Eigenvalue of angular momentum operator

1. Jan 7, 2014

leonmate

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

I'm running through practice papers for my 3rd year physics exam on atomic and nuclear physics:

This is the operator we found in the previous part of the question
L = -i*(hbar)*d/dθ

Next, we need to find the eigenvalues and normalised wavefunctions of L

3. The attempt at a solution

So I know that operator * eigenstate = eigenvalue * eigenstate and have been trying to use this in an attempt to find a solution but to no avail. I think the first thing I need to do is to find an eigenstate:

I also know the eigenstate is found to be: U = K*exp(i*l*θ/hbar) where K is a constant (There's a very unhelpful solution posted by the lecturer that skips the working to this step and I just can't work out how to get there!)

I can find the eigenvalue from that to be l = m*hbar

Any help getting me to the eigenstate would be very appreciated. I've been searching the net trying to find something useful but I can't!

2. Jan 7, 2014

vanhees71

First of all, you are considering only one component of the orbital angular-momentum operator (usually the $z$ component).

Further, the wave function for a scalar particle is unchanged by rotations around an arbitrary axis with rotation angle $2 \pi$, and rotations are generated by the total angular momentum. Since the spin is 0 for scalar particles that's identical with the orbital angular momentum. So, what value must your eigenvalue take to make the found solution reproduce under rotations around the $z$ axis by an angle of $2 \pi$!

PS: The above assumptions are not as trivial as they look. You need a bit of representation theory of Lie groups and Lie algebras to prove them! This is usually not covered in a first quantum mechanics lecture.

3. Jan 7, 2014

leonmate

The question considers a particle free to move around a circle of radius R

I figured that as this is a 1d problem with θ being the only angular coordinate. I'm not sure if I'm looking at the z component or just the angular component?

I follow what you said as the angle can only be between 0-2pi but I haven't done much work with eigenvalues outside of matrix notation so their use in this topic has really confused me.

I'm still unsure of what step I should take after setting up: Operator (L) * Eigenstate (U[θ]) = Eigenvalue (l) * Eigenstate (U[θ])

If i use the operator I found I get the equation:

i*hbar*d/dθ*U(θ) = l * U(θ)

The solution posted by my lecturer shows a step from that to:

U(θ) = K*exp(i*l*θ / hbar) where K is a constant

I have no idea what to do to get there

4. Jan 7, 2014

vanhees71

Ok, then it's simpler. Your Hilbert space is given by wave functions on the circle, $\psi(\theta)$ with periodic boundary conditions $\psi(\theta+2 \pi)=\psi(\theta)$. The corresponding momentum operator (physically indeed an angular momentum operator) is then given by the differential operator
$$\hat{L}=-\frac{\mathrm{i}}{\hbar} \frac{\mathrm{d}}{\mathrm{d} \theta}.$$
The Hilbert space here is $L^[(0,2\pi)]$, i.e., the set of all square-integrable functions. The eigenvector of the linear operator $\hat{L}$ is thus a function,
$u_l(\theta)$ such that
$$\hat{L} u_l(\theta)=-\mathrm{i} u_l'(\theta)=l u_l(\theta),$$
where $l$ is the eigenvalue of $\hat{L}$. To find now the possible values for $l$ you have to solve for this differential equation and find the values that fulfill the periodic boundary condition.

5. Jan 7, 2014

strangerep

1) Do you know enough calculus to verify that the solution given above does indeed satisfy the differential equation? If "yes", then do so and show your verification here.

2) Do you know enough calculus to solve the (slightly simpler) differential equation:
$$\frac{df(x)}{dx} ~=~ f(x)$$ to find the function $f(x)$ ?

6. Jan 11, 2014

leonmate

no not really, at least i dont think ive covered it before.

well i assume the only function f(x) that would be equal to f'(x) is e^x??

7. Jan 11, 2014

leonmate

ok i think i've solved this problem but i'm a little confused as to why the solution is:

U(θ) = K*exp(i*l*θ / hbar)

and not:

U(θ) = K*exp(-l*θ / i*hbar)

As when i rearrange to make dU(θ)/dθ the subject i have = (l/-i*hbar)*U(θ)

8. Jan 11, 2014

strangerep

You're doing 3rd yr physics, without being proficient in such basic calculus? OK,... something has gone wrong in your selection of math courses up to now. But that's not something I can fix in a just a few posts. Maybe you need to ask in the Academic Guidance forum about the best way to catch up on the necessary calculus.

For a N-th order ordinary differential equation, there's N constants in the general solution. Here, we have a 1st order O.D.E., so there should be 1 constant. I'll give you the next step of the solution to the ODE I posed in my previous post #5. You rearrange it to:
$$\frac{df}{f} ~=~ dx ~,$$ and then integrate both sides:
$$\int \frac{df}{f} ~=~ \int dx ~.$$Can you perform these integrals?

Last edited: Jan 11, 2014
9. Jan 11, 2014

strangerep

It looks like the only difference is that in the 1st exponent you have
$$\frac{i\ell\theta}{\hbar}$$ and in the 2nd exponent you have
$$\frac{-\ell\theta}{i\hbar}$$Do you not know that $\frac{1}{i} = -i$ ?

10. Jan 11, 2014

leonmate

No I didn't realise that 1/i = -i that could be handy

And yeah I think I can do that calculus I've done all the math modules but they were in first year in not as sharp as I'd like to be

So x = ln(f(x))?

F(x) = exp(x)

Yeah?

11. Jan 11, 2014

strangerep

What about the "constant" I mentioned? Do the integrations on both sides more carefully.

12. Jan 11, 2014

leonmate

Ok well I would say that

X = ln(f(x)) + ln(k)
Exp(x) = k*f(x)
F(x) = 1/k * exp(x)

Then as k is a constant we can write it as k instead of 1/k

Am I on the correct lines?

13. Jan 12, 2014

strangerep

Yes.

Then you need to invoke some other condition on $f(x)$ to determine the value of $k$.

But first, can you redo the integration if the original differential equation is
$$\frac{df}{dx} ~=~ a f ~,$$ where $a$ is some constant, independent of $x$ ?

If you can do that, then this is probably a good point to switch back to the angular variable $\theta$ instead of $x$. Then, if $f(\theta)$ is a wavefunction, what condition do we usually impose on wavefunctions in QM?