review of early …antum mechanics formalization of …antum

Review of Early�antum MechanicsFormalization of �antum Mechanics

Physics 280�antum Mechanics LectureSpring 2015

Dr. Jones1

1Department of PhysicsDrexel University

August 3, 2016

Dr. Jones Physics 280�antum Mechanics Lecture

Objectives

Review Early�antum Mechanics

Schrodinger’s Wave Equation

Heisenberg Uncertainty

Strange Consquences

Strange Contradictions

Objectives

Strange Consquences

Objectives

Strange Consquences

Objectives

Strange Consquences

Objectives

Strange Consquences

So far..

Electrical charges are quantized in the form of electrons andprotons, nothing is smaller than ±e.

Classical Physics predicts that the intensity of radiation fromblackbody-like sources goes to infinity with decreasingwavelength, which is experimentally shown to be untrue and isimpossible anyway.

Planck uses a mathematical trick for the first time ever–energyquantization: E = hf to ”fix” the equations.

Einstein takes this result at face value and shows that photonsare quantized in the same way, explaining the photoelectrice�ect.

Bohr adapts the idea of quantization to ”explain” the Rydbergequation and why the emission/absorption spectrum isquantized. He quantizes angular momentum.

So far..

Electrical charges are quantized in the form of electrons andprotons, nothing is smaller than ±e.Classical Physics predicts that the intensity of radiation fromblackbody-like sources goes to infinity with decreasingwavelength, which is experimentally shown to be untrue and isimpossible anyway.

So far..

A new truthEnergy is quantized.

The Ultraviolet Catastrophe

Thermal radiation of abody depends upon thetemperature andwavelength.

We use blackbodyradiation as a proxy for themore complex thermalradiation.

Wien’s law can explainthe shi� of the peakle�wards withtemperature.

λmaxT =(0.2898× 10−2

Why we can approximate some glowing bodies asblackbody radiation

Stefan’s law predicts thepower emi�ed byblackbody radiation:

P = σAeT 4

Combining these lawsyields the Rayleigh-Jeansprediction:

I(λ, T ) =2πckBTλ4

Classical physics couldtake us no further, and itwas wrong.

P = σAeT 4

The Ultraviolet Catastrophe: Resolved Reluctantly

Planck’s leap: Supposing that blackbodyradiation was emi�ed by ’resonators’, theseresonators could only have energy in discreetquantity:

En = nhf

n is a positive integer, the quantum number, fis frequency of vibration, h is a new constant hefit to experimental results:

h = 6.626× 10−34J · s

I(λ, T ) =2πhc2

hcλkBT − 1

En = nhf

h = 6.626× 10−34J · s

I(λ, T ) =2πhc2

hcλkBT − 1

En = nhf

h = 6.626× 10−34J · s

I(λ, T ) =2πhc2

hcλkBT − 1

)Dr. Jones Physics 280�antum Mechanics Lecture

Atomic mystery

�antization of energy solved the Blackbody Radiationproblem and the Photoelectric E�ect. It solved the sca�eringproblem (Compton e�ect) and although nobody could quitemake sense of it,

Another mystery dominated atomic physics–nobody couldexplain the spectra of gasses.

We might expect a continuous distribution of wavelengths, butinstead we find discrete line spectrum called the emissionspectrum.

Passing white light through gasses result in discrete missingwave- lengths, this is called the absorption spectrum.

Atomic mystery

Example emission (hydrogen, mercury, and neon) andabsorption spectrum for hydrogen:

19th century physicists can’t explain these spectra.

Atomic mystery

Example emission (hydrogen, mercury, and neon) andabsorption spectrum for hydrogen:

19th century physicists can’t explain these spectra.

Atomic mystery

Balmer found an empirical equation that correctly predictedthe wavelengths of four of the visible emission lines; Rydbergexpanded this equation to find all emission lines:

(122− 1

), n = 3, 4, 5, · · ·

RH is a constant called the Rydberg constant and is equal to1.0973732 ×107m−1.The shortest wavelength is found when n→∞ is called theseries limit with wavelength 364.6 nm (ultraviolet).

Atomic mystery

(122− 1

), n = 3, 4, 5, · · ·

RH is a constant called the Rydberg constant and is equal to1.0973732 ×107m−1.

The shortest wavelength is found when n→∞ is called theseries limit with wavelength 364.6 nm (ultraviolet).

Atomic mystery

(122− 1

), n = 3, 4, 5, · · ·

RH is a constant called the Rydberg constant and is equal to1.0973732 ×107m−1.The shortest wavelength is found when n→∞ is called theseries limit with wavelength 364.6 nm (ultraviolet).

Atomic mystery

Other physicists started experimenting with these numbers andfound similar equations that described other lines in the spectrum:

Lyman series:1λ

(1− 1

), n = 2, 3, 4, · · ·

Paschen series:1λ

(132− 1

), n = 4, 5, 6, · · ·

Bracket series:1λ

(142− 1

), n = 5, 6, 7, · · ·

It all works very well, but nobody knows why.

Atomic mystery

�antum solution

A physicists named Niels Bohr saw that energy quantization hadsolved the blackbody and photoelectric problem, and he wondered ifit could solve the mystery of the atomic spectrum lines as well. Hereis a basic outline of his reasoning.

�antum solution

Assume the classical model of the electron orbiting the nucleusof the hydrogen atom under electrical forces. In this case thetotal energy is

E = K + U =12mev2 − ke

and since we assume the electron is going in a circle, theelectric force must act as a centripetal force:

r=⇒ v2 =

Using this value for velocity, we have kinetic energy:

K =12mev2 =

2rand E = −kee2

So far, so classical.

�antum solution

Assume the classical model of the electron orbiting the nucleusof the hydrogen atom under electrical forces. In this case thetotal energy is

E = K + U =12mev2 − ke

and since we assume the electron is going in a circle, theelectric force must act as a centripetal force:

r=⇒ v2 =

Using this value for velocity, we have kinetic energy:

K =12mev2 =

2rand E = −kee2

2rSo far, so classical.

�antum solution

Next comes the quantum step. Assume that only certain orbitsare stable (called stationary states, which validates are previousassumption of using classical physics). The atom emitsradiation when it makes a quantum leap from one state toanother. The change in its energy a�er making this leap isquantized:

Ef−Ei = hf , positive value means absorbed, negative value means emi�ed

�antum solution

Bohr next a new leap: quantizing the electron’s orbital angularmomentum:

mevr = nh2π

where ~ = h/2π. The energy quantization was enough tofluster some, but this la�er assumption was a radically newexpansion of the concept of quantization.

Now lets apply this radical concept:

v2 =n2~2

mer=⇒ rn =

mekee2, n = 1, 2, 3

�antum solution

Bohr next a new leap: quantizing the electron’s orbital angularmomentum:

mevr = nh2π

where ~ = h/2π. The energy quantization was enough tofluster some, but this la�er assumption was a radically newexpansion of the concept of quantization.

Now lets apply this radical concept:

v2 =n2~2

mer=⇒ rn =

mekee2, n = 1, 2, 3

�antum solution

Call the orbit with the smallest radius the Bohr radius (whenn = 1):

a0 =~2

mekee2= 0.0529nm

Plug this all back into the energy equation to find:

En = −kee2

), n = 1, 2, 3, · · ·

or numerically:

En = −13.606eVn2

, n = 1, 2, 3, · · ·

Now let’s find out the frequency of light emi�ed from aquantum leap:

f =Ei − Ef

(1n2f− 1

)Dr. Jones Physics 280�antum Mechanics Lecture

�antum solution

That looks familiar…

(1n2f− 1

)Plugging in all of our numbers:

2a0hc= 1.0973732× 107m−1 = RH

�antum mechanics had theoretically predicted the value ofRH . At this point, a�empts to rescue classical physics werebeginning to look hopeless.We see it again: integers describing physical reality–one of thekey signatures of quantum mechanics. This made mostphysicists uncomfortable, but also energized by therevolutionary spirit in the air.

�antum solution

(1n2f− 1

2a0hc= 1.0973732× 107m−1 = RH

�antum mechanics had theoretically predicted the value ofRH . At this point, a�empts to rescue classical physics werebeginning to look hopeless.

We see it again: integers describing physical reality–one of thekey signatures of quantum mechanics. This made mostphysicists uncomfortable, but also energized by therevolutionary spirit in the air.

�antum solution

(1n2f− 1

2a0hc= 1.0973732× 107m−1 = RH

�antum mechanics had theoretically predicted the value ofRH . At this point, a�empts to rescue classical physics werebeginning to look hopeless.We see it again: integers describing physical reality–one of thekey signatures of quantum mechanics. This made mostphysicists uncomfortable, but also energized by therevolutionary spirit in the air.

Summary:

Planck predicts quantization (1900): E = nhf , solves ultravioletcatastrophe

Einstein proposes that light is quantized (1905), solvedphoto-electric e�ect

Bohr proposes electron orbital angular momentum is quantized(1913), solves hydrogen spectrum problem, En = −13.606eV

de Broglie (1923/24) proposes that if photons are waves andparticles, so are massive particles like electrons:

p = mv =Ec

=hλ→ λ =

Summary:

p = mv =Ec

=hλ→ λ =

Summary:

p = mv =Ec

=hλ→ λ =

Summary:

p = mv =Ec

=hλ→ λ =

Schrodinger’s formalization

If particles are waves, how do we discuss their amplitudes anddisplacements like we do for classical waves? We need somesort of wave function to do so, which we write as Ψ.

For light, the intensity of the wave is proportional to the squareof the electric field. The more intense the light, the morephotons we can find, and so the number of photons isproportional to the intensity is proportional to the square ofthe electric field (the EM wave amplitude).

Put another way, the square of the electric field amplitude isproportional to the probability that we will find a photon at acertain region.

We seek a ma�er wave such that Ψ2 is proportional theprobability of finding an electron at a certain place and time.

Schrodinger found the equation that properly describes howthese ma�er waves behave:

− ~2

2md2ψdx2

+ Uψ = Eψ

This is called the Schrodinger equation.For most particles, we can separate the time-dependence sothat

Ψ(r, t) = ψ(r)e−iωt

This is a probabilistic equation, where the probability offinding a particle whose wave is ψ within any given region ofspace is P(x, y, z)dV =| ψ |2 dV .What is ma�er then? A particle? A wave? Is ψ the truemanifestation of ma�er?ψ is the true manifestation of ma�er.

− ~2

2md2ψdx2

+ Uψ = Eψ

This is called the Schrodinger equation.

For most particles, we can separate the time-dependence sothat

− ~2

2md2ψdx2

+ Uψ = Eψ

− ~2

2md2ψdx2

+ Uψ = Eψ

This is a probabilistic equation, where the probability offinding a particle whose wave is ψ within any given region ofspace is P(x, y, z)dV =| ψ |2 dV .

What is ma�er then? A particle? A wave? Is ψ the truemanifestation of ma�er?ψ is the true manifestation of ma�er.

− ~2

2md2ψdx2

+ Uψ = Eψ

This is a probabilistic equation, where the probability offinding a particle whose wave is ψ within any given region ofspace is P(x, y, z)dV =| ψ |2 dV .What is ma�er then? A particle? A wave? Is ψ the truemanifestation of ma�er?

ψ is the true manifestation of ma�er.

− ~2

2md2ψdx2

+ Uψ = Eψ

Probability

For any probability function, the integral of the entire function overinfinity must be equal to one:∫ +∞

−∞|ψ(x)|2 dx = 1

It can be show that the ”standard deviation in the results ofrepeated measurements on identically prepared systems”, or the”uncertainty”, has the following relation:

∆x∆p ≥ ~2

This is the Heisenberg Uncertainty Principal.

Probability

−∞|ψ(x)|2 dx = 1

∆x∆p ≥ ~2

Probability

−∞|ψ(x)|2 dx = 1

∆x∆p ≥ ~2

Probability

When there is a barrierwhich has too muchenergy for a particle topass through in terms ofclassical physics, quantummechanics correctlypredicts that there is still aprobability it will passthrough. This is calledtunneling.

review of early …antum mechanics formalization of …antum

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