Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

So I am given a 2-dimensional harmonic oscillator with $H=H_1+H_2$ where $$H_i=\frac{p_i^2}{2m}+\frac{1}{2}m\omega^2x_i^2$$ Additionally, $$L=x_1p_2-x_2p_1$$ If we define $$A=\frac{1}{2\omega}[H_1-H_2]$$ $$B=\frac{1}{2}L$$ $$C=\frac{-i}{\hbar}[A,B]$$ Where [A,B] is the commuatator of A with B. We are asked for the explicit form of C, but isnt it just $$[H_1-H_2,L] = [H_1,L]-[H_2,L]=0$$ Due to the isotropy of space. It just does not make sense that C would be 0, because then the three would not be closed under commutation (which I am supposed to show).

share|improve this question
add comment

1 Answer 1

up vote 3 down vote accepted

When I compute the commutator explicitly, I don't get $0$. Use the canonical commutation relations \begin{align} [x_j, p_k] = i\hbar I\delta_{jk} \end{align} where $I$ is the identity operator, and recall that the harmonic oscillator components are independent which means; \begin{align} [x_k, x_j] = 0, \qquad [p_i, p_j] = 0 \end{align} to compute: \begin{align} [H_1-H_2, L] &= [H_1, L] - [H_2, L] \\ &= [H_1, x_1p_2 - x_2p_1] - [H_2, x_1p_2 - x_2p_1] \\ &= [H_1, x_1]p_2 -x_2[H_1, p_1] - x_1[H_2, p_2] + [H_2, x_2]p_1 \\ &= \frac{1}{2m}[p_1^2, x_1]p_2 - \frac{1}{2}m\omega^2 x_2[x_1^2, p_1] - \frac{1}{2} m\omega^2 x_1[x_2^2,p_2] + \frac{1}{2m} [p_2^2, x_2]p_1 \\ &= \frac{1}{2m} (-2i\hbar)(p_1p_2 + p_2p_1) - \frac{1}{2}m\omega^2(2i\hbar)(x_2x_1+x_1x_2) \\ &= -\frac{2i\hbar}{m}p_1p_2 - 2im\omega^2\hbar x_1x_2\\ &\neq 0 \end{align}

share|improve this answer
    
Is this result hermitian? Because it has i in the result doesnt that negate, making adj(c)=-C, not hermitian –  yankeefan11 Feb 6 at 14:21
    
@yankeefan11 Well the commutator I computed above is not hermitian, but $C$ is given by $-i$ times the commutator, so the $i's$ go away making $C$ hermitian. –  joshphysics Feb 6 at 16:18
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.