Non-Commutative Angular Momentum, and Muonium

[YouWish]

We are doing several things in my physics class that I really do not understand, and I would really appreciate it if someone could help me.

First:

I do not understand the non-commutivity of angular momentum in the case of [Lx,Ly]. I understand how to arrive at the values of Lx = -ih(y * d/dz - z * d/dy) and Ly = -ih(z * d/dx - x * d/dz), but I do not understand how these values do not commute.

For instance, as far as I understand, [Lx,Ly] = (Lx* Ly) - (Ly *Lx) = -h2( (y * d/dz - z * d/dy)(z * d/dx - x * d/dz) - (z * d/dx - x * d/dz)(y * d/dz - z * d/dy). However, whenever I multiply all these values out, I get an answer of zero, meaning that the values do commute, but I know that they are not supposed to. Wikipedia lists the final answer as ihLz, but the steps are not shown and I have no idea how they got that answer. All other online searches have proven equally fruitless for the steps involved. I'm just lost as to how to get that answer.

Second:

One of our problems involved finding the center of mass in a muonium atom. This atom is comprised of an electron, which has a mass of .511 MeV/c2, and a muon, which has a mass of 105.6 MeV/c2. I am utterly lost as to how to find the center of mass of this system, because it seems to me that you would have to know the radius between these two particles to find the center of mass. I realize this is not much information to go on, but I would greatly appreciate any help with these two problems.

 

[Ben Niehoff]

First: Remember that derivatives don't simply "multiply"; you have to use the product rule:

$$\left( y \frac{\partial}{\partial z} \right) \left( z \frac{\partial}{\partial x}\right) = y \frac{\partial z}{\partial z} \frac{\partial}{\partial x} + yz \frac{\partial^2}{\partial z \partial x} = y \frac{\partial}{\partial x} + yz \frac{\partial^2}{\partial z \partial x}$$

So, the terms with mixed partials will cancel, but other terms will not.

Second:

Try looking up the derivation of the Bohr radius (can't remember it off the top of my head). You will probably find how it depends upon the masses of the proton and electron.

 

[denian]

Hello,

I can definitely help you with your questions about non-commutative angular momentum and muonium. Let's start with the first question about non-commutativity.

To understand why [Lx,Ly] does not equal zero, we need to look at the mathematical definition of the commutator. The commutator of two operators A and B is defined as [A,B] = AB - BA. This means that we need to multiply A and B in both orders and then take the difference between the two products.

In the case of angular momentum, Lx and Ly are operators that represent the components of angular momentum in the x and y directions, respectively. When we multiply Lx and Ly, we get:

Lx * Ly = -ih(y * d/dz - z * d/dy)(z * d/dx - x * d/dz)

= -ih(y * d/dz * z * d/dx - y * d/dz * x * d/dz - z * d/dy * z * d/dx + z * d/dy * x * d/dz)

= -ih(y * (z * d/dx) * d/dz - (y * x) * (d/dz)^2 - (z * d/dy) * (z * d/dx) + (z * x) * (d/dz)^2)

= -ih(y * (z * d/dx) * d/dz + (z * x) * (d/dz)^2 - (z * d/dy) * (z * d/dx) )

Similarly, when we multiply Ly and Lx, we get:

Ly * Lx = -ih(z * (x * d/dy) * d/dz + (x * y) * (d/dz)^2 - (x * d/dx) * (y * d/dz) )

Now, let's subtract Ly * Lx from Lx * Ly:

Lx * Ly - Ly * Lx = -ih(y * (z * d/dx) * d/dz + (z * x) * (d/dz)^2 - (z * d/dy) * (z * d/dx) ) - (-ih(z * (x * d/dy) * d/dz + (x * y) * (d/dz)^2 - (