Calculating Commutator of Differential Angular Momentum

[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:

<br /> [\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)<br />

The angular momenta are defined as

<br /> \hat{L}_x = y \frac{\partial}{\partial z} - z \frac{\partial}{\partial y} \\<br /> \hat{L}_y = z \frac{\partial}{\partial x} - x \frac{\partial}{\partial z} \\<br /> \hat{L}_z = x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x}<br />

Sets up the commutator expression, and writes it out:

<br /> [\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] <br />

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

<br /> \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] <br /> - \left[z \frac{\hbar}{i} \frac{\partial}{\partial y}, z \frac{\hbar}{i} \frac{\partial}{\partial x} \right] <br /> + \left[z \frac{\hbar}{i} \frac{\partial}{\partial y}, x \frac{\hbar}{i} \frac{\partial}{\partial z} \right]<br />

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

<br /> \left[y \frac{\hbar}{i} \frac{\partial}{\partial z}, z \frac{\hbar}{i} \frac{\partial}{\partial x} \right]<br /> + \left[z \frac{\hbar}{i} \frac{\partial}{\partial y}, x \frac{\hbar}{i} \frac{\partial}{\partial z} \right] \phantom{...} [1]<br />

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:

<br /> \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)<br />

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:

<br /> \frac{\hbar}{i} y \psi \frac{\partial \psi}{\partial x}<br />

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

<br /> -\frac{\hbar}{i} x \psi \frac{\partial \psi}{\partial y}<br />

Substituting into [1] yields

<br /> [\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<br />

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

 

[Bill_K]

First I factored the complex fraction out. Then I wrote out the commutator and let each differential fraction act on \psi:
<br /> \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)<br />

[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!)

 

[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}]

 

[Jilang]

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

 

[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}

 

[Chem.Stud.]

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