Calculating Commutator of Differential Angular Momentum

Question

I have tried for hours to calculate the commutator of angular momentum in the differential form, but I cannot get the correct answer. This is my first experience with actually checking if two operators commutes, so there may be some beginner's misunderstandings that causes the problem.

I know from my textbook the correct answer is:

$$ [\hat{L}_x,\hat{L}_y] = \frac{\hbar}{\mathrm{i}}\hat{L}_z = \frac{\hbar}{\mathrm{i}} \left( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right) $$

The angular momenta are defined as

$$ \hat{L}_x = y \frac{\partial}{\partial z} - z \frac{\partial}{\partial y} \\ \hat{L}_y = z \frac{\partial}{\partial x} - x \frac{\partial}{\partial z} \\ \hat{L}_z = x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} $$

Sets up the commutator expression, and writes it out:

$$ [\hat{L}_x,\hat{L}_y] = \left[ y \frac{\partial}{\partial z} - z \frac{\partial}{\partial y} , z \frac{\partial}{\partial x} - x \frac{\partial}{\partial z} \right] $$

From what I assume is a well-known algebraic rule for handling commutators, I get the following:

$$ \left[y \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z}, z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial x} \right] - \left[y \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z}, x \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z} \right] - \left[z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial y}, z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial x} \right] + \left[z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial y}, x \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z} \right] $$

It was straight forward to find that the two middle terms commute, i.e. they equal zero. Hence, we are left with:

$$ \left[y \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z}, z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial x} \right] + \left[z \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial y}, x \frac{\hbar}{\mathrm{i}} \frac{\partial}{\partial z} \right] \tag{1} $$

Upon evaluating these terms, I let them act on a function $ \psi $. Looking at the first term: First I factored the complex fraction out. Then I wrote out the commutator and let each differential fraction act on $ \psi $:

$$ \frac{\hbar}{\mathrm{i}} \left(y \frac{\partial \psi z}{\partial z} \frac{\partial \psi}{\partial x} - yz \frac{\partial \psi }{\partial x} \frac{\partial \psi}{\partial z} \right) $$

The y in the second term above can be factored out from the differentials, because the differential is not done with respect to y. In the first term, however, the z must be differentiated along with psi by using the product rule. Doing this, and cancelling equal terms of opposite sign, yields:

$$ \frac{\hbar}{\mathrm{i}} y \psi \frac{\partial \psi}{\partial x} $$

Repeating this calculation for the second term in equation $(1)$ yields

$$ -\frac{\hbar}{\mathrm{i}} x \psi \frac{\partial \psi}{\partial y} $$

Substituting into $(1)$ yields

$$ [\hat{L}_x,\hat{L}_y] \psi = \frac{\hbar}{\mathrm{i}} \left( y \frac{\partial \psi}{\partial x} - x \frac{\partial \psi}{\partial y} \right) \psi $$

Now, comparing with the correct answer:

i) My signs are opposite of what they should be

ii) More importantly, I have a $ \psi $ in the differentials which should not be there. My question is, "why, and how do I rid my answer of it?".

I hope I have sufficiently showed what I have done. I have checked my notes several times, and I cannot get rid of those extra psi's in my answer.

Answer 1

Your problem with the sign stems from two things: Firstly, you use a wrong equation for the commutator $[\hat{L}_x,\hat{L}_y]$ - your prefactor is wrong by a minus sign: The correct equation (see here) is

\begin{equation} [\hat{L}_x,\hat{L}_y] = i \hbar \hat{L}_z \end{equation}

Secondly, your equations for the angular momentum operators are slightly wrong because you leave out the prefactor $\frac{\hbar}{i}$ in them, although you use the correct versions in your equation [1]. So, substituting the differential operator expression for $\hat{L}_z$

\begin{equation} \hat{L}_z = \frac{\hbar}{i} \left( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right) \end{equation}

into the commutator equation you should get:

\begin{equation} [\hat{L}_x,\hat{L}_y] = i \hbar \frac{\hbar}{i} \left( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right) = \hbar^{2} \left( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right) \end{equation}

As for your problem with the additional $\psi$: Everything is fine until you let equation [1] act on a wavefunction $\psi$.

\begin{equation} [\hat{L}_x,\hat{L}_y] \psi = \left( \left[y \frac{\hbar}{i} \frac{\partial}{\partial z}, z \frac{\hbar}{i} \frac{\partial}{\partial x} \right] + \left[z \frac{\hbar}{i} \frac{\partial}{\partial y}, x \frac{\hbar}{i} \frac{\partial}{\partial z} \right] \right) \psi \end{equation}

You seem to make the following mistake which then leads to the appearance of an additional $\psi$ in your equations: When you let the commutator $ [\hat{A},\hat{B}]$ act on a wavefunction $\psi$ you did the following

\begin{equation} [\hat{A},\hat{B}] \psi = \hat{A} \psi \hat{B} \psi - \hat{B}\psi \hat{A}\psi \qquad \qquad \text{(wrong)} \end{equation}

which is incorrect. The correct way to do this is

\begin{equation} [\hat{A},\hat{B}] \psi = \hat{A} \hat{B} \psi - \hat{B} \hat{A}\psi \qquad \qquad \text{(correct)} \end{equation}

where $\hat{A} \hat{B} \psi$ means that first $\hat{B}$ acts on $\psi$ and only then $\hat{A}$ acts on the result. So, in the case at hand this leads to

\begin{align} \left[y \frac{\hbar}{i} \frac{\partial}{\partial z}, z \frac{\hbar}{i} \frac{\partial}{\partial x} \right] \psi &= \underbrace{\left(\frac{\hbar}{i}\right)^{2}}_{= \, -\hbar^{2}} \Biggl( \underbrace{ y \frac{\partial}{\partial z} \biggl( z \frac{\partial \psi}{\partial x} \biggr)}_{= \, y \frac{\partial \psi}{\partial x} + yz \frac{\partial^{2} \psi}{\partial z \partial x}} - z y \frac{\partial^{2} \psi}{\partial x \partial z} \Biggr) \\ &= -\hbar^{2} y \frac{\partial \psi}{\partial x} \end{align}

and

\begin{align} \left[z \frac{\hbar}{i} \frac{\partial}{\partial y}, x \frac{\hbar}{i} \frac{\partial}{\partial z} \right] \psi &= -\hbar^{2} \Biggl( z x \frac{\partial^{2} \psi}{\partial y \partial z} - \underbrace{ x \frac{\partial}{\partial z} \biggl( z \frac{\partial \psi}{\partial y} \biggr)}_{= \, x \frac{\partial \psi}{\partial y} + xz \frac{\partial^{2} \psi}{\partial z \partial y}} \Biggr) \\ &= \hbar^{2} x \frac{\partial \psi}{\partial y} \end{align}

If you substitute these equations into the commutator equation you get

\begin{equation} [\hat{L}_x,\hat{L}_y] \psi = \hbar^{2} \left(x \frac{\partial \psi}{\partial y} - y \frac{\partial \psi}{\partial x} \right) \end{equation}

which is exactly the result you should expect.