Download Solutions - Fernando Brandao

Ph125: Problem Set 5 Solutions
December 1, 2016
Problem 1
(a) We require a two-particle state which satisfies three properties: it should be invariant under
particle exchange, it should be conisistent with finding the particles in states |φi and ψi, and it
should be normalized. All but the normalization requirement are satisfied by |φi|ψi + |ψi|φi. We
can ensure the state is normalized by dividing by dividing by an appropriate factor:
=
|φi|ψi + |ψi|φi
p
(hφ |hψ | + hψ |hφ ||)(|φi|ψi + |ψi|φi)
|φi|ψi + |ψi|φi
p
2 + 2|hφ |ψi|2
(1)
(2)
(b) No, momentum is not conserved by this system. The reason is that the Hamiltonian does
not commute with the momentum operator, and therefore the momentum will change with time
in general under time evolution (by Ehrenfest’s theorem). To show directly that the Hamiltonian
does not commute with the momentum operator, observe that for an arbitrary function f (x):
i ∂
[H, p]f (x) = [p2 /2m + V (x), p]f (x) = [V (x), p]f (x) = [V0 sin(2πx/a), −
]f (x)
~ ∂x
iV0
∂f (x)
∂
= −
sin(2πx/a)
−
[sin(2πx/a)f (x)]
~
∂x
∂x
iV0 2π
= −
cos(2πx/a)f (x) 6= 0
~
a
(3)
(4)
(5)
Problem 2
(a) Substituting the following relations into |ΨAB i =
√1
2
(|ẑ; ↑iA |ẑ; ↑iB + |ẑ; ↓iA |ẑ; ↓iB ),
1
|x̂; ↓i = √ (|ẑ; ↑i − |ẑ; ↓i)
2
1
|ẑ; ↑i = √12 (|x̂; ↑i + |x̂; ↓i) |ẑ; ↓i = √ (|x̂; ↑i − |x̂; ↓i) ,
2
|x̂; ↑i
=
√1
2
(|ẑ; ↑i + |ẑ; ↓i)
(6)
(7)
we obtain,
|ΨAB i =
=
=
1
√ (|ẑ; ↑iA |ẑ; ↑iB + |ẑ; ↓iA |ẑ; ↓iB )
(8)
2
1
1
1
1
1
√
√ (|x̂; ↑i + |x̂; ↓i)A √ (|x̂; ↑i + |x̂; ↓i)B + √ (|x̂; ↑i − |x̂; ↓i)A √ (|x̂; ↑i − |x̂; ↓i)B
2
2
2
2
2
1
√ (|x̂; ↑iA |x̂; ↑iB + |x̂; ↓iA |x̂; ↓iB ) ,
(9)
2
1
as required.
Alice and Bob are friends. Bob has the bright idea to use this state to do faster than light
communication: he prepares an ensemble with many copies of the state |ΨAB i. and then carefully
separates the particles A and B in each pair, keeping particles B in Pasadena and giving particles
A to Alice who travels to Pluto. When he wants to send the message Yes he measures his particles
spins along the z direction, which immediately prepares Alice’s particles in an ensemble of the
states |ẑ; ↑i and |ẑ; ↓i. For the message No he measures his particles spins along the x direction,
which immediately prepares Alice’s particles in an ensemble of the states |x̂ :↑i and |x̂; ↓i. Alice
then measures her spins to see which state Bob been prepared and hence deduce his message. Let
us suppose that Alice measures her spins along the z axis. If Bob has sent the message Yes or No
show that Alice will measure spin up half the time and spin down half the time. Similarly if Alice
measures her spins along the x axis. Silly Bob, well at least he isn’t stuck on Pluto.
(b) If Bob sends a message Yes, then he measures in the z basis. There is a 0.5 chance the
resulting state will be |ẑ; ↑iA |ẑ; ↑iB which means alice will measure |ẑ; ↑iA , and a 0.5 chance the
resulting state will be |ẑ; ↓iA |ẑ; ↓iB , in which case Alice will measure |ẑ; ↓iA (Bob has no control
over which).
Alternatively, if Bob sends a message No, then he measures in the x basis. There is a 0.5
chance the resulting state will be |x̂; ↑iA |x̂; ↑iB . Given this occurs, Alice’s state is |x̂; ↑iA =
√1 (|ẑ; ↑i + |ẑ; ↓i), so she will measure |ẑ; ↑iA with probability 0.5, and |ẑ; ↓iA with probability
2
0.5. Similarly, there is a 0.5 chance the resulting state will be |x̂; ↓iA |x̂; ↓iB . Given this occurs,
Aalice’s state is |x̂; ↑iA = √12 (|ẑ; ↑i − |ẑ; ↓i), so (as in the other case of Bob’s measurement) she
will measure |ẑ; ↑iA with probability 0.5, and |ẑ; ↓iA with probability 0.5. Therefore, irrespective
of Bob’s outcome, Alice meaures |ẑ; ↑iA and |ẑ; ↓iA each with probability 0.5.
Hence, irrespective of whether Bob measures in the x or z basis, the outcome statistics of Alice’s
measurements are the same, so she receives no information about what Bob chose to measure, and
therefore does not receive any message, even if they use many entangled pairs.
Problem 3
Show that hJx i = hJy i = 0 in the states |j, mi and that in these states
1 hJx2 i = hJy2 i = ~2 j(j + 1) − m2 .
2
transform operators Jx and Jy into J+ and J− , we get
J+ + J−
2
J+ − J−
Jy =
2i
Jx =
1 2
2
Jx2 = (J+
+ J+ J− + J− J+ + J−
)
4
1 2
2
Jy2 = − (J+
− J+ J− − J− J+ + J−
)
4
(10)
(11)
(12)
(13)
then it is obvious that hJx i = hJy i = 0, and
1
1 hJx2 i = hJy2 i = (hJ 2 i − hJz2 i) = ~2 j(j + 1) − m2
2
2
2
(14)
Problem 4
Consider a spinless particle in a state represented by the wave-function
ψ(x, y, z) = C(x + y + 2z)exp(−γr)
p
where r = x2 + y 2 + z 2 .
(a) the angular part of the wave equation x + y + 2z can be expressed in terms of spherical
harmonics Y1± and Y10 . explicitly we have
√
x + y + 2z ∼ (1 + i)Y1−1 + (−1 + i)Y11 + 2 2Y10
(15)
since we only have spherical harmonics for l = 1, the wave function is an eigenfunction for L2 with
l = 1.
(b) the upper index of Ylm denotes its eigenvalue of Lz , in unit of ~
√
|1 + i|2 ∗ (−1) + | − 1 + i|2 ∗ 1 + |2 2|2 ∗ 0
√
=0
(16)
hLz i = ~
|1 + i|2 + | − 1 + i|2 + |2 2|2
(c) from the explicit construction of (a) part
P (Lz = ~) =
| − 1 + i|2
1
√
=
2
2
2
6
|1 + i| + | − 1 + i| + |2 2|
3
(17)