Homework Solution 3
PHYS234 Quantum Mechanics
Prof. Liantao Wang
by Chiu Yu Hung
Problem 1
a) For |n+i, eigenstate of ~σ · ~n = sin θ cos φσ1 + sin θ sin φσ2 + cos θσ3 ,
If ~σ · ~n 6= ±σ1 , i.e. θ 6= π/2, 3π/2 and φ 6= 0, π, then |n+i is an eigenstate of a spin 1/2 in
~n direction, but not an eigenstate of σ1 (in x-direction). And the vector space is of dimension
2, the state |n+i must be a mixed state of both ±1 eigenstate of σ1 .
Hence, the possible outcomes are ±1. And after measurement, the state would collapse
into the eigenstate of σ1 corresponding the eigenvalue measured, i.e. √12 (|+i ± |−i), where
|±i are the eigenstates of σ3 .
If ~σ · ~n = ±σ1 , then |n+i is an eigenstate of σ1 . Hence there would be only one possible
outcome. For ~σ · ~n = σ1 , the only possible outcome is 1, and the state would stay as before
measurement. For ~σ · ~n = −σ1 , the outcome is -1.
b) The density matrix is given by ρ = |n+ihn + |, and to obtain it in the basis of |±i,
one computes h±|ρ|±i. First, express |n+i in the basis of |±i:
|n+i = α|+i + β|−i
(1)
And by ~σ · ~n|n+i = |n+i:
sin θ cos φ(α|−i + β|+i) + i sin θ sin φ(α|−i − β|+i) + cos θ(α|+i − β|−i) = α|+i + β|−i (2)
Solving with normalization condition gives:
θ
θ
|n+i = cos( )e−iφ/2 |+i + sin( )eiφ/2 |−i
2
2
(3)
Hence, ρ is given by:

θ
 cos ( 2 )
ρ=
θ
θ
sin( ) cos( )eiφ
2
2
Note 1:
but this
Note 2:
without
2

θ
θ −iφ
sin( ) cos( )e

2
2

2 θ
sin ( )
2
(4)
In solving |n+i in the basis |±i, one could have an arbitrary phase added to |n+i,
phase doesn’t appear in ρ.
Solving α, β is actually equivalent to solving h±|n+i, one can actually solve h±|n+i
writing |n+i explicitly in basis of |±i
c) Tr(ρ) is always 1 for any state, this is due to the normalization condition of the probability distribution. One can check with the result in part (b). Indeed, the trace of the density
matrix means adding up the probability of the state being in all the eigensates, hence it must
be 1. And for pure state(i.e. ρ = |ψihψ|),
ρ2= ρ, hence, Tr(ρ2 ) = 1. If one work in the basis
1 0
of |n±i, it would be obvious as ρ =
, and trace does not depend on the basis chosen.
0 0
d) hσ1 i = α∗ β + β ∗ α = 2 sin(θ/2) cos(θ/2) cos φ = sin θ cos φ
Note that if ~σ · ~n = ±σ1 , the result is ±1, which is just the eigenvalue obviously.
1
Homework Solution 3
PHYS234 Quantum Mechanics
Prof. Liantao Wang
by Chiu Yu Hung
Problem 2
a) To obtain eignevalue, solve det(H − λI) = 0:
(a − λ)(a − λ) − |c|2 = 0
(5)
λ± = a ± |c|
(6)
H|λ± i = λ± |λ± i
(7)
which gives:
To solve for eigensates, solve:
For λ+ = a + |c|,
−|c| c
c∗
−|c|
α
β
=0
(8)
which gives
1
1 c
|λ+ i = √ ( |+i + |−i) = √ (e−iφ |+i + |−i)
2 |c|
2
(9)
for Arg(c) = −φ, i.e. c/|c| = e−iφ .
For λ− = a − |c|,
|c| c
c∗ |c|
α
β
=0
(10)
which gives
1
1 c
|λ− i = √ ( |+i − |−i) = √ (e−iφ |+i − |−i)
2 |c|
2
(11)
b) To obtain state at time t, one evolves the state by the evolution operator U :
|ψ(t)i = e−iHt/h̄ |ψ(0)i
(12)
And now |ψ(0)i = |+i. We want to decompose it into eigenstate of H, which can be done
by an identity operator 1 = |λ+ ihλ+ | + |λ− ihλ− |:
1
|+i = (|λ+ ihλ+ | + |λ− ihλ− |)|+i = √ eiφ (|λ+ i + |λ− i)
2
(13)
Then, the state at time t is given by:
|ψ(t)i = e−iHt/h̄ |+i =
√1 eiφ e−iHt/h̄ (|λ+ i
2
=
√1 eiφ (e−iλ+ t/h̄ |λ+ i
2
=
√1 ei(φ−at/h̄) (e−i|c|t/h̄ √1 (e−iφ |+i
2
2
+ |λ− i)
+ e−iλ− t/h̄ |λ− i)
(14)
+ |−i) + ei|c|t/h̄ √12 (e−iφ |+i − |−i))
|c|t iφ
= e−iat/h̄ (cos |c|t
h̄ |+i − i sin h̄ e |−i)
2
Homework Solution 3
PHYS234 Quantum Mechanics
Prof. Liantao Wang
by Chiu Yu Hung
c) The probability of getting a +1 for measurement of σ2 is given by |h2+ |ψ(t)i|2 , where
|2+ i = √12 (|+i + i|−i) could be easily obtained.
|h2+ |ψ(t)i|2
|c|t iφ 2
|c|t
|c|t iφ 2
1
= | √12 e−iat/h̄ (cos |c|t
h̄ − sin h̄ e )| = 2 | cos h̄ − sin h̄ e |
=
1
|c|t
|c|t
|c|t iφ
|c|t
(cos2
+ sin2
− cos
sin
(e + e−iφ ))
2
h̄
h̄
h̄
h̄
=
1
2|c|t
(1 − sin
cos φ)
2
h̄
(15)
Note: The result is always between 0 and 1, which we expect by physics.
Also, note that when |c| = 0, H is a diagonal one, the state would remain |+i with a
phase, then its component in |2+ i would be time-independent, which is shown in the result.
When φ = nπ/2, the Hamiltonian would only have component in the identity and σ2 , representing a pure rotation about y-axis on the state. This rotation would not change the
component of the state in y-axis, hence the result would be time independent, as shown by
the result.
d) To obtain the expectation value, one calculates hψ(t)|σ2 |ψ(t)i
|c|t iφ
σ2 |ψ(t)i = e−iat/h̄ σ2 (cos |c|t
h̄ |+i − i sin h̄ e |−i)
|c|t iφ
= e−iat/h̄ (cos |c|t
h̄ (i|−i) − i sin h̄ e (−i|+i))
(16)
|c|t
iφ
= −e−iat/h̄ (sin |c|t
h̄ e |+i − i cos h̄ |−i)
Hence,
|c|t iφ
|c|t
|c|t −iφ
hψ(t)|σ2 |ψ(t)i = − cos |c|t
h̄ sin h̄ e − cos h̄ sin h̄ e
(17)
= − sin 2|c|t
h̄ cos φ
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