Commutation Relations, 2D Harmonic Oscillator

  • Thread starter BOAS
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  • #1
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Homework Statement


Consider a two-dimensional harmonic oscillator, described by the Hamiltonian

##\hat H_0 = \hbar \omega (\hat a_x \hat a_x ^{\dagger} + \hat a_y \hat a_y^{\dagger} + 1)##

Calculate ##\hat H_0 \hat L | n_1, n_2 \rangle## and ##\hat L \hat H_0 |n_1, n_2 \rangle##. What does this imply for ##[\hat H_0 , \hat L]##

Homework Equations




The Attempt at a Solution


[/B]
I can solve this problem by "brute force", having shown that ##\hat L = \hat x \hat P_y - \hat y \hat P_x = i \hbar (\hat a_y^{\dagger} \hat a_x - \hat a_x ^{\dagger} \hat a_y)##. I can apply one operator and then the other, but the results are messy making me think this can be done in a smarter manner.

Trying to combine these operators into a single simplified operator gives me

##H_0 \hat L = i \hbar^2 \omega (\hat a_x ^{\dagger} \hat a_x (\hat a_y ^{\dagger} \hat a_x - \hat a_x ^{\dagger} \hat a_y) + \hat a_y^{\dagger} \hat a_y (\hat a_y^{\dagger} \hat a_x - \hat a_x^{\dagger} \hat a_y) + (\hat a_y^{\dagger} \hat a_x - \hat a_x^{\dagger} \hat a_y))##

I need to deal with this ##(\hat a_y ^{\dagger} \hat a_x - \hat a_x ^{\dagger} \hat a_y)## term but I am unsure of how to do it. I know that ##[\hat a_i, \hat a_i^{\dagger}] = 1## and that the x and y operators are mutually commuting.

If I set ##\hat N = \hat a_y^{\dagger} \hat a_x##, then ##(\hat a_y ^{\dagger} \hat a_x - \hat a_x ^{\dagger} \hat a_y) = [\hat N, \hat N^{\dagger}]## but the numbered states or not eigenstates of my invented operator...

So, is there anything I can realise to make this problem less of an accounting exercise?

I hope it's clear what i'm asking; I can solve this problem, but it does not feel efficient.

Thank you for your help!
 

Answers and Replies

  • #2
763
71
Do you realize that the Hamiltonian on a number state ##\mid n_1,n_2\rangle ## has eigenvalue related to the total ##n_1+n_2##?
\begin{align*} H& = \hbar \omega (a_x ^{\dagger}a_x + a_y^{\dagger}a_y \ + 1) \\
H &= \hbar\omega(N_x + N_y + 1)\\
H|n_1, n_2> &= \hbar\omega(n_1 + n_2+1)|n_1, n_2>\end{align*}
So a question you can ask is "does ##L## change the total ##n_1 + n_2## ?"
You can examine the operator ##L## or try it out.
 

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