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Harmonic oscillator with ladder operators - proof for sum rule

Physics Asked by Nad on January 13, 2021

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 $H|nrangle=hbar(n+frac 12)|nrangle$ and I know that $H|nrangle=E|nrangle$

One Answer

Just a tip: $$ sum_{l}langle n|p|lrangle E_l langle l|p|nrangle = langle n|pHp|nrangle,\ sum_{l}langle n|p|lrangle langle l|p|nrangle = langle n|p^2|nrangle, $$ where I used the expansion of identity $$ 1 = sum_l |lranglelangle l| $$

Update
Perhaps more in the spirit of the exercise is to use the properties of the raising and lowering operators: $$ b|nrangle = sqrt{n}|n-1rangle,\ b^dagger|nrangle = sqrt{n+1}|n+1rangle, $$ which allows easily to calculate th matrix elements (remember that $langle m|nrangle = delta_{n,m}$).

Answered by Vadim on January 13, 2021

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