Physics 215 – Quantum Mechanics 1 Assignment...

Physics 215 – Quantum Mechanics 1Assignment 1

Logan A. Morrison

January 19, 2016

Problem 1

Prove via the dual correspondence definition that the hermitian conjugate of |α〉〈β| is |β〉〈α|.

Solution

By definition, the hermitian conjugate of an operator A is defined by

A |n〉 → 〈n| A† (1)

under the dual correspondence. Let |n〉 be and arbitrary vector. Then, we know that

(|α〉〈β|) |n〉 → 〈β|n〉∗ 〈α| (2)= 〈n|β〉〈α| (3)= 〈n| (|β〉〈α|) (4)

(5)

Since this hold for all |n〉, we conclude that

(|α〉〈β|)† = |β〉〈α| (6)

Logan A. Morrison Assignment 2– Problem 2 / 9

Problem 2

Prove that, in the absence of degeneracy, a sufficient condition for the following to be true

∑b′〈c′|b′〉〈b′|a′〉〈a′|b′〉〈b′|c′〉 =

∑b′b′′〈c′|b′〉〈b′|a′〉〈a′|b′′〉〈b′′|c′〉 (7)

(where as usual |a′〉 is and eigen-ket of A etc.) is that [A, B] = 0 or that [B,C] = 0.

Solution

Assume that both A and B are self-adjoint. Additionally, assume that [A, B] = 0. Then,

AB |a′〉 = BA |a′〉 (8)= a′B |a′〉 (9)

Therefore, B |a′〉 is an eigenket of A with eigenvalue a′. Since there is no degeneracy, there is only one eigenvectorof A corresponding to the eigenvalue a′: |a′〉. Therefore, B |a′〉 is proportional to |a′〉. That is, B |a′〉 = b′ |a′〉. Inothere words, the eigenkets of B are proportional to the eigenkets of A. Thus, we can write

|b′〉 = γ′ |a′〉 (10)

Since the states are properly normalized, (〈b′|b′〉 = 1 and 〈a′|a′〉 = 1), we require that |γ′| = 1. Note that sincethe operators are self-adjoint, the eigenvectors are orthogonal. Therefore, 〈b′|a′〉 = 0 unless 〈b′| = γ′∗ 〈b′| Thisyields: ∑

b′〈c′|b′〉〈b′|a′〉〈a′|b′〉〈b′|c′〉 = 〈c′| γ′ |a′〉〈a′| γ′∗ |a′〉〈a′| γ′ |a′〉〈b′| γ′∗ |c′〉 (11)

= |γ′|4 〈c′|a′〉〈a′|c′〉 (12)= 〈c′|a′〉〈a′|c′〉 (13)

Since B is hermitian, ∑b′|b′〉〈b′| =

∑b′′|b′′〉〈b′′| = I (14)

where I is the identity operator.∑b′b′′〈c′|b′〉〈b′|a′〉〈a′|b′′〉〈b′′|c′〉 = 〈c′| I |a′〉〈a′| I |c′〉 (15)

= 〈c′|a′〉〈a′|c′〉 (16)

Thus, the two expressions are equal and we have proved the equality in equation 7.


Problem 3

Show that for the |S z; +〉 state of a spin 12 system

⟨S 2

x

⟩− 〈S x〉

2 = ~2/4

Solution

First, from class, we know that

|S x,+〉 =|S z,+〉 + |S z,−〉

√2

(17)

|S x,−〉 =|S z,+〉 − |S z,−〉

√2

(18)

We can invert these equations to obtain

|S z,+〉 =|S x,+〉 + |S x,−〉

√2

(19)

(20)

Using this, we find that ⟨S 2

x

⟩= 〈S z; +| S 2

x |S z; +〉 (21)

=12

(〈S x; +| + 〈S x;−|) S xS x (|S x; +〉 + |S x;−〉) (22)

=12

(〈S x; +| + 〈S x;−|) S x

(~

2|S x; +〉 −

~

2|S x;−〉

)(23)

=12

(〈S x; +| + 〈S x;−|)(~2

4|S x; +〉 +

~2

4|S x;−〉

)(24)

=12

(~2

4+~2

4

)(25)

=~2

4(26)

and

〈S x〉 = 〈S z; +| S x |S z; +〉 (27)

=12

(〈S x; +| + 〈S x;−|) S x (|S x; +〉 + |S x;−〉) (28)

=12

(〈S x; +| + 〈S x;−|)(~

2|S x; +〉 −

~

2|S x;−〉

)(29)

=12

(~

2−~

2

)(30)

= 0 (31)

and hence,⟨S 2

x

⟩−〈S x〉

2 = ~2/4. Physically, we know thatσS x =

√⟨S 2

x⟩− 〈S x〉

2 = ~/2 whereσS x is the uncertaintyof the value of S x for a particle in the state |S x; +〉. Since there is a probability of 1/2 that the value ~/2 will bemeasured and a probability of 1/2 that the value −~/2 will be measured.


Problem 4

A two-state system is characterized by the Hamiltonian

H = H11 |1〉〈1| + H22 |2〉〈2| + H12 (|1〉〈2| + |2〉〈1|) (32)

where H11,H12 and H22 are real numbers with the dimension of energy, and |1〉 and |2〉 are eigenkets of someobservable , H. FInd the energy eigenkets and corresponding energy eigenvalues.

Solution

To solve this problem, we should choose a representation for our states. Let

|1〉 =

(10

)(33)

and

|2〉 =

(01

)(34)

In this representation, we have that

H =

(H11 H12

H12 H22

)(35)

Now, let’s find the eigenvalues:

0 = det(H − λI) (36)

=

∣∣∣∣∣∣H11 − λ H12

H12 H22 − λ

∣∣∣∣∣∣ (37)

= (H11 − λ) (H22 − λ) − H212 (38)

= H11H22 − (H11 + H22) λ + λ2 − H212 (39)

Therefore,

λ =12

(H11 + H22 ±

√(H11 + H22)2

− 4H11H22 + 4H212

)(40)

=12

(H11 + H22 ±

√(H11 − H22)2 + 4H2

12

)(41)

(42)

Thus, the two possible energies are:

E+ =12

(H11 + H22 +

√(H11 − H22)2 + 4H2

12

)(43)

and

E− =12

(H11 + H22 −

√(H11 − H22)2 + 4H2

12

)(44)


Now let’s find the energy eigenstate corresponding to E+. To do so, we will determine the null space of H −E+I:

0 = (H − E+I) |E+〉 (45)

=

(H11 − E+ H12

H12 H22 − E+

) (ab

)(46)

12

(H11 − H22 −

√(H11 − H22)2 + 4H2

12

)H12

H1212

(H22 − H11 −

√(H11 − H22)2 + 4H2

12

)(ab

)(47)

Thus,

0 = a(12

(H11 − H22 −

√(H11 − H22)2 + 4H2

12

))+ bH12 (48)

a = −2H12(

H11 − H22 −

√(H11 − H22)2 + 4H2

12

)b (49)

and therefore, choosing b =

(H11 − H22 −

√(H11 − H22)2 + 4H2

12

), we find

|E+〉 =

2H12(H11 − H22 −

√(H11 − H22)2 + 4H2

12

) (50)

And now let’s find the energy eigenstate corresponding to E−. To do so, we will determine the null space ofH − E−I:

0 = (H − E−I) |E−〉 (51)

=

(H11 − E− H12

H12 H22 − E−

) (ab

)(52)

12

(H11 − H22 +

√(H11 − H22)2 + 4H2

12

)H12

H1212

(H22 − H11 +

√(H11 − H22)2 + 4H2

12

)(ab

)(53)

Thus,

0 = a(12

(H11 − H22 +

√(H11 − H22)2 + 4H2

12

))+ bH12 (54)

a = −2H12(

H11 − H22 +

√(H11 − H22)2 + 4H2

12

)b (55)

and therefore, choosing b =

(H11 − H22 −

√(H11 − H22)2 + 4H2

12

), we find

|E−〉 =

2H12(H11 − H22 +

√(H11 − H22)2 + 4H2

12

) (56)

Clearly these states are not normalized to 1. We could do so by defining∣∣∣E′+⟩ =

1〈E+|E+〉

|E+〉 and a similar

expression for∣∣∣E′−⟩.


Problem 5

Construct the transformation matric that connects the S z diagonal basis to the S x diagonal basis and show thatyour result is consistent with the general relation

U =∑

r

∣∣∣b(r)⟩ ⟨

a(r)∣∣∣ . (57)

Solution

From class, we know that:

|S x; +〉 =1√

2(|S z; +〉 + |S z;−〉) (58)

|S x;−〉 =1√

2(|S z; +〉 − |S z;−〉) (59)

(60)

Thus, in the usual representation of |S z; +〉 and |S z;−〉, i.e.

|S z; +〉 =

(10

)and |S z;−〉 =

(01

)(61)

one can clearly see that the transformation matrix is:

U =1√

2

(1 11 −1

)(62)

In other words,

|S x; +〉 = U |S z; +〉 (63)|S x;−〉 = U |S z;−〉 (64)

In a representation-free form, U is:

U =1√

2(|S z; +〉〈S z; +| + |S z;−〉〈S z; +| + |S z; +〉〈S z;−| − |S z;−〉〈S z;−|) (65)

Let’s check that this transformation matrix yields the appropriate transformation:

U |S z; +〉 =1√

2(|S z; +〉〈S z; +| + |S z;−〉〈S z; +| + |S z; +〉〈S z;−| − |S z;−〉〈S z;−|) |S z; +〉 (66)

=1√

2(|S z; +〉〈S z; +|S z; +〉 + |S z;−〉〈S z; +|S z; +〉 + |S z; +〉〈S z;−|S z; +〉 − |S z;−〉〈S z;−|S z; +〉) (67)

=1√

2(|S z; +〉 + |S z;−〉) (68)

= |S x; +〉 (69)


and

U |S z;−〉 =1√

2(|S z; +〉〈S z; +| + |S z;−〉〈S z; +| + |S z; +〉〈S z;−| − |S z;−〉〈S z;−|) |S z;−〉 (70)

=1√

2(|S z; +〉〈S z; +|S z;−〉 + |S z;−〉〈S z; +|S z;−〉 + |S z; +〉〈S z;−|S z;−〉 − |S z;−〉〈S z;−|S z;−〉) (71)

=1√

2(|S z; +〉 − |S z;−〉) (72)

= |S x;−〉 (73)

Now, we can write our transformation matrix in a slightly different form:

U =1√

2(|S z; +〉〈S z; +| + |S z;−〉〈S z; +| + |S z; +〉〈S z;−| − |S z;−〉〈S z;−|) |S z; +〉 (74)

=1√

2(|S z; +〉 + |S z;−〉) 〈S z; +| +

1√

2(|S z; +〉 − |S z;−〉) 〈S z;−| (75)

= |S x; +〉〈S z; +| + |S x;−〉〈S z;−| (76)

which is of the form∑

r

∣∣∣b(r)⟩ ⟨

a(r)∣∣∣.


Problem 6

An operator A, corresponding to an observable α, has two normalized eigenstates |φ1〉 and |φ2〉, with eigenvalues a1

and a2. An operator B, corresponding to an observable β, has normalized eigenstates |χ1〉 and |χ2〉. The eigenstatesare related by

|φ1〉 =2 |χ1〉 + 3 |χ2〉

√13

, |φ1〉 =3 |χ1〉 − 2 |χ2〉

√13

. (77)

α is measured and the value a1 is obtained. If β is then measured and then α again, find the probability of obtaininga1 a second time.

Solution

First off, let’s invert equation 78:

|χ1〉 =2 |φ1〉 + 3 |φ2〉√

13, |χ1〉 =

3 |φ1〉 − 2 |φ2〉√

13. (78)

Assume that α is measured and a1 is obtained. Then we know that the state of our system is |φ1〉. Now if wemeasure β and then α, there are two ways of obtaining a1 again. The first case is that we measure β and obtain|χ1〉 and then measure α and obtain a1. The second case is we measure β and obtain |χ2〉 and then measure α andobtain a1. The probability of obtaining a1 after the two measurements will be

Prob(|φ1〉 → |χ1〉 → |φ1〉) + Prob(|φ1〉 → |χ2〉 → |φ1〉) (79)

The probability of obtaining |χ1〉 is:

Prob(|φ1〉 → |χ1〉) = | 〈χ1|φ1〉 |2 (80)

=

∣∣∣∣∣∣〈χ1|

(2 |χ1〉 + 3 |χ2〉

√13

)∣∣∣∣∣∣2 (81)

=4

13(82)

The probability of obtaining |φ1〉 after realizing |χ1〉 is:

Prob(|χ1〉 → |φ1〉) = | 〈φ1|χ1〉 |2 (83)

=

∣∣∣∣∣∣〈φ1|

(2 |φ1〉 + 3 |φ2〉√

13

)∣∣∣∣∣∣2 (84)

=4

13(85)

Thus,Prob(|φ1〉 → |χ1〉 → |φ1〉) = (4/13)(4/13) = 16/169 (86)

The probability of obtaining |χ2〉 after measuring β is:

Prob(|φ1〉 → |χ2〉) = | 〈χ2|φ1〉 |2 (87)

=

∣∣∣∣∣∣〈χ1|

(2 |χ1〉 + 3 |χ2〉

√13

)∣∣∣∣∣∣2 (88)

=9

13(89)


The probability of obtaining |φ1〉 after realizing |χ2〉 is:

Prob(|χ2〉 → |φ1〉) = | 〈φ1|χ2〉 |2 (90)

=

∣∣∣∣∣∣〈φ1|

(3 |φ1〉 − 2 |φ2〉√

13

)∣∣∣∣∣∣2 (91)

=9

13(92)

Therefore,Prob(|φ1〉 → |χ2〉 → |φ1〉) = (9/13)(9/13) = 81/169 (93)

Therefore, the probability of obtaining a1 after measuring β then α is

Prob(|φ1〉 → |χ1〉 → |φ1〉) + Prob(|φ1〉 → |χ2〉 → |φ1〉) =16169

+81

169=

97169

(94)

Physics 215 – Quantum Mechanics 1 Assignment...

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