University of California at San Diego – Department of Physics – Prof. John McGreevy TA:
Shauna Kravec
Quantum Mechanics (Physics 212A) Fall 2015
Assignment 2 –
Solutions
Due 12:30pm Wednesday, October 14, 2015
1. Paulis
(a) Show (from the matrix representations) that
σ i σ j = iijk σ k + δij 1
(1)
where ijk is the completely antisymmetric tensor in three indices, with
123 = 312 = 231 = +1, 231 = 132 = 321 = −1
and ijk = 0 if any two indices are equal. (On the right hand side of (1), we are
using the Einstein summation convention.)
First note that (σ i )2 = 1 for any i. This is the source of the δ ij 1 term. The ijk
term will vanish if i = j because by definition it is 0 on repeated indicies.
The iijk σ k piece one can find by exhaustively multiplying pairs, shown not here.
You always get the third Pauli matrix with a factor of i and sign which depends
on whether the permutation (the order of the product) is even or odd.
(b) Use (1) to show that
~ )2 = 1
(ň · σ
if ň is a unit vector.
It’s actually easy to show (~
σ · ~a)(~
σ · ~b) = ~a · ~b + i~
σ · ~a × ~b .
P
(~σ · ~a)(~σ · ~b) = ij ai bj σ i σ j
P
P
P
= ij ai bj (iijk σ k + δ ij 1) = ij ai bj δ ij + i ij ijk ai bj σ k
P ijk
= ~a · ~b + i~σ · (~a × ~b) as (~a × ~b)k =
ai b j
ij
2
(~σ · n̂) = 1 is then simply a special case.
(c) Show by series expansion that
eiň·~σθ = cos θ1 + sin θň · σ.
θ
ei 2 ~σ·n̂ = 1 + (i 2θ ~σ · n̂) + 21 (i 2θ ~σ · n̂)2 + 16 (i 2θ ~σ · n̂)3 + 4!1 (i 2θ ~σ · n̂)4 · · ·
= (1 − 21 ( 2θ )2 + 4!1 ( 2θ )4 + · · · )1 + i~σ · n̂(1 − 16 ( 2θ )3 + · · · )
= cos 2θ 1 + i~σ · n̂ sin 2θ
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2. Projector onto spin-up along ň.
Show that the projector onto the state of a qbit | ↑n̂ i with spin up along an arbitrary
~ | ↑n̂ i = ||ii↑n̂ )1 can be written as
unit vector n̂ (defined by the relation n̂ · σ
| ↑n̂ ih ↑n̂ | =
1
(1 + n̂ · σ) .
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That is: check that the operator on the RHS is hermitian, squares to itself, and acts
correctly on a basis.
The operator is hermitian as each individual Pauli (and 1) are hermitian and the
coefficients of n̂ are real.
2
n
n
−
in
z
x
y
~ )2 = (nx σx + ny σy + nz σz )2 =
Note that (n̂ · σ
=1
nx + iny
−nz
~ )2 = 41 (1 + 2n̂ · σ
~ + 1) = 12 (1 + n̂ · σ
~ ) so P 2 = P
Now 12 (1 + n̂ · σ
Finally you should check that the operator annihilates | ↓n̂ i which is clear by inspection
~ | ↓n̂ i = −| ↓n̂ i
n̂ · σ
3. Functions of operators.
Given a linear operator A, define the ‘superoperator’ adA , which implements the adjoint
action of A, by
adA (B) = [A, B].
(a) Show that this operator is a derivation,
adA (BC) = (adA B)C + B(adA C) .
(2)
That is: it satisfies a product rule, like a derivative.
This follows from the usual Leibniz rule for commutators.
(b) Show that (for finite-dimensional Hilbert spaces) one can “integrate by parts”
under the trace:
Tr [BadA C] = −Tr [(adA B) C] .
From above [A, BC] = [A, B]C + B[A, C] = adA BC + BadA C and take trace of
both hand side using Tr [., .] = 0
(c) We are going to derive the formula
eA+B = eA eĈA (B) ,
(3)
where
Z
ds e
ĈA =
0
1
∞
1
−sadA
1
1
eadA − 1 X (−1)n
=
adnA = 1 − adA + ad2A + . . . .
=
adA
(n + 1)!
2
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n=0
Sometimes it is called |+n̂ i ≡ | ↑n̂ i.
2
To do this, consider the operator-valued function
G(s) = e−sA es(A+B) .
Derive a first-order differential equation for G(s) and solve it.
Then
∂s G(s) = −AG(s) + e−sA (A + B)esA e−sA es(A+B) = B(s)G(s)
with
B(s) ≡ e−sA BesA = e−sadA B.
Therefore
G(s) = e
Rs
0
dtB(t)
G(0)
Since G(0) = 1 we get
e−A eA+B = G(1) = eĈ(B)
which implies (3).
(d) For the special case where [A, B] = c is a c-number show that (3) implies
1
eA+B = eA eB e− 2 c .
(4)
Look at the series expanded form of ĈA in (3). For a c-number commutator
ĈA = 1 − 21 adA which plugging this in gives (4)
4. Time evolution of a two-level system. [Le Bellac]
Consider the Hamiltonian
A B
H=~
B −A
in the basis |0i ≡ | ↑i, |1i ≡ | ↓i, where A, B are real numbers. Set ~ = 1. You know
the eigensystem of this matrix from our discussion of Pauli gymnastics. It is useful to
write it in terms of Ω, θ defined by
√
Ω = 2 A2 + B 2 ,
tan θ =
B
.
A
(a) The state vector at time t can be written as
|ϕ(t)i = c+ (t)|0i + c− (t)|1i.
Find a system of ODEs for the components c± (t) that follow from the Schrödinger
equation.
ċ+ (t)
cos θ c+ (t) + sin θ c− (t)
i∂t |ϕ(t)i = H|φ(t)i → i
=
ċ− (t)
sin θ c+ (t) − cos θ c− (t)
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(b) Decompose the initial state |ϕ(t = 0)i in the energy eigenbasis (H|± i = ± |± i):
X
|ϕ(t = 0)i =
a± |± i .
±
Express c+ (t) in terms of the coefficients a± .
± = ± Ω2 and |+ i = cos 2θ |0i + sin 2θ |1i and |− i = − sin 2θ |0i + cos 2θ |1i
This gives us an expression for |ϕ0 i which we can time evolve as follows: |ϕ(t)i =
Ω
Ω
e−iHt |ϕ0 i = a+ e−i 2 t |+ i + a− ei 2 t |− i
Ω
Ω
To evaluate c+ (t) we can compute h0|ϕ(t)i = a+ e−i 2 t cos 2θ − a− ei 2 t sin 2θ
(c) Suppose c+ (0) = 0. Find a± (up to a phase) in terms of A, B (or rather, Ω and
θ). Find the probability of finding the system in the state | ↑i at time t.
If c+ (0) = 0 then aa−+ = tan 2θ which we can use in our evaluation of |c+ (t)|2 , the
probability of being in state |0i at time t.
Note that by normalization : c− (0) = 1 = a+ sin 2θ + a− cos 2θ → a− = cos 2θ
c+ (t)c∗+ (t) = 12 (a2+ + a2− + (a+ − a− )(a+ + a− ) cos θ − 2a+ a− cos Ωt sin θ)
= 12 (a2− (1 + tan2 2θ ) + a2− (tan2 2θ − 1) cos θ − 2a2− tan θ cos Ωt sin θ)
= 4a2− sin2 2θ sin2 Ωt
= 4 cos2 2θ sin2 2θ sin2 Ωt
= sin2 θ sin2 Ωt
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5. The variational method. [Le Bellac]
(a) Let |ϕi be a vector (not necessarily normalized) in the Hilbert space of states,
and let H be a Hamiltonian of interest. The expectation value hHiϕ is
hHiϕ =
hϕ|H|ϕi
.
hϕ|ϕi
Show that if the minimum and maximum of this quantity on the space of states are
obtained for |ϕi = |ϕm i and |ϕi = |ϕM i respectively, then |ϕm,M i are eigenvectors
of H with the smallest and largest eigenvalues in the spectrum, Em , EM .
First normalize the wavefunctions : |Φi = |ϕi 1 so hĤiϕ = hΦ|Ĥ|Φi
hϕ|ϕi 2
P
P
P
Next insert 1 = i |Ei ihEi | so hĤiϕ = i hΦ|Ei ihEi |Ĥ|Φi = i Ei |hEi |Φi|2
Where all the coefficients |hEi |Φi|2 are positive. Let E0 be the groundstate energy
and EM be the highest.
P
P
P
2
2
2
2
2
i Ei |hEi |Φi| = E0 hE0 |Φi| +
i6=0 Ei |hEi |Φi| ≥ E0 hE0 |Φi| +
i6=0 E0 |hEi |Φi| =
E0 as Ei6=0 > E0 this proves the minimization criterion.
A similar rearrangement proves the maximization.
(b) Suppose the vector |ϕi depends on a parameter α: |ϕi = |ϕ(α)i. Show that if
∂α hHiφ(α) |α=α0 = 0
then Em ≤ hHiϕ(α0 ) if α0 is a minimum of hHiϕ(α) . Similarly, show that EM ≥
hHiϕ(α0 ) if α0 is a maximum of hHiϕ(α) .
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The above is proven if we can show that ∂α hĤiα |α=α0 = 0 around energy eigenstates.
Let’s call out paramaterized state |ϕ(α)i = |E + αφi for some φ and |Ei being
an energy eigenstate. With this parameterization α = 0 gives |Ei
hĤiα =
hE+αφ|Ĥ|E+αφi
hE+αφ|E+αφi
Ĥ|Ei(hφ|Ei+hE|φi)
∂α hĤiα |α=0 = hE|Ei(hφ|Ĥ|Ei+hE|Ĥ|φi)−hE|
= E(hφ|Ei+hE|φi−hφ|Ei−
hE|Ei2
hE|φi) = 0
So minima of this expectation value are eigenvectors of Ĥ. Then the variational
procedure in the first part implies the rest.
This fact forms the basis of an approximation method called the variational
method. If you have a good guess for the form of the groundstate, you can
find the best approximation within that family of states, and it produces a bound
on the correct groundstate energy.
(c) Consider the special case of dim H = 2, with Hamiltonian
a+c
b
H=
b
a−c
with a, b, c real. Parametrizing the variational state as
cos α/2
|ϕ(α)i =
,
sin α/2
find the values of α = α0 which extremize the expectation value of the Hamiltonian. Show that this reproduces the eigenstates:
cos θ/2
− sin θ/2
|χ+ i =
, |χ− i =
.
sin θ/2
cos θ/2
where tan θ = b/c.
(a + c) cos α2 + b sin α2
(a − c) sin α2 + b cos α2
hĤiα = a + b sin(α) + c cos(α) which has extremum at tan(α) =
This is the answer we get from exact diagonalization as well.
b
c
Note that Ĥ = a1 + bX + cZ so Ĥ|αi =
6. The Feynman-Hellmann theorem.
Suppose the Hamiltonian depends on a parameter s, H = H(s). Suppose E(s) is a
non-degenerate eigenvalue with normalized eigenvector |φ(s)i. Show that
∂s E(s) = hφ(s)|∂s H(s)|φ(s)i.
First we note that E(s) = hφ(s)|Ĥ(s)|φ(s)i and 1 = hφ(s)|φ(s)i
Using the chain rule: ∂s E(s) = hφ(s)|∂s Ĥ(s)|φ(s)i+E(s)[h∂s φ(s)|φ(s)i+hφ(s)|∂s φ(s)i]
But by normalization and the chain rule:[h∂s φ(s)|φ(s)i + hφ(s)|∂s φ(s)i] = ∂s 1 = 0
This proves the theorem.
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