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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/
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Momentum and Position Commutation Relation:ˆ)3 xp
)()(2
d
)(dxdxˆ1ˆˆ
ppi
xxex
xxxpxxxpxpxppxi
:ˆ)4 pp
)(
ˆˆˆ ***
pppppppppppp
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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?
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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:
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Heisenberg Uncertainty Relations and Measurements
The greater the accuracy in determining A, the bigger the uncertainty in B just after the measurement
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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:
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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:
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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!
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
Dynamics of a Two Level System
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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!
Dynamics of a Two Level System
21 sincos)( eUteieUtettiti
Solution is:
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
Dynamics of a Two Level System
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Dynamics of a Two Level System
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)(
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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:
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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:
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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!
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Dynamics of a Two Level System in the Heisenberg Picture
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
ˆ)(ˆ)(ˆˆ),(ˆ)(ˆ
12
21
1
tN1ˆ
Notice that:
212121 ˆˆ)(ˆ)(ˆ0)(ˆ)(ˆ NNtNtNtNtNdtd
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Dynamics of a Two Level System in the Heisenberg Picture
)(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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Dynamics of a Two Level System in the Heisenberg Picture
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!
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
The greater the accuracy in determining A, the bigger the uncertainty in B just after the measurement
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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)
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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)
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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??
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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)
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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:
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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?
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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!
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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!
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Dynamics of a Two Level System via the Density Matrix in the Schrodinger Picture
12212211ˆ eeUeeUeeeeH
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
12212211ˆ eeUeeUeeeeH
ttUtdtdi 211211
tUt
eeeUeeUeeeeteetHe
tUteteeUeeUeeeeeetHe
1211
112212211111
2111
112212211111
ˆˆˆ
ˆˆˆ
Note:
tttt
tHttHtti
2221
1211ˆˆˆˆˆˆ
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Dynamics of a Two Level System via the Density Matrix
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Dynamics of a Two Level System via the Density Matrix
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
Boundary conditions:
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Dynamics of a Two Level System via the Density Matrix
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!
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Dynamics via the Density Matrix with Decoherence
02d 2
2
2
tUt
dtt
dtd
ddd
0
1000
0
1122
td
d
tdtd
ttt
Boundary conditions:
Solution is:
22
2
22
sin2
cos
U
ttett
d
ttet
ttet
t
t
sin2
cos121
sin2
cos121
222
211
Therefore:
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Dynamics via the Density Matrix with Decoherence
As time t →∞:
2121
22
11
t
t
0
0
21
12
t
t
Therefore as t →∞:
210
021
ˆttp
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Dynamics of a Two Level System via the Density Matrix in the Heisenberg Picture
12212211ˆ eeUeeUeeeeH
222
111ˆˆ
eeN
eeN
ˆˆˆ
ˆ
12
21
ee
ee
ˆˆˆˆˆ 21 UNNHThe Hamiltonian is:
dt
tditNtNUHtdt
tdi
dttNdittUHtN
dttNdi
ˆ)(ˆ)(ˆˆ),(ˆ)(ˆ
ˆ)(ˆ)(ˆˆ),(ˆ)(ˆ
12
21
1
The Heisenberg equations are:
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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!
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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:
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Joint or Poduct Hilbert Spaces
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Joint or Poduct Hilbert Spaces
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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)
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
Meas. of energy of a
1
2
ba ee 11 3/4
1/4ba ee 12
Meas. of energy of a
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
Meas. of energy of a
21
ba ee 12
Meas. of energy of a
11
Meas. of energy of a
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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:
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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!
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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:
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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!
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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
ECE 407 – Spring 2009 – Farhan Rana – Cornell University