15.3wave-particle duality - light and matter · the correct interpretation of wave-particle duality...

15.3 Wave-particle dualityHow can light be both a particle and a wave? We are now readyto resolve this seeming contradiction. Often in science when some-thing seems paradoxical, it’s because we (1) don’t define our termscarefully, or (2) don’t test our ideas against any specific real-worldsituation. Let’s define particles and waves as follows:

• Waves exhibit superposition, and specifically interference phe-nomena.

• Particles can only exist in whole numbers, not fractions.

As a real-world check on our philosophizing, there is one partic-ular experiment that works perfectly. We set up a double-slit inter-ference experiment that we know will produce a diffraction patternif light is an honest-to-goodness wave, but we detect the light witha detector that is capable of sensing individual photons, e.g., a dig-ital camera. To make it possible to pick out individual dots due toindividual photons, we must use filters to cut down the intensity ofthe light to a very low level, just as in the photos by Prof. Pageon p. 344. The whole thing is sealed inside a light-tight box. Theresults are shown in figure i. (In fact, the similar figures in on page344 are simply cutouts from these figures.)

i / Wave interference patternsphotographed by Prof. LymanPage with a digital camera. Laserlight with a single well-definedwavelength passed through aseries of absorbers to cut downits intensity, then through a set ofslits to produce interference, andfinally into a digital camera chip.(A triple slit was actually used,but for conceptual simplicity wediscuss the results in the maintext as if it was a double slit.) Inpanel 2 the intensity has beenreduced relative to 1, and evenmore so for panel 3.

Neither the pure wave theory nor the pure particle theory canexplain the results. If light was only a particle and not a wave, therewould be no interference effect. The result of the experiment wouldbe like firing a hail of bullets through a double slit, j. Only twospots directly behind the slits would be hit.

If, on the other hand, light was only a wave and not a particle,we would get the same kind of diffraction pattern that would happen

Section 15.3 Wave-particle duality 351

j / Bullets pass through a doubleslit.

k / A water wave passes througha double slit.

l / A single photon can gothrough both slits.

with a water wave, k. There would be no discrete dots in the photo,only a diffraction pattern that shaded smoothly between light anddark.

Applying the definitions to this experiment, light must be botha particle and a wave. It is a wave because it exhibits interferenceeffects. At the same time, the fact that the photographs containdiscrete dots is a direct demonstration that light refuses to be splitinto units of less than a single photon. There can only be wholenumbers of photons: four photons in figure i/3, for example.

15.3.1 A wrong interpretation: photons interfering with eachother

One possible interpretation of wave-particle duality that occurredto physicists early in the game was that perhaps the interference ef-fects came from photons interacting with each other. By analogy,a water wave consists of moving water molecules, and interferenceof water waves results ultimately from all the mutual pushes andpulls of the molecules. This interpretation has been conclusivelydisproved by forming interference patterns with light so dim thatno more than one photon is in flight at a time. In figure i/3, forexample, the intensity of the light has been cut down so much by theabsorbers that if it was in the open, the average separation betweenphotons would be on the order of a kilometer! Although most lightsources tend to emit photons in bunches, experiments have beendone with light sources that really do emit single photons at widetime intervals, and the same type of interference pattern is observed,showing that a single photon can interfere with itself.

The concept of a photon’s path is undefined.

If a single photon can demonstrate double-slit interference, thenwhich slit did it pass through? The unavoidable answer must be thatit passes through both! This might not seem so strange if we thinkof the photon as a wave, but it is highly counterintuitive if we tryto visualize it as a particle. The moral is that we should not thinkin terms of the path of a photon. Like the fully human and fullydivine Jesus of Christian theology, a photon is supposed to be 100%wave and 100% particle. If a photon had a well defined path, then itwould not demonstrate wave superposition and interference effects,contradicting its wave nature. (In sec. section 16.4 we will discussthe Heisenberg uncertainty principle, which gives a numerical wayof approaching this issue.)

15.3.2 The probability interpretation

The correct interpretation of wave-particle duality is suggestedby the random nature of the experiment we’ve been discussing: eventhough every photon wave/particle is prepared and released in thesame way, the location at which it is eventually detected by the

352 Chapter 15 Light as a particle

m / Example 4.

digital camera is different every time. The idea of the probabilityinterpretation of wave-particle duality is that the location of thephoton-particle is random, but the probability that it is in a certainlocation is higher where the photon-wave’s amplitude is greater.

More specifically, the probability distribution of the particle mustbe proportional to the square of the wave’s amplitude,

(probability distribution) ∝ (amplitude)2.

This follows from the correspondence principle and from the factthat a wave’s energy density is proportional to the square of its am-plitude. If we run the double-slit experiment for a long enough time,the pattern of dots fills in and becomes very smooth as would havebeen expected in classical physics. To preserve the correspondencebetween classical and quantum physics, the amount of energy de-posited in a given region of the picture over the long run must beproportional to the square of the wave’s amplitude. The amount ofenergy deposited in a certain area depends on the number of pho-tons picked up, which is proportional to the probability of findingany given photon there.

A microwave oven example 4. The figure shows two-dimensional (top) and one-dimensional(bottom) representations of the standing wave inside a microwaveoven. Gray represents zero field, and white and black signify thestrongest fields, with white being a field that is in the opposite di-rection compared to black. Compare the probabilities of detectinga microwave photon at points A, B, and C.

. A and C are both extremes of the wave, so the probabilities ofdetecting a photon at A and C are equal. It doesn’t matter that wehave represented C as negative and A as positive, because it isthe square of the amplitude that is relevant. The amplitude at B isabout 1/2 as much as the others, so the probability of detecting aphoton there is about 1/4 as much.

Discussion questions

A Referring back to the example of the carrot in the microwave oven,show that it would be nonsensical to have probability be proportional tothe field itself, rather than the square of the field.

B Einstein did not try to reconcile the wave and particle theories oflight, and did not say much about their apparent inconsistency. Einsteinbasically visualized a beam of light as a stream of bullets coming froma machine gun. In the photoelectric effect, a photon “bullet” would onlyhit one atom, just as a real bullet would only hit one person. Supposesomeone reading his 1905 paper wanted to interpret it by saying thatEinstein’s so-called particles of light are simply short wave-trains that onlyoccupy a small region of space. Comparing the wavelength of visible light(a few hundred nm) to the size of an atom (on the order of 0.1 nm), explainwhy this poses a difficulty for reconciling the particle and wave theories.

C Can a white photon exist?

Section 15.3 Wave-particle duality 353

n / A photon hits a piece ofglass that reflects half of the lightand transmits the other half.

D In double-slit diffraction of photons, would you get the same patternof dots on the digital camera image if you covered one slit? Why should itmatter whether you give the photon two choices or only one?

15.4 Nonlocality and entanglement15.4.1 Nonlocality

People sometimes say that quantum mechanics is the set of rulesfor describing the world of the very small, but this is a false general-ization, like saying that terriers are untrainable. How do we defineour measure of how small is small? The only distance scales we’vediscussed have been wavelengths, and there is no upper limit onwavelengths. The wavelength of an FM radio photon is bigger thanmy terrier, who is very obedient to Newton’s laws. The only scalebuilt in to the structure of quantum mechanics is Planck’s constant,and Planck’s constant has units of joules per hertz, not meters, so itcan’t be converted into a distance. Quantum mechanics is, as far aswe can tell, a valid tool for describing systems at scales from quarksto galaxies.

So quantum behavior can occur at any scale, even large ones. Foran example that may be a little disturbing, consider the arrangementshown in figure n. A single photon comes in from the left andencounters a diagonal piece of glass. The glass reflects half the lightand transmits half of it. The photon is a wave, and this is expectedwave behavior. But the photon is also a particle, and we can’t havehalf a particle. Therefore either camera A will detect a whole photonand B will see none, or it will be the other way around. If we repeatthe experiment many times times, we might come up with a list ofresults like this:

A B

no yesyes noyes nono yesno yesyes nono yesyes no

An instant before the moment of detection, the photon is a wavepattern that just happens to consist of two widely separated pieces,each carrying half the energy. The situation seems perfectly sym-metric, but then a moment later we find that B has detected thephoton and A hasn’t. If B’s detection of the photon is random, thenhow does the information get to A that it had better not detect it?This seems as though there is some sort of conspiracy being carriedout over arbitrarily large distances and with no time delay. It’s asthough the two parts of the wave are a pair of criminal suspects who


would like to line up their stories but are being kept in separate jailcells so that they can’t communicate. If the part of the wave at B isgoing to be detected (at full strength, carrying 100% of the energyE = hf), how does the part at A get the message that it should fadeaway like the Cheshire cat? This coordination would have to occurover very large distances — real-world experiments of this type havebeen done over distances of a thousand kilometers, with the photonstraveling either through outer space or through fiber-optic cables.Einstein derisively referred to this apparent coordination as “spookyaction at a distance.”

Niels Bohr and two collaborators proposed in 1924 the seem-ingly reasonable solution that there can’t be any such coordination.Then the random detection of the photon by camera A and cam-era B would be independent. Independent probabilities multiply, sothere would be a probability of (1/2)(1/2) = 1/4 that both cam-eras would see photons. This would violate conservation of energy,since the original energy E = hf would have been detected twice,and the universe would have gained 1hf worth of total energy. ButBohr pointed out that there would also be the same probability thatneither camera would detect a photon, in which case the change inthe universe’s energy would be −1hf . On the average, energy wouldbe conserved. According to Bohr’s theory, conservation of energyand momentum would not be absolute laws of physics but only rulesthat would be true on the average.

The experimentalists Geiger and Bothe immediately set out totest this prediction. They performed an experiment analogous tothe one in figure n, but with x-rays rather than visible light. Theirresults, published in 1926, showed that if one detector saw the x-rayphoton, the other did not, so that energy was always conserved atthe microscopic level, not just on the average. We never observe anoutcome in which both A and B detect a photon, or one in whichneither detects it. That is, the occurrence of event A (camera Asees a photon) and event B (camera B sees one) are both random,but they are not independent.

15.4.2 Entanglement

At a 1927 conference in Brussels, Einstein protested that this wasa problem, because the two detectors could in principle make theirobservations simultaneously, and it would then seem that some influ-ence or communication was being transmitted between them fasterthan the speed of light. “It seems to me,” he complained, “that thisdifficulty cannot be overcome unless the description of the processin terms of the . . . wave is supplemented by some detailed specifi-cation of the [trajectory of the particle]. . . . If one works only with. . . waves, the interpretation . . . , I think, contradicts the postulateof relativity.”

The experimental fact ends up being that the spooky action at a

Section 15.4 Nonlocality and entanglement 355

distance exists, and it does go faster than light. In 2012, Guerreiroet al.1 carried out a very direct and conceptually simple enactmentof exactly the experiment in figure n, with electronic timing preciseenough to prove that the detection events at A and B were sepa-rated from each other by too great a distance to have been linkedby any influence traveling at ≤ c. These findings are summarized bysaying that quantum mechanics is nonlocal. A single wave-particlecan be spread out over an arbitrarily large region of space, but itsinteractions that transfer energy and momentum are always corre-lated over these distances in such a way that the conservation lawsare maintained.

What Einstein had not originally appreciated was that thesecorrelations do not violate relativity because they do not actuallytransport any energy, or even any information, between A and B.For example, if Alice is at detector A, and Bob is at B, a millionkilometers away, Alice can detect the photon and know immediatelythat Bob did not detect it. She learns something seemingly instan-taneously about Bob — Bob is probably sad and disappointed rightnow. But because Bob does not have any control over the result,he cannot use this fact to send a message to Alice, so there is notransmission of information. Alice and Bob’s states are said to beentangled.

o / Entanglement is like findingthat you only have your left glove,so that you must have left yourright glove at home. There is again in information, but no suddentransmission of information fromthe dog to you.

By analogy, suppose that you head off to work on a winter day inNew York. As you step out of the subway station into the cold air,you reach into your pockets for your gloves, but you find that youonly have your left glove. Oh, you must have dropped your rightglove on the floor while you were petting your adorable terrier onthe way out the door. The presence of your left glove tells you thatyour right glove must be at home. But there has been no spookyaction at a distance. You have simply recovered some informationabout a region of space that lies at some distance from you.

Einstein and Bohr had strong physical intuitions that led them

1arxiv.org/abs/1204.1712. The paper is very readable.


to incorrect predictions about experiments, and these predictionswere the fruits of inappropriate mental pictures of what was goingon. If we take the principles of quantum mechanics seriously, thenthe correct picture is the following. Before the photon in figure nhits the glass diagonal, the state of things is the following.

A photon is headed to the right.

Our photon is then partially reflected and partially transmitted.Now we have a superposition of two wave patterns:

cThe photon has beenreflected upward.

+ c′The photon has con-tinue to the right.

,

where the amplitudes c and c′ are equal in absolute value.2

Let’s say that the cameras are at equal distances from the glassdiagonal, so that their chances to detect the photon occur simulta-neously.3 After detection, we have this:

cCamera A detected aphoton and B didn’t.

+ c′B detected a photonand A didn’t.

,

Here we have made the nontrivial assumption that material objectslike cameras obey the same wave-superposition rules as photons.This turns out to be true. Cameras are made out of things likeelectrons, and as we’ll see in chapter 16, things like electrons arealso wave-particles, and they obey all the same wave-particle rulesas photons. The states of the two cameras are now entangled.

You can see where this is going. Alice had been standing bycamera A, watching anxiously, while Bob, a million kilometers away,was breathlessly observing camera B.

c

Alice saw a photonand Bob didn’t. Theyconsider this result tohave been random.

+ c′

Bob saw a photon andAlice didn’t. Theyconsider this result tohave been random.

,

It doesn’t seem to Alice and Bob as though their brains are in asuperposition of two states. They feel as though they have only

2Conservation of energy requires c2 = 1/2 and c′2 = 1/2, even in classicalphysics. We could have, for example, c = 1/

√2 and c′ = −1/

√2. Such a possible

difference in signs wouldn’t concern us in this example. It would only be relevantif there were some later opportunity for the two parts of the wave to recombineand superimpose on one another, producing interference effects.

3According to special relativity, this simultaneity holds only in one frameof reference, say the lab frame. But if simultaneity does hold in one frame,then we can also say that in all frames, the distance between the two events is“spacelike,” i.e., they are too far apart to have been connected by any causalinfluence propagating at ≤ c.

Section 15.4 Nonlocality and entanglement 357

p / Schrodinger’s cat.

experienced the one possibility that actually happened, not a mix-ture of both at the same time. And yet this picture of the physicsexplains very nicely how the deterministic laws of physics producea result that seems to them to have been random.

If Alice and Bob have been split into two ghostlike halves ofthemselves, then conceivably these half-selves could undergo inter-ference, as in the double-slit experiment. But there are practicalreasons why we cannot actually detect such interference effects. Forone thing, Alice and Bob are macroscopic objects, with energies E onthe order of many joules. Because Planck’s constant is small, theirwave frequencies f = E/h are extremely high, and their wavelengthsincredibly short (on the order of 10−34 m!). We have seen thatdiffraction becomes undetectable when wavelengths are too short.Furthermore, there is a phenomenon called decoherence, discussedfurther in sec. 16.5.1, p. 381, in which interactions with the environ-ment tend to rapidly randomize the wave-phases of large objects.When phases are randomized, interference and diffraction effectsbecome undetectable.

Historically, it seemed absurd to the originators of quantum me-chanics to imagine a macroscopic object in a superposition of states.The most celebrated example is called the Schrodinger’s cat experi-ment. Luckily for the cat, there probably was no actual experiment— it was simply a “thought experiment” that the German theoristSchrodinger discussed with his colleagues. Schrodinger wrote:

One can even construct quite burlesque cases. A cat isshut up in a steel container, together with the followingdiabolical apparatus (which one must keep out of thedirect clutches of the cat): In a Geiger tube [radiationdetector] there is a tiny mass of radioactive substance, solittle that in the course of an hour perhaps one atom ofit disintegrates, but also with equal probability not evenone; if it does happen, the counter [detector] respondsand ... activates a hammer that shatters a little flaskof prussic acid [filling the chamber with poison gas]. Ifone has left this entire system to itself for an hour, thenone will say to himself that the cat is still living, if inthat time no atom has disintegrated. The first atomicdisintegration would have poisoned it.

It seemed ridiculous to Schrodinger that at the end of the hour,“The uncertainty originally restricted to the atomic domain has beentransformed into a macroscopic uncertainty...,” and the cat wouldbe in a superposed state.

In modern language, people like Einstein and Schrodinger didn’tfeel comfortable with nonlocality, or with entanglement of subatomicparticles, and they felt even less comfortable with applying these


q / Probability is the volumeunder a surface defined byD(x , y ).

concepts to macroscopic objects. Today, entanglement has beendemonstrated using objects that clearly deserve to be called macro-scopic. For example, in 2012, K.C. Lee et al. created a version ofthe experiment in figure n in which the cameras were replaced bysmall diamonds, about 1 mm in size. They were separated by 15cm, which is a macroscopic distance. When a photon hit one of thediamonds, it produced a vibration in the crystal lattice. This vibra-tion was localized to a relatively small region within the diamond,but this region was still large enough that one has to admit that itqualifies as macroscopic. Its atoms had a total weight of about 0.1nanograms, which is a quantity big enough to weigh on a state-of-the-art balance, and the region was about 0.01 mm in size, whichwould make it visible with a magnifying glass.

The quantum states of the two diamonds became entangled: ifone had detected the photon, the other hadn’t. This entangledstate was maintained for only about 7 picoseconds before decoher-ence destroyed the phase relationship between one diamond and theother. But Lee was able to use additional photons to “read out” thequantum states in only 0.5 ps, before decoherence occurred, and ver-ify that there were wave interference effects in which one diamond’squantum-mechanical wave had a definite phase relationship with theother’s. Although these experiments are difficult, they suggest thatthere is no obstruction in principle to observing quantum-mechanicaleffects such as superposition in arbitrarily large objects.

15.5 Photons in three dimensionsUp until now I’ve been sneaky and avoided a full discussion of thethree-dimensional aspects of the probability interpretation. The ex-ample of the carrot in the microwave oven, for example, reduced to aone-dimensional situation because we were considering three pointsalong the same line and because we were only comparing ratios ofprobabilities.

A typical example of a probability distribution in section 14.1was the distribution of heights of human beings. The thing thatvaried randomly, height, h, had units of meters, and the probabil-ity distribution was a graph of a function D(h). The units of theprobability distribution had to be m−1 (inverse meters) so that ar-eas under the curve, interpreted as probabilities, would be unitless:(area) = (height)(width) = m−1 ·m.

Now suppose we have a two-dimensional problem, e.g., the prob-ability distribution for the place on the surface of a digital camerachip where a photon will be detected. The point where it is detectedwould be described with two variables, x and y, each having unitsof meters. The probability distribution will be a function of bothvariables, D(x, y). A probability is now visualized as the volume

Section 15.5 Photons in three dimensions 359

under the surface described by the function D(x, y), as shown infigure q. The units of D must be m−2 so that probabilities will beunitless: (probability) = (depth)(length)(width) = m−2 ·m ·m. Interms of calculus, we have P =

∫D dx dy.

Generalizing finally to three dimensions, we find by analogy thatthe probability distribution will be a function of all three coordi-nates, D(x, y, z), and will have units of m−3. It is unfortunatelyimpossible to visualize the graph unless you are a mutant with anatural feel for life in four dimensions. If the probability distri-bution is nearly constant within a certain volume of space v, theprobability that the photon is in that volume is simply vD. If not,then we can use an integral, P =

∫D dx dy dz.


Problem 4.

ProblemsKey√

A computerized answer check is available online.? A difficult problem.

1 When light is reflected from a mirror, perhaps only 80% of theenergy comes back. One could try to explain this in two differentways: (1) 80% of the photons are reflected, or (2) all the photons arereflected, but each loses 20% of its energy. Based on your everydayknowledge about mirrors, how can you tell which interpretation iscorrect? [Based on a problem from PSSC Physics.]

2 Suppose we want to build an electronic light sensor using anapparatus like the one described in section 15.2 on p. 346. Howwould its ability to detect different parts of the spectrum depend onthe type of metal used in the capacitor plates?

3 The photoelectric effect can occur not just for metal cathodesbut for any substance, including living tissue. Ionization of DNAmolecules can cause cancer or birth defects. If the energy requiredto ionize DNA is on the same order of magnitude as the energyrequired to produce the photoelectric effect in a metal, which of thefollowing types of electromagnetic waves might pose such a hazard?Explain.

60 Hz waves from power lines100 MHz FM radiomicrowaves from a microwave ovenvisible lightultraviolet lightx-rays

4 (a) Rank-order the photons according to their wavelengths,frequencies, and energies. If two are equal, say so. Explain all youranswers.(b) Photon 3 was emitted by a xenon atom going from its second-lowest-energy state to its lowest-energy state. Which of photons 1,2, and 4 are capable of exciting a xenon atom from its lowest-energystate to its second-lowest-energy state? Explain.

5 The beam of a 100 W overhead projector covers an area of1 m× 1 m when it hits the screen 3 m away. Estimate the numberof photons that are in flight at any given time. (Since this is only anestimate, we can ignore the fact that the beam is not parallel.)

√

Problems 361

6 In the photoelectric effect, electrons are observed with virtuallyno time delay (∼ 10 ns), even when the light source is very weak.(A weak light source does however only produce a small numberof ejected electrons.) The purpose of this problem is to show thatthe lack of a significant time delay contradicted the classical wavetheory of light, so throughout this problem you should put yourselfin the shoes of a classical physicist and pretend you don’t knowabout photons at all. At that time, it was thought that the electronmight have a radius on the order of 10−15 m. (Recent experimentshave shown that if the electron has any finite size at all, it is farsmaller.)(a) Estimate the power that would be soaked up by a single electronin a beam of light with an intensity of 1 mW/m2.

√

(b) The energy, Es, required for the electron to escape through thesurface of the cathode is on the order of 10−19 J. Find how long itwould take the electron to absorb this amount of energy, and explainwhy your result constitutes strong evidence that there is somethingwrong with the classical theory.

√

7 As far as we know, the mass of the photon is zero. However,it’s not possible to prove by experiments that anything is zero; allwe can do is put an upper limit on the number. As of 2008, thebest experimental upper limit on the mass of the photon is about1× 10−52 kg. Suppose that the photon’s mass really isn’t zero, andthat the value is at the top of the range that is consistent withthe present experimental evidence. In this case, the c occurring inrelativity would no longer be interpreted as the speed of light. Aswith material particles, the speed v of a photon would depend onits energy, and could never be as great as c. Estimate the relativesize (c− v)/c of the discrepancy in speed, in the case of a photon ofvisible light.

. Answer, p. 455

8 Give a numerical comparison of the number of photons persecond emitted by a hundred-watt FM radio transmitter and ahundred-watt lightbulb.

√

9 (a) A radio transmitter radiates power P in all directions, sothat the energy spreads out spherically. Find the energy density ata distance r.

√

(b) Let the wavelength be λ. As described in example 2 on p. 349,find the number of photons in a volume λ3 at this distance r.

√

(c) For a 1000 kHz AM radio transmitting station, assuming rea-sonable values of P and r, verify, as claimed in the example, thatthe result from part b is very large.


Problem 10.

Problem 11.

Problem 12.

10 The two diffraction patterns were made by sending a flash oflight through the same double slit. Give a numerical comparison ofthe amounts of energy in the two flashes.

√

11 Three of the four graphs are consistently normalized to rep-resent a wave consisting of the same number of photons. Which oneisn’t? Explain.

12 Many radio antennas are designed so that they preferen-tially emit or receive electromagnetic waves in a certain direction.However, no antenna is perfectly directional. The wave shown in thefigure represents a single photon being emitted by an antenna at thecenter. The antenna is directional, so there is a stronger wave onthe right than on the left. What is the probability that the photonwill be observed on the right?

Problems 363

Dorothy melts the Wicked Witchof the West.

Chapter 16

Matter as a wave

[In] a few minutes I shall be all melted... I have been wickedin my day, but I never thought a little girl like you would everbe able to melt me and end my wicked deeds. Look out —here I go!

The Wicked Witch of the West

As the Wicked Witch learned the hard way, losing molecularcohesion can be unpleasant. That’s why we should be very grate-ful that the concepts of quantum physics apply to matter as wellas light. If matter obeyed the laws of classical physics, moleculeswouldn’t exist.

Consider, for example, the simplest atom, hydrogen. Why doesone hydrogen atom form a chemical bond with another hydrogenatom? Roughly speaking, we’d expect a neighboring pair of hy-drogen atoms, A and B, to exert no force on each other at all,attractive or repulsive: there are two repulsive interactions (protonA with proton B and electron A with electron B) and two attractiveinteractions (proton A with electron B and electron A with protonB). Thinking a little more precisely, we should even expect that oncethe two atoms got close enough, the interaction would be repulsive.For instance, if you squeezed them so close together that the twoprotons were almost on top of each other, there would be a tremen-dously strong repulsion between them due to the 1/r2 nature of the

365

electrical force. A more detailed calculation using classical physicsgives an extremely weak binding, about 1/17 the strength of whatwe actually observe (☡388), which is far too weak to make the bondhold together.

Quantum physics to the rescue! As we’ll see shortly, the wholeproblem is solved by applying the same quantum concepts to elec-trons that we have already used for photons.

16.1 Electrons as waves16.1.1 Wavelength related to momentum

We started our journey into quantum physics by studying therandom behavior of matter in radioactive decay, and then asked howrandomness could be linked to the basic laws of nature governinglight. The probability interpretation of wave-particle duality wasstrange and hard to accept, but it provided such a link. It is nownatural to ask whether the same explanation could be applied tomatter. If the fundamental building block of light, the photon, isa particle as well as a wave, is it possible that the basic units ofmatter, such as electrons, are waves as well as particles?

A young French aristocrat studying physics, Louis de Broglie(pronounced “broylee”), made exactly this suggestion in his 1923Ph.D. thesis. His idea had seemed so farfetched that there wasserious doubt about whether to grant him the degree. Einstein wasasked for his opinion, and with his strong support, de Broglie gothis degree.

Only two years later, American physicists C.J. Davisson and L.Germer confirmed de Broglie’s idea by accident. They had beenstudying the scattering of electrons from the surface of a sampleof nickel, made of many small crystals. (One can often see such acrystalline pattern on a brass doorknob that has been polished byrepeated handling.) An accidental explosion occurred, and whenthey put their apparatus back together they observed somethingentirely different: the scattered electrons were now creating an in-terference pattern! This dramatic proof of the wave nature of mattercame about because the nickel sample had been melted by the ex-plosion and then resolidified as a single crystal. The nickel atoms,now nicely arranged in the regular rows and columns of a crystallinelattice, were acting as the lines of a diffraction grating. The newcrystal was analogous to the type of ordinary diffraction grating inwhich the lines are etched on the surface of a mirror (a reflectiongrating) rather than the kind in which the light passes through thetransparent gaps between the lines (a transmission grating).

Although we will concentrate on the wave-particle duality of elec-

366 Chapter 16 Matter as a wave

a / A double-slit interference pattern made with neutrons. (A. Zeilinger,R. Gahler, C.G. Shull, W. Treimer, and W. Mampe, Reviews of ModernPhysics, Vol. 60, 1988.)

trons because it is important in chemistry and the physics of atoms,all the other “particles” of matter you’ve learned about show waveproperties as well. Figure a, for instance, shows a wave interferencepattern of neutrons.

It might seem as though all our work was already done for us,and there would be nothing new to understand about electrons:they have the same kind of funny wave-particle duality as photons.That’s almost true, but not quite. There are some important waysin which electrons differ significantly from photons:

1. Electrons have mass, and photons don’t.

2. Photons always move at the speed of light, but electrons canmove at any speed less than c.

3. Photons don’t have electric charge, but electrons do, so electricforces can act on them. The most important example is theatom, in which the electrons are held by the electric force ofthe nucleus.

4. Electrons cannot be absorbed or emitted as photons are. De-stroying an electron or creating one out of nothing would vio-late conservation of charge.

(In section 18.6 we will learn of one more fundamental way in whichelectrons differ from photons, for a total of five.)

Because electrons are different from photons, it is not immedi-ately obvious which of the photon equations from chapter 15 can be

Section 16.1 Electrons as waves 367

• The wavenumber k =2π/λ is inversely relatedto the wavelength λ.

• It has units of radians permeter.

• In three dimensions, k isthe magnitude of a thewave-vector k (p. 51).

• The wavevector k is tothe wavelength as the fre-quency ω is to the period.

b / Review of the wavenumber k

applied to electrons as well. With hindsight, we know that there isa simple and consistent way of putting this all together. As a pre-liminary in order to make the notation simpler, we define a versionof Planck’s constant ~ = h/2π, read as “h-bar.” We also recall fromsec. 2.4.6, p. 45, the definition of the wavenumber k = 2π/λ alongwith a few of its properties, table b. With these definitions, our twofundamental quantum-mechanical relations are

E = ~ω [the same as E = hf ]

p = ~k [the same as p = h/λ]

It’s only a slight exaggeration to say that these two equations sum-marize all of quantum mechanics. They are true for all the ordinarybuilding blocks of light and matter in everyday life, and also for everyweird creature in the particle-physics zoo. Each of these equationshas a particle thing on the left and a wave thing on the right.

The second equation, although true for photons, takes on agreater importance for electrons. This is first of all because themomentum of matter is more likely to be significant than the mo-mentum of light under ordinary conditions, and also because forceis the transfer of momentum, and electrons are affected by electricalforces.

The wavelength of a cat example 1. What is the wavelength of a trotting cat?

. One may doubt whether the equation p = ~k = h/λ should beapplied to a cat, which is not just a single particle but a ratherlarge collection of them. Throwing caution to the wind, however,we estimate the cat’s mass at 1 kg and its trotting speed at 1 m/s.Its wavelength is therefore roughly

λ =hp

=h

mv

=6.63× 10−34 J·s

(1 kg)(1 m/s)

∼ 10−33

(kg·m2/s2)·s

kg·m/s= 10−33 m.

The wavelength found in this example is so fantastically smallthat we can be sure we will never observe any measurable wavephenomena with cats or any other human-scale objects. For exam-ple, if we sent the cat through a pair of double slits separated by10 cm, then the angular spacing of the diffraction pattern wouldbe 10−32 radians, which would make the fringes too close together


to be distinguished. This is in agreement with the correspondenceprinciple.

Although a smaller mass in the equation λ = h/mv does re-sult in a longer wavelength, the wavelength is still quite short evenfor individual electrons under typical conditions, as shown in thefollowing example.

The typical wavelength of an electron example 2. Electrons in circuits and in atoms are typically moving throughvoltage differences on the order of 1 V, so that a typical energy is(e)(1 V), which is on the order of 10−19 J. What is the wavelengthof an electron with this amount of kinetic energy?

. This energy is nonrelativistic, since it is much less than mc2.Momentum and energy are therefore related by the nonrelativisticequation K = p2/2m. Solving for p and substituting in to theequation for the wavelength, we find

λ =h√

2mK= 1.6× 10−9 m.

This is on the same order of magnitude as the size of an atom,which is no accident: as we will discuss in the next chapter inmore detail, an electron in an atom can be interpreted as a stand-ing wave. The smallness of the wavelength of a typical electronalso helps to explain why the wave nature of electrons wasn’t dis-covered until a hundred years after the wave nature of light. Toscale the usual wave-optics devices such as diffraction gratingsdown to the size needed to work with electrons at ordinary ener-gies, we need to make them so small that their parts are compa-rable in size to individual atoms. This is essentially what Davissonand Germer did with their nickel crystal.

self-check A

These remarks about the inconvenient smallness of electron wavelengthsapply only under the assumption that the electrons have typical ener-gies. What kind of energy would an electron have to have in order tohave a longer wavelength that might be more convenient to work with?. Answer, p. 454

16.1.2 What kind of wave is it?

If a sound wave is a vibration of matter, and a photon is avibration of electric and magnetic fields, what kind of a wave isan electron made of? The disconcerting answer is that there isno experimental “observable,” i.e., directly measurable quantity, tocorrespond to the electron wave itself. In other words, there aredevices like microphones that detect the oscillations of air pressurein a sound wave, and devices such as radio receivers that measure


c / These two electron wavesare not distinguishable by anymeasuring device.

the oscillation of the electric and magnetic fields in a light wave,but nobody has ever found any way to measure the electron wavedirectly.

We can of course detect the energy (or momentum) possessed byan electron just as we could detect the energy of a photon using adigital camera. (In fact I’d imagine that an unmodified digital cam-era chip placed in a vacuum chamber would detect electrons just ashandily as photons.) But this only allows us to determine where thewave carries high probability and where it carries low probability.Probability is proportional to the square of the wave’s amplitude,but measuring its square is not the same as measuring the waveitself. In particular, we get the same result by squaring either apositive number or its negative, so there is no way to determine thepositive or negative sign of an electron wave. In general, the phaseof the wavefunction is not an observable thing.

This meaninglessness of phase only applies to absolute phases.Relative phases do have real-world consequences. For example, fig-ure a on p. 367 shows a double-slit interference pattern made byneutrons. There are places where we observe a lot of neutrons, andplaces where we don’t. The places where we don’t see many neu-trons are the ones where waves are (at least partially) canceling,which means that they’re out of phase — relative to each other. Asa political metaphor, my wife and I would sometimes, before 2016,cancel each other out at the polls by voting for different politicalparties. It was a matter of opinion who was right, but it was anobjective fact that we canceled.

Most physicists tend toward the school of philosophy known asoperationalism, which says that a concept is only meaningful if wecan define some set of operations for observing, measuring, or testingit. According to a strict operationalist, then, the electron wave itselfis a meaningless concept, because we can’t measure its absolutephase. Nevertheless, it turns out to be one of those concepts likelove or humor that is impossible to measure and yet very useful tohave around. We therefore give it a symbol, Ψ (the capital Greekletter psi), and a special name, the electron wavefunction (becauseit is a function of the coordinates x, y, and z that specify where youare in space). It would be impossible, for example, to calculate theshape of the electron wave in a hydrogen atom without having somesymbol for the wave. But when the calculation produces a resultthat can be compared directly to experiment, the final algebraicresult will turn out to involve only Ψ2, which is what is observable,not Ψ itself.

Since Ψ, unlike E and B, is not directly measurable, we are freeto make the probability equations have a simple form: instead ofhaving the probability density equal to some funny constant multi-plied by Ψ2, we simply define Ψ so that the constant of proportion-


ality is one:

(probability distribution) = |Ψ|2.

Since the probability distribution has units of m−3, the units of Ψmust be m−3/2. The square of a negative number is still positive,so the absolute value signs may seem unnecessary, but as we’ll insec. 17.5, p. 403, the wavefunction may in general be a complexnumber. In fact, only standing waves, not traveling waves, canreally be represented by real numbers, although we will often cheatand draw pictures of traveling waves as if they were real-valuedfunctions.

We have defined a wave as something that can superpose, anda particle as something that you can’t have a fraction of (p. 351).Since you can’t have a fraction of an electron, we conclude that if Ψis a valid wavefunction for an electron, then Ψ/2 should be somehowillegal. We can see this from the requirement that the probabilitydistribution be normalized (p. 326).

Normalization and phase of a particle in a box example 3A certain electron confined to a box of length L has a wavefunc-tion

Ψ(x) =

{A sin(2πx/L) if 0 ≤ x ≤ L0 elsewhere.

This quantity inside the sine function varies from 0 to 2π, so this isa standing wave with one wavelength fitting inside the box, . Ifwe pick positive or negative values of A we can have the two func-tions in figure c, p. 370, but this is a difference in phase, which wedon’t expect to be physically observable. Normalization requiresthat ∫ L

0|Ψ|2 dx = 1

or

|A|2∫ L

0sin2(2πx/L) dx = 1.

This integral can be done using straightforward calculus, but ashortcut is to recognize that the sine oscillates symmetrically be-tween 0 and 1, and therefore over any whole number of half-cycles its average value is 1/2. We can therefore find the sameresult for the definite integral by replacing the sine with 1/2, andthis results in

|A|2 =2L

.

Two possible values of A are√

2/L and −√

2/L, but these rep-resent physically indistinguishable states. We could also have acomplex value of A, as long as |A|2 had the correct value.


16.1.3 Quantum numbers and bra-ket notation

When we want to describe a wavefunction, it can be cumbersometo write out its equation as in example 3, the particle in a box. Inthat particular example, it’s much easier just to draw a picture

or to define some simple numerical label like N = 2 as we didin problem 12 on p. 59. A number like this is called a quantumnumber, and in this particular example the N quantum number isa wavelength or energy label. (It isn’t a momentum label, becausethe standing wave is a superposition of traveling waves going in bothdirections.)

When we want to refer specifically to the wavefunction that islabeled in these ways, there is a handy notation involving anglebrackets, called the “bra-ket” notation. In our example, we couldwrite something like

| 〉

or|N = 2〉.

These are called kets. We can also have notation where the barand angle bracket are flipped the other way, like 〈 | or 〈N = 2|.These are called bras. For real-valued wavefunctions, which are goodenough for standing waves, there is no important distinction to bemade between bras and kets. When we sandwich together a bra anda ket, both of which represent the same state, this is a shorthandway of notating the type of integral that we did in example 3 fornormalization. That is, if we want to express the requirement thatthe state be normalized, we can write something like

〈 | 〉 = 1,

which is a lot easier than writing out the whole integral as∫ L0 |A|

2 sin2(2πx/L) dx = 1. The bra-ket notation is generalized insec. 17.7.1, p. 409.

16.1.4 “Same state” versus “same wavefunction”

We’ve already seen that phases (or at least absolute phases) arenot physically meaningful. We saw this principle at work in exam-ple 3, where | 〉 and | 〉 would give identical predictions aboutprobabilities. In general two different wavefunctions can actuallyrepresent the same state, if they only differ by a constant factor.It doesn’t matter whether we multiply by −1, or for that matterby 1/2, forming | 〉. The latter would not be normalized, but itwouldn’t be a different state, just an unnormalized version of thesame state. We could of course rule out unnormalized wavefunc-tions, but that’s not a good idea, for the following reasons.

Suppose you do the Schrodinger’s cat experiment (p. 358). Atthe end of the experiment, your brain is in a superposition of twostates that really are different: a state in which your brain saw a


live cat, and a state in which it saw a dead one. Let’s label thesestates as | L〉 and | D〉. The state of your brain is

c| L〉+ c′| D〉.

Now as far as | L〉 is concerned, a live cat is the only possibilitythat has happened. That brain doesn’t know about the alternateuniverse in which | D〉 saw a dead cat.

In a certain calculation, we might say that c was a negativenumber, but the brain doesn’t say, “Huh, I feel negative today,” or“The universe seems to have flipped its phase now.”

Similarly, the brain has no way of knowing whether c is bigor small. Suppose that the Quantum Police did decree that allwavefunctions had to be normalized. Then we would have |c| < 1.(If the L and D outcomes each have probability 1/2, then |c| = |c′| =1/√

2.) But in | L〉’s world, it’s meaningless to talk about this.That brain doesn’t say, “I feel weak, as though my normalizationwas sagging,” or “The universe seems gray and boring today — Ithink its normalization has gotten lower.”

For these reasons, we adopt the convention that if | . . .〉 is awavefunction and c is any nonzero number (either real or complex),then any wavefunction c| . . .〉 represents the same state.

As a matter of mathematical elegance, waves are things thatsuperpose, and this means that they inhabit a vector space (p. 155).One of the axioms defining a vector space is that we have well-defined operations of addition and multiplication by a scalar. Ifwe ruled out unnormalized wavefunctions, we would be violatingthis requirement. And physically, it is sometimes OK to talk aboutan unnormalized state as representing more than one particle, or,in the case of a traveling wave, a steady flux of particles like thebeam of a cathode ray tube. We don’t actually want to stop talkingcompletely about unnormalized wavefunctions, like someone fromthe 1950’s who’s afraid to use the word “toilet” in polite company.

None of this is meant to imply that the c’s are never important.Relative phases are meaningful. For example, | 〉 + | 〉 will pro-duce a different interference pattern than | 〉 − | 〉. And we doneed to keep track of normalization in order to say anything aboutprobabilities. For example, it’s not true that in the state describedby the unnormalized wavefunction 5| 〉+ 7| 〉, the probability of

is 25; probabilities have to be between 0 and 1.

16.2 Dispersive wavesA colleague of mine who teaches chemistry loves to tell the storyabout an exceptionally bright student who, when told of the equa-tion p = h/λ, protested, “But when I derived it, it had a factor of2!” The issue that’s involved is a real one, albeit one that could be

Section 16.2 Dispersive waves 373

glossed over (and is, in most textbooks) without raising any alarmsin the mind of the average student. The present optional sectionaddresses this point; it is intended for the student who wishes todelve a little deeper.

Here’s how the now-legendary student was presumably reason-ing. We start with the equation v = fλ, which is valid for any sinewave, whether it’s quantum or classical. Let’s assume we alreadyknow E = hf , and are trying to derive the relationship betweenwavelength and momentum:

λ =v

f

=vh

E

=vh

12mv

2

=2h

mv

=2h

p.

The reasoning seems valid, but the result does contradict theaccepted one, which is after all solidly based on experiment.

The mistaken assumption is that we can figure everything out interms of pure sine waves. Mathematically, the only wave that hasa perfectly well defined wavelength and frequency is a sine wave,and not just any sine wave but an infinitely long sine wave, d. Theunphysical thing about such a wave is that it has no leading ortrailing edge, so it can never be said to enter or leave any particularregion of space. Our derivation made use of the velocity, v, and ifvelocity is to be a meaningful concept, it must tell us how quicklystuff (mass, energy, momentum, . . . ) is transported from one regionof space to another. Since an infinitely long sine wave doesn’t removeany stuff from one region and take it to another, the “velocity of itsstuff” is not a well defined concept.

Of course the individual wave peaks do travel through space, andone might think that it would make sense to associate their speedwith the “speed of stuff,” but as we will see, the two velocities arein general unequal when a wave’s velocity depends on wavelength.Such a wave is called a dispersive wave, because a wave pulse consist-ing of a superposition of waves of different wavelengths will separate(disperse) into its separate wavelengths as the waves move throughspace at different speeds. Nearly all the waves we have encounteredhave been nondispersive. For instance, sound waves and light waves(in a vacuum) have speeds independent of wavelength. A water waveis one good example of a dispersive wave. Long-wavelength waterwaves travel faster, so a ship at sea that encounters a storm typi-cally sees the long-wavelength parts of the wave first. When dealing


d / Part of an infinite sine wave.

e / A finite-length sine wave.

f / A beat pattern created bysuperimposing two sine waveswith slightly different wave-lengths.

with dispersive waves, we need symbols and words to distinguishthe two speeds. The speed at which wave peaks move is called thephase velocity, vp, and the speed at which “stuff” moves is calledthe group velocity, vg.

An infinite sine wave can only tell us about the phase velocity,not the group velocity, which is really what we would be talkingabout when we refer to the speed of an electron. If an infinitesine wave is the simplest possible wave, what’s the next best thing?We might think the runner up in simplicity would be a wave trainconsisting of a chopped-off segment of a sine wave, e. However, thiskind of wave has kinks in it at the end. A simple wave should beone that we can build by superposing a small number of infinitesine waves, but a kink can never be produced by superposing anynumber of infinitely long sine waves.

Actually the simplest wave that transports stuff from place toplace is the pattern shown in figure f. Called a beat pattern, it isformed by superposing two sine waves whose wavelengths are similarbut not quite the same. If you have ever heard the pulsating howlingsound of musicians in the process of tuning their instruments to eachother, you have heard a beat pattern. The beat pattern gets strongerand weaker as the two sine waves go in and out of phase with eachother. The beat pattern has more “stuff” (energy, for example)in the areas where constructive interference occurs, and less in theregions of cancellation. As the whole pattern moves through space,stuff is transported from some regions and into other ones.

If the frequency of the two sine waves differs by 10%, for in-stance, then ten periods will be occur between times when they arein phase. Another way of saying it is that the sinusoidal “envelope”(the dashed lines in figure f) has a frequency equal to the differencein frequency between the two waves. For instance, if the waves hadfrequencies of 100 Hz and 110 Hz, the frequency of the envelopewould be 10 Hz.

Applying similar reasoning to the wavelength, the wavenumberk of the envelope equals the difference between the wavenumbers ofthe two sine waves.

The group velocity is the speed at which the envelope movesthrough space. Let ∆ω and ∆k be the differences between the fre-quencies and wavenumbers of the two sine waves, which means thatthey equal the frequency and wavenumber of the envelope. Thegroup velocity is vg = ωenvelope/kenvelope = ∆ω/∆k. If ∆ω and ∆kare small, we can approximate this expression as a derivative,

vg =dω

dk.

This expression is usually taken as the definition of the group veloc-ity for wave patterns that consist of a superposition of sine waves

Section 16.2 Dispersive waves 375

g / Three possible standing-wave patterns for a particle in abox.

having a narrow range of frequencies and wavelengths. In quan-tum mechanics, with ω = E/~ and k = p/~, we have vg = dE/dp.In the case of a nonrelativistic electron the relationship betweenenergy and momentum is E = p2/2m, so the group velocity isdE/dp = p/m = v, exactly what it should be. It is only the phasevelocity that differs by a factor of two from what we would haveexpected, but the phase velocity is not the physically importantthing.

16.3 Bound statesElectrons are at their most interesting when they’re in atoms, thatis, when they are bound within a small region of space. We canunderstand a great deal about atoms and molecules based on simplearguments about such bound states, without going into any of therealistic details of atom. The simplest model of a bound state isknown as the particle in a box: like a ball on a pool table, theelectron feels zero force while in the interior, but when it reachesan edge it encounters a wall that pushes back inward on it witha large force. In particle language, we would describe the electronas bouncing off of the wall, but this incorrectly assumes that theelectron has a certain path through space. It is more correct todescribe the electron as a wave that undergoes 100% reflection atthe boundaries of the box.

Like generations of physics students before me, I rolled my eyeswhen initially introduced to the unrealistic idea of putting a particlein a box. It seemed completely impractical, an artificial textbookinvention. Today, however, it has become routine to study elec-trons in rectangular boxes in actual laboratory experiments. The“box” is actually just an empty cavity within a solid piece of silicon,amounting in volume to a few hundred atoms. The methods forcreating these electron-in-a-box setups (known as “quantum dots”)were a by-product of the development of technologies for fabricatingcomputer chips.

For simplicity let’s imagine a one-dimensional electron in a box,i.e., we assume that the electron is only free to move along a line.The resulting standing wave patterns, of which the first three areshown in the figure, are just like some of the patterns we encoun-tered with sound waves in musical instruments. The wave patternsmust be zero at the ends of the box, because we are assuming thewalls are impenetrable, and there should therefore be zero proba-bility of finding the electron outside the box. Each wave pattern islabeled according to n, the number of peaks and valleys it has. Inquantum physics, these wave patterns are referred to as “states” ofthe particle-in-the-box system.

The following seemingly innocuous observations about the par-ticle in the box lead us directly to the solutions to some of the most


h / The spectrum of the lightfrom the star Sirius.

vexing failures of classical physics:

The particle’s energy is quantized (can only have certain values).Each wavelength corresponds to a certain momentum, and a givenmomentum implies a definite kinetic energy, E = p2/2m. (This isthe second type of energy quantization we have encountered. Thetype we studied previously had to do with restricting the numberof particles to a whole number, while assuming some specific wave-length and energy for each particle. This type of quantization refersto the energies that a single particle can have. Both photons andmatter particles demonstrate both types of quantization under theappropriate circumstances.)

The particle has a minimum kinetic energy. Long wavelengths cor-respond to low momenta and low energies. There can be no statewith an energy lower than that of the n = 1 state, called the groundstate.

The smaller the space in which the particle is confined, the higherits kinetic energy must be. Again, this is because long wavelengthsgive lower energies.

Spectra of thin gases example 4A fact that was inexplicable by classical physics was that thin

gases absorb and emit light only at certain wavelengths. Thiswas observed both in earthbound laboratories and in the spectraof stars. The figure on the left shows the example of the spec-trum of the star Sirius, in which there are “gap teeth” at certainwavelengths. Taking this spectrum as an example, we can give astraightforward explanation using quantum physics.

Energy is released in the dense interior of the star, but the outerlayers of the star are thin, so the atoms are far apart and electronsare confined within individual atoms. Although their standing-wave patterns are not as simple as those of the particle in thebox, their energies are quantized.

When a photon is on its way out through the outer layers, it can beabsorbed by an electron in an atom, but only if the amount of en-ergy it carries happens to be the right amount to kick the electronfrom one of the allowed energy levels to one of the higher lev-els. The photon energies that are missing from the spectrum arethe ones that equal the difference in energy between two elec-tron energy levels. (The most prominent of the absorption lines inSirius’s spectrum are absorption lines of the hydrogen atom.)

The stability of atoms example 5In many Star Trek episodes the Enterprise, in orbit around a planet,suddenly lost engine power and began spiraling down toward theplanet’s surface. This was utter nonsense, of course, due to con-servation of energy: the ship had no way of getting rid of energy,so it did not need the engines to replenish it.

Section 16.3 Bound states 377

i / Two hydrogen atoms bondto form an H2 molecule. In themolecule, the two electrons’ wavepatterns overlap, and are abouttwice as wide.

Consider, however, the electron in an atom as it orbits the nu-cleus. The electron does have a way to release energy: it has anacceleration due to its continuously changing direction of motion,and according to classical physics, any accelerating charged par-ticle emits electromagnetic waves. According to classical physics,atoms should collapse!

The solution lies in the observation that a bound state has a min-imum energy. An electron in one of the higher-energy atomicstates can and does emit photons and hop down step by step inenergy. But once it is in the ground state, it cannot emit a photonbecause there is no lower-energy state for it to go to.

Chemical bonds example 6I began this section with a classical argument that chemical

bonds, as in an H2 molecule, should not exist. Quantum physicsexplains why this type of bonding does in fact occur. There areactually two effects going on, one due to kinetic energy and onedue to electrical energy. We’ll concentrate on the kinetic energyeffect in this example. Example 6 on page 402 revisits the H2bond in more detail. (There is also a qualitatively different typeof bonding called ionic bonding that occurs when an atom of oneelement steals the electron from an atom of another element.)

The kinetic energy effect is pretty simple. When the atoms arenext to each other, the electrons are shared between them. The“box” is about twice as wide, and a larger box allows a smaller ki-netic energy. Energy is required in order to separate the atoms.


A Neutrons attract each other via the strong nuclear force, so accordingto classical physics it should be possible to form nuclei out of clusters oftwo or more neutrons, with no protons at all. Experimental searches,however, have failed to turn up evidence of a stable two-neutron system(dineutron) or larger stable clusters. These systems are apparently notjust unstable in the sense of being able to beta decay but unstable inthe sense that they don’t hold together at all. Explain based on quantumphysics why a dineutron might spontaneously fly apart.

B The following table shows the energy gap between the ground stateand the first excited state for four nuclei, in units of picojoules. (The nucleiwere chosen to be ones that have similar structures, e.g., they are allspherical in shape.)

nucleus energy gap (picojoules)4He 3.23416O 0.96840Ca 0.536208Pb 0.418

Explain the trend in the data.


j / Werner Heisenberg (1901-1976). Heisenberg helped todevelop the foundations of quan-tum mechanics, including theHeisenberg uncertainty principle.He was the scientific leader ofthe Nazi atomic-bomb programup until its cancellation in 1942,when the military decided that itwas too ambitious a project toundertake in wartime, and toounlikely to produce results.

16.4 The uncertainty principle16.4.1 Eliminating randomness through measurement?

A common reaction to quantum physics, among both early-twentieth-century physicists and modern students, is that we shouldbe able to get rid of randomness through accurate measurement. IfI say, for example, that it is meaningless to discuss the path of aphoton or an electron, one might suggest that we simply measurethe particle’s position and velocity many times in a row. This seriesof snapshots would amount to a description of its path.

A practical objection to this plan is that the process of measure-ment will have an effect on the thing we are trying to measure. Thismay not be of much concern, for example, when a traffic cop mea-sure’s your car’s motion with a radar gun, because the energy andmomentum of the radar pulses are insufficient to change the car’smotion significantly. But on the subatomic scale it is a very realproblem. Making a videotape through a microscope of an electronorbiting a nucleus is not just difficult, it is theoretically impossi-ble. The video camera makes pictures of things using light that hasbounced off them and come into the camera. If even a single photonof visible light was to bounce off of the electron we were trying tostudy, the electron’s recoil would be enough to change its behaviorsignificantly.

16.4.2 The Heisenberg uncertainty principle

This insight, that measurement changes the thing being mea-sured, is the kind of idea that clove-cigarette-smoking intellectualsoutside of the physical sciences like to claim they knew all along. Ifonly, they say, the physicists had made more of a habit of readingliterary journals, they could have saved a lot of work. The anthro-pologist Margaret Mead has recently been accused of inadvertentlyencouraging her teenaged Samoan informants to exaggerate the free-dom of youthful sexual experimentation in their society. If this isconsidered a damning critique of her work, it is because she couldhave done better: other anthropologists claim to have been able toeliminate the observer-as-participant problem and collect untainteddata.

The German physicist Werner Heisenberg, however, showed thatin quantum physics, any measuring technique runs into a brick wallwhen we try to improve its accuracy beyond a certain point. Heisen-berg showed that the limitation is a question of what there is to beknown, even in principle, about the system itself, not of the abilityor inability of a specific measuring device to ferret out informationthat is knowable but not previously hidden.

Suppose, for example, that we have constructed an electron in abox (quantum dot) setup in our laboratory, and we are able to adjust

Section 16.4 The uncertainty principle 379

the length L of the box as desired. All the standing wave patternspretty much fill the box, so our knowledge of the electron’s positionis of limited accuracy. If we write ∆x for the range of uncertaintyin our knowledge of its position, then ∆x is roughly the same as thelength of the box:

∆x ≈ L

If we wish to know its position more accurately, we can certainlysqueeze it into a smaller space by reducing L, but this has an unin-tended side-effect. A standing wave is really a superposition of twotraveling waves going in opposite directions. The equation p = h/λreally only gives the magnitude of the momentum vector, not itsdirection, so we should really interpret the wave as a 50/50 mixtureof a right-going wave with momentum p = h/λ and a left-going onewith momentum p = −h/λ. The uncertainty in our knowledge ofthe electron’s momentum is ∆p = 2h/λ, covering the range betweenthese two values. Even if we make sure the electron is in the groundstate, whose wavelength λ = 2L is the longest possible, we have anuncertainty in momentum of ∆p = h/L. In general, we find

∆p & h/L,

with equality for the ground state and inequality for the higher-energy states. Thus if we reduce L to improve our knowledge of theelectron’s position, we do so at the cost of knowing less about itsmomentum. This trade-off is neatly summarized by multiplying thetwo equations to give

∆p∆x & h.

Although we have derived this in the special case of a particle in abox, it is an example of a principle of more general validity:

The Heisenberg uncertainty principleIt is not possible, even in principle, to know the momentum and theposition of a particle simultaneously and with perfect accuracy. Theuncertainties in these two quantities are always such that ∆p∆x &h.

(To make this approximate inequality into an exact one, we wouldneed to pick a mathematical definition of our measure of uncertainty.If we use the standard deviation, p. 327, then it can be shown that∆p∆x ≥ h/4π.)

Note that although I encouraged you to think of this deriva-tion in terms of a specific real-world system, the quantum dot, noreference was ever made to any specific laboratory equipment or pro-cedures. The argument is simply that we cannot know the particle’sposition very accurately unless it has a very well defined position,it cannot have a very well defined position unless its wave-pattern


covers only a very small amount of space, and its wave-pattern can-not be thus compressed without giving it a short wavelength anda correspondingly uncertain momentum. The uncertainty princi-ple is therefore a restriction on how much there is to know abouta particle, not just on what we can know about it with a certaintechnique.

An estimate for electrons in atoms example 7. A typical energy for an electron in an atom is on the order of(1 volt)·e, which corresponds to a speed of about 1% of the speedof light. If a typical atom has a size on the order of 0.1 nm, howclose are the electrons to the limit imposed by the uncertaintyprinciple?

. If we assume the electron moves in all directions with equalprobability, the uncertainty in its momentum is roughly twice itstypical momentum. This is only an order-of-magnitude estimate,so we take ∆p to be the same as a typical momentum:

∆p∆x = ptypical∆x

= (melectron)(0.01c)(0.1× 10−9 m)

= 3× 10−34 J·s

This is on the same order of magnitude as Planck’s constant, soevidently the electron is “right up against the wall.” (The fact thatit is somewhat less than h is of no concern since this was only anestimate, and we have not stated the uncertainty principle in itsmost exact form.)

self-check B

If we were to apply the uncertainty principle to human-scale objects,what would be the significance of the small numerical value of Planck’sconstant? . Answer, p. 454


A Compare ∆p and ∆x for the two lowest energy levels of the one-dimensional particle in a box, and discuss how this relates to the uncer-tainty principle.

B On a graph of ∆p versus ∆x, sketch the regions that are allowed andforbidden by the Heisenberg uncertainty principle. Interpret the graph:Where does an atom lie on it? An elephant? Can either p or x be mea-sured with perfect accuracy if we don’t care about the other?

16.5 Decoherence and quantum computing16.5.1 Decoherence

Starting around 1970, physicists began to realize that ideas in-volving a loss of coherence, or “decoherence,” could help to explainsome things about quantum mechanics that had previously seemed

Section 16.5 Decoherence and quantum computing 381

mysterious. The classical notions of coherence and coherence lengthwere described in sec. 12.8, p. 273, and quantum-mechanical deco-herence was briefly introduced on p. 358.

One mystery was the fact that it is difficult to demonstrate waveinterference effects with large objects. This is partly because thewavelength λ = h/p = h/mv tends to be small for an object with alarge mass (example 1, p. 368). But even taking this into account,we do not seem to have much luck observing, for example, double-slit diffraction of very large molecules, even when we use slits withappropriate dimensions and a detector with a good enough angularresolution.

In the early days of quantum mechanics, people like Bohr andHeisenberg imagined that there was simply a clear division betweenthe macroscopic and microscopic worlds. Big things and small thingsjust had different rules: Newton’s laws in one case, quantum me-chanics in the other. But this is no longer a tenable position, becausewe now know that there is no limit on the distance scales over whichquantum-mechanical behavior can occur. For example, a commu-nication satellite carried out a demonstration in 2017 in which acoherence length of 1200 km was demonstrated using photons.1

k / A large molecule such asthe one in the Eibenberger ex-periment is represented by itswavepacket. As the moleculestarts out, its coherence length,shown by the arrows, is quitelong. As it flies to the right,it is bombarded by infrared pho-tons, which randomize its phase,causing its coherence length toshorten exponentially: by a fac-tor of two in the second panel,and by a further factor of two inthe final one. When the packetenters the double slit, its coher-ence length is on the same orderof magnitude as the slits’ spacingd , which will worsen but not en-tirely eliminate the observability ofinterference fringes. (This is onlya schematic representation, withthe wavepacket shown as beingmany orders of magnitude biggerthan its actual size in relation tothe vacuum chamber. Also, thereal experiment used a reflectinggrating, not a transmitting doubleslit.)

The insight about decoherence was the following. Consider themost massive material object that has so far been successfully dif-fracted through a grating, which was a molecule consisting of about810 atoms in an experiment by Eibenberger et al. in 2013.2 Whilethis molecule was propagating through the apparatus as a wave,the experimenters needed to keep it from simply being stopped bya collision with an air molecule. For this reason, they had to dothe experiment inside a vacuum chamber, with an extremely goodvacuum. But even then, the molecule was being bombarded by

1Yin et al., arxiv.org/abs/1707.013392arxiv.org/abs/1310.8343


photons of infrared light emitted from the walls of the chamber. Theeffect of this bombardment is to disrupt the molecule’s wavefunctionand reduce its coherence length (p. 274).

This decoherence effect was the reason that the experiment waslimited to molecules of the size they used. Even though the moleculestook only about 400 nanoseconds to fly through the apparatus, therewas a significant amount of decoherence. A larger molecule wouldhave been a bigger target for photons and would have undergonedecoherence more quickly, making interference unobservable.

16.5.2 Brains, classical computers, and quantum computers

Decoherence helps to explain why it seems as though measure-ments of quantum systems only produce one result, chosen at ran-dom. Suppose you do the Schrodinger’s cat experiment (p. 358),after which, in the notation of sec. 16.1.4, p. 372, your brain is inthe superposition of states

c| L〉+ c′| D〉.

Although absolute phases and normalizations are not physically ob-servable, relative phases are. So in principle, you could tell that yourbrain was in a superposed state, because, for example, the state

| L〉+ | D〉

is different from| L〉 − | D〉,

which has a different relative phase. This really does work withelectrons, since | 〉+ | 〉 gives a different interference pattern than| 〉−| 〉. But because your brain is a macroscopic object exposed tolots of subatomic collisions, decoherence affects it on an incrediblyshort timescale. It randomizes the relative phases of the L and

D parts, so that we can never observe any interference patternbetween them.

For these reasons, the brain is a classical computer. But there isa completely different type of computing device, the quantum com-puter. Rather than bits, a quantum computer operates on qubits. Aqubit, or quantum bit, can exist in a mixture of states like c|0〉+c′|1〉.You might think that a quantum computer would be worse than aclassical one, since it would probably act like a classical computerwith some randomness thrown in, making it unreliable. But the ran-domness of quantum mechanics is only apparent. It crops up onlywhen decoherence happens, as often occurs in measurements. Aquantum computer has to be isolated cleanly from its environment,so that there will be no decoherence. In such an environment, therereally is an observable difference between |0〉+ |1〉 and |0〉−|1〉. Fur-thermore, the different qubits in a quantum computer can becomeentangled in complicated ways. A chunk of memory in a quan-tum computer is therefore a much richer structure than a same-size

Section 16.5 Decoherence and quantum computing 383

chunk of classical bits. This leads to a concept of quantum informa-tion, which is a higher grade of information than classical informa-tion. That is, we can losslessly convert from classical informationto quantum information, but not the other way around.

Quantum computers can efficiently solve certain types of prob-lems that a classical computer finds intractable, such as factoringlarge numbers into their prime factors. For this reason, the crypto-graphic infrastructure that lets you safely transmit your credit cardnumber to amazon.com will become completely broken if large-scalequantum computing ever becomes a reality. Luckily for our privacy,quantum computers are hard to build, and the biggest factoringproblems they can currently solve are examples like 15 = 3× 5.

16.6 A crude model of the hydrogen atom16.6.1 Modeling

Probably the most important reason for wanting to understandquantum physics is in order to understand the atom. As tiny asthey may be, atoms can be pretty complicated. Let’s think aboutways to simplify them — even at the risk of oversimplifying. Startwith hydrogen, because it’s the simplest of all atoms, consistingof only one electron and one proton. Because the proton is 1800times more massive than the electron, it’s a good approximationto imagine that it stays at rest while the electron moves around,influenced by the proton’s static electric field. This is good: we’vereduced our problem to the discussion of just one particle.

Classically, the electron moves around in the proton’s electricfield, speeding up, slowing down, and never straying farther thana certain distance r because it doesn’t have enough energy. As itdoes all these things, its momentum changes, so if we mix in a littlequantum mechanics, p = ~k then tells us that its wavenumber andwavelength change. So its wavefunction can’t be a nice, simple sinewave with a single well-defined wavelength. We expect it to lookmore like this: . When the electron is close to the proton,its electrical energy is low, so its kinetic energy is high, and itswavelength is short. When it’s farther out, its wavelength is longer.In chapter 17 we’ll talk in more detail about how to handle this sortof thing.

But for now we’re going to do something much more crude, whichis to approximate the electron’s wavefunction as a sine wave like thisone . We take the nodes and antinodes and make them niceand regular: all evenly spaced and equal in amplitude, and fillingthe whole space available to the electron based on the amount ofenergy it has.

Also, by drawing the wavefunctions on a piece of paper in thisway, I’ve snuck yet another crude approximation past you. We’re


l / The energy levels of a particlein a box, contrasted with those ofthe hydrogen atom.

treating the hydrogen atom as one-dimensional, Ψ(x) rather thanΨ(x, y, z). The wave fills in the line segment −r ≤ x ≤ r, where ris the maximum distance. The drawings also implicitly imply thatΨ is a function whose outputs are real rather than complex, but wecan get away with this because this is a standing wave.

16.6.2 Estimation of the energy levels

Let n be the same quantum number defined in problem 12, p. 59,and in sec. 16.1.3, p. 372, i.e., it’s the number of antinodes or thenumber of half-wavelengths. In the examples drawn above, n = 9.The ground state has n = 1. Because n is the number of half-wavelengths, the wavelength is 4r/n, but because this whole calcu-lation is so crude, we’re going to throw away factors like the 4, andjust say that

λ ∼ r/n.

If r had the same value for every standing-wave pattern, thenwe’d essentially be solving the particle-in-a-box problem, but thereis a fundamental difference here, which is that we don’t have a boxof a fixed size with impenetrable walls. If we did, then it would beimpossible to separate the electron from the proton, whereas in reallife we know that this can be done by putting in a certain amountof energy to ionize the atom. The force keeping the electron boundisn’t an infinite force encountered when it bounces off of a wall,it’s the attractive electrical force from the nucleus. If we put moreenergy into the electron, it’s like throwing a ball upward with ahigher energy — it will get farther out before coming back down.

Figure l shows how we expect this to turn out. In the hydrogenatom, we expect r to increase as we go to states of higher energy.This tends to keep the wavelengths of the high energy states fromgetting too short, reducing their kinetic energy. The closer andcloser crowding of the energy levels in hydrogen also makes sensebecause we know that there is a certain energy that would be enoughto make the electron escape completely, and therefore the sequenceof bound states cannot extend above that energy.

Now let’s crunch some numbers. When the electron is at themaximum classically allowed distance r from the proton, it has zerokinetic energy. Thus when the electron is at distance r, its energyis purely electrical,

[1] E = −ke2

r,

where k is the Coulomb constant (not the wavenumber). The zero-level of the electrical energy scale is chosen to be the energy of anelectron and a proton that are infinitely far apart. With this choice,negative energies correspond to bound states and positive energiesto unbound ones.

Section 16.6 A crude model of the hydrogen atom 385

Finally we assume that the typical kinetic energy of the electronis on the same order of magnitude as the absolute value of its totalenergy. (This is true to within a factor of two for a typical classicalsystem like a planet in a circular orbit around the sun.) We thenhave

absolute value of total energy[2]

=ke2

r∼ K= p2/2m

= (h/λ)2/2m

∼ h2n2/2mr2

We now solve the equation ke2/r ∼ h2n2/2mr2 for r and throwaway numerical factors we can’t hope to have gotten right, yielding

[3] r ∼ h2n2

mke2.

Plugging n = 1 into this equation gives r = 2 nm, which isindeed on the right order of magnitude compared to the observedsizes of atoms. Finally we combine equations [3] and [1] to find

E ∼ −mk2e4

h2· 1

n2[result of crude treatment]

which turns out to be correct except for the numerical factors wenever aimed to find. The exact result is

E = −mk2e4

2~2· 1

n2, [exact result]

which is different by a factor of 2π2. It might seem incredible thatsuch a crude approximation would provide a result so close to thecorrect one. However, the factor in front is constrained by the re-quirement that the result have units of energy. All of the exponentsin this factor have to be as they are just based on units. For theground state, n = 1, we will show in example 5, p. 401, that theexact result is correct.

16.6.3 Comparison with experiment

The experimental technique for measuring the energy levels ofan atom accurately is spectroscopy: the study of the spectrum oflight emitted (or absorbed) by the atom. Only photons with certainenergies can be emitted or absorbed by a hydrogen atom, for ex-ample, since the amount of energy gained or lost by the atom mustequal the difference in energy between the atom’s initial and finalstates. Spectroscopy had become a highly developed art several


decades before Einstein even proposed the photon, and the Swissspectroscopist Johann Balmer determined in 1885 that there was asimple equation that gave all the wavelengths emitted by hydrogen.In modern terms, we think of the photon wavelengths merely as in-direct evidence about the underlying energy levels of the atom, andwe rework Balmer’s result into an equation for these atomic energylevels:

En = − An2

,

where A = 2.2× 10−18 J. About 30 years later, the constant A wasexplained to be the constant factor in the equation at the end of theprevious section.


A Sketch graphs of r and E versus n for the hydrogen atom, and com-pare with analogous graphs for the one-dimensional particle in a box.

B States of hydrogen with n greater than about 10 are never observedin the sun. Why might this be?

Section 16.6 A crude model of the hydrogen atom 387

Notes for chapter 16☡366 Failure of the classical H2

molecule

We use classical physics to make an esti-mate of the electrical binding energy of the H2

molecule. It is much too small compared to re-ality, and much to small to make the moleculehold together.

To give classical physics a fair chance, weshould not use any classical model with mov-ing parts, such as the planetary model, be-cause such models are always unstable due totheir ability to radiate away energy as electro-magnetic waves. We also want a model thatis fairly easy to calculate with. Let’s use amodel in which the proton is a point chargefixed at the center of a rigid electron cloudwhose density of charge is e−r, where r isin units of some basic distance scale (about0.05 nm in SI units), and the charge is inunits of the fundamental charge. This is ac-tually somewhat realistic as a time-averagedpicture of the charge of a hydrogen atom, andit turns out that the electrical potential energywe need has been calculated by A. Mukherji,arxiv.org/abs/0903.3304. We assume thatthe two clouds simply interpenetrate one an-other. The resulting electrical energy, in-cluding electron-electron, proton-proton, andelectron-proton interactions, turns out to be

U = e−r(

1

r+

5

16− 3

16r − 1

48r2

).

This function has a very shallow minimum atr ≈ 3.8 (which is about triple the real-worldvalue for the bond length), attaining a mini-mum value of about −0.01 in our energy units,which converts to −0.3 eV. The negative signdoes indicate an equilibrium, but the bind-ing energy is far too small compared to thereal-world value of −4.7 eV. Not only is thecalculated binding energy an order of magni-tude too small compared to experiment, butit is of a size that would probably cause themolecule to break apart spontaneously due tothe Heisenberg uncertainty principle.


Problem 2.

ProblemsKey√

A computerized answer check is available online.? A difficult problem.

1 In a television, suppose the electrons are accelerated from restthrough a voltage difference of 104 V. What is their final wavelength?√

2 The figures show the wavefunction of an electron as a functionof position. Which one could represent an electron speeding up asit moves to the right? Explain.

3 Use the Heisenberg uncertainty principle to estimate the mini-mum velocity of a proton or neutron in a 208Pb nucleus, which has adiameter of about 13 fm (1 fm=10−15 m). Assume that the speed isnonrelativistic, and then check at the end whether this assumptionwas warranted.

√

4 Suppose that an electron, in one dimension, is confined to acertain region of space so that its wavefunction is given by

Ψ =

0 if x < 0

A sin(2πx/L) if 0 ≤ x ≤ L0 if x > L

Determine the constant A from normalization.√

5 Find the energy of a nonrelativistic particle in a one-dimensionalbox of length L, expressing your result in terms of L, the particle’smass m, the number of peaks and valleys n in the wavefunction, andfundamental constants.

√

6 A free electron that contributes to the current in an ohmicmaterial typically has a speed of 105 m/s (much greater than thedrift velocity).(a) Estimate its de Broglie wavelength, in nm.

√

(b) If a computer memory chip contains 108 electric circuits in a 1cm2 area, estimate the linear size, in nm, of one such circuit.

√

(c) Based on your answers from parts a and b, does an electricalengineer designing such a chip need to worry about wave effectssuch as diffraction?(d) Estimate the maximum number of electric circuits that can fit ona 1 cm2 computer chip before quantum-mechanical effects becomeimportant.

Problems 389

Problem 7.

7 The figure shows eight of the possible ways in which an electronin a hydrogen atom could drop from a higher energy state to astate of lower energy, releasing the difference in energy as a photon.Of these eight transitions, only D, E, and F produce photons withwavelengths in the visible spectrum.(a) Which of the visible transitions would be closest to the violetend of the spectrum, and which would be closest to the red end?Explain.(b) In what part of the electromagnetic spectrum would the photonsfrom transitions A, B, and C lie? What about G and H? Explain.(c) Is there an upper limit to the wavelengths that could be emittedby a hydrogen atom going from one bound state to another boundstate? Is there a lower limit? Explain.

8 Find an equation for the wavelength of the photon emittedwhen the electron in a hydrogen atom makes a transition from en-ergy level n1 to level n2.

√

9 Hydrogen is the only element whose energy levels can beexpressed exactly in an equation. Calculate the ratio λE/λF of thewavelengths of the transitions labeled E and F in problem 7 onp. 390. Express your answer as an exact fraction, not a decimalapproximation. In an experiment in which atomic wavelengths arebeing measured, this ratio provides a natural, stringent check on theprecision of the results.

√

10 Assume that the kinetic energy of an electron in the n = 1state of a hydrogen atom is on the same order of magnitude as theabsolute value of its total energy, and estimate a typical speed atwhich it would be moving. (It cannot really have a single, definitespeed, because its kinetic and interaction energy trade off at differentdistances from the proton, but this is just a rough estimate of atypical speed.) Based on this speed, were we justified in assumingthat the electron could be described nonrelativistically?

11 Use physical reasoning to explain how the equation for theenergy levels of hydrogen,

En = −mk2e4

2~2· 1

n2,

should be generalized to the case of an atom with atomic number Zthat has had all its electrons removed except for one.


Problem 14.

12 A muon is a subatomic particle that acts exactly like anelectron except that its mass is 207 times greater. Muons can becreated by cosmic rays, and it can happen that one of an atom’selectrons is displaced by a muon, forming a muonic atom. If thishappens to a hydrogen atom, the resulting system consists simplyof a proton plus a muon.(a) How would the size of a muonic hydrogen atom in its groundstate compare with the size of the normal atom?(b) If you were searching for muonic atoms in the sun or in theearth’s atmosphere by spectroscopy, in what part of the electromag-netic spectrum would you expect to find the absorption lines?

13 Before the quantum theory, experimentalists noted that inmany cases, they would find three lines in the spectrum of the sameatom that satisfied the following mysterious rule: 1/λ1 = 1/λ2 +1/λ3. Explain why this would occur. Do not use reasoning thatonly works for hydrogen — such combinations occur in the spectraof all elements. [Hint: Restate the equation in terms of the energiesof photons.]

14 In sec. 16.6.2, p. 385, we derived an approximate expressionfor the energies of states in hydrogen. As input to the calculation,we used the the proportionality U ∝ r−1, which is a characteristicof the electrical interaction. The result for the energy of the nthstanding wave pattern was En ∝ n−2.

There are other systems of physical interest in which we have U ∝ rkfor values of k besides −1. Problem 4 discusses the ground state ofthe harmonic oscillator, with k = 2 (and a positive constant of pro-portionality). In particle physics, systems called charmonium andbottomonium are made out of pairs of subatomic particles calledquarks, which interact according to k = 1, i.e., a force that is in-dependent of distance. (Here we have a positive constant of pro-portionality, and r > 0 by definition. The motion turns out notto be too relativistic, so the Schrodinger equation is a reasonableapproximation.) The figure shows actual energy levels for thesethree systems, drawn with different energy scales so that they canall be shown side by side. The sequence of energies in hydrogenapproaches a limit, which is the energy required to ionize the atom.In charmonium, only the first three levels are known.3

Generalize the method used for k = −1 to any value of k, and findthe exponent j in the resulting proportionality En ∝ nj . Comparethe theoretical calculation with the behavior of the actual energiesshown in the figure. Comment on the limit k →∞.

√

3See Barnes et al., “The XYZs of Charmonium at BES,” arxiv.org/abs/

hep-ph/0608103. To avoid complication, the levels shown are only those in thegroup known for historical reasons as the Ψ and J/Ψ.

Problems 391

15 The electron, proton, and neutron were discovered, respec-tively, in 1897, 1919, and 1932. The neutron was late to the party,and some physicists felt that it was unnecessary to consider it asfundamental. Maybe it could be explained as simply a proton withan electron trapped inside it. The charges would cancel out, givingthe composite particle the correct neutral charge, and the massesat least approximately made sense (a neutron is heavier than a pro-ton). (a) Given that the diameter of a proton is on the order of10−15 m, use the Heisenberg uncertainty principle to estimate thetrapped electron’s minimum momentum.

√

(b) Find the electron’s minimum kinetic energy.√

(c) Show via E = mc2 that the proposed explanation may have aproblem, because the contribution to the neutron’s mass from theelectron’s kinetic energy would be comparable to the neutron’s en-tire mass.


a / An electron in a gentleelectric field gradually shortensits wavelength as it gains energy.(As discussed on p. 371, it isactually not quite correct to graphthe wavefunction of an electronas a real number unless it is astanding wave, which isn’t thecase here.)

Chapter 17

The Schrodinger equation

17.1 Electrons in electric fields17.1.1 Defining a wavelength when the wavelength is varying

So far the only electron wave patterns we’ve considered havebeen simple sine waves, but whenever an electron finds itself in anelectric field, it must have a more complicated wave pattern. Let’sconsider the example of an electron being accelerated by the elec-tron gun at the back of a TV tube. Newton’s laws are not useful,because they implicitly assume that the path taken by the particle isa meaningful concept. Conservation of energy is still valid in quan-tum physics, however. In terms of energy, the electron is movingfrom a region of low voltage into a region of higher voltage. Sinceits charge is negative, it loses electrical energy by moving to a highervoltage, so its kinetic energy increases. As its electrical energy goesdown, its kinetic energy goes up by an equal amount, keeping thetotal energy constant. Increasing kinetic energy implies a growingmomentum, and therefore a shortening wavelength, a.

The wavefunction as a whole does not have a single well-definedwavelength, but the wave changes so gradually that if you only lookat a small part of it you can still pick out a wavelength and relateit to the momentum and energy. (The picture actually exagger-ates by many orders of magnitude the rate at which the wavelengthchanges.)

But what if the electric field was stronger? The electric field inan old-fashioned vacuum tube TV screen is only ∼ 105 N/C, but theelectric field within an atom is more like 1012 N/C. In figure b, thewavelength changes so rapidly that there is nothing that looks likea sine wave at all. We could get a rough idea of the wavelength in agiven region by measuring the distance between two peaks, but thatwould only be a rough approximation. Suppose we want to knowthe wavelength at point P . The trick is to construct a sine wave, likethe one shown with the dashed line, which matches the curvature ofthe actual wavefunction as closely as possible near P . The sine wavethat matches as well as possible is called the “osculating” curve, froma Latin word meaning “to kiss.” The wavelength of the osculatingcurve is the wavelength that will relate correctly to conservation ofenergy.

393

c / The wavefunction’s tailsgo where classical physics saysthey shouldn’t.

b / A typical wavefunction of an electron in an atom (heavy curve)and the osculating sine wave (dashed curve) that matches its curvatureat point P.

17.1.2 Tunneling

We implicitly assumed that the particle-in-a-box wavefunctionwould cut off abruptly at the sides of the box, c/1, but that would beunphysical. A kink has infinite curvature, and curvature is relatedto energy, so it can’t be infinite. A physically realistic wavefunctionmust always “tail off” gradually, c/2. In classical physics, a particlecan never enter a region in which its interaction energy U would begreater than the amount of energy it has available. But in quantumphysics the wavefunction will always have a tail that reaches intothe classically forbidden region. If it was not for this effect, calledtunneling, the fusion reactions that power the sun would not occurdue to the high electrical energy nuclei need in order to get closetogether! Tunneling is discussed in more detail on p. 397.

17.2 The Schrodinger equationIn section 17.1 we were able to apply conservation of energy toan electron’s wavefunction, but only by using the clumsy graphicaltechnique of osculating sine waves as a measure of the wave’s cur-vature. You have learned a more convenient measure of curvaturein calculus: the second derivative. To relate the two approaches, wetake the second derivative of a sine wave:

d2

dx2sin kx =

d

dx(k cos kx)

= −k2 sin kx.

Taking the second derivative gives us back the same function,but with a minus sign and a constant out in front that is relatedto the wavelength. We can thus relate the second derivative to theosculating wavelength:

[1]d2 Ψ

dx2= −k2Ψ,

where k = 2π/λ. This could be solved for k or λ in terms of Ψ, butit will turn out below to be more convenient to leave it in this form.

394 Chapter 17 The Schrodinger equation

Applying this to conservation of energy, we have

E = K + U

=p2

2m+ U

=(~k)2

2m+ U

[2]

We can simplify our algebra by multiplying both sides of equation[2] by Ψ to make it look more like equation [1]:

E ·Ψ =(~k)2

2mΨ + U ·Ψ,

which leads to the important result known as the Schrodingerequation:

E ·Ψ = − ~2

2m

d2 Ψ

dx2+ U ·Ψ

(Actually this is a simplified version of the Schrodinger equation,applying only to standing waves in one dimension.) Physically itis a statement of conservation of energy. The total energy E mustbe constant, so the equation tells us that a change in interactionenergy U must be accompanied by a change in the curvature ofthe wavefunction. This change in curvature relates to a changein wavelength, which corresponds to a change in momentum andkinetic energy.

self-check AConsidering the assumptions that were made in deriving the Schrodingerequation, would it be correct to apply it to a photon? To an electron mov-ing at relativistic speeds? . Answer, p.454

Usually we know right off the bat how U depends on x, so thebasic mathematical problem of quantum physics is to find a functionΨ(x) that satisfies the Schrodinger equation for a given interaction-energy function U(x). An equation, such as the Schrodinger equa-tion, that specifies a relationship between a function and its deriva-tives is known as a differential equation.

Section 17.2 The Schrodinger equation 395

Discussion question

A The figure shows a skateboarder tipping over into a swimmingpool with zero initial kinetic energy. There is no friction, the corners aresmooth enough to allow the skater to pass over the smoothly, and thevertical distances are small enough so that negligible time is required forthe vertical parts of the motion. The pool is divided into a deep end and ashallow end. Their widths are equal. The deep end is four times deeper.(1) Classically, compare the skater’s velocity in the left and right regions,and infer the probability of finding the skater in either of the two halves ifan observer peeks at a random moment. (2) Quantum-mechanically, thiscould be a one-dimensional model of an electron shared between twoatoms in a diatomic molecule. Compare the electron’s kinetic energies,momenta, and wavelengths in the two sides. For simplicity, let’s assumethat there is no tunneling into the classically forbidden regions. What isthe simplest standing-wave pattern that you can draw, and what are theprobabilities of finding the electron in one side or the other? Does thisobey the correspondence principle?

17.3 Solutions when U is constant; tunnelingThe detailed study of the solution of the Schrodinger equation isbeyond the scope of this book, but we can gain some importantinsights by considering the easiest version of the Schrodinger equa-tion, in which the interaction energy U is constant. We can thenrearrange the Schrodinger equation as follows:

d2 Ψ

dx2=

2m(U − E)

~2Ψ,

which boils down to

d2 Ψ

dx2= aΨ,


d / Tunneling through a bar-rier. (As discussed on p. 371, it isactually not quite correct to graphthe wavefunction of an electronas a real number unless it is astanding wave, which isn’t thecase here.)

where, according to our assumptions, a is independent of x. We needto find a function whose second derivative is the same as the originalfunction except for a multiplicative constant. The only functionswith this property are sine waves and exponentials:

d2

dx2[ q sin(rx+ s) ] = −qr2 sin(rx+ s)

d2

dx2

[qerx+s

]= qr2erx+s

The sine wave gives negative values of a, a = −r2, and theexponential gives positive ones, a = r2. The former applies to theclassically allowed region with U < E.

This leads us to a quantitative calculation of the tunneling effectdiscussed briefly in sec. 17.1.2, p. 394. The wavefunction evidentlytails off exponentially in the classically forbidden region. Suppose,as shown in figure d, a wave-particle traveling to the right encountersa barrier that it is classically forbidden to enter. Although the formof the Schrodinger equation we’re using technically does not applyto traveling waves (because it makes no reference to time), it turnsout that we can still use it to make a reasonable calculation of theprobability that the particle will make it through the barrier. If welet the barrier’s width be w, then the ratio of the wavefunction onthe left side of the barrier to the wavefunction on the right is

qerx+s

qer(x+w)+s= e−rw.

Probabilities are proportional to the squares of wavefunctions, sothe probability of making it through the barrier is

P = e−2rw

= exp

(−2w

~√

2m(U − E)

).

self-check BIf we were to apply this equation to find the probability that a person canwalk through a wall, what would the small value of Planck’s constantimply? . Answer, p. 454

Tunneling in alpha decay example 1Naively, we would expect alpha decay to be a very fast process.The typical speeds of neutrons and protons inside a nucleus areextremely high (see problem 3). If we imagine an alpha particlecoalescing out of neutrons and protons inside the nucleus, thenat the typical speeds we’re talking about, it takes a ridiculouslysmall amount of time for them to reach the surface and try toescape. Clattering back and forth inside the nucleus, we could

Section 17.3 Solutions when U is constant; tunneling 397

e / The electrical, nuclear,and total interaction energies foran alpha particle escaping from anucleus.

imagine them making a vast number of these “escape attempts”every second.

Consider figure e, however, which shows the interaction energyfor an alpha particle escaping from a nucleus. The electrical en-ergy is kq1q2/r when the alpha is outside the nucleus, while itsvariation inside the nucleus has the shape of a parabola, as aconsequence of the shell theorem. The nuclear energy is con-stant when the alpha is inside the nucleus, because the forcesfrom all the neighboring neutrons and protons cancel out; it risessharply near the surface, and flattens out to zero over a distanceof ∼ 1 fm, which is the maximum distance scale at which thestrong force can operate. There is a classically forbidden regionimmediately outside the nucleus, so the alpha particle can onlyescape by quantum mechanical tunneling. (It’s true, but some-what counterintuitive, that a repulsive electrical force can make itmore difficult for the alpha to get out.)

In reality, alpha-decay half-lives are often extremely long — some-times billions of years — because the tunneling probability is sosmall. Although the shape of the barrier is not a rectangle, theequation for the tunneling probability on page 397 can still beused as a rough guide to our thinking. Essentially the tunnelingprobability is so small because U − E is fairly big, typically about30 MeV at the peak of the barrier.

A marble tunneling out of a box example 2On p. 334, I introduced Gell-Mann’s whimsically named totalitar-ian principle, that any process not forbidden by a conservationlaw will happen with some nonzero probability or rate. One of theexamples I used there was a marble locked in a box. As a sillyexample, let’s make a crude numerical estimate of the marble’sprobability of tunneling out through the box. As in self-check Bon p. 397, we expect based on the correspondence principle thatthis probability will be very small.

A typical marble has a radius on the order of a centimeter, andlet’s say the box has about that thickness. We’ll say that the mar-ble has a mass of 5 g and contains something like n = 1023 atoms.The atomic energy scale is roughly 1 eV, so that if the marble isto be inside the wall of the box, its electrical energy U probablyhas to be something on the order of n multiplied by 1 eV, givingU ∼ 104 J, or about an order of magnitude greater than the ki-netic energy of a bullet. We could of course give the marble thatmuch energy, and then it would blast through the box easily, butthe point here is to estimate its probability of just tunneling out,so we’ll say that the kinetic energy is negligible, and thereforeU − E is basically the same as U. Plugging in numbers, we findP ∼ e−1033

. When you have to express a small number usinga stack of exponents like this, you know that it’s very small. For


f / β+ and β− spectra of 64Cu.

example, this number is many, many orders of magnitude smallerthan 10−1000.

Beta decay: a push or pull on the way out the door example 3The nucleus 64Cu undergoes β+ and β− decay with similar prob-abilities and energies. Each of these decays releases a fixedamount of energy Q due to the difference in mass between theparent nucleus and the decay products. This energy is sharedrandomly between the beta and the neutrino. In experiments, thebeta’s energy is easily measured, while the neutrino flies off with-out interacting. Figure f shows the energy spectrum of the β+ andβ− in these decays.1 There is a relatively high probability for thebeta and neutrino each to carry off roughly half the kinetic energy,the reason being identical to the kind of phase-space argumentdiscussed in sec. 9.4, p. 205. Therefore in each case we get abell-shaped curve stretching from 0 up to the energy Q, with Qbeing slightly different in the two cases.

So we expect the two bell curves to look almost the same exceptfor a slight rescaling of the horizontal axis. Yes — but we also seemarkedly different behavior at low energies. At very low energies,there is almost no chance to see a β+ with very low energy, butquite a high probability for a β−.

We could try to explain this difference in terms of the release ofelectrical energy. The β+ is repelled by the nucleus, so it getsan extra push on the way out the door. A β− should be heldback as it exits, and so should lose some energy. The bell curvesshould be shifted up and down in energy relative to one another,as observed.

But if we try to estimate this energy shift using classical physics,we come out with a wildly incorrect answer. This would be aprocess in which the beta and neutrino are released in a point-like event inside the nucleus. The radius r of the 64Cu nucleusis on the order of 4 fm (1 fm = 10−15 m). Therefore the en-ergy lost or gained by the β+ or β− on the way out would beU ∼ kZe2/r ∼ 10 MeV. The actual shift is much smaller.

To understand what’s really going on, we need quantum mechan-ics. A beta in the observed energy range has a wavelength ofabout 2000 fm, which is hundreds of times greater than the sizeof the nucleus. Therefore the beta cannot be much better local-ized than that when it is emitted. This means that we should reallyuse something more like r ∼ 500 fm (a quarter of a wavelength) inour calculation of the electrical energy. This gives U ∼ 0.08 MeV,which is about the right order of magnitude compared to obser-vation.

A byproduct of this analysis is that a β+ is always emitted within

1Redrawn from Cook and Langer, 1948.

Section 17.3 Solutions when U is constant; tunneling 399

the classically forbidden region, and then has to tunnel out throughthe barrier. As in example 1, we have the counterintuitive factabout quantum mechanics that a repulsive force can hinder theescape of a particle.

17.4 Three dimensionsFor simplicity, we’ve been considering the Schrodinger equation inone dimension, so that Ψ is only a function of x, and has units ofm−1/2 rather than m−3/2. Since the Schrodinger equation is a state-ment of conservation of energy, and energy is a scalar, the general-ization to three dimensions isn’t particularly complicated. The totalenergy term E ·Ψ and the interaction energy term U ·Ψ involve noth-ing but scalars, and don’t need to be changed at all. In the kineticenergy term, however, we’re essentially basing our computation ofthe kinetic energy on the squared magnitude of the momentum, p2

x,and in three dimensions this would clearly have to be generalized top2x+p2

y+p2z. The obvious way to achieve this is to replace the second

derivative d2 Ψ/ dx2 with the sum ∂2Ψ/∂x2 + ∂2Ψ/∂y2 + ∂2Ψ/∂z2.In other words, we replace the second derivative with the Laplacian(sec. 2.7.4, p. 53),

∇2 =∂2

∂x2+

∂2

∂y2+

∂2

∂z2,

just as we did for waves such as sound. We recall from sec. 2.7.4that:

• The partial derivative symbol ∂, introduced on page 40, in-dicates that when differentiating with respect to a particularvariable, the other variables are to be considered as constants.

• Like the second derivative, the Laplacian is essentially a mea-sure of curvature.

• As shown in figure ag, p. 55, we can also think of the Laplacianas a measure of how much the value of a function at a certainpoint differs from the average of its value on nearby points.

A classically allowed region with constant U example 4In a classically allowed region with constant U, we expect thesolutions to the Schrodinger equation to be sine waves. A sinewave in three dimensions has the form

Ψ = sin(kxx + kyy + kzz

).

When we compute ∂2Ψ/∂x2, double differentiation of sin gives− sin, and the chain rule brings out a factor of k2

x . Applying allthree second derivative operators, we get

∇2Ψ =(−k2

x − k2y − k2

z

)sin(kxx + kyy + kzz

)= −

(k2

x + k2y + k2

z

)Ψ.


15.3wave-particle duality - light and matter · the correct interpretation of wave-particle duality...

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