PHY4604 HW 1 Solutions

Prof. Hebin Li

September 17, 2017

Problem 1 solution:

a = 4eiπ/2 = 4(cosπ/2 + i sinπ/2) = 4i;

ab = 4i(3 + 2i) = 12i− 8;

|ab| =√

122 + (−8)2 = 4√

13.

Problem 2 solution:

|ψ1〉 = C[3|+〉+ 4|−〉];〈ψ1|ψ1〉 = 1;

⇒ C∗[3〈+|+ 4〈−|]C[3|+〉+ 4|−〉] = 1;

⇒ C∗C(9 + 25) = 1 :

⇒ |C|2 = 1/25;

We choose C = 1/5, so the normalized state is |Ψ1〉 = 35|+〉+ 4

5|−〉.

Likewise, we have

|ψ2〉 =1√5|+〉+

2i√5|−〉

|ψ3〉 =3√10|+〉 − 1√

10eiπ/3|−〉

Problem 3 solution:(a)

For state |ψ1〉 = 1√3|+〉+ i

√2√3|−〉, we assume the orthogonal vector is 〈φ1| = a〈+|+ b〈−| and they

have to satisfy 〈φ1|ψ1〉 = 0, so we have

[a〈+|+ b〈−|][ 1√3|+〉+ i

√2√3|−〉] = 0;

⇒ a√3〈+|+〉+ i

√2√3b〈−|−〉 = 0;

⇒ a√3

+ i

√2√3b = 0;

⇒ a = −i√

2b

1

〈φ1| should also be normalized, i.e. |a|2 + |b|2 = 1.We can then solve for a and b (we choose a to be real): a =

√2√3, b = i√

3, so

〈φ1| =√

2√3〈+|+ i√

3〈−|;

i.e. |φ1〉 =

√2√3|+〉 − i√

3|−〉;

Likewise, we can have

|φ2〉 =2√5|+〉+

1√5|−〉

|φ3〉 =1√2|+〉 − eiπ/4 1√

2|−〉

(b)

〈ψ1|ψ2〉 = [1√3〈+| − i

√2√3〈−|][ 1√

5|+〉 −

√2√5|−〉]]

=1√3× 1√

5+ i

√2√3×√

2√5

=1√15

+ i2√

2√15

;

〈ψ1|ψ3〉 = [1√3〈+| − i

√2√3〈−|][ 1√

2|+〉+ eiπ/4

1√2|−〉]]

=1√3× 1√

2− i√

2√3× eiπ/4 1√

2

=

√2√3− i 1√

6;

〈ψ2|ψ3〉 = [1√5〈+| −

√2√5〈−|][ 1√

2|+〉+ eiπ/4

1√2|−〉]

=1√5× 1√

2−√

2√5× eiπ/4 1√

2

=1−√

2√10− i 1√

5;

〈ψ2|ψ1〉 = 〈ψ1|ψ2〉∗ =1√15− i2√

2√15

;

〈ψ2|ψ1〉 =

√2√3

+ i1√6

;

〈ψ3|ψ2〉 =1−√

2√10

+ i1√5

;

〈ψ1|ψ1〉 = 〈ψ2|ψ2〉 = 〈ψ3|ψ3〉 = 1.

2

Problem 4 solution:

(a)Heads or tails.

(b)The probability for heads is 50% (or 1/2), PH = 1/2.The probability for tails is 50% (or 1/2), PT = 1/2.

(c)

Problem 5 solution:

(a)Possible results: 1, 2, 3, 4, 5, 6 dots

(b)Each result has a probability of 1/6.P1 = P2 = P3 = P4 = P5 = P6 = 1/6.

(c)

3

Problem 6 solution:There are four possible measurement results: 2 eV, 4 eV, 7 eV, and 9 eV. Note that |ψ〉 is alreadynormalized.

P2eV = |〈2eV |ψ〉|2 = |〈2eV | 1√39

[3|2eV 〉 − i|4eV 〉+ 2eiπ/7|7eV 〉+ 5|9eV 〉]|2

=3

13;

P4eV = |〈4eV |ψ〉|2 =1

39;

P7eV = |〈7eV |ψ〉|2 =4

39;

P9eV = |〈9eV |ψ〉|2 =25

39.

Problem 7 solution:

P = |〈ψf |ψi〉|2

= |[1− i√3〈a1|+

1√6〈a2|+

1√6〈a3|][

i√3|a1〉+

√2

3|a2〉]|2

= |1− i√3× i√

3+

1√6×√

2

3|2

=5

9

4