Quantum mechanics - creation and annihilation operators

[Anabelle37]

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> ?

 

[sgd37]

use the fact that \left| n \right\rangle = \frac{1}{\sqrt{n!}} (a^{\dagger})^n \left| 0 \right\rangle and then use the commutation relations for the a's

note you will need to workout [a^n, a^{\dagger}] and other similar ones write if you have trouble

 

[vela]

What do you get if you apply p to |n>?

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

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

\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

You could also do it the way sgd37 suggested, but it's a bit more tedious for this particular problem.

 

[Anabelle37]

Thank you both! Vela I had done it a longer way and then just realized 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!?

 

[vela]

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.

 

[grey_earl]

A self-adjoint operator \hat{A} is an operator for which
\hat{A}^\dagger = \hat{A} (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 \hat{x} and the momentum operator \hat{p} are postulated to be self-adjoint.