# Measurement of angular momentum

1. Aug 28, 2007

### noospace

This problem was inspired by my thread in the quantum physics room

1. The problem statement, all variables and given/known data

Suppose the wavefunction is $\psi = aY_{00} + bY_{10}$, where $Y_{lm}$ are the spherical harmonics. I want to find the possible measurements of the $x$-component of angular momentum.

3. The attempt at a solution

Since the operators $L_x$ and $\mathbf{L}^2$ do not commute (ie there is uncertainty relation between them), the operator $L_x$ has a different set of eigenfunctions than $\mathbf{L}^2$, so we must express $\psi$ in terms of its eigenfunctions. To do this, we presumably solve the eigenvalue equation

$L_x |X\rangle = \lambda |X\rangle$

where $|X>$ are the yet to be determined eigenfunctions. If we let $|X\rangle = \sum_{l,m} a_{lm} |Y_{lm}\rangle$ then we get an infinite matrix equation and an infinite number of eigenvalues and corresponding eigenvectors. Only a small subset of these will be necessary to generate the originial function $\psi$. So how do we know what restriction to place on the test function $|X\rangle$ for general operators? Is it just clever thinking or is there a systematic approach?

2. Aug 28, 2007

### dextercioby

What do you mean by "possible measurements" ? I guess you mean "possible outcomes" of L_{x} measurements. If so, then i suggest you reread the 3-rd postulate. What does it say ?

3. Aug 28, 2007

### noospace

Hi dextercioby,

I don't see what L_x being a Hermitian operator has to do with my question.

The postulate I'm using here is that the eigenfunctions of L_x form a complete set so that we can express the wavefunction as a linear combination of these eigenfunctions. I understand that to do this we must obtain the eigenfunctions of L_x in terms of the spherical harmonics. But how do we do this without solving infinite matrix equations?

4. Aug 28, 2007

### genneth

If you regard finding a Fourier series as solving "infinite matrix equations" then you can't. Understanding how the formula for Fourier coefficients come about will help you see what is needed. Symmetry will help you to rule out the obviously zero coefficients. The rest might fall into place. Try it.

Edit: actually, ignore this. See below.

Last edited: Aug 28, 2007
5. Aug 28, 2007

### genneth

Btw, L_x and L^2 do commute...

6. Aug 29, 2007

### noospace

I understand how do do this now. Here it is for future use.

A sensible choice for the test function is $|X> = aY_{00} + bY_{11}+ cY_{10} + dY_{1(-1)}$ because the L_x operator can only change the value of the quantum number m and not the l number. If you plug this test function into the eigenvalue equation $L_x | X \rangle = \lambda | X \rangle$ you immediately see that $a=0$. You then get a 3 x 3 matrix equation

$A (b,c,d)^T = (0,0,0)^T$.

Nontrivial solutions iff det A = 0 from which you can deduce the eigenvalues $0,\hbar,-\hbar$ and the corresponding eigenvectors. Rewritting the original wavefunction in terms of these eigenfunctions gives the allowable states.

Note also that if the eigenfunctions of two operators coincide, this implies that they commute, but the converse need not be true.

Last edited: Aug 29, 2007