# Calculating Commutator of Differential Angular Momentum

1. May 9, 2014

### Chem.Stud.

Hi there!

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:

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

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:

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

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

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

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 $\psi$:

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

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:

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

Repeating this calculation for the second term in [1] yields

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

Substituting into [1] yields

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

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.

I would appreciate any help!

Regards,
Anders

2. May 9, 2014

### Bill_K

[A,B] ψ = A(Bψ) - B(Aψ)

Looks like you forgot the parentheses. Also there should be two factors of ħ/i in front, not one. (And just one ψ!)

3. May 9, 2014

### ChrisVer

For such cases, because it is a bit difficult not to make a mistake during your calculations, it's better to write some things in a more compact, yet known form....
For example, you could avoid so many $\frac{h}{i} \frac{d}{dx_{i}}$ by substituting them with $p_{i}$, for which you know its commutation relations with the coordinates $x_{i}$

So for example the first parenthesis is:
$[y p_{z},z p_{x}]= y [p_{z},zp_{x}]= y [p_{z},z] p_{x} = -ih y p_{x}$
The 2nd is:
$[z p_{y},x p_{z}]= [z, xp_{z}] p_{y}= x [z,p_{z}] p_{y}= ih x p_{y}$

Also I don't understand why you have h/i twice in the commutators...where did they come from?
$a [\hat{A},\hat{B}] \ne [a \hat{A},a \hat{B}]$

Last edited: May 9, 2014
4. May 9, 2014

### Jilang

Your main problem is that you are operating too soon on the wavefunction. Leave it right until the end.

5. May 9, 2014

### ChrisVer

You need to show
$L_{z}= \frac{i}{h} [L_{x},L_{y}]=\frac{i}{h}[ y \frac{\partial}{\partial z}-z\frac{\partial}{\partial y}, z \frac{\partial}{\partial x}-x\frac{\partial}{\partial z}]$
Now you should replace $\frac{\partial}{\partial x_{j}} = \frac{i}{h} p_{j}$

$\frac{i}{h} [L_{x},L_{y}]=\frac{i}{h}[ y \frac{i}{h} p_{z}-z\frac{i}{h} p_{y}, z \frac{i}{h} p_{x}-x\frac{i}{h} p_{z}]$

$\frac{i}{h} [L_{x},L_{y}]=(\frac{i}{h})^{3}[ y p_{z}-z p_{y}, z p_{x}-x p_{z}]$

$\frac{i}{h} [L_{x},L_{y}]=(\frac{i}{h})^{3}([ y p_{z}, z p_{x}]+ [ z p_{y}, x p_{z}])$

$\frac{i}{h} [L_{x},L_{y}]=(\frac{i}{h})^{3}(-ih y p_{x}+ ih x p_{y})$

$\frac{i}{h} [L_{x},L_{y}]=(\frac{i}{h})^{3}(-ih y \frac{h}{i} \frac{\partial}{\partial x}+ ih x \frac{h}{i} \frac{\partial}{\partial y})$

$\frac{i}{h} [L_{x},L_{y}]=-\frac{i}{h}(- y \frac{\partial}{\partial x}+ x \frac{\partial}{\partial y})=\frac{i}{h} L_{z}$

But obviously your angular momenta are not well defined.... For example, you know that angular momenta have dimensions of $h$, also you show partial derivatives in their definition (reminding me of momenta) which is not correct....
Another way to see that is by remembering that angular momenta are:
$\vec{L}=\vec{r} \times \vec{p}$
which after the quantization of the momenta will get:
$\vec{L}=\frac{h}{i} \vec{r} \times \vec{∇}$
with the cross giving the things you give as: xd_y - y d_x etc....
In fact the definitions you have given correspond to:
$\frac{i L_{i}}{h}$ in order to make them quantum operators...
with that definition:
$\frac{i}{h} [\frac{ i L_{x}}{h}, \frac{i L_{y}}{h}]= \frac{i}{h} \frac{i L_{z}}{h}$
which will give you the
$[L_{x},L_{y}]= \frac{h}{i}L_{z}$

Last edited: May 9, 2014
6. May 11, 2014

### Chem.Stud.

I see where I went wrong. I have managed to get the correct answer. Thank you for your help!