University of California at San Diego – Department of Physics – Prof. John McGreevy TA:
Shauna Kravec
Quantum Mechanics (Physics 212A) Fall 2015
Assignment 3 –
Solutions
Due 12:30pm Wednesday, October 21, 2015
1. Neutrino oscillations [This problem has some overlap with a problem from the
previous homework, but let’s celebrate this year’s Nobel Prize in Physics.]
Neutrinos are subatomic particles which come in several types; different types interact
differently with matter. Consider a neutrino which can be an electron-neutrino νe
or a muon-neutrino νµ . We can describe its wave function as |ψi = a|νe i + b|νµ i,
where |νe i, |νµ i are orthonormal. A neutrino in the state |νe i or |νµ i is detected as an
electron-neutrino or muon-neutrino, respectively, with certainty. The Hamiltonian is
H=p+
M2
2p
where p is the (magnitude of the) momentum, and M2 is a 2×2 matrix with eigenvectors
|ν1 i = cos θ|νe i + sin θ|νµ i|ν2 i = − sin θ|νe i + cos θ|νµ i
(1)
with respective eigenvalues m21 and m22 . (We work in units with c = 1 for this problem.)
(a) Evaluate
cei ≡ hνi |νe i and cµi ≡ hνi |νµ i,
i = 1, 2
as functions of θ.
From the eigenvectors above, we have
|ν1 i
|νe i
=R
|ν2 i
|νµ i
which means
|νe i
cos θ − sin θ
|ν1 i
−1 |ν1 i
=R
=
|νµ i
|ν2 i
sin θ cos θ
|ν2 i
so that e.g.
|νe i = cos θ|ν1 i − sin θ|ν2 i
and
cei ≡ hνi |νe i = (cos θ, − sin θ)i , cµi ≡ hνi |νµ i = (sin θ, cos θ)i
1
(b) At t = 0 a neutrino with momentum p is in the state |νe i. If we were to measure
the observable M2 , what are the possible outcomes and with what probabilities
will they occur?
Possible answers are m21 , m22 . They occur with probability
Prob(m2i ) = hνe | (|νi ihνi |) |νe i = |cei |2
So
Prob(m21 ) = cos2 θ,
Prob(m22 ) = sin2 θ.
(c) At t = 0 a neutrino with momentum p is in the state |νe i. At time t it reacts
with matter inside of a detector. What is the probability that it is detected as a
muon-neutrino?
m2
[Hint: I found it useful to introduce the notation αi ≡ t p + 2pi .]
The state evolves to
|ψ(t)i = e−iHt |νe i = e
2
−it p+ M
2p
|νe i.
Note that we can treat p as a number. To proceed we should insert the identity
in the form of a spectral resolution of M2 :
X
X
1=
|νi ihνi |, M2 =
|νi ihνi |m2i ,
i=1,2
i=1,2
or better, for the time evolution operator:
e−iHt =
X
m2
−it p+ 2pi
|νi ihνi |e
.
i=1,2
X
|ψ(t)i =
m2
−it p+ 2pi
e
|νi ihνi |νe i
i=1,2
From the previous part, we have
hν1 |νe i = cos θ, hν2 |νe i = − sin θ
which gives
|ψ(t)i = cos θe
m2
−it p+ 2p1
|ν1 i − sin θe
m2
−it p+ 2p2
|ν2 i ≡
X
eiαi ci |νi i.
i=1,2
The probability that a neutrino in this state is detected as a νµ is
X
X
+iαi e
2
Probµ (t) = hψ(t)|νµ ihνµ |ψ(t)i =
hνj |e−iαj ce?
|ν
ihν
|e
c
|ν
i
=
|
e−iαi cei cµ?
µ
µ
i
j
i
i |
ij=1,2
2
i=1,2
µ
e
Here
ci 2 ≡
hνi |νe i = (cos θ, − sin θ)i , ci ≡ hνi |νµ i = (sin θ, cos θ)i and αi ≡
m
t p + 2pi .
−iα2 2
−iα1
| = | sin θ cos θe−iα1 − cos θ sin θe−iα2 |2
+ cµ2 ce?
Probµ (t) = |cµ1 ce?
2 e
1 e
= | sin θ cos θ ei(α2 −α1 )/2 − ei(α1 −α2 )/2 |2 = | sin θ cos θ2 sin (α2 − α1 ) /2|2
2
m1 − m22
2
2
= sin (2θ) sin
ct .
4p
In the last step I restored the speed of light. Note that ct is the approximate
distance travelled by the neutrinos (if m1,2 are small).
(d) [bonus] Restore the factors of c.
See the last expression from the previous part. In the Hamiltonian, it would be
H = pc +
M2 c4
+ ...
pc
Note that these are the leading terms in an expansion in the limit M2 p2 .
2. Density operator is non-negative.
Show that even if the states |αi i are not orthogonal, the density operator ρ =
is non-negative, in the sense that
P
i
pi |αi ihαi |
hψ|ρ|ψi ≥ 0 ∀ψ.
Check that this condition implies that all eigenvalues of ρ are non-negative.
Note that the pi are probabilities bounded in [0, 1]
P
hψ|ρ|ψi = i pi |hαi |ψi|2 which is ≥ 0 because |hαi |ψi|2 ≥ 0
This implies the eigenvalues are non-negative by considering |ψi an eigenvector of ρ
3. Density operators for two-state systems.
For the case of a single qbit, the most general density matrix is a hermitian 2 × 2
matrix, so can be written
1
ρ = n0 1 + ni σ i .
2
Find the restrictions on the na from the properties of the density matrix. What is the
geometry of the space of allowed ~n ? Which ones are the pure states?
We must have 1 = Tr ρ = 2n0 (recall that the σ i are traceless), so
1
1 1 + n3 n1 − in2
~) =
ρ = (1 + ~n · σ
.
2
2 n1 + in2 1 − n3
(2)
Its determinant is det ρ = 41 (1 − ~n2 ). No negative eigenvalues requires 0 ≤ ρ1 ρ2 =
det ρ =⇒ ~n2 ≤ 1. (This is in fact sufficient since Tr ρ = 1 means at most one
3
negative eigenvalue for the density matrix of a qbit.) So: possible density matrices of a
qbit correspond to points inside the unit ball {~n|0 ≤ ~n2 ≤ 1}. This is called the Bloch
ball.
Its boundary |~n| = 1 is the Bloch sphere: it corresponds to density matrices whose
determinant vanishes; since Tr ρ = 1, the nonzero eigenvalue of such an operator must
be 1, and these are projectors onto pure states. (We can see directly that ρ in (2)
~ · n̂2 = 1.) With a little effort, we can figure out
is a projector when ~n2 = 1 using σ
which pure state; you won’t be too surprised that n̂ is the direction in which the spin
is pointing up.
4. A density operator is a probability distribution on quantum states. [from A.
Manohar]
An electron gun produces electrons randomly polarized with spins up or down along
one of the three possible radomly selected orthogonal axes 1, 2, 3 (i.e. x, y, z), with
probabilities
1
1
pi,↑ = di + δi
2
2
1
1
pi,↓ = di − δi
2
2
i = 1, 2, 3
(3)
Probabilites must be non-negative, so di ≥ 0 and |δi | ≤ di .
(a) Write down the resultant density matrix ρ describing our knowledge of the electron
spin, in the basis | ↑i, | ↓i with respect to the z axis.
ρ is a sum of six projectors onto the eigenstates of X, Y , and Z weighted by the
appropriate probabilities. Using the Z basis I’ll write the associated matrices:
ρ = ρX + ρY + ρZ
py,↑ + py,↓ −i(py,↑ − py,↓ )
px,↑ + px,↓ px,↑ − px,↓
pz,↑ 0
1
1
ρX = 2
ρZ =
ρY = 2
i(py,↑ − py,↓ )
py,↑ + py,↓
px,↑ − px,↓ px,↑ + px,↓
0 pz,↓
~ · n̂) and | ↓n ih ↓n | = 12 (1 − σ
~ · n̂)
These are given by recalling | ↑n ih ↑n | = 21 (1 + σ
P
1
i
i
So ρ = 2 i pi,↑ (1 + σ ) + pi,↓ (1 − σ )
Summing these and using the form of the probabilities above one finds:
1 (dx + dy + dz ) + δz
δx − iδy
ρ=
δx + iδy
(dx + dy + dz ) − δz
2
(b) Any 2 × 2 matrix ρ can be written as
~
ρ = a 1 + ~b · σ
(4)
in terms of the unit matrix and 3 Pauli matrices. Determine a and ~b for ρ from
part (a).
P
Using i di = 1 we can determine by inspection that a = 12 and ~b = 12 ~δ
4
(c) A second electron gun produces electrons with spins up or down along a single
axis in the direction ň with probabilities (1 ± ∆)/2. Find ň and ∆ so that the
electron ensemble produced by the second gun is the same as that produced by
the first gun.
Such a gun corresponds to a density matrix which can also be written in the above
form ρ = 21 (1 + ~n · ~σ ) so by inspection we should choose n̂ along ~δ and ∆ = |~δ|
5. Purity test. [from Chuang and Nielsen]
Show that for any density matrix ρ:
(a) Tr ρ2 ≤ 1
(b) the inequality is saturated only if ρ is a pure state.
[Hint: don’t forget that the trace operation is basis-independent.]
P
Evaluate the trace in the eigenbasis of ρ: Tr ρ2 = i λ2i . The eigenvalues are positive
real numbers whose sum
That means each one is less than one. That means
P is2 1. P
2
λi ≤ λi ∀i. Therefore i λi ≤ i λi = 1. And at most one of these inequalities can
be saturated, and that only if only one of the λi is nonzero, in which case the density
matrix is a 1d projector, i.e. a pure state.
6. Polarization. [from Boccio]
In the expression for the general density matrix of a qbit
1
~
ρ=
1 + P~ · σ
2
the vector P~ is called the polarization.
(a) To justify this name, show that it gives the expectation for the spin operator in
the state ρ:
P~ = h~
σ iρ .
~ σi ] = 12 Tr [Pi 1] = Pi
hσi i = Tr [ρσi ] = 12 Tr [σi + P~ · σ
(b) Subject the qbit to a external magnetic field, which couples by
~
~ = − γ σ · B.
H = −~
µ·B
2
(γ is called the gyromagnetic ratio.) Assuming the qbit is isolated, so that its
time evolution is unitary, what is the time evolution of the polarization, ∂t P~ ?
∂t ρ = −i[H, ρ(t)] = −i
−γ
γ~
~.
Bi Pj [σ i , σ j ] = − B
× P~ · σ
| {z }
2
2
=2iijk σ k
and
1
~
∂t ρ = ∂t P~ · σ
2
5
So
~ × P~ .
∂t P~ = −γ B
7. von Neumann entropy.
Consider again the general density matrix for a qbit,
ρ~n =
1
(1 + ~n · σ) .
2
Recall that ~n, ~n · ~n ≤ 1, specifies a point on the Bloch ball.
Compute the von Neumann entropy S(ρ~n ) of this operator; it is defined in general as
S(ρ) ≡ −Tr ρ log ρ.
Express the answer as a function of a single variable. Interpret this variable in terms
of the geometry of the space of states.
[Hint: use the fact that the von Neumann entropy is independent of the choice of basis.]
Because it is basis-independent,
S(ρ~n ) does not depend on the direction of ~n, only on
√
the magnitude r ≡ ~n · ~n. WLOG, then, we can compute the entropy of
1
1 1+r
0
z
ρrẑ = (1 + rσ ) =
.
0
1−r
2
2
Then the entropy is
S(r) = −Tr ρ log ρ = −
1+r
log
2
1+r
2
1−r
log
+
2
1−r
2
So r determines how deep inside the Bloch ball you are with r = 0 corresponding to
the center. This is the point of maximum entropy and thus uncertainty about the
quantum state.
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