Harmonic oscillator with ladder operators - proof using the Sum Rule

[chocopanda]

Homework Statement

Verify the proof of the sum rule for the one-dimensional harmonic oscillator:
$$\sum_l^\infty (E_l-E_n)\ | \langle l \ |p| \ n \rangle |^2 = \frac {mh^2w^2}{2} $$

Relevant Equations

The exercise explicitly says to use laddle operators and to express $p$ with
$$b=\sqrt{\frac {mw}{2 \hbar}}-\frac {ip}{\sqrt{2 \hbar mw}} $$
$$b^\dagger =\sqrt{\frac {mw}{2 \hbar}}+\frac {ip}{\sqrt{2 \hbar mw}} $$

I'm trying verify the proof of the sum rule for the one-dimensional harmonic oscillator:
$$\sum_l^\infty (E_l-E_n)\ | \langle l \ |p| \ n \rangle |^2 = \frac {mh^2w^2}{2} $$
The exercise explicitly says to use laddle operators and to express $p$ with
$$b=\sqrt{\frac {mw}{2 \hbar}}-\frac {ip}{\sqrt{2 \hbar mw}} $$
$$b^\dagger =\sqrt{\frac {mw}{2 \hbar}}+\frac {ip}{\sqrt{2 \hbar mw}} $$

For p I get $$p=i \sqrt{\frac{\hbar}{2mw}} (b-b^\dagger) $$

To solve the exercise, we need to calculate the left side. I'm still very much a novice and am not very sure how to use the ladder operators... To start, I at least tried to expand the bra-ket:
$$\sum_l^\infty (E_l-E_n)\ \langle l \ |p| \ n \rangle \langle n \ |p| \ l \rangle $$
and tried to insert the p I solved:
$$\sum_l^\infty (E_l-E_n)\ (-\frac{\hbar}{2mw}) \langle l \ |b-b^\dagger| \ n \rangle \langle n \ |b-b^\dagger| \ l \rangle $$
is this correct? If yes, how do I continue? The hint says to probably use $$H|n\rangle=\hbar(n+\frac 12)|n\rangle$$ and I know that $$H|n\rangle=E|n\rangle$$

 

[DrClaude]

To start, I at least tried to expand the bra-ket:
$$\sum_l^\infty (E_l-E_n)\ \langle l \ |p| \ n \rangle \langle n \ |p| \ l \rangle $$

I wouldn't do that right away. I would start by calculating $\langle l \ |p| \ n \rangle$ first, and then consider the absolute value squared.

and tried to insert the p I solved:
$$\sum_l^\infty (E_l-E_n)\ (-\frac{\hbar}{2mw}) \langle l \ |b-b^\dagger| \ n \rangle \langle n \ |b-b^\dagger| \ l \rangle $$
is this correct? If yes, how do I continue?

Calculate $(b-b^\dagger) |n\rangle$.

 

[chocopanda]

I wouldn't do that right away. I would start by calculating $\langle l \ |p| \ n \rangle$ first, and then consider the absolute value squared.Calculate $(b-b^\dagger) |n\rangle$.

Hello DrClaude, thank you for replying. I tried to do what you suggested:

$$| \langle l|p|n \rangle |^2 = \langle n|p^2|n\rangle = \frac{h}{2mw} (2n+1) $$
That's my result. How would I continue?

 

[DrClaude]

Hello DrClaude, thank you for replying. I tried to do what you suggested:

$$| \langle l|p|n \rangle |^2 = \langle n|p^2|n\rangle = \frac{h}{2mw} (2n+1) $$
That's my result. How would I continue?

That's not correct. How can the bra $\langle l |$ even become $\langle n |$?

As I said, forget the absolute value squared for now. Start by calculating $(b-b^\dagger) |n\rangle$ and then apply that to $\langle l|p|n \rangle$.

 

[vanhees71]

Are you forced to do it in that complicated way? It's simpler to use the commutation relations, $[\hat{x},\hat{p}]=?$, $[\hat{H},\hat{x}]=?$ as well as to think about what's
$$\sum_{i} |\langle n|\hat{O}|i \rangle|^2=?$$
for a general operator $\hat{O}$.