Cohen - Quantum Mechanics (solutions H)

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QM 2 Homework
Sebastian Requena
Fall 2011
1 Cohen-Tannoudji HII Exercise 2
In a two dimensional vector space, consider the operator whose matrix, in an
orthonormal basis {|1〉 , |2〉}, is written:
σy =
(
0 −i
i 0
)
(1)
a) Is σy Hermitian? Calculate its eigenvalues and eigenvectors, (Give their nor-
malized expansion in terms of the {|1〉 , |2〉} basis.)
It is easy to see that σy is in fact hermitian.
σ†y =
(
0 −i
i 0
)†
=
(
0 −i
i 0
)
= σy (2)
The eigenvalues are found by,
det
(
0− λ −i
i 0− λ
)
= λ2 − 1 = (λ− 1)(λ + 1) = 0 (3)
The eigenvalues are 1 and -1. Now to find the eigenvectors. For λ = 1,
(
0− 1 −i
i 0− 1
)(
v1
v2
)
=
(
0
0
)
(4)
The normalized vector satisfying this equation.
1√
2
(−i
1
)
=
−i√
2
|1〉+ 1√
2
|2〉 (5)
Similarly for λ = −1, the eigenvector is,
1√
2
(
i
1
)
=
i√
2
|1〉+ 1√
2
|2〉 (6)
b) Calculate the matrices which represent the projectors onto these eigen vec-
tors. Then verify that they satisfy the orthogonality and closure relations.
1
To find the Projectors, I take the outer product of the eigenvectors.
|λ1〉 〈λ1| = 12
(−i
1
) (
i 1
)
=
1
2
(
1 −i
i 1
)
(7)
|λ−1〉 〈λ−1| = 12
(
i
1
) (−i 1) = 1
2
(
1 i
−i 1
)
(8)
Test for orthogonality,
|λ1〉 〈λ1|λ−1〉 〈λ−1| = 14
(
1 i
−i 1
)(
1 −i
i 1
)
=
(
0 0
0 0
)
(9)
Check for closure,
|λ1〉 〈λ1|+ |λ−1〉 〈λ−1| = 12
(
1 i
−i 1
)
1
2
(
1 −i
i 1
)
=
(
1 0
0 1
)
(10)
2 Cohen-Tannoudji HII Exercise 3
The state space of a certain physical system is three-dimensional. Let {|u1〉 , |u2〉 , |u3〉}
be an orthonormal basis of this space. Th kets |φ0〉 and |φ1〉 are defined by:
|φ0〉 = 1√
2
|u1〉+ i2 |u1〉+
1
2
|u3〉 (11)
|φ1〉 = 1√
3
|u1〉+ i√
3
|u3〉 (12)
a) Are these kets normalized?
〈φ0|φ0〉 = 12 +
1
4
1
4
= 1 (13)
〈φ1|φ1〉 = 13 +
1
3
=
2
3
(14)
Therefore, |φ0〉 is normalized and |φ1〉 is not.
b) Calculate the matrices ρ0 and ρ1 representing the projection operators onto
the state |φ0〉 and onto the state |φ1〉. Verify these matrices are Hermitian.
|φ0〉 〈φ0| =


1√
2
i
2
1
2


(
1√
2
− i2 12
)
=


1
2 − i2√2
1
2
√
2
i
2
√
2
1
4
i
4
1
2
√
2
− i4 14

 (15)
Check if this is Hermitian, Yep.
(|φ0〉 〈φ0|)† =


1
2 − i2√2
1
2
√
2
i
2
√
2
1
4
i
4
1
2
√
2
− i4 14


†
= |φ0〉 〈φ0| (16)
2
Similarly for |φ1〉,
|φ1〉 〈φ1| =


1√
3
0
i√
3


(
1√
3
0 − i√
3
)
=


1
3 0 − i3
0 0 0
i√
3
0 13

 (17)
It is also Hermitian.
(|φ1〉 〈φ1|)† =


1
3 0 − i3
0 0 0
i√
3
0 13


†
= |φ1〉 〈φ1| (18)
3 Cohen-Tannoudji HII Exercise 6
The matrix σx is defined by:
σx =
(
0 1
1 0
)
(19)
Prove the relation:
eiασx = Icos(α) + iσxsin(α) (20)
Where I is the 2x2 identity matrix.
It’s clear that this is using Euler’s Identity. To derive Euler’s Identity use
the Taylor expansion of left side of equation 19.
eiασx =
∑ 1
n!
(iασx)n (21)
We can split this apart to be two summations.
eiασx =
∑ 1
2n!
(iασx)2n +
∑ 1
(2n + 1)!
(iασx)2n+1 (22)
Since the matrix σx is idempotent we can assign even and odd forms.
eiασx = I
∑
(−1)n 1
2n!
(α)2n + iσx
∑
(−1)n 1
(2n + 1)!
(α)2n+1 (23)
These are simply the Taylor expansions for sine and cosine.
eiασx = Icos(α) + iσxsin(α) (24)
3