Let's say I have two operators, $\hat{x}^k$ and $\hat{p}_x^l$, where $\hat{x}$ and $\hat{p}_x$ are the ordinary position and momentum operators. It seems fairly straight forward to show that $\hat{x}^k$ and $\hat{p}_x^l$ are Hermitian. And the product of two Hermitian operators is Hermitian if they commute. However, it seems like something funny happens with $\hat{x}^k$ and $\hat{p}_x^l$, when I try and work out the commutator. I end up with:

$$ [\hat{x}^k,\hat{p}_x^l] = -\left( \frac{\hbar}{i} \right)^l\frac{\partial^l}{\partial x^l}x^k $$

If $l=k=1$, then this is just the plain old-fashioned commutator for position and moment. No problems. However, because of the integer powers:

$$ -\left( \frac{\hbar}{i} \right)^l\frac{\partial^l}{\partial x^l}x^k=-\left( \frac{\hbar}{i} \right)^l\frac{\partial^{l-1}}{\partial x^{l-1}}kx^{k-1} = -\left( \frac{\hbar}{i} \right)^l\frac{\partial^{l-2}}{\partial x^{l-2}}(k-1)kx^{k-2} $$


So this is the first step that I'm a little unsure of. So my first question is: is the above correct?

Now if $k>l$, then I get:

$$ [\hat{x}^k,\hat{p}_x^l] =-\left( \frac{\hbar}{i} \right)^lk(k-1)(k-2)...(k-l)x^{k-l} $$ so clearly the two operators do not commute. If $l=k$:

$$ [\hat{x}^k,\hat{p}_x^l] =-\left( \frac{\hbar}{i} \right)^lk! $$ And again, they don't commute. However, if I let $l>k$, then:

$$ \begin{align} [\hat{x}^k,\hat{p}_x^l] &= -\left( \frac{\hbar}{i} \right)^{l-k}\frac{\partial^{l-k}}{\partial x^{l-k}}x^k!x^0\\ \\ &=-\left( \frac{\hbar}{i} \right)^{l-k}\frac{\partial^{l-k-1}}{\partial x^{l-k-1}}\frac{\partial}{\partial x}k!\\ \\ &=0 \end{align} $$

So now they commute and $\hat{x}^k\hat{p}_x^l$ is Hermitian! What's happened here? Have I made some trivial error somewhere?

  • $\begingroup$ The first equation (v1) is incorrect. $\endgroup$ – Qmechanic Mar 24 '19 at 11:31
  • $\begingroup$ Ah yes, thank you. Doing it again, I can see that it should be quite different, as I have only applied the product rule once, when it should be applied $l$ times. However, my situation is now worse, because I end up with a long expression that I can't simplify. $\endgroup$ – monkeyofscience Mar 24 '19 at 20:30

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