# Given a wave function, predict the result

1. May 3, 2017

### spacetimedude

1. The problem statement, all variables and given/known data
The wave function of a particle is known to have the form $$u(r,\theta,\phi)=AR(r)f(\theta)\cos(2\phi)$$ where $f$ is an unknown function of $\theta$. What can be predicted about the results of measuring
(a) the z-component of angular momentum;
(b) the square of the angular momentum?

2. Relevant equations
$cos(2\phi)={exp(2i\phi)+exp(-2i\phi)}/2$
Also, I know the spherical harmonics and the corresponding eigenvalues.

The solutions are:
a) $L_z=+/- 2$ with equal probability.
b) $l\geq 2$ but nothing can be said if we don't know $f(\theta)$

where l denotes the angular momentum quantum number.
3. The attempt at a solution
For the first one, I am trying to make sense of what the physical interpretation of the two exponential functions is in the wave function. Why is $L_z=+/- 2$? Is u a linear combination of two states, $exp(2i\phi)$ and $exp(-2i\phi)$ and do the value -2 and 2 indicate the angular momentum? If I act some operator to u, do we only consider one of the exponential state? I am a bit rusty with the meaning of a wave function.

The second one, I just know that $$\hat{L^2} |l,m>=\hbar^2 l(l+1)|l,m>$$.

2. May 3, 2017

### Staff: Mentor

Can't you write u as a linear combination?

No, they represent the eigenvalues of the $\hat{L}_z$ operator, which correspond to the projection of the angular momentum on the z axis.

No, you would have to act on the full wave function. But you don't actually need to do that, as the probability of measuring the system in a given $|l, m \rangle$ is
$$\langle l, m | u \rangle$$
and you need to figure out for which $|l ,m \rangle$ the result would be something other than 0.

Since the $|l ,m \rangle$ are spherical harmonics, it would probably help for you to look at a table of spherical harmonic functions to familiarize yourself with their form in terms of l and m.

3. May 3, 2017

### spacetimedude

Yes, by expanding the two exponential terms. But I thought $$u(r,\theta,\phi)=AR(r)f(\theta)\exp(2i\phi)/2+AR(r)f(\theta)\exp(-2i\phi)/2$$ represents just one of all the possible state. Or is each term in the right hand side representing a different state (is this the superposition?)? I don't see how one term vanish while the other one stays when the wave function collapses.

And thank you. I understand the rest.

EDIT: Sorry, I still don't see why Lz=+/- 2. They come from reading the exponential terms, correct? How are they the eigenvalues of Lz?

4. May 4, 2017

### Staff: Mentor

The angular momentum operators in spherical coordinates are
$$\hat{L}_z = -i \hbar \frac{\partial}{\partial \phi} \\ \hat{L}^2 = - \hbar^2 \left[ \frac{1}{\sin \theta}\frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial}{\partial \theta} \right) + \frac{1}{\sin^2 \theta} \frac{\partial^2}{\partial \phi^2} \right]$$
So it is obvious that for $\hat{L}_z$, one needs only look at the $\phi$ part of the wave function. The eigenstates of $\hat{L}_z$ are $e^{i m \phi} / \sqrt{2\pi}$ (with $m$ an integer). One can make use of the orthogonality property
$$\frac{1}{2\pi} \int_0^{2 \pi} e^{-i m' \phi} e^{i m \phi} d\phi = \delta_{m,m'}$$

So you you make a measurement of $\hat{L}_z$, after the measurement the wave function will be in an eigenstate of $\hat{L}_z$, so the $\phi$ part will be $e^{i m \phi}$. To see which value of $m$ are possible, you make use of the orthogonality property, and thus the only possibilities will be $m=2$ or $m=-2$.

This is the state of the system before measurement. It represents a single state, but it is a superposition from the point of view of the eigenstates of $\hat{L}_z$. In other words, since it is not an eigenstate of $\hat{L}_z$, it has to be a superposition of those eigenstates.

That's just the way it is. If you measure $L_z = 2 \hbar$, then you know that after measurement the wave function is $\propto e^{2i\phi}$.