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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).

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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}

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Is this result hermitian? Because it has i in the result doesnt that negate, making adj(c)=-C, not hermitian – yankeefan11 Feb 6 '14 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 '14 at 16:18

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