11. Interpretation of Quantum Mechanics
Why Discuss Philosophy?
• It’s fun!
• Different ways of thinking about quantum mechanics sometimes involve
different calculation techniques
• Sometimes, these techniques make problems easier
My goals:
• Teach you useful techniques
• Show you how to use them to solve difficult problems
• Explode your brains
11A. The Time Evolution Operator
Definition
• The state vector |(t) is a linear function of the initial state vector |(t0)
• Call the operator that performs this function: U  t , t0    t   U  t , t    t 
0
0
• This operator must preserve probability, so it must be unitary U †  t , t U  t , t   1
0
0
• Some other easy-to-prove identities:
U  t0 , t 0   1 U  t , t  U  t , t   U  t , t 
2 1
1 0
2 0
Schrödinger’s Equation for U:
d
i
 t   H t   t 
• We know that
dt
• Therefore


i
U  t , t0    t0   H  t  U  t , t 0    t 0 
i
U  t , t0   H  t  U  t , t 0 
t
t
• This, together with the boundary condition U(t0,t0) = 1 defines U(t,t0)
Linearity of Time Evolution operator
  t   U  t , t0    t0 
• Suppose we have two solutions of Schrödinger’s Equation, |1 and |2
• We know Schrödinger’s equation is linear, so that
  c1 1  c2  2
is also a solution
• We therefore
U  t , t0    t0     t   c1 1  t   c2  2  t 
know that
 c1U  t , t0  1  t0   c2U  t , t0   2  t 
U  t , t0  c1 1  t0   c2  2  t0    c1U  t , t0  1  t0   c2U  t , t0   2  t 
• We see that U is a linear operator
• It is also reversible
  t0   U †  t , t 0    t   U  t 0 , t    t 
• Measurement, in contrast, is neither linear nor reversible
• Clearly, the time evolution operator only applies when not performing
measurements
Finding the Time Evolution Operator

i
U  t , t0   H  t  U  t , t 0 
t
U  t0 , t 0   1
• If H is independent of time, easy to solve these equations:
U  t , t0   exp  i H  t  t0 

• Suppose we have a complete set of orthonormal eigenstates of H: H n  En n
• Then insert these states into expression for U:
iE t t
U  t , t0    n e n  0  n
U  t , t0   exp  i H  t  t0    n n
n
n
• If H depends on time, expression gets complicated
U  t , t0   1   i
 t
1
t
H  t   dt    i
0
 i
t
0
 t
3
 t
2
t
0
t
H  t   dt  H  t   dt 
t0
t
t 
t0
t0
H  t   dt  H  t   dt  H  t   dt  
Sample Problem
A harmonic oscillator with mass m and angular frequency  in 1D is initially
in the state |(t0) at time t0. At a later time t, the energy is measured. What is
the probability that it will be measured to have the minimum value ½?
• The probability is just P  0   t 
• We need to evolve the initial state to the final state:
2
P  0 U  t , t0    t0 
2
 0 exp   iH  t  t0 
   t0 
2
• Now we get clever: let H act to the left:
P 0 e
iE0  t t0 
  t0 
P
m

 0   t0 
2



e
 m x2 2
2
e
 12 i  t t0 
  x, t0  dx
2
2
 0   t0 
2
Sample Problem
A spin-1/2 particle is in one of the states |+ or |- at time t0.
Find a Hamiltonian that evolves it to the state |+ by time t.
U  t , t0     and U t , t0    
• We need:
• Take the inner product of
1      U †  t , t0 U  t , t0       0
the first with the second
• This is impossible, so NO SOLUTION
• Does this mean one can never create a spin + particle?
• Yes, if this particle is the only particle in the universe, and all it has is spin
• If we have some other particle, it is always possible to do something like
U  t , t0  , e0  , e
and U  t , t0  , e0  , e
Sample Problem
Consider a superposition detector. This device is initially in the state |S0, but such that
when it interacts with a spin |, it will change into state |S-,? when faced with a pure
spin state | +  or | –  , and state |S+,? when presented with a superposition state, where
“?” means that it may represent any quantum state. Show such a device is impossible
U  t , t0   S 0     S  , a
• What we want:
• We will make no assumptions
about a, b, c, other than that the
S+ and S- states are orthogonal
S , a S  , c  0  S  , b S  , c

U  t , t0   S 0     S  , b

U  t , t0  S 0 

U  t , t0  S0  12   12 
• By linearity:
• So impossible
S , c  12  S , a  S  , b 
1  S , c S  , c 
1
2

 

1
2
1
2
 
1
2

  S , c

U  t , t0   S 0    S 0  
S , c S , a  S , c S , b
 0

11B. The Propagator
It’s Reason for Existence
• Consider a single spinless particle (in 3D)
– This can be generalized
• Given (r,t0), what is (r,t)?   r,t   r   t   r U  t , t0    t0 
• Insert a complete set of states |r0
3
  r, t    d 3r0 r U  t , t0  r0 r0   t0    d r0 r U  t , t0  r0   r0 , t0 
• Now define the propagator, also called the kernel: K  r, t; r0 , t0   r U  t , t0  r0
• Then:
  r, t    d 3r0 K  r, t ; r0 , t0    r0 , t0 
• We can find the propagator, and use it to get the wave function later in
one step
Schrödinger’s Equation for the Propagator
  r, t    d 3r0 K  r, t ; r0 , t0    r0 , t0 
• From this equation, easy to see
K  r, t0 ; r0 , t0    3  r  r0 
• Assume we have Hamiltonian
H  P2 2m  V  r, t 
• Schrödinger’s Equation
 2 2


i
  r, t    
  V  r, t     r, t 
t
 2m

2

 3

3
2
i
d r0 K  r, t; r0 , t0    r0 , t0   
  V  r, t   d r0 K  r, t ; r0 , t0    r0 , t0 

t
 2m

2





3
3
2
 d r0 r0 , t0  i t K r, t; r0 , t0    d r0 r0 , t0   2m   V r, t K r, t; r0 , t0 
• Since true for all
2

i
K  r, t ; r0 , t0   
 2 K  r, t; r0 , t0   V  r, t  K  r, t ; r0 , t0 
(r0,t0), we
t
2m
must have
Propagator for Constant H
• By definition,
K  r, t; r0 , t0   r U  t , t0  r0
• If H is constant, recall
iE t t
U  t , t0    n e n  0 
n
n
• It follows that
K  r, t ; r0 , t0    r n e
iEn  t t0 
n r0
n
• Therefore
K  r, t ; r0 , t0    n  r  e
n
iEn  t t0 
n*  r0 
Propagator for No Potential
K  r, t ; r0 , t0    n  r  e
iEn  t t0 
n*  r0 
n
• Let’s work it out for a free particle in one dimension V(x,t) = 0
2 2
k
ikx
1
k  x   2 e
• The eigenstates are plane waves
Ek 
2m
• The propagator is then
2


dk
i
k
dk ikx i k 2 t t0   2 m  ikx0 
exp ikx  ikx0 
 t  t0  
K  x, t ; x0 , t0   
e e
e

2
2m
2


• Each of these ki integrals can be done using:
1
K  x, t ; x0 , t0  
2



e
 12 Ax2  Bx
2
2


i
x

x

2m
0 m
exp 

i  t  t0 
 2i  t  t0  
2

im  x  x0  
m
K  x, t ; x0 , t0  
exp 

2 i  t  t0 
 2  t  t0  
dx  2 Ae
B 2  2 A
Sample Problem
At t = 0, the wave function of a free particle is given by   x, 0   Nxe
Find (x,t) at arbitrary time
 Ax 2 2
2
• In 1D, the final wave function

im  x  x0  
m
K  x, t ; x0 , t0  
exp 

will just be
2 i  t  t0 
 2  t  t0  
  x, t    dx0 K  x, t ; x0 , t0    x0 , t0 
2

im  x  x0  
m 
 Ax02 2
x0e
exp 
dx0

• Now we just substitute in:   x, t   N

i 2 t 
2 t


m 
2
2
1

  x, t   N
x
exp

A

im
t
x

imxx
t

imx
2 t  dx0


0
0
0
2


i 2 t 
• Use the identity:



xe
 12 Ax2  Bx
dx  2 BA
3/2 B2 2 A
  x, t   2 N   imx t   A  im t 
3/2
e
2



imx
t
 imx 2 


m
exp 
 exp 

i 2 t
2
A

im
t
2
t
  
 

Sample Problem
At t = 0, the wave function of a free particle is given by   x, 0   Nxe
Find (x,t) at arbitrary time
 Ax 2 2
2

imx t  
 imx 2 

m
3/2
  x, t   2 N   imx t   A  im t 
exp 
 exp 

i 2 t
2
A

im
t
2
t
  
 

3/2

Nx  m i t 
m2 x 2
imx 2 

exp 


3/2
 A  im t 
 2 t  A t  im  2 t 
 imx 2 A 
 m2 x 2  imx 2 A t  m2 x 2 
Nx
exp 

exp 


3/2
3/2
2
A
t

im

2 t  A t  im 
1  iA t m 
 

 1  iA t m 
Nx


 Ax 2
  x, t  
exp 

3/2
1  iA t m 
 2 1  iA t m  
Nx
11C. The Feynman Path Integral Formalism
The Idea Behind it
•
•
•
•
It is hard to find K for large time differences, but easy for small
We can build up large ones out of many small ones
2
P
Consider the Hamiltonian in 1D:
H
 V  x, t 
2m
We wish to solve:
2

d2
i
K  x, t ; x0 , t0   
K  x, t ; x0 , t0   V  x, t  K  x, t ; x0 , t0 
2
t
2m dx
• For short enough times, we expect V to change relatively little, and K to
be non-zero only near x = x0
• Estimate V(x,t) = V(x0,t0)
2

d2
i
K  x, t; x0 , t0   
K  x, t ; x0 , t0   V  x0 , t0  K  x, t ; x0 , t0 
2
t
2m dx
Propagator for Constant H
•
•
•
•
•
2

d2
i
K  x, t; x0 , t0   
K  x, t ; x0 , t0   V  x0 , t0  K  x, t ; x0 , t0 
2
t
2m dx
Multiply by eit t0 V  x0 ,t0 
Cleverly write this as
2
2

d
i  t t0 V  x0 ,t0 
i  t t0 V  x0 ,t0 



i
K  x, t; x0 , t0  e
K
x
,
t
;
x
,
t
e


0
0
2


t 
2m dx 
This is same as equation for free propagator,
K  x, t0 ; x0 , t0     x  x0 
and has the same boundary condition
It therefore has same solution, at time t1 slightly after t0, at position x1:
2

im  x1  x0  
m
i  t1 t0 V  x0 ,t0 
K  x1 , t1 ; x0 , t0  e

exp 

i 2  t1  t0 
 2  t1  t0  
Let t = t1 – t0, then we have
2



x

x
m
i

t
m


 1 0
K  x1 , t1; x0 , t0  
exp 

V
x
,
t
 0 0  
 

i 2 t
 2  t 
 



Wave Function at Time tN
2



m
 it m  x1  x0 

K  x1 , t1; x0 , t0  
exp 
 
  V  x0 , t0   
i 2 t
 2  t 
 



• Since U(t2,t0) = U(t2,t1)U(t1,t0), we can get it at t2 = t0 + 2t
K  x2 , t2 ; x0 , t0   x2 U  t2 , t1 U t1 , t0  x0   dx1 x2 U  t2 , t1  x1 x1 U  t1 , t0  x0
  dx1 K  x2 , t2 ; x1 , t1  K  x1 , t1 ; x0 , t0 
2
2



m
m  x2  x1 
 it m  x1  x0 


dx1 exp 
 
  V  x1 , t1  
  V  x0 , t0   

2 i t
2  t 
 2  t 
 



• Iterate it N times to get it at time tN = t0 + Nt
 m 
K  xN , t N ; x0 , t0   

 2 i t 
N
2
 dx
N 1
 it N 1  m  x  x  2


i 1
i
dx
exp




  V  xi , ti   
 1
i 0 
 
 2  t 

Functional Integrals
N
2
 it N 1  m  x  x
 m 
i 1
i
K  xN , t N ; x0 , t0   
 

  dxN 1  dx1 exp 
 2 i t 
i 0 
 2  t

• In limit t  0, we are considering all
possible functions xi(t) that start at x0
x0
and end at xN
• Define the functional integral:
xN  t N 

x0  t0 
 m 
D  x  t    lim 

N  i 2 t


N 2
 dx
N 1
t0 t 1 t2 t3 t4 t5
 dx1
 

  V  xi , ti   

 
2
xN
…
…
tN-1 tN
• The propagator is now:
K  xN , t N ; x0 , t0  
xN  t N 

x0  t0 

i
D  x  t  exp 


 m  xi 1  xi 2



t

V
x
,
t


 

i i 

i 0
 2  t 
 

N 1
The Lagrangian and the Action
xN  t N 
2
N 1



m  xi 1  xi 
i

K  xN , t N ; x0 , t0    D  x  t  exp   t  
  V  xi , ti   
 
x0  t0 

 i 0  2  t 

• In the limit t  0, the term in round parentheses is a derivative
xN  t N 
 i N 1

K  xN , t N ; x0 , t0    D  x  t   exp   t  12 mx 2  ti   V  xi , ti   
 i 0

x0  t0 
• The inner sum is the value of a function at various times, added up, and multiplied
by the time step
xN  t N 
t
1 N 1
2

K
x
,
t
;
x
,
t

D
x
t
exp
i
mx






N N
0 0
x0 t0   
t0  2  V  x, t  dt
– An integral



F
F
• That thing in []’s
K  xF , tF ; xI , t I    D  x  t   exp i 1  L  x, x, t  dt
xI
tI
is the Lagrangian
xF
• The integral of the
K  xF , t F ; xI , t I    D  x  t   exp i 1S  x  t  
xI
Lagrangian is the action
x
t



Second Postulate Rewritten:

K  xF , t F ; xI , t I    D  x  t   exp i
xI
xF
1

S  x  t  
• The propagator acts on the wave function to make a new wave function
• This can be generalized completely to rewrite the second postulate:
Postulate 2: When you do not perform a measurement, the state
vector evolves according to

  t    D  x  t   exp i
t0
t
1

S  x  t     t0  ,
where S[x(t)] is the classical action associated with the path x(t)
• I am being deliberately vague because we won’t ever actually use this version
• It is identical with the previous one
Why This Version of the Postulate?

  t    D  x  t   exp i
t0
t
1

S  x  t     t0 
• The Lagrangian and action are considered more fundamental then the
Hamiltonian
– Hamiltonian is normally derived from the Lagrangian
• The action is relativistically invariant, the Hamiltonian is not
• In quantum field theory, it is far easier to work with the Lagrangian
• For some problems in quantum chromodynamics, it is actually the only known
way to do the computation
Why Not This Version of the Postulate?
• To do any problem, you must do infinity integrals – hard even for a computer
• I know of no doable problem with this approach
Connection with Classical Physics

K  xF , t F ; xI , t I    D  x  t   exp i
xI
xF
1
• According to this postulate,
to go from xI to xF, the particle
xI
takes all possible paths – pretty cool
• But which ones contribute the most?
• If we consider  small, then almost
everywhere, the phase is constantly
changing for even a slight change of path
• Unless small changes in path leave the action stationary
– Stationary phase approximation
• This is the same as the classical path!

S  x  t  
xF
 S  x  t  
0
 x t 
11D. The Heisenberg Picture
Rearranging Where the Work is Done
• Quantum mechanics makes predictions about outcomes of measurements
• Can be shown: All we need to do is predict
A   t  A  t 
expectation values of operators at arbitrary time
• Using the time evolution operator, we relate this to time t0:
  t   U  t , t0    t0 
A    t0  U †  t , t0  AU  t , t0    t0 
• In Schrödinger picture, the state vector changes and the operator is constant
• Why not try it the other way?
– Let the state vector be constant and the operator changes
• Define the Heisenberg picture:
 H   S  t0 
AH  t   U †  t , t0  ASU t , t0 
• Then we have:
A   S  t  AS  S  t    H AH  t   H
Evolution of Operators in Heisenberg
•
•
•
•
Assume an operator in the Schrödinger picture has no time dependance
In the Heisenberg picture, it evolves according to: AH  t   U †  t , t0  ASU t , t0 
Recall Schrödinger’s equation for U:

U  t , t0   H S  t  U  t , t 0 
Hermitian Conjugate of this expression: i
t
 †
†
i
U  t , t0   U  t , t0  H S  t 
t
• Take time derivative of AH:
i †
i †
d

  †
†
AH  t    U  ASU  U AS U  U H S  t  ASU  U AS H S  t U
dt
t
 t 
i
i
 U H S  t  UU ASU  U † ASUU † H S  t U

i
†
†
H H  t  AH  t  
i
AH  t  H H  t 
d
i
AH  t    H H  t  , AH  t  
dt
Second Postulate In Heisenberg
d
i
AH  t    H H  t  , AH  t  
dt
Postulate 2: All observables A(t) evolve according to
d
i

A  t    H  t  , A  t    A  t 
dt
t
where H(t) is another observable.
• Note that if A has explicit time dependence, another term must be added
• If the Hamiltonian has no explicit time dependence, then H will not evolve, so
d
i
H H  t   H H  t0   H S
A  t    H , A  t  
dt
• Other postulates must be changed slightly as well
• State vector does change, but only during measurement
Heisenberg vs. Schrödinger
• Schrödinger says the state vector is constant, but the operators change
• To me, this is counterintuitive, since, for example, it is only in measurement
that a particle changes
• Since the two have identical predictions, there is no way to know which one
is “right”
• I think in Schrödinger but will do calculations in whatever is convenient
Commutation of Operators in Heisenberg
•
•
•
•
•
Suppose we have a commutation relation in Schrödinger:  AS , BS   CS
What is the corresponding commutation in Heisenberg?
†
A
t

U


t , t0  ASU t , t0 
Recall:
H
Abbreviate this: A  t   U † ASU
We therefore have
 A  t  , B  t    A  t  B  t   B  t  A  t   U † ASUU † BSU  U † BSUU † ASU
†
 U † AS BSU  U † BS ASU  U †  AS , BS U  U CSU
 X  t  , P  t    i
• For example, in 1D, we have
• Note that in unequal times, there
 X  t  , P  t     i
is no comparable relationship
• At unequal times, many operators
 X  t  , X  t     0
don’t commute with themselves
 A  t  , B  t    C  t 
Example of Operator Evolution
• Consider a free particle in 1D
H  P 2 2m
dP
dP i
0
• Let’s find evolution of momentum operator first
  H , P
dt
dt
• And now for position:
dX i
i
i
i  i 
2
 P , X  
 H , X  
P  P, X    P, X  P  

P  P

dt
2
m
2m
2m
• Need to solve these two equations simultaneously
dX 1
• Momentum one is easy:
 P
P  t   P  0
dt m
• Then we solve position one
1
X
t

X
0

t
P  0

 
• Note that
m
t
 X  0  , X  t     X  0  , P  0    i t m
m
• Recall generalized uncertainty principle:  A B   12 i  A, B 
• This implies, in this case,
t

X
0

X
t


    
2m
Sample Problem
How long can you balance a pencil on its tip before it falls over?
• Not exactly a real problem, but we’ll tackle it anyway
• We need expressions for the
2
1
kinetic and potential energy E  2 I  d dt   mgh
• If we treat pencil as a uniform
2
1
1
h

I

mL
2 L cos 
3
rod of mass m and length L, then
• In a manner similar to how this is handled classically, you define
the momentum corresponding to , and call it P P  I  d dt 
1 2 1
P  2 mgL cos 
• Then the Hamiltonian will be H 
2I
• For small angles cos   1  12  2
• So
H
1 2 1
P  2 mgL  14 mgL 2
2I

Sample Problem (2)
How long can you balance a pencil on its tip before it falls over?
1 2 1
d
i
H
P  2 mgL  14 mgL 2
A   H , A
2I
dt
• The angle variable  and its
 , P   i
corresponding momentum satisfy:
• Now work out time derivatives of operators:
i
d
i
i 1
   H ,     P2 ,  
dt
 2I
 2I
 P  P ,    P ,  P  
P
I
d
i
imgL
i
2
1
P   H , P     4 mgL , P  
  , P    , P    12 mgL

dt
4
Sample Problem (3)
How long can you balance a pencil on its tip before it falls over?
d
d
1
P  12 mgL
  P
I  13 mL2

, P   i

dt
dt
I
• Take second derivative
of  with respect to t:
• The solution to this is:
3g
mgL
d2
1 d
  

P 
2
2L
2I
dt
I dt
  t   A cosh  t   B sinh  t 
• This has boundary values:
  0  A and   0   B
• Rewrite second equation in terms of P:
• Write (t) in terms of (0) and P(0):
  t     0  cosh  t  
P  0   I B
1
P  0  sinh  t 
I
3g

2L
Sample Problem (4)
How long can you balance a pencil on its tip before it falls over?
1
  t     0  cosh  t  
P  0  sinh  t 
I
3g

2L
I  13 mL2
• Look at commutator of  at different times:
1
i
sinh  t    0  , P  0   
  t  ,  0   
sinh  t 
I
I
• Generalized uncertainty relationship:  A B   12 i  A, B 
• Therefore:
sinh  t 
   t      0   
2 I

sinh  t
   t     0  
m 2 gL3

3
3g 

2L 
Sample Problem (5)
How long can you balance a pencil on its tip before it falls over?

sinh  t
   t     0  
3
m 2 gL

3
3g 

2L 
• Typical pencil: m  0.0050 kg , L  0.170 m , g  9.80 m/s
• Substitute in:   t    0   1.18 1031 sinh t 0.1075 s


     
2
• Order of magnitude: if (0) = 1 or (t) = 1, then the pencil has
tipped over
31
1.18

10
sinh  tmax 0.1075 s   1
• Maximum time for it to balance:
• Solve for tmax:
tmax
1


  0.1075 s  sinh 
31    0.1075 s  71.9 
 1.18 10 
1
tmax  7.7 s

The Interaction Picture
•
•
•
•
Half way between Schrödinger and Heisenberg
Divide the Hamiltonian into two pieces, H0 and H1(t): H  H 0  H1  t 
Normally, H0 is chosen time-independent and easy to find the eigenstates of
Then operators evolve due to H0 and state vectors due to H1:
d
i
d
A   H 0 , A
i
  H1  t  
dt
dt
Why would we do this?
• It is a useful way to do time-dependent perturbation theory
– We will ultimately use this approach, but not use this notation
• It is a useful way to think about things
– Very common in particle physics
     
• Think of the state as “unchanging” until the pion decays
• We will, in this class, nonetheless always work in Schrödinger picture
11E. The Trace
Definition
• Let {|i} be a complete orthonormal basis of a vector space 
• Let A be any operator in that vector space
• Define the trace of A as Tr  A   i A i
Tr  A   Aii
i
i
• In components:
• Can be shown: trace is independent of choice of basis:
Tr  A   i A i   i A  j  j i    j i i A  j    j A  j
i
i
i
j
j
j
1
• Consider trace of a product of operators:
Tr  AB    i AB i
i
   j BA  j
j
  i A  j  j B i
i
j
Tr  AB   Tr  BA
 Tr  A
   j B i i A  j
i
j
1
Tr  ABC   Tr  BCA  Tr  CAB 
Partial Trace
• A trace reduces an operator in vector space  to a number
• If we have an operator A in a product space of vector spaces,  , we can
do a trace over just one of them, say , to get an operator on vector space 
• Suppose the vector spaces  and  have basis vectors {|i} and {|j}
respectively
• Basis vectors of   look like {|i,j}
• Define the partial trace as TrW  A   i i , k A  j , k  j
i
j
• This makes Tr(A) an operator on 
• In components, this is
Tr A
W
k
 ij   Aik , jk
k
11F. The State Operator / Density Matrix
Two Types of Probability
• There is a classical sense of probability that has nothing to do with quantum
mechanics:
– If I pull a card from a deck of cards, the probability of getting a heart is 25%
– We don’t believe it is truly indeterminate, just that we are ignorant
• Quantum mechanics introduces another kind of probability
– If a particle has spin + in the x-direction, and we measure the spin in the zdirection, the probability that it comes out + is 50%
• Up to now, we assumed that the quantum state is completely known
• What if there are multiple possible quantum states?
i  t  , fi
• Quantum states |i(t) each with probability fi
• The probabilities fi are
fi  0 ,  fi  1
non-negative and add to one
i
• The quantum states will be normalized, but not generally orthogonal


The State Operator
  t  , f  f  0 ,  f  1
i
i
i
i
i
• In principle, this list of possible states/probabilities could be very complicated
• Define the state operator  as
  t    fi  i  t   i  t 
i
Properties of the state operator:
• Trace: Tr      fi  j i  t  i  t   j   fi i  t   j  j i  t 
j
i
  fi  i  t   i  t 
i
  fi
j
i
Tr     1
i
†
• Hermitian (obvious)   
• Positive semi-definite: for any state vector |:
     fi   i  t   i  t     fi   i  t 
i
i
2
0
   0
Sample Problem
An electron is in the spin state + as measured along an axis at an
angle randomly chosen in the xy-plane. What is the state operator?
• We are in the normalized positive eigenstate of the operator
 i


0
e
S  cos  S x  sin  S y  12  cos   x  sin   y   12 

i
0 
e
1 1
• The normalize eigenstate is:  
 i 
2 e 
• If we knew what
 i


1
e
1
1
1 
 i
the angle  was,
      i  1 e    i

2 e
1 
2 e 
the state operator would be
• Since we don’t, and all angles are equally likely, we have to average over all angles:

2
0
d

2
  
2
0
d  1
 i
4  e
• Let’s check we got the trace right:
i
2
e 
ie 
 12 0 
1  

 i
 
1
1  4  ie
0
 0 
2
Tr     12  12  1
i
Eigenvectors and Eigenvalues of State Operator
Tr     1
†  
 
   0
i
 i  i i ,
• Like any Hermitian operator, we can find a complete,
orthonormal set of eigenstates of  with real eigenvalues
• Because it is positive semi-definite, we have
0  i  i  i i i  i
i  0
• Trace condition: 1  Tr      i  i   i
i
i  j   ij

i
1
i
i
•  written in terms of its eigenvectors:     i i   i i i
i
i
   fi  i  i
• Compare to the definition of :
i
Conclusions:
• We can pretend  is a combination of orthonormal states, even if it isn’t
• Any positive semi-definite Hermitian operator with trace 1 can form a valid
state operator
Pure States and Mixed States
   i i i

i  0
i
i
1
i
i   in
• If there is only one non-zero i, then we have a pure state
– If it isn’t a pure state, it’s a mixed state
You can prove something is pure in a variety of ways:
• You can find the eigenvalues (homework), and show they are 0 and 1
• You can find 2 and compare it to :
 2   i i i   j  j  j   i  j i i  j  j   i  j i  ij  j
i
  i2 i i
i
j
i
j
i
j
 2   for pure state
• For a pure state, i2 = i (because it is zero
Tr   2   1 for a pure state
or one), for a mixed state it isn’t
• One measure of how mixed
S  kB Tr   ln    k B  i ln i
the state is is the quantum
i
mechanical entropy
• Pure states have S = 0, mixed states have S > 0
Time Evolution of the State Operator
  t  , f   t    f  t   t 
i
i
i
i
i
i
• Whichever state vector it is in, it satisfies Schrödinger’s Equation:
d
d
i
 t   H  t 
i
 t    t  H
dt
dt
• It follows that:
d
d
d
d

i t  i t   i t 
i t  
  t    fi
 i  t   i  t    fi 
dt
dt
dt
 dt

i
i
1
1
  fi
H i t  i t   i t  i t  H   H   t     t  H 
i
i
i


d
1
  t    H ,   t  
dt
i


d
i
A t   H H  t  , AH  t  
• Don’t confuse this with the Heisenberg picture: dt H  
Expectation Values of Operators
  t  , f   t    f  t   t 
i
i
i
i
i
i
• If we knew which state we were in, the expectation value of an operator A is:
A i  i A i
• Since we don’t know which state it is, we must weight it by the probabilities:
A   fi A i   fi  i A  i   fi i A  j  j i
i
i
i
j
  fi  j i i A  j    j  A  j  Tr   A
i
j
j
A  Tr   A
Sample Problem
d
1
A  Tr   A Prove, using the
A  i  H , A
state operator, that:
dt
for operators A that have no time dependence
d
1
  t    H ,   t  
dt
i
d
d

 d  1
A  Tr    A    Tr 
A   Tr  i  H ,   A   1 Tr  iH  A  i  HA 
dt
 dt

 dt 
1
 Tr  i  HA  i  AH   1 Tr  i   H , A  1 i  H , A
Sample Problem
Show that Tr   2  is constant
d
d 
 d
 d 
Tr   2   Tr 


  2Tr 
dt
dt 
 dt
 dt 
1
1
1
 Tr  H ,      Tr  H    H    Tr  H   H    0
i
i
i
Comments on Entropy
• We showed in the previous problem that Tr(2) is constant
– Easily could have generalized it to any power
– Generalizes to trace of any function of 
• The quantum mechanical entropy is not changed by Schrödinger’s Equation
• Pure states should always evolve into pure states
• It is disputed whether this applies to gravity
– We don’t have a quantum theory of gravity
• In principle, if you put something in a black hole, it eventually comes back
out as black body radiation with a lot of entropy
Postulates in Terms of the State Operator
d
1
  t    H ,   t  
dt
i
• Note that to get (t), you don’t need |i(t) and fi, you just need (0)
• This suggests we could write postulates in terms of the state operator
Postulate 1: The state of a quantum mechanical system at time t can
be described as a positive semi-definite Hermitian operator (t) in a
complex vector space with positive definite inner product
• The equation above becomes our second postulate
Postulate 2: When you do not perform a measurement, the
state operator evolves according to
d
i
  t    H ,   t  
dt
where H(t) is an observable.
Measurement Postulates for the State Operator
 t    f  t   t 
  t  , f 
i
i
i
i
i
i
• Postulate 3 (measurements correspond to observables) doesn’t need changing
• Postulate 4 concerns the probability of getting a result a if you measure A.
P  a    P  i  P  a | i    f i  a, n  i
i
i
n
2
  f i  a, n  i  i a, n
i
n


  a , n   f i  i  i  a , n   a, n  a, n
n
 i

n
Postulate 4: Let {|a,n} be a complete orthonmormal basis of
the observable A, with A|a,n = a|a,n, and let (t) be the state
operator at time t. Then the probability of getting the result a
at time t will be
P  a    a, n   t  a, n
n
Post-Measurement State Operator
 t    f  t 
  t  , f 
This one is tricky
i t 
i
i
i
i
•
i
• Need to figure out what the probability that it was in the state i given that the
measurement produced a
– Requires a good understanding of conditional probabilities
P i  P  a | i 
P a & i
P a | i


fi  P  i | a  
 fi
P a
P a
P a
1
• Need to figure out the state vector if it
 i 
a, n a , n  i

was in the state i after the measurement
P a | i n
• Then find the new state operator
P a | i 1




   fi  i  i   fi
a, n a, n  i   i a, m a, m

P a P a | i n
i
i
m
1

a, n a, n  fi  i  i

P a n
i
 a, m
m
a, m
Post-Measurement State Operator (2)
  t    fi  i  t   i  t 
i
1
 
a, n a , n  f i  i  i  a , m a , m

P a n
i
m
1

 
a, n a, n  a, m a, m

P a n m

• This now serves as our final postulate
Postulate 5: If the results of a measurement of the observable
A at time t yields the result a, the state operator immediately
afterwards will be given by
1
 t  
a, n a , n   t  a , m a , m

P a n m

Comments on Postulates using State Operator
• The postulates in terms of the state operator are equivalent to those in terms of
the state vector
Pros and cons of using the state operator approach:
• The irrelevant overall phase in | is cancelled out in 
• The formalism simultaneously deals with both quantum and ordinary
probability
• You have to work with matrices (more complicated) rather than vectors
• Requires greater mathematical complexity
• Postulates are slightly more complicated
• Whether you believe the postulates should be stated in terms of state operators
or not, they are useful anyway
11G. Working With the State Operator
It’s Useful
• The state operator can be used even if we don’t write our postulates this way
• It allows us to prove powerful theorems, and simplify what would otherwise be
complicated calculations
• Example 1: How do you calculate scattering if the polarization of a spin-1/2
particle is random?
• Example 2: How do you calculate interactions of a particle which is produced
in a particular state, but at an unknown time
Sample Problem
Suppose a spin- ½ particle is polarized in the |+ state 50% of the
time, or the |– state 50% of the time, as measured along an arbitrary
axis described by the angles  and . What is the state operator ?
• We need to find the normalized eigenstates of the spin operator in an arbitrary
direction
S  ,    nˆ  S  12 nˆ  σ 
 12
 cos 

i
sin

e

1
2
 sin  cos 
x
 sin  sin  y  cos  z 
sin  ei 

 cos  
• Now we find the eigenstates
of this matrix:
 cos  12   
  
,
i 
1

 sin  2   e 
  sin  12   
  
i 
1

cos

e


2


Sample Problem (2)
Suppose a spin- ½ particle is polarized in the |+ state 50% of the
time, or the |– state 50% of the time, as measured along an arbitrary
axis described by the angles  and . What is the state operator ?
• Now find state operator:
  12    12  
1



sin

1  cos  12   
2 

i

 i
1
1
1
1
1
 
cos

sin

e


sin

cos

e
2 
 2   2 
2 
2  


i 
i 
1
1


2  sin  2   e 
 cos  2   e 
2 1
 i
1
1


cos

cos

sin

e






1
2
2
2
 

i
2 1
1
1
sin  2  
2  cos  2   sin  2   e

sin 2  12  
 cos  12   sin  12   ei  1  1 0 
1
 
  2  0 1 
i
2 1
1
1
cos  2  
2   cos  2   sin  2   e



• Interestingly, result is independent of angle!
Calculations for Unpolarized Particles
• Suppose the spin of a particle is completely random and uncontrolled
• It could be spin up or down on any axis
1 1 0
• As demonstrated, no matter what axis it is on, the state operator is   

2 0 1
• Since it doesn’t matter, pick any axis, say the z-axis
• Calculate interaction, say a cross section, for:
– Spin up
 , 
– Spin down
 unpol  12      
• The unpolarized cross-section is then
• In the Large Hadron collider, for instance, unpolarized protons are collided with
unpolarized protons:
 unpol  12            
Sample Problem
A particle is produced in a quantum state (t0) = 0 at a completely
unknown time t0. At time t, what does the state operator look like?
• If it is in the state |0 at time t0,
  t   U  t , t0   0
then at time t it will be in the state
• Write this in terms of eigenstates of the Hamiltonian: H n  En n
U  t , t0    n e
 iEn  t t0 
n
n
  t    n e
• If we knew the time t0, we would have
iEn  t t0 


e
 n
   t   t 
n
 iEn  t t0 
n
n  0  0 m eiE
m
 t  t0 
m
• Since we don’t, average over the time t0, from time –½T to +½T
1
   n
T n m

1T
2
 12 T
e
i  En  Em  t t0 
• We will then take the limit T  
n  0
dt0 n  0  0 m m
m
Sample Problem (2)
A particle is produced in a quantum state (t0) = 0 at a completely
unknown time t0. At time t, what does the state operator look like?
1T
1
2
i  En  Em  t t0 
   n  1 e
dt0 n  0  0 m m

T
2
T n m
• Do the integral:
1
i  En  Em t0
e
T  12 T
1T
2
 2 sin  En  Em  T 2 

dt0  
T  En  Em 

1

if En  Em
if En  Em
• In the limit T  , first expression numerator is never larger than 2, and
1 12 T i En  Em t0
denominator goes to 
lim  1 e
dt0   En , Em   nm

T
T  T
2
– So it vanishes
• If we assume that different states have different energies, expression even simpler
   n  nm n  0  0 m m
n
m
   n  0
n
2
n n
The Spectrum is (Almost) All That Matters
• For a particle in a known state emitted at an unknown time
   n  0
2
n n
n
• If it is emitted on one of several states |i at unknown time with probability fi:

   fi n  i
i
n
2

n
n
• If we measure the energy of the resulting particle, the expression in parentheses
is the probability of getting the energy En:

P  En    fi n  i
i
2

• Hence we know the state operator , i.e., we know everything we can know, if
we know the spectrum
• This might be modified if we have degenerate energy states
– But only connecting states with the same energy
– For example, for photons, polarization information
11H. Separability
Entangled States
• Consider two spin ½ particles in a state of total spin 0:   12      
• These particles can be physically separated by a large distance
– We call these pair of particles an Einstein-Podolsky-Rosen (EPR) pair
• We measure one or both spins with Stern-Gerlach devices, at arbitrary angles
What happens when we measure the first one’s spin?
• The spin of the first particle is indefinite: 50% spin up, 50% spin down
• The state afterwards will depend
50% :    
on the result of the first measurement
50% :    
N
N
Source
S
S

Instantaneous Quantum Communication?
• According to the postulates (as stated), the system,

including the second particle, changes instantaneously

when we preform a measurement

– Faster than light
• We call states like this entangled states
Can we somehow use this to communicate faster than light?
• Reduce information to bits
• Produce a pair of entangled particles, one at sender, one at receiver
• Based on each bit, decide to measure or not measure one particle
• Measure the other particle and see if the state vector has changed
N
N
sender
Source
S
S
receiver
Why It Doesn’t Work in This Case

If the sender didn’t perform a measurement, then:
1
• State is still in a superposition
    
2
• When receiver measures, result is uncertain

– 50% spin up, 50% spin down
If the sender did perform a measurement, then:
• State is not in a superposition
• But we don’t know which state: 50% spin up, 50% spin down
• There is no way the receiver can tell the state changed
Conclusion: The measurement by the sender just converts quantum uncertainty to
classical uncertainty
N
N
sender
Source
S
S
receiver
Faster Than Light Communication
• Suppose we have a quantum system distributed in two regions, one for the
sender, one for the receiver
• The quantum states for the sender live in vectors space , and for the receiver 
• The vector space for the whole system is  
• Let basis of  be |vi and of  be |wj, so those of   will be |vi,wj
Can we communicate from sender to receiver?
• We could change the Hamiltonian for the sender
• We could measure the state for the sender
• Would either of these affect a measurement for the receiver?
sender
V
receiver
W
Local Measurements and Hamiltonian
VW
• We will imagine doing a measurement A by the sender or B by the receiver
A  A 1
• These measurements will be assumed to be local,
that is, A acts on vector space  and B acts only on 
B  1 B
• The Hamiltonian, similarly, will consist of two pieces,
HV that acts only on , and HW that acts only on  H  HV 1  1  HW
• Note that these automatically guarantee that:
 A, B  AB  BA   A 11 B   1 B  A 1  A  B  A  B  A, B  0
 H , B   HV 1  1 HW ,1 B  0  1 HW ,1 B  1  HW B  1  BHW
 H , B  1  HW , B
sender
A
V
HV
receiver
B
W
HW
Can the Hamiltonian Communicate Instantly?
VW
 H , B  1  HW , B
• We can change the state vector by
d
letting it evolve under Schrödinger’s equation i dt   t    H ,   t  
• This causes the expectation
value of B to change:
1
d
1
B  i  H , B   i1   HW , B 
dt
• However, the sender can only adjust HV, not HW, so he can’t affect what the
receiver measures
• You can’t communicate instantly by modifying the Hamiltonian
sender
A
V
HV
receiver
B
W
HW
Can Measurement Communicate Instantly? (1)
VW
 A, B  0
A  A 1
B  1 B
• The state operator changes when we perform a measurement
• Would measuring A cause a change in the expectation value of B?
• Since A and B commute, we can simultaneously diagonalize both of them
– Assume that |vi and |wj are eigenstates of A and B respectively
A vi , w j  ai vi , w j
• To make the argument simpler, assume no degeneracies
First, assume we don’t measure A first
B vi , w j  b j vi , w j
B  Tr   B    vi , w j  B vi , w j
i, j
B   b j vi , w j  vi , w j
i, j
sender
V
A
receiver
B
W
Can Measurement Communicate Instantly? (2)
B   b j vi , w j  vi , w j
VW
 A, B  0
A vi , w j  ai vi , w j
B vi , w j  b j vi , w j
i, j
This time, let’s assume we do measure A first
• The probability of getting the result ai is now P  ai    vi , w j  vi , w j
j
• If we knew the result, then the
1

i 
vi , w j vi , w j  vi , wk vi , wk
state vector afterwards is

P  ai  j ,k
• But we don’t, so we
must take a weighted
    P  ai  i   vi , w j vi , w j  vi , wk vi , wk
average
i , j ,k
i
• Then we can calculate the average measurement result from B:

B  Tr    B    vl , wm vi , w j vi , w j  vi , wk vi , wk B vl , wm
l ,m i , j ,k
   il jm vi , w j  vi , wk bm vi , wk vl , wm   vi , w j  vi , wk b j vi , wk vi , w j
l ,m i , j ,k
B

  b j vi , w j  vi , w j
i, j
i , j ,k
Announcements
ASSIGNMENTS
Day
Read
Homework
Today
11I
11.3
Monday
11J
11.4, 11.5
Wednesday 11K
11.6
1/23
Instantaneous Communication?
• Bottom line: Neither measurement nor modification of the Hamiltonian
VW
can yield instantaneous communication from the sender to the receiver
• In particular, if you are never going to perform measurements in the sender’s
region, then you can ignore all particles and interactions there
• Measurements of objects in vector space  have expectation values:
• Note we are summing on the basis |vi
B   b j vi , w j  vi , w j
i, j
– This is effectively a partial trace over 
W  TrV   
• We can, in fact, work with the simplified state operator:
• Note: This means when we do an experiment with an
electron, we don’t have to worry about what it has interacted with before
sender
A
V
HV
receiver
B
W
HW
Instantaneous Communication?
• Have we proven instantaneous (or fast) practical communication is impossible?
Yes, assuming:
• Particles that carry information are no faster than light
• All terms in the Hamiltonian are local (no action at a distance)
• All measurements are local
• Quantum mechanics is valid
• Note, however, that the hypotheses of quantum mechanics do transmit quantum
information instantaneously
– Hypotheses 4 and 5 (about measurement)
sender
A
V
HV
receiver
B
W
HW
Sample Problem
A pair of spin-1/2 particles are in the pure spin state   12      
(a) What is the state operator, as a matrix?
(b) Suppose the second particle is lost forever. What is the effective state
operator for the first particle?
(c) What is the entropy of the initial and final state operators?
• We have two spin-1/2 particles. First
thing we need to do is pick a basis:
• The state vector in this basis is then:
• The corresponding
state operator is:
 
0 0

1
0
2
    
 0  12

0 0
,  ,  , 
0
 12
1
2
0

0

0
0

0
• To find the effective state operator, trace over the second spin:

0
 
1 1
 
2  1
 
0
Sample Problem (2)
A pair of spin-1/2 particles are in the pure spin state   12      
(a) What is the state operator, as a matrix?
(b) Suppose the second particle is lost forever. What is the effective state
operator for the first particle?
(c) What is the entropy of the initial and final state operators?
• To find the effective state operator, trace over the second spin:

0 0
 0 0
0 0
0 0
Tr 
Tr  1



1


1
1
0 2
 2 0


0

0

2
2
W  TrV    

 
 0  1 

1
0
 0  12 12 0 


2
2


Tr
Tr








0
0
0
0
0
0
0
0








• The initial state was a pure state, so it has entropy S = 0
1
0

2
• The final state’s entropy can be found from:
W  
1
0

2
SW  kB  i ln i  2k 1 ln 1
S  k ln 2
i
B 2
2
W
B

Quantum Decoherence
• Note that according to Schrödinger’s Equation, entropy is preserved
– Pure states evolve into pure states
• However, if particles escape the system, entropy is effectively generated
• Hence Schrödinger’s equation can result in effective irreversibility in any
situation where information is lost
• This process is sometimes called quantum decoherence
• Practically, any measurement causes the quantum system to become hopelessly
entangled with the environment
• This causes effective entropy, even without the measurement hypotheses
11I. Hidden Variables and Bell’s Inequality
Can We Get around Probabilities?
• Consider measuring a particle’s spin in a Stern-Gerlach device
• According to quantum, the outcome is probabilistic
• Could there be secret additional information
S
that actually determines the outcome?
– Details of the particle
– Details of the measuring apparatus
– Etc.
• Philosophy of hidden variables – there is some sort of additional information
that makes the outcome actually certain
– Einstein: “God does not play with dice”
– Additional information is called hidden variables
• No assumption that this additional information is accessible
• In this picture, quantum uncertainty is just hidden classical uncertainty
N
Einstein Separability and Hidden Variables
• We will study the results of an EPR experiment with
two Stern-Gerlach devices at arbitrary angles 1 and 2
 
1
2
 
 

Let’s make some reasonable-sounding assumptions about Hidden Variables:
• All quantum outcomes are actually certain
– The uncertainties are all classical, representing hidden variables
• The results of experiments on one side can’t affect the outcome on the other side
– Can be arranged by doing experiments simultaneously far apart
– Einstein Separability
• We will discover that this leads to predictions that contradict quantum mechanics
N
Source
S
Using the Assumptions of Hidden Variables
• Each of the SG devices will measure either spin up or spin down
• They will either agree or disagree with each other,
P  a  b
so we have some probability that they disagree
– According to hidden variables, due to unknown hidden variables
• According to hidden variables, we can talk about what the measurement would
have given, even if we didn’t make that measurement
– Not true in quantum mechanics!
• According to Einstein separability, performing the measurement on one side
can’t effect the outcome on the other side
– Not true in Copenhagen interpretation!
N
Source
S
Bell’s Inequality (Carlson’s Version):
• Imagine measuring spin 1 in one of the two directions a or c
• Imagine measuring spin 2 in one of the two direction b or d
• I now make the following logical claim:
a
d
c
If a = b and b = c and c = d, then a = d
• This is logically equivalent to the following claim:
If a  d then a  b or b  c or c  d
• From which follows the following probability statement:
P  a  d   P  a  b   P b  c   P  c  d 
N
Source
S
b
ab
bc
cd
Bell’s Inequality vs. Quantum Mechanics
P  a  d   P  a  b   P b  c   P  c  d 
a
d
45
• Hidden variables plus separability implies Bell’s Inequality
90
• What does quantum mechanics predict?
90
135
– Homework problem
P  a  b   cos 2  12  a  b  
b
• Assume particular angles
• Then we have: P a  b  P b  c  P c  d  cos 2 67.5  0.1464 ,

 
 



P  a  d   cos 2  22.5   0.8536
0.8536  3 0.1464  0.4393
• Quantum mechanics predicts Bell’s Inequality is violated!
c
Testing Hidden Variables
Quantum mechanics contradicts Bell’s Inequality
Three possibilities:
1. Hidden variables is incorrect
2. Information can move faster than light
3. Quantum mechanics makes incorrect predictions
• The third one can be, and has been, tested experimentally
– Not exactly with this setup
– Uses photons
• The results agree with predictions of quantum mechanics
N
Source
S
Salvaging Hidden Variables
Can we save hidden variables from extinction?
Experiment assumes all photons are measured
• Realistically, only a fraction are captured
• If this is the explanation, then quantum mechanics would become
obviously wrong if we had perfect detectors
Perhaps initial state is modified by setup of the detectors
• Solution: Adjust detector at the last moment
• Choose it randomly by computer
Counter-argument – computers aren’t random
• Use fast graduate students and large (astronomical?) scale instead
• Do graduate students have free will?
N
Source
S
11J. Measurement
Why is it Important?
• Three of our five postulates of quantum mechanics have to do with measurement,
and it was mentioned in one other
1

a, n a , n   t 
• These are among our most complicated   t  

P a n
and least elegant postulates:
• It is very different from our other postulates
– Irreversible
– Non-linear
– Time asymmetric
– Probabilistic
• Worst of all, It was never defined
– A theory with undefined expressions is not a theory
• 60% of our postulates concern measurement; 90% of physics concerns only
Schrödinger’s equation
What constitutes a measurement?
• Take a single spin-½ particle in the state
 x  12     
• Put it through a Stern-Gerlach device
• When does the measurement occur
S
– When it passes through the magnets?
– When it hits the screen?
– Later?
• We can test if state vector collapses when you split the beam
• Recombine the two beams
• Then measure the x-component of the spin
1


 
x
• If it is all deflected one way, then state is
2
• With photons, this has definitely been demonstrated
N
N
N
+
S
S
How Does Measuring Actually Work?
• Theorist picture:
x 
• Experimentalist picture:
1
2

 

Measuring
device


 50% 
 50% 
• Real detectors are built on principles of how particles interact
• Particles interact via the Hamiltonian/Schrödinger equation
• Maybe these small interactions don’t really cause collapse of the state vector
• Many books (at least those that discuss it), imply or state that it is when effects
get macroscopic that measurement occurs
– One electron doesn’t count, but one milliamp does
Quantum Measurement Devices
• Since measuring devices interact with particles via the Hamiltonian, we
should include these measuring devices in our state vectors
M 0  Not measured yet
• We might think of a simple measuring device M that
measures the spin state as having three states:
M   Measured, result 
• The state of the system will be described by the state
M   Measured, result 
of the spin and the state of the measuring device:
• Interactions cause the state of the measuring  , M
, M 0  , M 
device to change to reflect the measurement:
• Since this is due to some sort of interaction Hamiltonian, , M 0  , M 
there must be a corresponding time evolution operator:
U  tm , t0  , M 0  , M  , U  tm , t0  , M 0  , M 
• Recall that the time evolution operator is linear!
U  t m , t0 
1
2

    M 0  U  t m , t0 
1
2
 , M
0
 , M 0
   , M
1
2

 , M 

When Does Measurement Occurs?
• When does state vector collapse occur?
• Assume first it occurs just after the measurement occurs
, M 
1
1

,
M


,
M



 , M   , M  
0
0
2
2
, M 
 50% 
 50% 
• Now assume it occurs just before the measurement occurs
1
2
 , M
0
 , M 0

, M 0
 , M 
 50% 
, M 
 , M 
 50% 
• The final state is the same, whether you think of it occurring before or after
the measurement
Measurement Without Collapse (1)
N
N
S
Path Detector
+
S
x  
• We previously explained that we can demonstrate that
no measurement occurs, because the final state is still
• Measuring Sx then shows us that
 Sx   12
  x   14
      
x
  
1
4
 
1
2

     
 
     12
• Now add a measuring device
1

,
M

 , M 0  , M 0
0
• Before measurement, quantum state is
2
• After measurement, state is
  12  , M   , M  
• Assume no state vector collapse occurs. What is Sx?


Measurement Without Collapse (2)
Path Detector
N
S
 
1
2
N
+
S
x  
 , M
 , M 


• Assume no state vector collapse occurs. What is Sx?
 Sx   12

1
2
  x   12
 , M

 , M
 , M 

 , M
 , M    x  , M   , M 

 , M 
 0
• Interference is lost without collapse of the state vector!
• Due to particle becoming entangled with the measuring device

How Long Can We Delay Measurement?
• Consider a system consisting of a source, a measurement device, a computer,
and an experimenter

 

N
1
2
S
Path Detector
 
• Initial state, particle is in a spin state, but the measuring device, the computer,
and the experimenter don’t know what it is:
0   , M 0 , C0 , E0
• Now, we can assume the collapse occurs at any stage:
– Just before the first measurement
– Between measurement and computer
– Between computer and experimenter
– Just after the experimenter gets the data
• What is final state in each case?
How Long Can We Delay Measurement? (2)
• The steps involved depend on when we think collapse of the state vector
0
, M 0 , C0 , E0  , M  , C0 , E0  , M  , C , E0  , M  , C , E
, M 0 , C0 , E0  , M  , C0 , E0  , M  , C , E0  , M  , C , E
1  , M  , C0 , E0   , M  , C0 , E0  , M  , C , E0  , M  , C , E
0 


2  , M  , C0 , E0  , M  , C0 , E0  , M  , C , E0  , M  , C , E
1  , M  , C , E0   , M  , C , E0  , M  , C , E
0  


2  , M  , C , E0  , M  , C , E0  , M  , C , E
0 
1  , M  , C , E  



2  , M  , C , E 
, M  , C , E
, M  , C , E
• But the final situation does not
• Can we push it later, i.e., can we leave the experimenter in a superposition?
How Long Can We Delay Measurement?
• Let’s add a theorist:
Path Detector
N
S
• Before the experimenter
  12  , M  , C , E , T0  , M  , C , E , T0
communicates to the theorist,
, M  , C , E , T
• Then, when the theorist gets the result, 
0
, M  , C , E , T
• According to this hypothesis, the experimenter is in a quantum superposition
up to the moment the theorist reads the published paper
• Schrödinger’s cat
• Wouldn’t the experimenter know she was in a superposition?

A Slightly More Sophisticated View
Path Detector
N
S
• Let’s describe the experimenter as having two internal degrees of freedom
– The result of the measurement
E, D
– Whether they are in a superposition or not
• Each of these two degrees of freedom have at least three states:
E0 Haven't measured yet
D0 Haven't measured/thought about yet
E Measured came out 
E Measured came out 
D Thought about it, in a definite state
D Theought about it, in superposition
Thinking Quantum Thoughts
Path Detector
N
S
• Now, we imagine a process I call “introspection” where the experimenter
examines her thoughts and decides if she is in a superposition of thoughts
• In particular, if she is given a pure spin state, introspection will cause
U  tI , tE  E , D0  E , D
and U tI , tE  E , D0  E , D
• Because in each case, the experimenter is in a mental pure state
• But U is a linear operator, and therefore:
U  tI , tE 
1
2
 E ,D

0
 E , D0
   E ,D
1
2


 E , D

Getting Extreme
• Let’s put the theorist back in:
Path Detector
N
S
1


  
• Put in a particle in the state
2
• The detector detects, the computer records, the experimenter reads it
 

1
2
 , M

, C , E , D0 , T0  , M  , C , E , D0 , T0

• Now the experimenter incorrectly concludes she is not in a superposition
 

1
2
 , M

, C , E , D , T0  , M  , C , E , D , T0
• Only when the theorist asks does
the state vector collapse

, M  , C , E , D , T
, M  , C , E , D , T

Quantum Solipsism
• Solipsism, from dictionary.com:
Solipsism: Noun
• I think therefore I am, but I’m not so 1. Philosophy. the theory that only the
sure about you
self exists, or can be proved to exist.
• One can consistently take the attitude
that only I can collapse the state vector
• In this view, the entire universe was in a complicated
superposition until November 22, 1961
• It then collapsed. If not for me, you probably
wouldn’t exist
• As far as I know, there is nothing wrong with this
as a quantum philosophy
• This seems remarkably egotistical
• As far as I know, no one takes this seriously
Quantum
collapse
device, ca.
2000
11K. The Many Worlds Hypothesis
Let’s Take it One More Step
• We can delay the state vector collapse as much as possible:
• Treat all measuring devices, recording devices, and even experimenters as
complicated quantum objects interacting with the quantum particles
• These are governed by a Hamiltonian, which implies that we can only measure
observables
• We found no inconsistencies with observations
• What happens if we assume the state vector never collapses?
•
•
•
•
Postulate 2 (Schrödinger’s equation) always applies
Postulate 3 (measurement  observables) is built in already
Postulate 5 (state vector collapse after measurement) never applies
Postulate 4 will take some discussion …
Many Worlds Postulates of Quantum
• We are going to go from five postulates of quantum mechanics to two:
Postulate 1: The state vector of a quantum mechanical system
at time t can be described as a normalized ket |(t) in a
complex vector space with positive definite inner product
• This one is unchanged from before
Postulate 2: The state vector evolves according to

i
 t   H t   t  ,
t
where H(t) is an observable.
• This one used to start, “When you do not perform a measurement, …”
The Fourth Postulate Revisited
Path Detector
• The fifth postulate talks about the state vector changing due to measurement
• This can roughly be explained in terms of decoherence
– Information lost during measurement mimics effective change of state
vector (or state operator)
• The FOURTH postulate makes a specific prediction, however
• Suppose we take an electron in the spin state
• Put it through Stern-Gerlach: + x  1     
2
• Copenhagen predicts
– 50% takes upper path
S
– 50% takes lower path
N
• Many Worlds predicts:
 
1
2
 , M

 , M 

• There is no statement or implication of probability, so how can we reproduce
this correct prediction (50% each) in Many Worlds?
Can We Ever Predict Things In Many Worlds?
N
0

• We can make a definite prediction if it is in an
eigenstate of what is being measured
Path Detector
• It seems like, because of superposition, we might never predict
anything in many worlds
z  
• Imagine we have spin state
• Stern-Gerlach measurement:
S
, M  , M
• Consider a continuous variable, like the position x
• For a general wave function, we can’t predict the result:
• The position has uncertainty x
• But if x  0, the result will be definite
• We can predict the result for any variable with zero uncertainty
Probability: What Does it Mean?
• What does it mean when you say something has a 50% probability of being
spin up?
• It DOES NOT imply anything specific about a single spin
• Instead, it implies something about what happens if you repeat it many times
•
•
•
•
Perform the experiment N times
Keep track of N+ and N-, the number of times each outcome occurs
Prediction is only in the limit N  
N
N
lim

lim
 0.500
We have
N  N
N  N
• Define the average spin of many particles as
• Then the probability statement is just saying
1
Sz 
N
N
S
i 1
lim S z  0
N 
zi
lim S z  S z
N 
Quantum States for Repeated Measurements
•
•
•
•
Copenhagen does not make a prediction about the results of one experiment
It is making a prediction about an experiment that is repeated many times
The initial state is not
  + x  12     
Instead, it is an N
  +x  +x   +x
particle state:
• We are not measuring Sz
1 N
S z   S zi
• We are measuring:
N i 1
Our goal:
• Find the expectation value of this average
• Find the uncertainty of this average
• Take the limit N  
Sz   Sz 
 S 
z
2
 S
2
z
 Sz
We want to know:
• Does the average value match the prediction of Copenhagen?
?
• Does the uncertainty go to zero? S  0
z
2
?
Sz  Sz  0
Calculating with Many Particles
  +x  +x 
 +x
• Recall: S z    12
• Then we have
x 
1
2

 

x 
1
2

 


Sz  x 
1
2
1
2

 

1
2
x
• If we just let one Szi act on |, we have
Szi   Szi +x  +x 
• So we have
 +x 
 +x  12
+x  +x 
 x  +x   +x 


N
  x  x   +x 
1

S z    S zi  

N i 1
2N 



       
x
x
x 

 x 
 +x
Does the Expectation Value Equal Sz = 0 ?
  +x  +x 
1
Sz 
N
N
S
i 1
zi
• First note that:
 +x
 x  +x   +x 


  x  x   +x 
Sz  

2N 



       
x
x
x 

x  x 
1
2
     
• Now let’s find the expectation value:
Sz   Sz  


2N
1
2


2N
N  x x
0 1N 1  0
 
x
x
 x 
x

N 1
 
 0
x 
1
2

 

x 
1
2

 

 x  x  x x  1
 x  +x   +x 


   x  x   +x 
 x 




       
x
x
x 

Sz  Sz  0
Does the Uncertainty Vanish as N   ? (1)
  +x  +x 
 +x
 x  +x   +x 
x  x  0


N
1
  x  x   +x 
S z   S zi
Sz  
 x  x  x x  1


N i 1
2N



       
• We now need
x
x
x 

2
2
2
Sz   Sz   Sz 
  x   x    x   x  + x   + x 


2

    x   x    x    x   x   + x 

 


2
N



 


                
x
x
x 
x
x
x 

2
N 1
N 2
2



N





N

N















x
x
x
x
x
x
x
x
x
x

4N 2 

2
 N 1   N  N  0 
2 

4N
2
S z2 
2
4N
Does the Uncertainty Vanish as N   ? (2)
Sz  0
• The Uncertainty is given by
S
 S 
z
S z 
2
2
z

2
4N
 S z2  S z
2

2
4N
2 N
• Now take N  :
lim S z  0
N 
• Conclusion: In the limit N  , Many Worlds makes the same prediction as
Copenhagen
Occam’s Razor
• William of Ockham, (c. 1287-1347):
“Pluralitas non est ponenda sine necessitate”
• For those who don’t speak Latin:
“Plurality is not to be posited without necessity”
• Given a choice between five postulates and two, two is better
• However, it does imply a very complicated view of the universe
What the Universe Looks Like
• What led us here today, according to Copenhagen interpretation:
inflation field
Milky Way
No Milky Way
No Solar System
Solar System
• What let us here today, according
to many worlds:
 sin  Milky Way
inflation field  
  cos  No Milky Way
 sin  sin  sin  Earth

  sin  sin  cos  No Earth

  sin  cos  No Solar System
  cos  No Milky Way




Earth
No Earth
Eric
No Eric
 sin  sin  Solar System

   sin  cos  No Solar System

  cos  No Milky Way





   sin  sin  sin   cos  No Eric  sin  Eric  
 

   sin  sin  cos  No Earth

 

   sin  cos  No Solar System

 


cos

No
Milky
Way
 

Do People Take Many Worlds Seriously?
• Many famous physicists take Many Worlds Seriously
Murray
Gell-Mann
Steven
Weinberg
Postulated
Quarks
Stephen
Hawking
Wizard of
black holes
Electroweak
Theory
Richard
Feynman
Unknown
Name
Feynman
Diagrams
Expert on
11/11/11
So Many Choices
•
•
•
•
You have a variety of options, like ordering off a menu
You can mix and match, almost any combination works
My favorite combination, philosophically, marked in yellow
What we use in this class, marked in black
Appetizer:
State Vector State Operator
Salad:
Schrödinger’s Eqn.
Feynman Path Integral
Entrée:
Schrödinger
Heisenberg
Dessert:
Copenhagen
Many Worlds Others
Interaction
• Using Copenhagen means we don’t have to infinitely repeat experiments