Given a wave function, predict the result

[spacetimedude]

Homework Statement

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?

Homework 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.

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 $$\hat{L^2} |l,m>=\hbar^2 l(l+1)|l,m>$$.

Thanks in advance.

 

[DrClaude]

Is u a linear combination of two states, $exp(2i\phi)$ and $exp(-2i\phi)$

Can't you write u as a linear combination?

and do the value -2 and 2 indicate the angular momentum?

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

If I act some operator to u, do we only consider one of the exponential state?

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.

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

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.

 

[spacetimedude]

Can't you write u as a linear combination?

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?

 

[DrClaude]

Lets start with

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

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 $m=-2$.

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.

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.

I don't see how one term vanish while the other one stays when the wave function collapses.

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