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ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Chapter 1: Review of Quantum Mechanics

In this lecture you will learn (…..all that you might have forgotten):

• Postulates of quantum mechanics• Commutation relations• Schrodinger and Heisenberg pictures• Time development • Density operators and density matrices• Decoherence in quantum mechanics

To brush up your quantum physics, you can also see my ECE 4060 slides at:https://courses.cit.cornell.edu/ece4060/


Postulates of Quantum Mechanics: 1-3There are several postulates of quantum mechanics. These are postulates since they cannot be derived from some other deeper theory. They are known only from experiments

First three postulates are:

1) The state of physical system at time ‘ ’ is described by a vector (or a ket), denoted by , that belongs to a Hilbert space

2) Every measurable quantity (like position or momentum of a particle) is described by an operator that acts in the Hilbert space.

3) The only possible out come of a measurement of the quantity is one of the eigenvalues of the operator

t)(t

AA

A


Hilbert Spaces

Hilbert space is just a fancy name for a vector space with the following properties:

Vectors …… belong to a Hilbert space if and only if:wuv ,,

1) For some operation, denoted by ‘+’, if and

2) and

3)

4)

5) if

6) and

7) Inner product: for and , the inner product has the properties:

8) An operator in the Hilbert space has the property that if then

uv v u

vuuv wuvwuv

vv 0

0 vv

v v

uvuv vvv

v u*uvvu 0vv

O v vO


Properties of Hilbert SpacesExample of a Hilbert Space: Column vectors

dc

uba

v vu

ba

dc **

hgfe

O

Eigenvectors and Eigenvalues:

vvO ˆ

Eigenvalue

Eigenvector

Adjoint Operators: The adjoint operator of is defined by:

Let: then,

O O*ˆˆ uOvvOu

vOw ˆ*uwwu This follows from the definition

of inner productTherefore: Ovw ˆ

Hermitian Operators: An operator is Hermitian or self-adjoint if,Hermitian operators have real eigenvalues:

O OO ˆˆ

**

**ˆˆ

ˆ

vvvv

vvvOvvvvOv

vvO

**

****

bdbcadacdc

ba

uv


Properties of Hilbert Spaces

Basis Vectors: A set of vectors that “span” a Hilbert space is a called a basis

If “span” a Hilbert space then any vector in the Hilbert space can be written as:

The minimum numbers of vectors needed to span a Hilbert space is called the dimensionality of the Hilbert space

Orthogonal and Orthonormal Basis: A basis set is orthogonal if

A basis set is orthonormal if,

For an orthonormal basis set, if then the coefficients of expansion are:

nvvvv ......,, 321 w

n

kkk vaw

1

kjvv kj if0

jkkj vv

n

kkk vaw

1

wva jj


Properties of Hilbert Spaces

Complete Basis: An orthormal basis is considered complete if:

11

kn

kk vv

kn

kk

n

kkk

n

kkk

vwv

wvv

wvvww

1

1

1

1

Eigenvectors of Hermitian Operators: i) All eigenvectors of a Hermitian operator form a complete setii) Eigenvectors can be chosen to be orthonormal

0

0ˆ

ˆˆ

ˆˆ

*

*

vuvuvuuOv

vuuOvvuvOu

uuOvvO

Completeness relation


Postulate 4: Measurement and Probabilities in Quantum MechanicsSuppose the quantum state of a system is known:

Question: if a physical observable A is measured, what is the result?

We know from postulate 3 that the result has to be one of the eigenvalues of the operator

Akkk vvA ˆ

4) We cannot know for certain the result of a measurement before it is made, but the probability of getting as a result is given by: k

2kvDiscussion: we can write:

The probability of getting as result is then:

kn

kk va

1 The state is a linear superposition

of the eigenvectors of A

k22 kk va

The probabilities must all add up to unity:

n

kk

n

kkk

n

kkk

avv

vv

1

2

1

1

11


Postulate 5: Collapse of the Quantum State upon MeasurementSuppose the quantum state of a system is:

Suppose a physical observable A is measured

And the result is:

kkk vvA ˆ

j

Question: What is the state of the system just after the measurement?

i) The quantum state represents knowledge all that is knowable about the systemii) The quantum state post measurement must reflect the result of the

measurementiii) If second measurement of A is made immediately after the first measurement,

the result must be obtained with probability 1

kn

kk va

1

j

Answer: The state immediately after the first measurement must be: jv

kn

kk va

1

measurementj

jv

5) This sudden change in the quantum state upon measurement is called the “collapse” of the quantum state


Postulate 5: Collapse of the Quantum State upon MeasurementSuppose the quantum state of a system is:

Suppose a physical observable A is measured

kkk vvA ˆ kn

kk va

1

But now A has two degenerate eigenvalues: 21

Suppose the result of the measurement is:

Question: What is the state of the system just after the measurement?

i) The measurement cannot distinguish between andii) The state right after the measurement must be in the eigensubspace

corresponding to the eigenvalue :

1v 2v

2

22

1

2211

aa

vava

Measurement projects the quantum state into the eigensubspacecorresponding to the measured eigenvalue


Some Common Observables in Quantum Mechanics

The Position Operator:xxxx ˆ

)'(' xxxx Orthogonality:

Completeness: 1

xxdx

The Momentum Operator:

Orthogonality:

Completeness:

pppp ˆ

)'(' pppp

1ppdp

Position Wavefunction:

xxdxxxdxxxdx

1 xx

Momentum Wavefunction:

ppp )(d pp


Momentum and Position Commutation RelationSuppose my quantum state is a position eigenstate:

And I want to expand it in momentum basis:

x

pxpdpppdpppdp

1

What is this?We need to know something more about the relationship between position and momentum in order to find xp

Commutation Relation:ˆˆ ,ˆ ˆˆ ˆ

x p ixp px i

Fundamental property

What does mean? What if it equals 1? What if it equals 0?

There is an intimate connection between position and momentum measurements and

This connection is most elegantly expressed by the operator commutation relation

xp

xp


Momentum and Position Commutation RelationCommutation Relation:

ipx ˆ,ˆ Fundamental property

xpixpxp

ipx

ˆ,ˆ

ˆ,ˆ

Can this give ?

xpxpxpxpixpixpxpxpxp

ˆˆ)(ˆˆ

xppppxpxpxpxpxpi ˆ'''dˆˆ1ˆ)( ˆ( ) d ' ' ' '

ˆ( ) d ' ' d ' ' ' ' '( ) d ' d ' ' ' ' ' ' '

i xp p x p p x p p p xi xp p x p p p x x x x p p xi xp p x p x p x p x x p p x

This is an integral equation for and the solution is:px

22

xpixpiepxexp

xp


Momentum and Position Commutation Relation

xxx )(d

d ( )p p p

pedpppxdpxx

pxi

2

It follows that:

2)(d)(

xpiexxp

There is a Fourier transform relation between the momentum and position wavefunctions!

:ˆ)1 xx

)(

ˆˆˆ ***

xxxxxxxxxxxx

:ˆ)2 px

xx

ipep

xipepp

ppxppppppxpxpx

pxipxi

)()(2

d)(2

d

ddˆ1ˆˆ


Momentum and Position Commutation Relation:ˆ)3 xp

)()(2

d

)(dxdxˆ1ˆˆ

ppi

xxex

xxxpxxxpxpxppxi

:ˆ)4 pp

)(

ˆˆˆ ***

pppppppppppp


Mean (or Expectation) Values and Standard Deviations of Operators

Mean Value: AA ˆˆ

Suppose:

kkk vvA ˆ k

kk va

k

kkaAA 2ˆˆThen:

Standard Deviation:

AAAAA

AAA

ˆˆ2ˆˆˆ

ˆˆˆ

222

Define:

222

22222

ˆˆˆ

ˆˆˆˆ2ˆˆˆ

AAA

AAAAAAA

Then:

Operator for the uncertainty or fluctuation in A


The Fourier Transform Relation Between Momentum and Position Wavefunctions

)(x

)(x

)(p

)(p

p

p

x

x

)(

2d)( pepx

xpi

4ˆˆ

222

px

2)(d)(

xpiexxp

Can we generalize this observation to all observables and operators?


Heisenberg Uncertainty Relations

If: CiBA ˆ,ˆ

Then for any quantum state:4

ˆˆ2

22 CBA

Example:

We know that:

Therefore for any quantum state:

ipx ˆ,ˆ

4ˆˆ

222

px

This is not a surprise since the same Fourier transform relation,

)(

2d)( pepx

xpi

4ˆˆ

222

px

)(x

)(x

)(p

)(p

p

p

x

x

already told us that:


Heisenberg Uncertainty Relations and Measurements

CiBA ˆ,ˆ4

ˆˆ2

22 CBA

Consider an experiment in which one measures the observable A and the state post measurement is:

Measurement of A 2

22

ˆˆ4

CBA

The greater the accuracy in determining A, the bigger the uncertainty in B just after the measurement

Measurements and Heisenberg Uncertainty Relation:

Turns out that the above inequality is not entirely true. Under the assumption that the state observables all commute with the observables of the measurement apparatus, which is a reasonable assumption, the correct relation is:

22

2ˆ

ˆCBA

LHS is 4 times larger!!!

22

2ˆ

ˆ4CB

A


Heisenberg Uncertainty Relations and Measurements


Example: Position and Momentum (Heisenberg Microscope)

Suppose one tries to measure the position of an electron with certainty xOne needs a photon of wavelength where, x~

xp

~ Need a photon of momentum:

~p

Resulting uncertainty in the particle momentum after scattering:

~

~~

pxx

pp

CiBA ˆ,ˆ4

ˆˆ2

22 CBA

Consider an experiment in which one measures the observable A and the state post measurement is:

Measurement of A 2

22

ˆˆ

CBA


Heisenberg Uncertainty Relations and Measurements ipx ˆ,ˆ

*2

pxexp

xpi

2)(d)(

xpiexxp

4ˆˆ

222

px

This course

His

toric

al d

evel

opm

ent


The Hamiltonian OperatorThe operator corresponding to energy (a physical observable) is the Hamiltonian operator

)ˆ(2ˆˆ

2xV

mpH

For a particle of mass m in a potantial V(x), the Hamiltonian operator is:


Postulate 6: Time Development in Quantum Mechanics

The time evolution of a quantum state is given by the equation: )(t

)(ˆ)( tHtt

i

If the Hamiltonian is time independent then the formal solution subject to the boundary condition that at time t=0 the state is is:

)0(.....ˆ!2

1ˆ1)0()(2ˆ

ttHitHitet

tHi

)0( t

Stationary states: Eigenstates of the Hamiltonian are called stationary states

Suppose:ˆ

k k kH e e k

kk ect )0(Then:

kk

ti

kkk

ktHi

eececetk

ˆ

)(

The probability of being in a particular energy eigenstate does not change with time:222

)0()( jjj ctete


Matrix Representation of OperatorsSuppose we have a complete basis set:

1k

kk

Then we can write any operator as:

kjjkkj

kjjkjk

k jjjkk

A

A

AAA

ˆ

ˆ1ˆ1ˆ

jkkj AA ˆ

Matrix representation:

.....100

010

001

321

2221

1211 ....ˆ AA

AAA

Matrix representation is a mapping between the Hilbert space of physical states and the Hilbert space of column vectors


Matrix Representation of Operators: The Hamiltonian Operator

Suppose:

kkk eeH ˆ

Then:

kkkk

j kjjkjk

jjj k

jkk

jj

jk

kk

eeH

ee

eeee

eeHee

HH

ˆ

ˆ

1ˆ1ˆ

Every operator is diagonal in its own eigenbasis

2

10

....0ˆ

H


The Hamiltonian Operator with Added PerturbationSuppose we had a Hamiltonian of a particle in a potential:

)ˆ(2ˆˆ

2xV

mpHo

And we had found all its eigenstates and eigenvalues:

kkko eeH ˆ

Now suppose an additional potential is added to the Hamiltonian:

)ˆ()ˆ(2ˆ)ˆ(ˆˆ

2xUxV

mpxUHH o

jkkjk

jk j

kkjkkk

jj

jok

kk

o

exUeUeeUeeH

eexUHee

xUHHH

ˆˆ

)ˆ(ˆ

1))ˆ(ˆ(11ˆ1ˆ

Lets write the new full Hamiltonian in matrix representation using the eigenbasis of the original Hamiltonian:


2211ˆ eeeeHo

The Hamiltonian for two completely isolated potential wells (when they are far away from each other) is approximately:

The Hamiltonian for two potential wells coupled via tunneling is:

1 1 2 2 1 2 2 1H e e e e U e e U e e

Dynamics of a Two Level System

d

(1) (2)

Only diagonal elements

Off-diagonal elements!


Initial Condition: 1)0( et

Question: What is ?

2211 )()()( etcetct

)(t

Write the Solution as:And plug into:

22112211 )()(ˆ)()( etcetcHetcetct

i

Multiply by bras and to get: 1e 2e

)()()(

)()()(

122

211

tcUtctct

i

tcUtctct

i

0)0(1)0(

2

1

tctc

Answer is:

UteitcUtetctiti

sin)(cos)( 21



What are the probabilities of finding the particle in well 1 and well 2 at time t ?

Uttcte

Uttcte

222

22

221

21

sin)()(

cos)()(

The particle oscillates between the two wells!


21 sincos)( eUteieUtettiti

Solution is:


12212211

)ˆ(ˆˆ

eeUeeUeeeexUHH o

Solution Using the Exact Eigenstates of the Complete Hamiltonian:

Let:

01

1e

10

2e

U

UH

The eigenstates of the Hamiltonian are:

11

21

11

21U U

211 21 eev 212 2

1 eev




211 21)0( vvet

Expand the initial state in the eigenstates:

Find the state at any later time:

2)(

1)(

ˆ

21)(

)0()(

vevet

tet

tUitUi

tHi

Uttvvte 22

212

1 cos)(21)(

What are the probabilities of finding the particle in well 1 and well 2 at time t ?

Utte 222 sin)(


Fermi’s Golden Rule

0e

kek0

100

1000ˆ

kkkkkk

kkk eeUeeUeeeeH

Consider a single state coupled to may states – a continuum of states:

1)(

kkEED

0)0( et Suppose:

Then find:2

0 )(te

100 )()()( 0

kk

ti

kti

eetbeetbt k Let:

)(ˆ)( tHtt

i

And then use:


Fermi’s Golden Rulekek

0

1)(

kkEED

t tikk

tik

k

kk

tik

tdetbUitb

etbUt

tbi

tbeUt

tb

k

k

k

0

)(

0

)(

0

1

)(0

0

0

0

)'()(

)()(

)()(

One gets the following equations:

Solve the second one:

,3,2,1001)0(0

ktbtb

k

And substitute in the first one to get:

t tti

kk tbetdU

ttb k

00

)()(2

120 )'(1)( 0

0e

Note the correspondence between summation and integration for any quantity A:

1

kk

A dE D E A E


Fermi’s Golden Rule: Another Math Routekek

0

)(2

)(

)'('21

)'(1

)'(1

)'(1)(

0

02

00

00

2002

00

)()(2

002

00

)()(2

2

00

)()(2

20

0

0

0

tb

tbUD

tbtttdUD

tbedEtdUD

tbeEUEDdEtd

tbetdEUEDdEt

tb

t

t ttEi

t ttEi

t ttEi

1)(

kkEED

0e

t tti

kk tbetdU

ttb k

00

)()(2

120 )'(1)( 0

)()(20

20 DU

Assume constant (energy independent) density of states and coupling

10

2 )(2k

kkU


A Time-Dependent Two-Level SystemConsider a two-level system consisting of a single potential well with two confined energy levels, as shown below

11e2e 2 1 1 1 2 2 2H e e e e

In the presence of a time dependent sinusoidal electric field the Hamiltonian is,

11e2e 2

Eo cos( t )

1 1 1 2 2 2ˆ ˆ cosoH t e e e e q x E t

1 1 1 2 2 2ˆ ˆ ,H t e e e e q x t

, cos, cos

o

x o

x t x E tx t

E t E tx


A Time-Dependent Two-Level System

In the presence of a time dependent sinusoidal electric field the Hamiltonian is,

11e2e 2

Eo cos( t )

1 1 1 2 2 2 1 2 2 1ˆ 2 cos 2 cos2

RH t e e e e t e e t e e

2 1 1 2ˆ ˆR o oq E e x e q E e x e Rabi energy

ˆ ˆ ˆ ˆ1 1H t H t

1 1 2 2 1j jj

e e e e e e

Resolution of the identity:

1 1 1 2 2 2ˆ ˆ cosoH t e e e e q x E t


Schrodinger Picture and Heisenberg PicturesSuppose, given an initial state , we need to find the mean value,

at some later time

)0( t

)(ˆ)( tAt

Schrodinger Picture:

1) Find using:)(t

)(ˆ)( tHtt

i

2) Then calculate:

)0()(ˆ

tettHi

)(ˆ)( tAt

Heisenberg Picture: Notice that,

)0(ˆ)0()0(ˆ)0()(ˆ)(ˆˆ

ttAtteAettAttHitHi

tHitHieAetA

ˆˆˆ)(ˆ

Define the time-dependent operator as:


Schrodinger Picture and Heisenberg Pictures

tHitHieAetA

ˆˆˆ)(ˆ

Differentiate the above w.r.t. to get:

HtAdt

tAdi ˆ),(ˆ)(ˆ

1) Find using,

HtAdt

tAdi ˆ),(ˆ)(ˆ

)(ˆ tABoundary condition:

AtA ˆ)0(ˆ

Heisenberg equation

2) Then calculate:

tAttAttAt ˆ)0(ˆ)0()(ˆ)(

In the Schrodinger picture the states evolve in time but the operators do not evolve

In the Heisenberg picture the operators evolve in time but the states do not evolve


Heisenberg Picture

The Hamiltonian is always time-independent:

HeHetHtHitHi ˆˆ)(ˆ

ˆˆ

The commutation relation do not change with time:

CBA ˆˆ,ˆ Suppose:

Then: )(ˆˆˆˆˆˆ)(ˆ),(ˆˆˆˆˆ

tCeCeeABBAetBtAtHitHitHitHi

More precisely, the equal-time commutation relations maintain their form!


Dynamics of a Two Level System in the Heisenberg Picture

12212211ˆ eeUeeUeeeeH

Define a few operators as follows:

222

111ˆˆ

eeN

eeN

ˆˆˆ

ˆ

12

21

ee

ee

ˆˆˆˆˆ 21 UNNHThe Hamiltonian is then:

Given we need to find1)0( et

)0()(ˆ)0()(

)0()(ˆ)0()(ˆ)()()()(

22

2

11112

1

ttNtte

ttNttNtteette

21 )(te



You can verify the following commutation relations: 12

21

21

ˆˆˆ,ˆˆˆ,ˆˆˆ,ˆˆˆ,ˆˆˆ,ˆ

NNNNNN

Example: ˆˆˆˆˆˆ,ˆ 2111212111111 eeeeeeeeeeNNN

Find using:

dt

tditNtNUHtdt

tdi

dttNdittUHtN

dttNdi

ˆ)(ˆ)(ˆˆ),(ˆ)(ˆ


12

21

1

tN1ˆ

Notice that:

212121 ˆˆ)(ˆ)(ˆ0)(ˆ)(ˆ NNtNtNtNtNdtd



)(ˆ)(ˆ)(ˆ 12 tNtNtNd

Define the population difference operator as:

Equation for it is:

)(ˆ4)(ˆ2

22tNU

dttNd

dd

Boundary conditions:

ˆˆ2)(ˆˆˆˆ)0(ˆ0

12 iU

dttNdNNNtN

t

ddd

Solution is:

tUitUNtN dd

2sinˆˆ2cosˆ)(ˆ

Finally:

tUitUNtUN

tNNNtNtNtNtN dd

2sinˆˆ

2sinˆcosˆ

2)(ˆˆˆ

2)(ˆ)(ˆ)(ˆ

)(ˆ

22

21

21211



We had:

tUitUNtUNtN

2sinˆˆ

2sinˆcosˆ)(ˆ 2

22

11

UtiUtNUtNtN 2sinˆˆ2

cosˆsinˆ)(ˆ 22

212

We needed to find:

Ut

etNe

ttNtte

2

11

12

1

cos

)(ˆ

)0()(ˆ)0()(

Similarly:

tUetNete

2121

22 sin)(ˆ)(

Same as obtained in the Schrodinger picture!


Heisenberg Uncertainty Relations and Measurementsˆ ˆ,A B i C 4

ˆˆ2

22 CBA

Consider an experiment in which one is able to measure the observable A with an accuracy A

A 22 ~ˆ AA Measurement of A

2

22

4ˆ

ACB


Example: Position and Momentum (Heisenberg Microscope)

Suppose one tries to measure the position of an electron with accuracy xOne needs a photon of wavelength where, x~

xp

~ Need a photon of momentum:

~p

Resulting uncertainty in the particle momentum after scattering:

~

~~

pxx

pp


Measurements and Commutators in Quantum MechanicsCommutation Relations and Common Eigenvectors: If two operators and commute, i.e.:

then they can have the same set of eigenvectors

A B

0ˆ,ˆ BA

Proof: Suppose:

kkk vvA ˆ 0ˆ,ˆ BAand

kkk

kkk

k

vBvBA

vBvBA

vABBAABBA

ˆˆˆ0ˆˆˆ

0ˆˆˆˆ0ˆˆˆˆ

Therefore, is also an eigenvector of with the same eigenvalue as kvB A kv

1) If has all distinct eigenvalues, then and therefore is also an eigenvector of

2) If has some degenerate eigenvalues, then at least lies in the same eigensubspace. In this case, the vectors in this eigensubspace can be chosen so that they are eigenvectors of both and

A kk vvB ˆ kvB

kvB

A B


Measurements and Commutators in Quantum MechanicsCommutation Relations and Simultaneous Measurements:

Consider two observables A and B:

kkkkkk uuBvvA ˆˆ

1) Suppose the observable A is measured for a state

A measuredj jvState collapse A measured

j

2) Suppose the observables A and B are measured for a state

A measuredj jvState collapse B measured

i iuState collapse

A measuredp

jvState remainsthe same

2jv 1

2jv

2ji vu

2ip uv

pvState collapse

0ˆ,ˆ BA

Measurement of B disturbs A and measurement of A disturbs B when A and B do not commute. Accurate measurements of A and B cannot be done simultaneously


Measurements and Commutators in Quantum MechanicsCommutation Relations and Simultaneous Measurements:

Consider two observables A and B:

kkkkkk uuBvvA ˆˆ 0ˆ,ˆ BA

Since A and B commute, they have a common set of eigenvectors

kkkA ˆ

kkkB ˆ

2) Suppose the observables A and B are measured for a state

A measuredj jState collapse B measured

j State remains the same

A measuredj

2 j

State remains the same

1

j

1

j

Accurate measurements of A and B can be done simultaneously and will not affect each other


Superposition, Measurements, and Decoherence

Consider the following linear superposition state of a particle made from eigenkets of A:

2211 vcvc

An observer makes a measurement to see if the particle was in state or 1v 2v

Depending on the result the state collapses:

2211 vcvc Measurementof A

1

2 2v

1v

kkk vvA ˆ

LESSON: If an observer “looks” at a quantum state, he destroys the linear superposition structure of the quantum state and collapses it

21c

22c


Superposition, Interaction, and DecoherenceConsider again the following linear superposition state of a particle made from eigenkets of A:

2211 vcvc

kkk vvA ˆ

2211 vcvc

LESSON: If environment degrees of freedom are changed by the interaction in a way that can let an observer determine the value of A by looking at the environment, then this is equivalent to a direct measurement of A and the quantum state collapses

2v

1v

Interaction with environment (atoms, radiation, phonons, etc)


Decoherence

Any interaction with the environment can destroy the linear superposition and collapse the quantum state

The products,

present a good measure of the degree of superposition in a quantum state

2211 vcvc

1*2cc and 2

*1cc

These products are generated by the operators and : 2 1ˆ v v 1 2ˆ v v

1221 ˆˆ cccc

One can expect that as time goes by, interaction with the environment can make these products go to zero:

0)()()(ˆ)( 21

ttctctt 0)()()(ˆ)( 12

ttctctt

This phenomenon which results in the destruction of quantum mechanical superpositions is called quantum mechanical decoherence


Superposition and Decoherence

2211 vcvc Suppose:

2211 vcvc

2v

1v

We said interaction with the environment tends to destroy linear superpositions

Wait a minute!!

Define a new basis: 2121 21

21 vvvvvv

vccvccvcvc22

21212211

v

v

Environment

EnvironmentLESSON: Whether or not a superposition is destroyed depends on the exact nature of interaction with the environment (i.e. on what information is extracted by the environment during the interaction)


Pure States and Statistical Mixtures

Set A Set B

2211 vcvc

A large number of identical copies of:

A large number of states and in the ratio

of: 1v 2v

22

21 : cc

Linear superposition states(pure states) Statistical mixture of states

kkk vvO ˆ

Consider two very different sets of states:

Measurement of O over the entire set Measurement of O over the entire set

222

211ˆ ccO 2

222

11 cc

Mean value obtained: Mean value obtained:

How does one represent and/or distinguish these two states then??


Density Operator in Quantum MechanicsDensity operators are a useful way to represent quantum statesMost generally, a quantum state is not represented by a state vector but by a density operator

Density Operator for Pure States (Set A):

For pure states the density operator is: ˆ

Example:

2211 vcvc

21121221222

2112

1ˆ vvccvvccvvcvvc

In matrix representation: 101 v 0

12 v

2221

122

1ˆccc

ccc

The diagonal elements indicate the occupation probabilities, and the off-diagonal elements represent coherences

Set A

2211 vcvc

A large number of identical copies of:

Linear superposition states(pure states)


Density Operator for Pure States

For pure states the density operator is: ˆ

How do we use the density operator to calculate mean values of observables?

AA ˆˆ

AA ˆˆTraceˆ

2 21 1 1 2 2 2

2 1 2 1 1 2 1 1 2

ˆ ˆˆTrace

ˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ ˆ

ˆ

n nn

n n n nn n

n nn

A v A v

v A v A v v

A v v A

c v A v c v A v

c c v A v c c v A v

A

The mean value of an observable is defined for a pure state as:


Density Operator for Statistical Mixtures

Set B

A large number of states and in the ratio

of: 1v 2v

22

21 : cc

Statistical mixture of states

Density operators for the statistical mixture (set B) is defined as:

222

2112

1ˆ vvcvvc

In matrix representation: 101 v 0

12 v

2

2

210

0ˆc

c

Off-diagonal elements are zero no coherences!

222

2112

1

222

2112

1

ˆ

ˆTrace

ˆˆTraceˆ

vAvcvAvc

Avvcvvc

AA

How do we use the density operator to calculate mean values of observables?


Decoherence and the Density Matrix

2221

122

1ˆccc

ccc

2211 vcvc

2v

1v

Environment

Decoherence

Environment

Decoherence

2

10

0ˆ

pp

Decoherence makes the off-diagonal components of the density matrix go to zero with time!


Density Operator for General Mixed States

Suppose we have a statistical mixture of two states:

pvvv 2121

pvvv 2121

vvpvvp

1 pp

The density operator is:

In matrix representation:

10v 0

1v

pp0

0

In matrix representation:

101 v 0

12 v

212221

ˆpp

pp

1221

2211

22

21

21

vvppvvpp

vvvv

Whether or not a density operator has off-diagonal components depends on the choice of basis!


Time Development: Schrodinger and Heisenberg Pictures for Density Operators

Given a state the mean value of an observable at time ‘t’ is: 0t

dependent) time is operator (the

Picture Heisenberg0ˆ0

dependent) time is state (the

Picture rSchrodingeˆˆ

ttAt

tAttA

dependent) time is operator (the

Picture Heisenbergˆ0ˆTrace

dependent) time is state (the Picture rSchrodingeˆˆTrace0ˆˆTraceˆ

tAt

AttAttA

Given a state the mean value of an observable at time ‘t’ is: 0ˆ t


Time Evolution of Density Operators in the Schrodinger Picture

tHitHi

eAetA ˆˆ

ˆˆ Heisenberg operators obey:

At

Aete

eAettAt

tHitHi

tHitHi

ˆˆTrace

ˆ0ˆTrace

ˆ0ˆTraceˆ0ˆTrace

ˆˆ

ˆˆ

ACB

BACCBA

ˆˆˆTrace

ˆˆˆTrace

ˆˆˆTrace

tHitHi

etet ˆˆ

0ˆˆ

Where:

Differentiate w.r.t. time to get:

tHtti ˆ,ˆˆ

Equation for the time development of the density operator in the Schrodinger picture


tHtti ˆ,ˆˆ

Given a state solve: 0ˆ t

AttAttA ˆˆTrace0ˆˆTraceˆ

And then calculate:

Schrodinger picture

HtAttAi ˆ,ˆˆ

This is not the same as the Heisenberg equation for other operators:

An advantage of the Heisenberg picture is that one can calculate correlations:

2121 ˆˆ0ˆTraceˆˆ tBtAttBtA

Quantum Mechanical Correlations of Observables:

Time Evolution of Density Operators in the Schrodinger Picture


Dynamics of a Two Level System via the Density Matrix in the Schrodinger Picture


10 et Suppose: 11000ˆ eettt

0001

0ˆ t

The goal is to find:

teeeteeeete

eetteette

11

21121111

11112

1

ˆˆ

ˆTrace

Start from:

tttt

tHttHtti

2221

1211ˆˆˆˆˆˆ


Dynamics of a Two Level System via the Density Matrix

Take the 11 matrix element of the above equation by multiplying from the left by and from the right by

1e1e


ttUtdtdi 211211

tUt

eeeUeeUeeeeteetHe

tUteteeUeeUeeeeeetHe

1211

112212211111

2111

112212211111

ˆˆˆ

ˆˆˆ

Note:

tttt

tHttHtti

2221

1211ˆˆˆˆˆˆ



tdtdittUt

ti 12112221d

d

ttUtdtdi 112212

tdtdittUt

dtdi 11211222

ttUtdtdi 211211

1 00

0

2211

2211

2211

tttt

ttdtd

tt *2112

One can obtain equations for all the diagonal and off-diagonal elements:

tttt

tHttHtti

2221

1211ˆˆˆˆˆˆ



ttt

ttt

s

d

2112

1122

tiUtdtd

sd

2

It follows that:

tiUtdtd

ds

2

And then one obtains:

tUtdtd

dd 2

2

2 2

0

002

021000

2112

0

1122

ttiU

tiUtdtd

ttt

st

d

d




Utt

UttU

tρttt

tUt

d

d

222

2221111

sin

cos2

2cos1

2

2cos

Solution is:

And therefore:

tUteette

21111

21 cosˆTrace

Same as obtained earlier!


Dynamics via the Density Matrix with Decoherence

ttUtdtd i

211211

tdtdittUit

dtd

11211222

The diagonal components are unaffected by decoherence:

The off-diagonal components are assumed to decay due to decoherence:

ttUittdtd

ttUittdtd

11222121

11221212

The equation we get now is:

02d 2

2

2

tUt

dtt

dtd

ddd



02d 2

2

2

tUt

dtt

dtd

ddd

0

1000

0

1122

td

d

tdtd

ttt


Solution is:

22

2

22

sin2

cos

U

ttett

d

ttet

ttet

t

t

sin2

cos121

sin2

cos121

222

211

Therefore:



As time t →∞:

2121

22

11

t

t

0

0

21

12

t

t

Therefore as t →∞:

210

021

ˆttp


Dynamics of a Two Level System via the Density Matrix in the Heisenberg Picture


222

111ˆˆ

eeN

eeN

ˆˆˆ

ˆ

12

21

ee

ee

ˆˆˆˆˆ 21 UNNHThe Hamiltonian is:

dt

tditNtNUHtdt

tdi

dttNdittUHtN

dttNdi



12

21

1

The Heisenberg equations are:


Dynamics of a Two Level System via the Density Matrix in the Heisenberg Picture

tUitUNtUN

tNNNtNtNtNtN dd

2sinˆˆ

2sinˆcosˆ

2)(ˆˆˆ

2)(ˆ)(ˆ)(ˆ

)(ˆ

22

21

21211

The goal is to find:

tNtNttN 111 ˆ0ˆTraceˆˆTraceˆ

11000ˆ eettt When:

Solution of the Heisenberg equation is:

tUtNttN

211 cosˆ0ˆTraceˆ

The result is:

Same as obtained earlier!


Dynamics of a Two Level System via the Density Matrix in the Heisenberg Picture with Decoherence

Consider the Heisenberg operator equations:

Their average w.r.t. the density operator will yield the familiar equations:

ttUitdtd

112221

ttUitdtd

112212

tt

tt

21ˆˆTraceˆ0ˆTrace

tt

tt

12ˆˆTraceˆ0ˆTrace

In the presence of decoherence the above equations got modified:

)(ˆ)(ˆˆ

)(ˆ)(ˆ)(ˆ

12

12

tNtNUidt

td

tNtNUidt

td

ttUittdtd

ttUittdtd

11222121

11221212


Dynamics of a Two Level System via the Density Matrix in the Heisenberg Picture with Decoherence

)(ˆ)(ˆ)(ˆˆ

)(ˆ)(ˆ)(ˆ)(ˆ

12

12

tNtNUitdt

td

tNtNUitdt

td

It would be tempting to introduce decoherence the following way:

PROBLEM:Commutation rules are always satisfied provided operator time evolution equations obey the Heisenberg equation:

CBA ˆˆ,ˆ tCeCeeBAetBtAtHitHitHitHi

ˆˆˆ,ˆˆ,ˆˆˆˆˆ

Added decoherence terms

Commutation rules will get violated when we add decoherence terms arbitrarily as done above!!

Later we will introduce quantum Langevin equations to get over this problem


Joint or Poduct Hilbert Spaces

The Hilbert space of two independent quantum systems is obtained by “sticking” together the Hilbert spaces of the individual systems:

ba

An operator in this enlarged Hilbert space is written as a tensor product: BA ˆˆ

baba BABA ˆˆˆˆ

Each operator acts in its own Hilbert space:

ba Or simply:

1

1

a b

Inner Product:

ba

ba

bbaa

Operators:



1

1

a b

Full Hilbert Space:

Consider two different two-level systems “a” and “b”The full Hilbert space consists of the following four states which form a complete set:

ba

ba

ba

ba

ee

ee

ee

ee

22

21

12

11

4321

Completeness:

111221122114

1

babbbbaaaak

eeeeeeeekk



1

1

a b

The Hamiltonian for two different two-level systems is:

baba HHH ˆ11ˆˆ

122111

122111

ˆ

ˆ

eeeeH

eeeeH

bbbbb

aaaaa

Where:bHHH ˆˆ ˆ a

Or with some abuse of notation:

An eigenstate of the combined system is, for example: ba ee 21

ba

baba

bbaabbaa

bababababa

ee

eeee

eHeeeH

eeHeeHeeH

2121

212211

2121

212121

ˆ11ˆ

ˆ11ˆˆ

Eigenvalue


Unentangled StatesStates belonging to a combined Hilbert space of two systems, “a” and “b”, are of two types:

1) Unentangled states2) Entangled states

Unentangled States:

These states can be written as:

ba b""systemofstateuniqueaa""systemofstateuniquea

Examples:

bbaa

bababba

bababaa

ba

eeee

eeeeeee

eeeeeee

ee

2121

2111211

1211121

21

21

21 iv)

21

21 iii)

21

21 ii)

i)


Entangled StatesEntangled States:

Entangled states cannot be factorized or separated in the same fashion, for example:

is an entangled state and it cannot be written as:

baba eeee 122121

ba x


Consider the complicated un-entangled state of two different two-level systems:

bbaa eeee 2121 2

121

23

Unentangled States and Measurements

Meas. of energy of b

1

2

baa eee 121 21

23

1/2

1/2

baa eee 221 21

23

Meas. of energy of a

1

2

ba ee 11 3/4

1/4ba ee 12


1

2

ba ee 11 3/4

1/4ba ee 12


1

2

3/4

1/4

bba eee 211 2

1

bba eee 212 2

1

LESSON: Measurement of system b does not affect the subsequent measurement results for system a


Entangled States and Measurements

baba eeee 1221 21

23

Consider the entangled state of two different two-level systems:

Meas. of energy of b

1

2

1/4

3/4


21

ba ee 12


11


1

2

3/4

1/4

ba ee 21

ba ee 12

LESSON: Measurement of system b affects the subsequent measurement results for system a(EPR Paradox, 1935)

ba ee 12

ba ee 21 ba ee 21


Density Operators for Joint Hilbert SpacesUnentangled States:

If the quantum state of a system consisting of two subsystems “a” and “b” is an unentangled state:

then the density operator is:

ba x

ba

bbaa

babaxx

xx

ˆˆ ˆ

Therefore, the density operator can be written as a tensor product of the density operators of the subsystems

Example:

ba

bbaa eeee

ˆˆˆ21

21

2121

22122111

22122111

21ˆ

21ˆ

eeeeeeee

eeeeeeee

bbbbbbbbb

aaaaaaaaa

Where:


Density Operators for Joint Hilbert SpacesEntangled States:

Consider the entangled state:

baba eeee 122121

21121221

1122221121

ˆ

eeeeeeee

eeeeeeee

bbaabbaa

bbaabbaa

The density operator is:

The density operator for entangled states cannot be written as a tensor product of the density operators of the subsystems:

ba ˆˆˆ


Density Operators for Joint Hilbert SpacesExample:

Suppose: ba ee 21

2211ˆ eeee bbaa

Calculation of average energy of the state:

21

22221122

21221121

12221112

11221111

22114

1

2211

toequal answernonzeroagivesabovelinethirdtheonly

ˆˆ

ˆˆ

ˆˆ

ˆˆ

ˆˆ

ˆˆTrace

ˆˆˆTraceˆˆ

bababbaaba

bababbaaba

bababbaaba

bababbaaba

babbaak

babbaa

baba

eeHHeeeeee

eeHHeeeeee

eeHHeeeeee

eeHHeeeeee

kHHeeeek

HHeeee

HHHH

ba

ba

ba

ba

ee

ee

ee

ee

22

21

12

11

4321


Density Operators of Subsystems: Partial Traces

Sometimes a density operator for two (or more) systems contains too much information

If one is interested in only system “a” but has the joint density operator for system “a” and system “b”, then one needs to “extract” a density operator for system “a”:

bbbb

aeeee 2211 ˆˆ

ˆTracepˆ

Unentangled States: For unentangled states, we know that: ba ˆˆˆ

a

ba

bbbbbba

bbbabbba

bbabbbab

eeee

eeee

eeee

ˆ ˆTracepˆ

ˆˆˆ ˆˆˆˆ

ˆˆˆˆˆTracep

2211

2211

2211

And this is exactly what we expected to get!


Density Operators of Subsystems: Partial Traces

Entangled States: Consider the entangled state:

baba eeee 1221 21

23

11222112

12212211

3

3341ˆ

eeeeeeee

eeeeeeee

bbaabbaa

bbaabbaa

Density operator of the subsystem ‘a’ is obtained as follows:

410

043

ˆ

41

43

ˆˆˆTracepˆ

2211

2211

a

aaaa

bbbb

a

eeee

eeee

Therefore, is a statistical mixture of states and

a

ae1 ae2


Entanglement and DecoherenceThere is an intimate connection between entanglement and decoherence

2221

122

1ˆccc

ccc

2211 vcvc

2v

1v

Environment

Decoherence

Environment

Decoherence

2

10

0ˆ

pp

Decoherence makes the off-diagonal components of the density matrix go to zero with time!

A Brief Review:


Entanglement and DecoherenceFirst, we need to make a model of the environmentSuppose the (mutually orthogonal) environment states are:

0 1 2E E EThe initial quantum state of the system is:

The initial joint state of the “system + environment” is:

2211 vcvc

02201100 EvcEvcEt

The environment “measures” the state of the system such that at some later time the environment state reflects the system state as follows:

222111 EvcEvct This is an entangled state

Interaction with the environment

00 Et 222111 EvbEvbt

There is no state collapse!


Entanglement and Decoherence

Interaction with the environment

00 Et

There is no state collapse!

Now we find the density operator for the system by taking the partial trace of the full density operator :

full

full 0 0 1 1 2 22 2

1 1 1 2 2 2

ˆ

ˆ ˆ ˆ ˆ ˆTrace

full full full

t t t

t t E t E E t E E t E

b v v b v v

2221

122

1ˆccc

ccc

Environment

Decoherence!!

2

2

210

0ˆb

b

222111 EvbEvbt

Interaction with the environment made the off-diagonal components of the system density operator go to zero

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