3
$\begingroup$

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).

$\endgroup$
3
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ Is this result hermitian? Because it has i in the result doesnt that negate, making adj(c)=-C, not hermitian $\endgroup$ – yankeefan11 Feb 6 '14 at 14:21
  • $\begingroup$ @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. $\endgroup$ – joshphysics Feb 6 '14 at 16:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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