〈ψ|p|ψ〉 = −i
√
m ̄hω
2
〈√
2
3
u 0 −i
√
1
3
u 1 |A−A†|
√
2
3
u 0 −i
√
1
3
u 1
〉
= −i
√
m ̄hω
2
〈√
2
3
u 0 −i
√
1
3
u 1 |−
√
2
3
u 1 −i
√
1
3
u 0
〉
= −i
√
m ̄hω
2
(−i
√
2
9
−i
√
2
9
)
= −
√
8
9
√
m ̄hω
2
=
2
3
√
m ̄hω
- Assumingunrepresents thenth1D harmonic oscillator energy eigenstate, calculate〈un|p|um〉.
Answer
p =
√
m ̄hω
2
A−A†
i
〈un|p|um〉 = −i
√
m ̄hω
2
〈un|A−A†|um〉
= −i
√
m ̄hω
2
(
√
mδn(m−1)−
√
m+ 1δn(m+1))
- Evaluate the “uncertainty” inxfor the 1D HO ground state
√
〈u 0 |(x− ̄x)^2 |u 0 〉where ̄x=
〈u 0 |x|u 0 〉. Similarly, evaluate the uncertainty inpfor the ground state. What is the product
∆p∆x?
Answer
Its easy to see that ̄x= 0 either from the integral or using operators. I’ll use operatorsto
compute the rest.
x =
√
̄h
2 mω
(A+A†)
〈u 0 |x^2 |u 0 〉 =
̄h
2 mω
〈u 0 |AA+AA†+A†A+A†A†)|u 0 〉
=
̄h
2 mω
〈u 0 |AA†|u 0 〉=
̄h
2 mω
1 =
̄h
2 mω
p =
1
i
√
m ̄hω
2
(A−A†)
〈u 0 |p^2 |u 0 〉 = −
m ̄hω
2
〈u 0 |−AA†|u 0 〉=
m ̄hω
2
∆x =
√
̄h
2 mω
∆p =
√
m ̄hω
2
∆p∆x =
̄h
2
