How Do Quantum Harmonic Oscillator Ladder Operators Affect State Vectors?

[glederfein]

Homework Statement

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

Homework Equations

Hamiltonian: H=\frac{P^2}{2m}+\frac{1}{2}m\omega^2X^2
Ladder operators:
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

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 = <br /> \sqrt{n} \left \langle n | n-1 \right \rangle = <br /> \sqrt{n} \left ( \left | n \right \rangle \right ) ^ \dagger \left | n-1 \right \rangle = <br /> \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 = <br /> \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...

 

[TSny]

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 = <br /> \sqrt{n} \left \langle n | n-1 \right \rangle = <br /> \sqrt{n} \left ( \left | n \right \rangle \right ) ^ \dagger \left | n-1 \right \rangle = <br /> \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 = <br /> \sqrt{\frac{n}{n!(n-1)!}} \left \langle 0 \right | a^n (a^\dagger)^{n-1} \left | 0 \right \rangle
.

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

 

[glederfein]

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

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?

 

[TSny]

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

And, yes, all 4 are zero