Transcript C. 11
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 eit 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
it 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 + 2t
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
it 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 + Nt
m
K xN , t N ; x0 , t0
2 i t
N
2
dx
N 1
it N 1 m x x 2
i 1
i
dx
exp
V xi , ti
1
i 0
2 t
Functional Integrals
N
2
it 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 , P2 ,
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 1031 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 ei
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 ei 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
VW
• 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 11 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?
VW
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)
VW
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
VW
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
VW
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
ab
bc
cd
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