quantum mechanics - notes to oxford lectures by … · quantum mechanics - notes to oxford lectures...

Quantum Mechanics - Notes to Oxford lectures by

Prof. James Binney Nov. 2009

Dipl.-Ing. Klemens Großmann

May 4, 2014

Contents

1 Preface 3

2 Quantum states 52.1 Spin of electrons . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 General definition of a quantum state in terms of a ket vector

living a vector space . . . . . . . . . . . . . . . . . . . . . . . 62.3 Two states and the complex conjugate rule . . . . . . . . . . 6

3 Basic/Important Operators in QM 73.1 The identity operator in quantum notation . . . . . . . . . . 73.2 The Hamiltonian operator H . . . . . . . . . . . . . . . . . . 7

4 Observables 74.1 Observables Q, their operators and their Eigenstates and Eigen-

values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74.2 Hermitian Adjoint of the matrix Q . . . . . . . . . . . . . . . 94.3 Functions of operators . . . . . . . . . . . . . . . . . . . . . . 94.4 Commutators of operators . . . . . . . . . . . . . . . . . . . . 9

5 Time Dependent Schrodinger Equation 115.1 Time Dependent SchrEq and Time Evolution . . . . . . . . . 115.2 Ehrenfest Theorem . . . . . . . . . . . . . . . . . . . . . . . . 11

6 Position and Momentum 136.1 Position Representation . . . . . . . . . . . . . . . . . . . . . 136.2 The Momentum Operator . . . . . . . . . . . . . . . . . . . . 146.3 The Commutator of the Position and the Momentum Operator 146.4 Rate of change of expectation value of position vs. that of

momentum. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156.5 Rate of change of expectation value of momentum vs. that

of change of potential engergy. . . . . . . . . . . . . . . . . . 16

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6.6 States of well-defined momentum . . . . . . . . . . . . . . . . 16

7 Free particle 187.1 Dynamics of a free particle . . . . . . . . . . . . . . . . . . . 187.2 Back to 2-slit interference . . . . . . . . . . . . . . . . . . . . 207.3 Generalisation to 3D . . . . . . . . . . . . . . . . . . . . . . . 22

8 Virial Theorem 23

9 Harmonic Oscillator 259.1 Normalization of the energy states . . . . . . . . . . . . . . . 289.2 Wave functions of stationary states . . . . . . . . . . . . . . . 299.3 Zero Point Energy . . . . . . . . . . . . . . . . . . . . . . . . 309.4 Expectation Value . . . . . . . . . . . . . . . . . . . . . . . . 319.5 Dynamics of Oscillators . . . . . . . . . . . . . . . . . . . . . 33

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1 Preface

The following lecture notes are a transcript with some further deliberationsand different or additional calculation steps of my own. I took them while“atttending“ Prof. James Binney’s Oxford lecture series on Quantum Me-chanics given online. They were recorded in November 2009 and uploadedto YouTube in April 2010 by a dilligent contemporary unkown to me. Youcan find the first lecture of the series here: http://youtu.be/AufmV0P6mA0

My motivation for watching these lectures was to acquire a mathemati-cal foundation for the University of Maryland MOOC ‘Exploring QuantumPhysics‘ on Cousera, which is self-contained, but having a prior solid under-standing of QM math is helpful for following the intuitive approach Prof.Galitsky often takes – it enchances the enjoyment a lot.

It comprises 27 roughly one-hour undergraduate lectures by Prof. Bin-ney’s. At least that is the number of thoese well-enumerated lectures Iwas able to find on YouTube. Interestingly in this time and age, they al-most completely forgo PowerPoint presentations, animations and the like,almost completely relying on scribbling equations, statements, derivationsand proofs on a blackboard with good old chalk.

What features this series of lectures – of course not specifically – seemsto me what might be considered by some a rather old-fangled style. I thinkit is absolutely not, surely not by judging by the means used – but rathera way of lecturing a not-that-young-anymore generation of past students isfamiliar with: a professor developing his equations on-the-spot, while you,the student, doing the same, transcribing the same steps into your notebookand thinking along with the lecturer, while you can literally see and hearhim think. From my own years at university I recall this, among other ways,as one of the most effective ways of learning.

I think it is in particular the consistent, precise and well-conceived com-ments on the concepts of QM accompanying Prof. Binney’s developing themathematical notations and giving them physical meaning, together withthe almost bottom-up structure of these lectures what accounts for the highbenefit one can gain from ”attending” them even over the Internet.

I think one does not have to be equipped with a substantial amount ofQM-specific mathematics in order to be able to follow the lectures and ben-efit from them. The basic knowledge is almost developed from scratch.Nevertheless, in not a few instances it would have been beneficial if the stu-dents had made comments or asked questions. Having said that, I felt thestudents in the lectures were, in terms of proactivity and interaction with

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the lecturer, astonishingly quiet and not as active as one would wish for –whatever the reasons might have been.

These lecture notes follow with high fidelity the order of the concepts asthey appear in the lectures. Some minor mathematical steps might havebeen skipped when they would have been straightforwardly to do for some-one with a good command of calculus. Others are depicted in almost gorydetail which to others some may appear worth skipping, but I did themanyway when I felt they could provide some deeper insight and/or learningimpact for others – or myself while writing the notes.

In any case - have fun!

Klemens Großmann, April 2014

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2 Quantum states

2.1 Spin of electrons

The angular momentum of half-spin particles, that is, fermions, in z direc-tion is

Jz = ±1

2~

Note that ~ has indeed unit of angular momentum J · s. The spin S of anelectron (of fermions) is dimensionless and also for sake of brevity the ~ isommitted. Hence, electrons as fermions have alway have a spin in z-directionof

Sz = ±1

2

It is important to note that this does not say anything about the spin in theother directions x and y.

The state notation (the ket notation) of the spin “along the axis n“

|n+〉 = sin(Φ

2)ei

Ψ2 |−〉+ cos(

Φ

2)e−i

Ψ2 |+〉

|n−〉 = cos(Φ

2)ei

Ψ2 |−〉 − sin(

Φ

2)e−i

Ψ2 |+〉

where Φ and Ψ are the polar coordinates as follows

Φ : angle measured from the z-axis in direction to the x-y-plane

Ψ : angle in the x-y-plance, emanating from the x-axis

and where n is expressed in term of Φ and Ψ, i.e. the y-axis is determinedas Φ = π

2 ,Ψ = π2 .

It is essential to keep in mind that if a particle has a certain spin “alongthe axis z“, that spin not really aligned along the z-axis, but there are stillsubstantial components of the spin in the x-y plane, of the direction is un-known. It is only for sure that if one measures the spin ”along the z axis”,he always gets either +1

2 or −12 .

Example: Probability of finding the particle’s spin in the “positive“ y-direction while is has +1

2 spin in z-direction:

|y+〉 = sin(π

4)ei

π4 |−〉+ cos(

π

4)e−i

π4 |+〉

=1√2

(eiπ4 |−〉+ e−i

π4 |+〉)

5

So the important result is

〈Ψ|Φ〉 = 〈Φ|Ψ〉∗ (1)

but also generally

Q|Ψ〉 = 〈I|QΨ〉 = (〈QΨ|I〉|)∗ = (Q〈Ψ|)∗ (2)

3 Basic/Important Operators in QM

3.1 The identity operator in quantum notation

The identity operator (without proof here) is expressed as [A,B2] = [A,BB] =[A,B]B +B[A,B]

I =∑

i |i〉〈i| (3)

where |i〉〈i| form a basis of kets.

Acting on a state |Φ〉 the identity opeator yields the same state |Φ〉

I|Φ〉 = |Φ〉

3.2 The Hamiltonian operator H

It is per defintion

H :=∑

iEi|Ei〉〈Ei| (4)

where Ei in the front is a real number and the possible engergies, the spec-trum of the engergy operator, and |Ei〉, 〈Ei| are the possible engergy states.

The expection value of the energy state of a system is simply obtained by

< Φ|H|Φ >=∑

i |ai|2E1 =∑

i PiE1 = 〈H〉 (5)

4 Observables

4.1 Observables Q, their operators and their Eigenstates andEigenvalues

We agree upon that

qi : the spectrum, the possible values that the measurment can return

|qi〉 : complete set of states for which the result of obeserving Q s certain

The form operator is defined as

Q :=∑

k qk|qk〉〈qk| (6)

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4.2 Hermitian Adjoint of the matrix Q

It is defined such that the ij-th element of the Hermitian adjoint of Q is thecomplex conjugate of ji-th element of Q:

(Q†)ij = Q∗ji

Following (1)

(Q†)ij = (qiδji)∗ = q∗i δji =ifqiisreal qi=j = Qij

Hence(Q†)ij = Qij

So we can say that if the eigenvalues of the observables a quantum systemare real, then the adjoint of the operator Q is Q itself, that is, it is Hermitionor ”self-adjoint”.

Side note:〈qi|qj〉

denotes the probability amplitude that the system which is in a state qj canbe measured in the state |qi〉, which is, of course, 0 for i 6= j.

Properties of the adjoint operator Q

〈Ψ1|QΨ2〉 = 〈Q†Ψ1|Ψ2〉 (8)

Some rule about the adjoint of a product of operators

(R · Q)† = Q† · R† (9)

4.3 Functions of operators

The definition isf(Q) =

∑f(qi)|qi〉〈qi| (10)

and with |qi〉〈qi| is has the same eigenstates/eigenkets as the operator Q,but the eigenvalues are a function of the eigenvalues of the operator Q.

4.4 Commutators of operators

We already know thatAB 6= BA

A definition of notation

[A,B] = AB −BA (11)

9

and from this follows

[A,B + C] = [A,B] + [A,C]

[A+B,C] = [A,C] + [B,C]

andAB = [A,B] +BA

and[AB,C] = [A,C]B +A[B,C]

which is mirror of the product rule of differentation:

dAB

dC=dA

dCB +A

dB

dC

Commutator with a product

[A,B2] = [A,BB] = [A,B]B +B[A,B] (12)

Important definiton: Commuting observables

if [A,B] = 0

then A and B are commuting obeservables.

If operators A and B are communting observables, then there is a com-plete set of mutual eigenkets/eigenstates |i〉 such that

A|i〉 = ai|i〉 and B|i〉 = bi|i〉

On the other hand, if[A,B] 6= 0

then there exists at least one state/ket |Ψ〉 such that

[A,B]|Ψ〉 6= 0

meaning: the commutator of A and B operating on the ket does not producenothing. It implies that there is no complete of state with certain value ofboth A and B. (but there actually CANbe some states, even a large numberof statees in which they do have certain values for both A and B).

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5 Time Dependent Schrodinger Equation

5.1 Time Dependent SchrEq and Time Evolution

Time Dependent Schrodinger Equation

i~∂|Ψ〉∂t = H|Ψ〉 (13)

where |Ψ〉 is the state of any system. Suppose we have a sate of well definedengery E, such that |Ψ〉 = |E〉, and, of course H|Ψ〉 = E|E〉, then

i~∂|E〉∂t

= E|E〉

which, obviously, has the solution

|E〉(t) = |E0〉e−iE~ t

where, naturally, a tremendously large frequency ω = E~ .

We know already that the state of a system is the sum of “weighted“ possibleeigenstates, and the time evolution of that states corresponds accordingly tothe time evolution of that engergy states by the eigenvalues and eigenstates:

|Ψ(t)〉 =∑i

ai(t)|Ei(t)〉

which, after plugging in the solution of the time-dep. SchrEq for the engergyyields a fabulous important equation :

|Ψ(t)〉 =∑n

an(t)e− iEn~ t|En0〉 (14)

As a reminder, the time-independent SchrEq:

H|En〉 = En|En〉

5.2 Ehrenfest Theorem

The Time Dep. SchrEq applied to the expetation value

i~∂

∂t〈Ψ|Q|Ψ〉 = i~

∂〈Ψ|∂t

Q|Ψ〉+ i~〈Ψ|∂Q∂t|Ψ〉+ 〈Ψ|Qi~∂Ψ

∂t

With (2), we have that i~∂〈Ψ|∂t = (i~∂|Ψ〉∂t )∗ = −i~∂|Ψ〉∂t , and with i~∂|Ψ〉∂t =H|Ψ〉 we obtain

i~∂

∂t〈Ψ|Q|Ψ〉 = −H|Ψ〉Q|Ψ〉+ 〈Ψ|QH|Ψ〉+ i~〈Ψ|∂Q

∂t|Ψ〉

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Since H is self-adjoint (Hermitian), −H|Ψ〉 = −〈Ψ|H

i~∂

∂t〈Ψ|Q|Ψ〉 = −〈Ψ|HQ|Ψ〉+ 〈Ψ|QH|Ψ〉+ i~〈Ψ|∂Q

∂t|Ψ〉

and applying the commutators, we obtain the Ehrenfest theorem :

i~ ∂∂t〈Ψ|Q|Ψ〉 = 〈Ψ|[Q, H]|Ψ〉+ i~〈Ψ|∂Q∂t |Ψ〉 (15)

The last term 〈Ψ|∂Q∂t |Ψ〉 is the expectation value of the rate of change of theobservable Q. which is expected to be zero, because the observable (positionor momentum) usually does not change over time, and the last term usuallyfalls away in most appplications. Hence, in normal states of affairs, that is,

if ∂Q∂t = 0, we have that the rate of change of expectation value of Q, which

is ∂∂t〈Ψ|Q|Ψ〉, is equal to the expectation value of the commutator [Q,H]:

i~ ∂∂t〈Q〉 = 〈[Q, H]〉 (16)

So, if Q commutes with H, the expectation value of Q is clearly constant:

〈Q〉 = const.

If at t = 0 〈Q〉 = qr, then 〈Q2〉 = q2r . The variance of Q becomes zero.

var(Q) = 〈Q2〉 − 〈Q〉2 = 0 (17)

Let Q = H, and since obviously H commutes with itself, then, since 〈H〉 = 0is must the case that

i~∂H

∂t= 0⇔ 〈H〉 = const

Another example of working out the time evolution, the rate of change ofthe amplitue of an observable given the system is a state |Ψ〉

i~∂

∂t〈qn|E〉

becomes, using the product rule

i~∂〈qn|∂t|E〉+ 〈qn|i~

∂|E〉∂t

Again, because of (2) i~∂〈qn|∂t = (i~∂|qn〉∂t )∗ = −i~∂|qn〉∂t = −H|qn〉 and with H

being Hermitian, −〈qn|H

−〈qn|H|E〉+ 〈qn|H||E〉 = 0

Which give the important result the the amplitude of 〈qn|E〉 is constantfor any observable Q. It translated to saying that any observable of ansystem in an engergy state of |E〉 are contant, and that sate |E〉 is called a”stationary state”.

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6 Position and Momentum

6.1 Position Representation

We define the position operator x, which produces states of well-definedpositions on the X-axis, and which spectrum is usually continous ]−∞,+∞[,so that their sum over a discrete spectrum becomes an integral

|Ψ〉 =

∫ +∞

−∞Ψ(x)|x〉dx

where Ψ(x) is the amplitude of finding the particle at postion x, and |x〉 isthe state of the system when the particle is at x. The identity operator inthe continuous case is

I =

∫|x〉〈x|dx

“Bra-ing“ is through with 〈Ψ|Ψ〉, that is, first with |Ψ〉

I|Ψ〉 =

∫|x〉〈x|Ψ〉dx

and then with 〈Ψ|〈Ψ|I|Ψ〉 =

∫〈Ψ|x〉〈x|Ψ〉dx

Since 〈x|Ψ〉 produces Ψ(x), and 〈Ψ|x〉, and 〈Ψ|x〉 is the complex conjugate,the integral produces something familiar, that is

〈Ψ|Ψ〉 =

∫Ψ(x)∗Ψ(x)dx =

∫|Ψ(x)|2dx = 1

A practise problem, that is the question expressed in the notation

〈x|x|Ψ〉

What is the amplitude to be at x for the state you obtain when the operatorx works on an arbritary state Ψ. We can solve that again by sliding in theidentity operator I =

∫dx′|x′〉〈x|

〈x|x|I|Ψ〉 =

∫dx′〈x|x|x′〉〈x′|Ψ〉 =

∫dx〈x|x|x′〉Ψ(x′)

x|x′〉 produced the eigenvalue and the eigenstate x′|x′〉, therefore

〈x|x|Ψ〉 =

∫dx x′〈x|x′〉Ψ(x′)

and 〈x|x′〉 is the Kronecker Delta δ(x− x′), hence

〈x|x|Ψ〉 =

∫dx′ x′δ(x− x′)Ψ(x′)

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Since δ(x−x′) becomes 1 when x′ = x (and has ”width” dx′), the integrandis non-zero only for that dx′, and the integral reduces to just one value tosumming up:

〈x|x|Ψ〉 = x ·Ψ(x)

This shows that the postition operator x acts on wavefunctions by ordinarymultiplication.

6.2 The Momentum Operator

Definition:

〈x|p|Ψ〉 = −i~∂〈x|Ψ〉∂x = −i~∂Ψ(x)∂x (18)

Is it Hermitian?

〈ΦI|p|Ψ〉 =

∫dx〈Φ|x〉〈x|p|Ψ〉

With 〈Φ|x〉 = Φ(x)∗ and 〈x|p|Ψ〉 = −i~∂Ψ(x)∂x

〈ΦI|p|Ψ〉 =

∫ +∞

−∞dx · Φ(x)∗ · (−i~∂Ψ(x)

∂x)

Integrating by parts

−i~ · (Φ(x)Ψ(x)|+∞−∞−∫ +∞

−∞

∂Φ(x)∗

∂xΨ(x))

∫dx ·Ψ(x)(i~

∂Φ(x)∗

∂x)

We know that z∗1z∗2 = (z1z2)∗, and therefore i~∂Φ(x)∗

∂x = (−i~∂Φ(x)∂x )∗ =

〈x|p|Φ〉∗. Hence

〈Φ|p|Ψ〉 =

∫dx〈Ψ∗|x〉〈x|p|Φ〉∗ = 〈Ψ|p|Φ〉∗

Stated again as important result

〈Φ|p|Ψ〉 = 〈Ψ|p|Φ〉∗ (19)

6.3 The Commutator of the Position and the MomentumOperator

〈x|[x, p]|Ψ〉 = 〈x|x, p|Ψ〉 − 〈x|p, x|Ψ〉 = 〈x|x(−i~∂Ψ

∂x)〉 − 〈x| − i~∂xΨ(x)

∂x〉

= −x〈x|i~∂Ψ(x)

∂x)〉 − 〈x| − i~∂xΨ(x)

∂x〉

= −i~x〈x|∂Ψ(x)

∂x)〉+ i~〈x|Ψ(x)〉+ i~x〈x|∂Ψ(x)

∂x〉

14

So finally〈x|[x, p]|Ψ〉 = i~〈x|Ψ〉

Hence we can state that[x, p] = i~ (20)

which is called a “canonical commutation relation“.

6.4 Rate of change of expectation value of position vs. thatof momentum.

Whe can show that applying the Ehrenfest Theorem 5.2 for general operatorsto the position operator yields, as have been expected, the expecation valueof the velocity, that is the momentum over the mass.

i~∂

∂x〈Ψ|x|Ψ〉 = 〈Ψ|[x, H]|Ψ〉 = i~〈Ψ| p

m|Ψ〉

The derivation of this result is as follows.

With the Hamiltonian as the engergy operator H = p2

2m − V (x) and lookingat the first term

〈Ψ|[x, H]|Ψ〉 = 〈Ψ|[x, ( p2

2m+ V (x))]|Ψ〉

Clearly, since V (x) is a function of the operator and gives a boring numnber,not an operator, V (x) commutes with x, that is, [x, V (x)] = 0, and we areleft with

〈Ψ|[x, H]|Ψ〉 = 〈Ψ|[x, p2

2m]|Ψ〉 =

1

2m〈Ψ|[x, pp]|Ψ〉

It can easily be shown that

[x, pp] = [x, p]p+ p[x, p]

and so1

2m〈Ψ|[x, pp]|Ψ〉 =

1

2m〈Ψ|[x, p]p+ p[x, p]|Ψ〉

From (11) we know that [x, p] = i~. Therefore

1

2m〈Ψ|[x, p]p+ p[x, p]|Ψ〉 =

i~2m〈Ψ|p+ p|Ψ〉 =

i~2m〈Ψ|2p|Ψ〉 = i~〈Ψ| p

m|Ψ〉

�

So we arrive at an equality analogous to what is known from classical physics.The rate of change of the expectation value of x is the expecation value ofthe velocity.

∂

∂x〈Ψ|x|Ψ〉 = 〈Ψ| p

m|Ψ〉 = v (21)

15

6.5 Rate of change of expectation value of momentum vs.that of change of potential engergy.

I can be shown that

∂

∂t〈Ψ|p|Ψ〉 = −〈Ψ|∂V (x)

∂x|Ψ〉 (22)

stating the the rate of change of the expection value momentum (which isthe acceleration) is equal to the expecation value of the derivative of thepotential engergy with respect to the position, which is the acceleration.This recovers Newton’s second law of motion.

6.6 States of well-defined momentum

The momentum operator p produces the eigenstates/eigenkets of possiblemomentum states along with the eigenvalues.

p|p〉 = p|p〉 (23)

This defines these states.

If we are interested in those states given a position, we bra-through with anx

〈x|p|p〉 = p〈x|p〉

where, of course, 〈x|p〉 is Ψp(x), and 〈x|p|p〉 we already know to be−i~∂〈x|p〉∂x =

−i~∂Ψp(x)∂x , so we the SchrEq

−i~∂Ψp(x)

∂x= pΨp(x) (24)

which has, obviously, a solution

〈x|p〉 = Ψp(x) = Ψp0eip~ x (25)

Normalization and a ”little bit“ of calculus and algebra (skipped here) yields

Ψp0 =1√h

So the state of well-defined momentum is a plain wave with the wave numberk = p

~ and the so-called ”de Broglie“ wavelength λ = 2πk = 2π~

p = hp , which

is, naturally, very small. It determines the characteristic size of an atom,say by the electron of a hydrogen atom.

The probability density |Ψp|2 is, obvious from the solution, a constant, in-dependent of x, which reflects the ultimate uncertainty principle: the prob-ability of finding a particle with a well-defined momentum is the same of

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every postion x over the whole universe.

〈p|x〉 = Ψx(p) = 〈x|p〉∗ = Ψ∗p(x) = Ψp0e−ip~ x

and, again, P (p, x) = |Ψx(p)|2 is a constant over p.

If we would like to calculate the amplitude of finding the particle at mo-mentum p given it is in a state Ψ, that is, 〈p|Ψ〉, we can do that with helpof the identity operator I =

∫|x〉〈x|dx by sticking it into it

〈p|I|Ψ〉 =

∫〈p|x〉〈x|Ψ〉dx

We have that 〈p|x〉 = 1√he−i

p~x, and we assume that Ψ(x) = e

− x2

4σ2

(2πσ2)4 is a

Gaussian distribution with dispersion in position σx = σ, so we get

〈p|I|Ψ〉 =

∫e−i

p~x

√h

e−x2

4σ2

(2πσ2)4dx

Solving the integral via bringing it into the Gaussian integral form andknowing that

∫∞−∞ e

−x2=√π, we obtain that

〈p|Ψ〉 =e−

σ2p2

~2

(2π~2

4σ2 )1/4

Expressing it as the probablitiy, which is P = |Ψ|2 and arranging the ex-ponent a little bit differntly in order to get the Gaussian form with thedispersion, that is

e−( x

σp)2

P (p) =e−( p

~2σ

)2

(2π~2

4σ2 )1/2

so we see that σp = ~2σx

, and, equivalently

σp · σx =~2

(26)

With is the Heisenberg Uncertainty Principle for the Gaussian distribution,meaning NOT if one measures the momentum of an electron he obtainsuncertainty σp, and then in a second measurment of the postion, he getsuncertainty σx, but instead, if one has a sample of electrons, all set up inthe same wavefunction, and he measures the momentum of half of them andthen measures the position of the other half of them, he obtains dispersions

17

of momentum σp and of position σx which satisfies the relationship above.

This is due to the fact that doing a measurement leads to a collapse ofthe wavefunction is a well-defined state, so if one measures the momen-tum or the position of an electron, the electron assumes a well-defined stateand the wavefunction with respect to other property be becomes highlychanged. The uncertainty principle as stated above is therefore to be ap-plied to a sample of different electrons setup with the same wavefunction,and the probabilities of the outcome of a measurement is the probabilitythat an electron of the sample is measured to have a certain momentum orposition, reflected in the distribution of the sample.

7 Free particle

7.1 Dynamics of a free particle

It is a free particle with respect to its potential energy, which meant to bezero at all positions. Then, the only engergy is possesses is the kinetic energy,and the Hamiltonian operator, which is an energy operators is reduced to

H =p2

2m

We conjecture that the complete set of amplitudes 〈x|p〉 of a free particlelocalized around the origin ±σ is characterized by the following function

〈p|Ψ〉 = Ψ(p) =1

(2π~2/4σ2)4e−

σ2

~2 (p−p0)2

(27)

where p0 is some constant, that is, it has a velocity that is at p0

m and islocalized around x = 0± σ.

What is the wavefunction in real space that corresponds to that function attime different time, as a function of time?

We notice that the Hamiltonian operator is confined to the mometum oper-ator, so the states of well-defined engergy En|En〉, obtained by the Hamil-tonian, are the states of well-defined momentum p|p〉.

We know the time evolution 5.1 of the wavefunction of well-defined energystates from the time-dependent SchrEq in the case of discrete energy states:

|Ψ(t)〉 =∑n

an(t) · e−iEn~ t · |En0〉

Analogous for the continuous case of, we setup the case of continuous mo-menta

|Ψ(t)〉 =

∫〈p|Ψ〉 · e−

ip2

2m~ t · |p〉dp

18

where 〈p|Ψ〉 is the amplitude of finding the particle with the momentum pgiven it is in a state Ψ, so it corresponds to the amplitudes an of the discretecase.

And, of course |p〉 are the states of well-defined momenta and thereforecorrespond to the sates of well-defined energies |En0〉 of the discrete case.

Now by bra-ing it through with 〈x|, we get the interference of time-evovledstates of well-defined momentum with real space:

〈x|Ψ(t)〉 =

∫〈p|Ψ〉 · e−

ip2

2m~ t · 〈x|p〉dp

We know from our conjecture 〈p|Ψ〉 to be e−σ

2p2

~2

( 2π~2

4σ2 )1/4, and 〈x|p〉 is the wave-

function for well-defined momenta with respect to x, and known from before

to be eip~ x√h

, so the johnny becomes

〈x|Ψ(t)〉 =

∫e−

ip2

2m~ t · e−σ2p2

~2

(2π~2

4σ2 )1/4· e

ip~ x

√h· dp

The result for the probability | · |2 comes out as

|〈x|Ψ〉|2 =σ√

2π~2|b|2· e−

(x− p0tm )2

2~4|b|4σ2

(28)

where b2 = σ2

~2 + it2m~ We notice p0

m = v and also that the dispersion is

σt =~2|b|2

σ=

~2(σ2

~2 + it2m~)

σ= σ +

i~t2mσ

From the uncertainty principle we know that uncertainty of the momentumis ∆p = ~

2σ and accordingly ∆v = ~2mσ . Therefore

σt = σ + i∆vt

(we remember that σ = σx). This makes physically perfect sense, as thedispersion with respect to time is the intial dispersion σx plus the dispersionstemming from the uncertainty of the velocity mutliplied by the time.

19

7.2 Back to 2-slit interference

Figure 1: 2-slit experiment with electrons

Pythogorean Theorem gives us for the two path lengths D+ and D− (thepath lenghts that actually differ, because the path lengths from the muzzleof the electron gun to either slit are symmetric):

D+ =√L2 + (x− s)2

D− =√L2 + (x+ s)2

The amplitude of finding a particle as position x give it is a state Ψ isreasonable going to be

〈x|Ψ〉 = eip~x

The probability to arrive at point x is the sum of the amplitudes of arrivingby the top slot and the bottom slot squared:

P (x) = |A+(x) +A−(x)|2

which is, multiplied out (and knowing that z2 = |z|2):

P (x) = |A+(x)|2 + |A−(x)|2 + 2Re{A∗+(x)A−(x)}

P (x) = P+ + P− + 2Re{A∗+A−}

The interference term 2Re{A∗+A−} is

2Re{A+A−e−i p~D+ei

p~D−}

= 2Re{A+A−e−i p~ (D+−D−)}

= 2|P±|cos(p

~(D+ −D−))

20

where |P±| is the probability to go through either slit.

We can rewrite the expression of for D+ and D− as

D+ = (L2 + (x− s)2)12 = (L2(1 +

(x− s)2

L2))

12 = L(1 +

(x− s)2

L2)

12

We can Newton-approximate it by (1 + x)a ≈ 1 + ax for x << 1, so itbecomes

D+ = L(1 +1

2

(x− s)2

L2) = L+

(x− s)2

2L

Correspondingly

D− = L+(x+ s)2

2L

Hence

D+ −D− = −2xs

L

With cos(−x) = cos(x)

2Re{A∗+A−} = 2|P±|cos(2p sL~x)

Therefore, the probability of the electron arriving at point x is

P (x) = P+ + P− + 2|P±|cos(2p sL~x)

Looking at the expression inside of the cosine, we can extract the period Xof the oscillation as

X = π~Lps

Assuming that P+ and P− are roughly the same and with the period men-tioned about, let’s express it differently

P (x) = 2|P±| · (1 + cos(2π

Xx))

1+cos(2πX x) is obviously oscillating between +2 and 0 with period X = π ~L

ps .

In the macroscopic world, say for a bullet with V = 300ms , m = 10g, L =1km, the period becomes, mostly because of the high momentum, relativelytoo small compared to the object in order to have an effect (X ≈ 10−29),so that the 1+cosine cancels itself out to 1, and the classical solution isrecovered:

P (x) = P+ + P−

21

Figure 2: P(x) = (P+) + (P-) + 2(P+/-)cos(2pi/X x)

7.3 Generalisation to 3D

We define 3-dimension for space and momentum, clearly with an orthogo-nality relationship and commute with each other

[xi, xj ] = 0

[pi, pj ] = 0

Furthermore, we already know that

[x, p] = i~δij

We denote the position in three dimensions, with x, y, z as as vector x.Correspondingly, the amplitude of finding the particle at position x becomesa function of x

〈x|Ψ〉 = Ψ(x)

Stating the analguous states of well-defined momentum

Ψp(x) = 〈x|p〉 =eip~ x

(~)32

where (~)32 comes from normalizing (6.6) with respect to three position vari-

ables instead of one.

The definition of the momentum operator we known from (??) is relativelystraightforward

〈x|p|Ψ〉 = −i~∇〈x|Ψ〉 = −i~∇Ψ(x)

22

8 Virial Theorem

For a stationary state

∀Q :d〈E|Q|E〉

dt= 0

Apply to Q = (x · p), and then using the Ehrenfest Theorem (5.2)

0 = i~∂〈E|x · p|E〉

∂t= 〈E|[x · p, H]|E〉

Take the familiar energy Hamiltonian p2

2m + V

0 = 〈E|[x · p, p2

2m+ V ]|E〉 (29)

Calculating this piece-wise, starting with the kinetic engergy operator:

[x · p, p2] =∑i,k

[xipi, p2k]

According to the known commutator rule that [AB,C] = [A,C]B−A[B,C],we consider xipi and obtain

[x · p, p2] =∑i,k

[xi, p2k]pj − xi[pi, p2

k]

where clearly pi commutes with pk, so what is left is

[x · p, p2] =∑i,k

[xi, p2k]pj

With the similiar rule [A,B2] = [A,B]B +B[A,B] and since that [xi, pk] =i~δik we get

[x·p, p2] =∑i,k

([xi, pk]pk+pk[xi, pk])pk =∑i,k

2i~δikp2k =

∑k

2i~p2k = 2i~

∑k

p2k = 2i~p2

[x · p, p2

2m] = 2i~

∑k

p2k = i~

p2

m

Taking the potential engergy, with is a function of an operator x, not anoperator itself, and since (remember 6.2 the momentum operator is p =−i~∇, the commutator becomes

[x · p, V (x)] = [−i~x∇, V (x)] = −i~(x∇V (x)− V (x)x∇1) = −i~ · x · ∇V (x)

The ”nabla”, the gradientg of 1 is clearly 0, so the equation reduces to

[x · p, V (x)] = −i~ · x · ∇V (x)

23

The potential energy with respect to x obeys the general equation

V (x) = A · |x|α

meaning that the potential engergy is proportional to some power of theradius, the distance from the origin. So what is the gradient of that?

∇V (x) = ∇(A|x|α) = A∇(

√∑i

x2i )α

= A∇(∑i

x2i )

α2

= A

∂(

∑i x

2i )α2 )

∂x1

∂(∑i x

2i )α2 )

∂x2

∂(∑i x

2i )α2 )

∂x3

= Aα

2

(∑

i x2i )

α2−1 · ∂(

∑i x

2i )

∂x1

...

...

= Aα

2

|x|α

|x|2 · 2x1

...

...

= Aα

|x|α

|x|2

x1

x2

x3

= Aα|x|α

|x|2x

Hence

x∇V (x) = Aα|x|αx · x|x|2

= Aα|x|α |x|2

|x|2= Aα|x|α = αV (x)

So luckily we can state that (22) is actually

0 = 〈E|[x · p, p2

2m+ V ]|E〉 = i~〈E| p

2

m− V |E〉

With leads to the conclusion that the sum of the expectation values of halfof kinetic energy times α and the potential energy has to be zero

〈E| p2

m|E〉 =

1

2〈K〉 = α〈V 〉 = α〈E|V (x)|E〉

In case of the Coloumb potential, in which α = −1, the potential energybecomes minus twice the kinetic engergy So we arrive at the Virial Theoremstating that for stationary states in a Coloumb potential

〈K〉 = −1

2〈V 〉 (30)

24

9 Harmonic Oscillator

Generally, for a force dependent on the the distance x to the position ofequilibrium, for small displacements the equation looks approximately likethis

F (x) = k · x+O(x2)

where, because x is small, we can neglect the 0-terms and the relationsshipbecomes proportional.

Integrating the force up over x becomes the potential energy V (x) = 12 ·k ·x

2,and the Hamiltonian becomes

H =p2

2m+

1

2kx2 =

p2 + (mωx)2

2m(31)

We want to find the states of well-defined energies, that is both the en-ergies and the states, so that we can then pretty do with them what wewant, like time-evolution. Of course, we obtain those states by applying theHamiltonian:

H|E〉 = E|E〉

Upfront, we betray a new operator, the so-called “Annihilation Opera-tor”, which reduces the exitation of the Harmonic Oscillator.

A = mωx+ip√2m~ω

(32)

The general ideal for this operator is to factorize the Hamiltonian under (24)(while the square root in the denominator is supposed to make the wholething dimensionless). Building the product with the complex conjugate ofsomething like A, just not as operators, but as boring number, usually bringsomething real and useful:

AA∗ =1

2m~ω(p2 + (mωx)2)

The Hermitian adjoint of A cam be found by substituting in the Hermitian

adjoing of each operator in A, that is, A† = mωx†+i†p†√2m~ω

, yielding the “Cre-

ation Operator”, which increases the exitation of the Harmonic Oscillator.

A† = mωx−ip√2m~ω

(33)

Note that the Hermitian adjoint of i is −i, because the Hermitian adjointof a complex number is the complex conjugate of that number.

In Quantum Field Theory, particles are exitations of the vacuum, there-fore in QFT the Creation Operator creates a particle and the Annihilation

25

Operator destroys it.

The product of the operator with its Hermitian adjoint is then, as almostalready predicted, but with remaining cross-terms because operators do notcommute neccessarily as in classical algebra, so they don’t in the case of xand p

A†A =1

2m~ω(p2 + (mωx)2) + imω[x, p])

A†A =p2 + (mωx)2

2m~ω+

imω

2m~ω[x, p]

By cancellation and knowing from 20 that [x, p] = i~, we obtain

A†A =p2 + (mωx)2

2m~ω− 1

2

Which is as we notice

A†A =H

~ω− 1

2

and we almost have a factorization of the Hamiltonian:

H = ~ω(A†A+ 12) (34)

Finding out what the commutator of both operators are

[A†, A] =1

2m~ω[mωx− ip,mωx+ ip]

[A†, A] =1

2m~ω([mωx,mωx] + [mωx, ip]− [ip,mωx] + [p, p])

Since [mωx,mωx] = 0 and [p, p] = 0

[A†, A] =1

2m~ωmω([x, ip]−[ip, x]) =

1

2~([x, ip]+[x, ip]) =

1

2~2[x, ip] =

1

~+ii~

So that

[A†, A] = −1 (35)

Now we are ready to apply the operator A† to our stationary state, and byoperating-through E|E〉 with A† we get

EA†|E〉 = A†H|E〉

Adding and substracting HA† to A†H, A†H can be expressed as

HA† + A†H − HA† = HA† + [A†, H]

and with H = ~ω(A†A− 12)

A†H|E〉 = HA† + ~ω[A†, A†A− 1

2]

26

Looking at the expression, A† clearly commutes with 12 , therefore the cor-

responding term vanishes. What is left from the commutator expression atthe end is [A†, A†A], which becomes according to the product rule [A,BC] =[A,B]C +B[A,C]

[A†, A†]A+ A†[A†, A] = 0 + A†(−1) = −A†

Hence,

[A†, A†A− 1

2] = −A†

Plugging that inEA†|E〉 = HA†|E〉 − ~ωA†|E〉

So the Hamiltonian acts like

HA†|E〉 = EA†|E〉+ ~ωA†|E〉

and for the Creation Operator A† we get the following action

HA†|E〉 = (E + ~ω)A†|E〉 (36)

This means that the Hamiltonian acting on an energy state on which theCreation Operator has acted upon generates a state of increased engergy|E + ~ω〉

Analogous for the Annihilation Operator A the result comes out as

HA|E〉 = (E − ~ω)A|E〉 (37)

This means that the Hamiltonian acting on an energy state on which theAnnihilation Operator has acted upon generates a state of decreased en-gergy |E − ~ω〉

So, the Creation Opereator A† generates states of discretely increasing en-ergies E + ~ω,E + 2~ω,E + 3~ω..., E + n~ω. The Annihilation OperatorA, for the Ground State, the state of lowest engergy, annihilates the energystate (i.e. A|E0〉 becomes 0), and the produced eigenvalue canactually notbee considered as an eigenvalue of the Hamiltonian H.

So |A|E0〉|2 = 〈E|A†A|E〉 = 0, which the expecation value 〈E0| H~ω −12 |E0〉 =

0. So we get that the ground state energy is, because H|E0〉 = ~ω2 |E0〉:

E0 = ~ω2 (38)

Hence, the n-th energies, obtained by the increasing it discretely by theCreation/Annihilation operators A† and A are

En = (n+ 12)~ω with n=0,1,2,... (39)

27

The usual notation and notion is that a state of energy E goes to state ofenergies (n+ 1

2)~ω, denoted as |n〉

|E〉 −→ |n〉

9.1 Normalization of the energy states

The normalization isA†|n〉 = K|n+ 1〉

where K is a constant. In order to find the value of K, we take the mod-squares, ||2 of both sides

〈n|AA†|n〉 = |K|2 = K2

Note that ||n + 1〉| = 1 because it is per definition normalized. In order toget the Hamiltonian into the game, we need to swap the order of AA† byremembering that AA† = A†A+ [AA†], and we obtain

〈n|A†A+ [AA†]|n〉 = 〈n| H~ω− 1

2+ [AA†]|n〉

Fortunately, [AA†] = 1, therefore

〈n| H~ω

+1

2|n〉

Note that |n〉 is a state obtain by having let A or A†, respectively, act onan energy state, therefore, H|n〉 is per defintion (n+ 1

2)~ω, and

H

~ω|n〉 = (n+

1

2)

Plus 12n, it becomes n+ 1, so

K2 = n+ 1

so we obtain a very important equation for normalized states:

|n+ 1〉 = A†|n〉√n+1

(40)

where the normalization constant is√n+ 1.

We would also obtain

|n− 1〉 = A|n〉√n

(41)

where the normalization constant is√n.

As general insight, the normalization constant is the square rootof the biggest number appearing in the equation.

28

9.2 Wave functions of stationary states

Ground-State Wavefunction:, that is, the amplitude of finding the par-ticle as x when it the groundstate is denoted as

U0(x) = 〈x, 0〉

where 0 refers to the numbering of states as noted before (|n〉). The Anni-hilation Operator A applied on the lowest energy state, the ground state,kills it (≡ A|0〉 = 0), so the amplitude of finding the particle at position xafter its lowest energy state has been annihilated is 0:

〈x|A|0〉 = 0

〈x|mωx+ ip|0〉 = 0

Let x act on 〈x, producing x〈x|

mωx〈x|0〉+ i〈x|p|0〉

where 〈x|0〉 is the wavefunction in the ground state U0(x), and 〈x|p|0〉 pro-

duces −i~∂U0(x)∂x

(mωxU0(x) + ~∂U0(x)

∂x= 0

mωxU0(x) + ~∂U0(x)

∂x= 0

∂U0(x)

∂x= −mωx

~U0(x)

or, in my favorite notation for the wavefunction

∂Ψ0(x)

∂x= −mωx

~Ψ0(x)

which obviously has as solution

Ψ0(x) = e−mω2~ x

2= e−( x

2l)2

where l =√

~2mω

The probability is

P (x) = |Ψ0(x)|2 = e−( x√

2l)2

Which we know to normalize, because its intregral over ] −∞,+∞[ is theGaussian integral. So, normalized to surely find the partical in that range,it is

P (x) =1√2πl2

e−( x√

2l)2

29

The normalized wave function is then, obviously, as the square root

Ψ0(x) =1

(2πl2)14

e−( x2l

)2

And also

P (p) =e−( p√

2σp)2

2πσ2p

The dispersions of both are l and σp, and by the Uncertainty Principle weknow σxσp = ~

2 , hence

σp =~2l

or

p =~2x

9.3 Zero Point Energy

One of the most important predictions in QM is the predictionof Zero-Pointenergy. If you try to minimize the energy of a particle, which is the ground-state, you strive to minimize its potential energy, that is, getting the particleas close as you can to the origin. Due to the uncertainty relationk confiningthe particle to a narrow space increases the uncertainty of its momentum,and you have kinetic energy, so you have a compromise between a smallpotential energy a relatively higher kinetic energy. Gluons and quarks, forexample, are in deep potential well, hence confined very narrowly in space,and therefore has a relatively high kinetic energy. and that kinetic energymakes for high mass of the matter.

In order to try to find the minimum of the energy with respect to x, let’ssetup the Hamiltonian only with repsect to x, using the Uncertainty Rela-tions

H =1

2m(p2 +m2ω2x2)

Plugging in the Uncertainty Relation p2 = ~2

4x2

=1

2m(~2

4x2+m2ω2x2)

and x2 ∼ l2

=1

2m(~2

4l2+m2ω2 l2)

First exited stateΨ1(x) = 〈x|1〉

|1〉 =1√1A†|0〉

30

Ψ1(x) = 〈x|A†|0〉 = 〈x|mωx† − i†p†√

2m~ω|0〉 (42)

l =√

~2mω therefore,

√2mω =

√~l , then

√2m~ω = ~

l So, a handy formular

for future reference

A† = x2l − l

∂∂x (43)

Hence (42) becomes

Ψ1(x) = (x

2l− l ∂

∂x)Ψ0(x)

= (x

2l− l ∂

∂x)e−( x

2l)2

(2πl2)14

= (x

2l− l ∂

∂x)e−( x

2l)2

(2πl2)14

and we have found the second exited state

Ψ1(x) = x

(2πl2)14e−( x

2l)2

(44)

and since we’re at it we betray the third exited state

Ψ2(x) = H2(x)e−( x2l

)2(45)

where H2(x) is a polynomial of second degree, to be found out as an exercise.The general n-th exited state is going to be

Ψn(x) = Hn(x)e−( x2l

)2(46)

We note that the ground-state is an even function of x, the 2nd exited state(n = 1) is an odd function x. Ψn is even, if n is even, odd otherwise, parity(−1)n.

9.4 Expectation Value

Adding the Annihilation and the Creation Operator A and A† together, weobtain that

l(A+ A†) = x

Thusly we can write the expection value of x2 in state |n〉 as

〈n|l2(A+ A†)2|n〉

= 〈n|l2(A2 + A†2 + AA† + A†A)|n〉

31

Now, A2|n〉 produces |n − 2〉 which is orthogonal to |n〉, similiarly A†2|n〉produces |n+ 2〉 orthogonal to |n〉 (why??), so it reduces to

= 〈n|l2(AA† + A†A)|n〉

= l2(〈n|A|n+ 1〉√n+ 1 + 〈n|A†|n− 1〉

√n)

= l2(〈n|n〉(n+ 1) + 〈n|n〉n)

= l2(1 · (n+ 1) + 1 · n) = 2En~

〈n|x2|n〉 = l2(2n+ 1) (47)

Since l2 = ~2mω and 2 · En = (2n+ 1)~ω, thusly 2n+ 1 = 2En

~ω

〈n|x2|n〉 =2En~ω

~2mω

=Enmω2

Dimensional analysis confirms it gives the right units [L]2.)

Classically, we expect the time average of x2 → x2 = 12x

2max, because

x = Xcos(ωt)⇒ x2 =1

2X2

(the average of cos2 is 12)

Classically, the energy is an harmonic oscillator with a spring constant kis

E =1

2kx2

For a harmonic oscillator, ω =√

km , hence

E =1

2mω2x2

With x2 = 2x2

x2 =E

mω2

confirming nicely the result of the quantum mechanical case.

For doing the opposite, that is, recovering classical physics from QM, wego to large n so that th energy becomes large with a decent, that is, a rea-sonable non-discrete spectrum, known as the Correspondence Principle:

QM → classical physics

for a large quantum number n.

32

9.5 Dynamics of Oscillators

Every stationary states is highly artificial. One way one can see their arti-ficiality is

|n, t〉 = e−i(n+ 12

)ωt|n, 0〉

where |n〉 is as known the state of energy En = (n + 12)~ω, and the phase

evolves as also known with En~ = (n+ 1

2)ω. Each one of these states have aphase which increments in time with a frequency (n+ 1

2)ω, but the oscillatoroscillates at frequency ω. So we have to explain why it is that none of thesestates evolves in time with ω. The oscillator we measure in the laboratoryhave n of humongous values, like 1028, so the frequency of the phase is stu-pendous.

In order to get back to reality, to classical physics, we calcuate the ex-pecation value of x, which has to be a constant, because a stationary statehas not time evolution whatsoever.

〈n|x|n〉 = const

Any state at time t can be written as a linear combination of states ofwell-defined energy:

|Ψ, t〉 =∑n

ane−i(n+ 1

2)ωt|n, 0〉

The expectation value of x is therefore

< Ψ, t, x,Ψ, t >=∑n,m

a∗ne+i(n+ 1

2)ωt〈n|x|m〉ame−i(m+ 1

2)ωt

=∑n,m

a∗name+i(n−m)ωt〈n|x|m〉

Since we can write x as l(A+ A†)

〈n|x|m〉 = l√m〈n|m− 1〉+ l

√m+ 1〈n|m+ 1〉

We are dealing with orthogonal states. In order for 〈n|m − 1〉 to producenon-zero, m has to be n + 1, and in order for 〈n|m + 1〉 6= 0, m has to ben− 1. Therefore, from the inner sum over m there remains∑

n

l√n+ 1 · a∗nan+1e

−iωt +∑n

l√n · a∗nan−1e

+iωt

Introducing a new notationn = n′ − 1

33

we can rewrite it as∑n′

l√n′ · a∗n′−1an′e

−iωt +∑n

l√n · a∗nan−1e

+iωt

Note that the sums are dominated by large values of n′ and n, respectively.With larger n and n′, there difference vanished, so that the second sum isjust the complex-conjugate of the first sum.

If we write√n · a∗nan−1, which is a complex number as Xne

jΦ, and con-jugate a∗n′−1an′ correspondinglly as Xne

−jΦn , then

√n′ · a∗n′−1an′e

−iωt = Xn′e−j(ωt+Φn′ )

and √n · a∗nan−1e

+iωt = Xne+j(ωt+Φn)

Hence, for the expecation value of x we obtain

〈x〉 = 2l∑n

Xncos(ωt+ Φn)

We have a sinusoidal oscillation with a frequency ν = ω2π , or a period T = 2π

ω .

Note that we have obtained these result by exitations states which differs inone unit of exitation (because the position can be written as the sum of theCreation and Annihilation Operators). Our result arose from interferencesof states with difference in energy by one unit, ~ω.

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