# Quantum Harmonic Oscillator

1. Dec 5, 2013

### glederfein

1. The problem statement, all variables and given/known data
Given a quantum harmonic oscillator, calculate the following values:
$\left \langle n \right | a \left | n \right \rangle, \left \langle n \right | a^\dagger \left | n \right \rangle, \left \langle n \right | X \left | n \right \rangle, \left \langle n \right | P \left | n \right \rangle$

2. Relevant equations
Hamiltonian: $H=\frac{P^2}{2m}+\frac{1}{2}m\omega^2X^2$
$a=\sqrt{\frac{m\omega}{2\hbar}}\left ( X + \frac{i}{m\omega}P \right )$
$a^\dagger=\sqrt{\frac{m\omega}{2\hbar}}\left(X-\frac{i}{m\omega}P \right )$
$\left [ a,a^\dagger \right ] = 1$
$\left [ a^\dagger a,a \right ] = -a$
N operator:
$N=a^\dagger a$
$N\left | n \right \rangle = n\left | n \right \rangle$
$a^\dagger \left | n \right \rangle = \sqrt{n+1} \left | n+1 \right \rangle$
$a \left | n \right \rangle = \sqrt{n} \left | n-1 \right \rangle$
$\left | n \right \rangle = \frac{(a^\dagger)^n}{\sqrt{n!}} \left | 0 \right \rangle$

3. The attempt at a solution
$\left \langle n \right | a \left | n \right \rangle = \left \langle n \right | \sqrt{n} \left | n-1 \right \rangle = \sqrt{n} \left \langle n | n-1 \right \rangle = \sqrt{n} \left ( \left | n \right \rangle \right ) ^ \dagger \left | n-1 \right \rangle = \sqrt{n} \left ( \frac{(a^\dagger)^n}{\sqrt{n!}} \left | 0 \right \rangle \right ) ^ \dagger \frac{(a^\dagger)^{n-1}}{\sqrt{(n-1)!}} \left | 0 \right \rangle = \sqrt{\frac{n}{n!(n-1)!}} \left \langle 0 \right | a^n (a^\dagger)^{n-1} \left | 0 \right \rangle$

Not sure how to continue from here...

2. Dec 5, 2013

### TSny

When you get to $\left \langle n \right | a \left | n \right \rangle = \left \langle n \right | \sqrt{n} \left | n-1 \right \rangle = \sqrt{n} \left \langle n | n-1 \right \rangle$ you should be able to see the answer.

3. Dec 5, 2013

### glederfein

Is the answer zero because eigenvectors are always perpendicular to one another?
Doesn't that mean that all the four values in the question are zero?

4. Dec 5, 2013

### TSny

Yes, eigenvectors of a Hermitian operator that correspond to different eigenvalues are orthogonal.

And, yes, all 4 are zero