• Support PF! Buy your school textbooks, materials and every day products Here!

Quantum mechanics - creation and annihilation operators

  • Thread starter Anabelle37
  • Start date
  • #1
39
0

Homework Statement



Evaluate <n|p^2|n>
where p is the momentum operator for the quantised harmonic oscillator.

Homework Equations



creation operator: a+|n>=sqrt(n+1)|n+1>
annihilation operator: a|n>=sqrt(n)|n-1>

The Attempt at a Solution



the operator p can be defined in terms of the creation and annihilation operators, ie. p = (-i/sqrt{2})(a-a+)

I also wrote down that for the quantised harmonic oscillator p = -(ihbar).(partial derivative wrt x) and so p^2= -(hbar)^2.(partial derivative wrt x)^2

I'm stuck on what to do next? How do I evaluate <n|p^2|n> ???
 
Last edited:

Answers and Replies

  • #2
213
8
use the fact that [tex] \left| n \right\rangle = \frac{1}{\sqrt{n!}} (a^{\dagger})^n \left| 0 \right\rangle [/tex] and then use the commutation relations for the a's

note you will need to workout [tex] [a^n, a^{\dagger}] [/tex] and other similar ones write if you have trouble
 
  • #3
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,639
1,276
What do you get if you apply p to |n>?

[tex]\vert\psi\rangle = \hat{p}\vert n\rangle = \frac{i}{\sqrt{2}}(\hat{a}^\dagger-\hat{a})\vert n\rangle = \cdots[/tex]

Once you have that, you can use the fact that p is self-adjoint so that

[tex]\langle \psi \vert \psi \rangle = \langle n \vert \hat{p}^\dagger \hat{p} \vert n \rangle = \langle n \vert \hat{p}^2 \vert n \rangle[/tex]

You could also do it the way sgd37 suggested, but it's a bit more tedious for this particular problem.
 
  • #4
39
0
Thank you both! Vela I had done it a longer way and then just realised your way would have been so much easier lol oh well. What does self adjoint mean again?

I got n+1/2 as my answer which i think is correct!?
 
  • #5
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,639
1,276
That's what I got. Don't forget to add the units back in if necessary. You're using the dimensionless version of the momentum operator.
 
  • #6
169
0
A self-adjoint operator [tex]\hat{A}[/tex] is an operator for which
[tex]\hat{A}^\dagger = \hat{A}[/tex] (there are also some technical requisites, but they shouldn't concern you now). Since all self-adjoint operators have real eigenvalues, they correspond to observables (which have real values when measured). On the other way round, an observable must therefore be self-adjoint, and so the position operator [tex]\hat{x}[/tex] and the momentum operator [tex]\hat{p}[/tex] are postulated to be self-adjoint.
 

Related Threads on Quantum mechanics - creation and annihilation operators

  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
5
Views
7K
Replies
5
Views
1K
Replies
2
Views
3K
Replies
0
Views
468
  • Last Post
Replies
1
Views
2K
Replies
1
Views
1K
Replies
3
Views
2K
Replies
1
Views
2K
Replies
1
Views
367
Top