Eigenvalue of angular momentum operator

In summary, the conversation discusses finding the eigenvalues and normalized wavefunctions of an operator in the context of a third year physics exam on atomic and nuclear physics. The conversation also touches on the use of eigenstates and eigenvalues in solving the problem and the potential need for representation theory of Lie groups and Lie algebras. Finally, there is a brief discussion on the necessary calculus skills needed to solve the problem.
  • #1
leonmate
84
1

Homework Statement



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


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!
 
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  • #2
First of all, you are considering only one component of the orbital angular-momentum operator (usually the [itex]z[/itex] component).

Further, the wave function for a scalar particle is unchanged by rotations around an arbitrary axis with rotation angle [itex]2 \pi[/itex], 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 [itex]z[/itex] axis by an angle of [itex]2 \pi[/itex]!

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
Hi, thanks for the reply

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
Ok, then it's simpler. Your Hilbert space is given by wave functions on the circle, [itex]\psi(\theta)[/itex] with periodic boundary conditions [itex]\psi(\theta+2 \pi)=\psi(\theta)[/itex]. The corresponding momentum operator (physically indeed an angular momentum operator) is then given by the differential operator
[tex]\hat{L}=-\frac{\mathrm{i}}{\hbar} \frac{\mathrm{d}}{\mathrm{d} \theta}.[/tex]
The Hilbert space here is [itex]L^[(0,2\pi)][/itex], i.e., the set of all square-integrable functions. The eigenvector of the linear operator [itex]\hat{L}[/itex] is thus a function,
[itex]u_l(\theta)[/itex] such that
[tex]\hat{L} u_l(\theta)=-\mathrm{i} u_l'(\theta)=l u_l(\theta),[/tex]
where [itex]l[/itex] is the eigenvalue of [itex]\hat{L}[/itex]. To find now the possible values for [itex]l[/itex] you have to solve for this differential equation and find the values that fulfill the periodic boundary condition.
 
  • #5
leonmate said:
[...] 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
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
no not really, at least i don't think I've covered it before.

well i assume the only function f(x) that would be equal to f'(x) is e^x??
 
  • #7
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
leonmate said:
no not really, at least i don't think I've covered it before.
:eek: 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.

well i assume the only function f(x) that would be equal to f'(x) is e^x??
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:
  • #9
leonmate said:
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)

[...]
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
No I didn't realize 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
leonmate said:
So x = ln(f(x))?
What about the "constant" I mentioned? Do the integrations on both sides more carefully.
 
  • #12
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
leonmate said:
Am I on the correct lines?
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?
 

1. What is the eigenvalue of the angular momentum operator?

The eigenvalue of the angular momentum operator is a physical quantity that represents the magnitude of angular momentum for a given quantum state. It is denoted by the symbol l and is a characteristic of the system being studied.

2. How is the eigenvalue of the angular momentum operator determined?

The eigenvalue of the angular momentum operator is determined by solving the corresponding eigenvalue equation, which is a mathematical equation that relates the operator to its eigenvalues and eigenvectors. This involves solving for the eigenvalues that satisfy the equation and then finding the corresponding eigenvectors.

3. What is the significance of the eigenvalue of the angular momentum operator?

The eigenvalue of the angular momentum operator is significant because it provides information about the properties of a quantum system. It helps to determine the allowed energy levels and the possible outcomes of measurements of angular momentum in a given state.

4. How does the eigenvalue of the angular momentum operator relate to the quantization of angular momentum?

The eigenvalue of the angular momentum operator is closely related to the quantization of angular momentum, as it represents the discrete values of angular momentum that a system can possess. This is due to the fact that angular momentum is a conserved quantity in quantum mechanics and can only take on certain values.

5. Can the eigenvalue of the angular momentum operator change over time?

No, the eigenvalue of the angular momentum operator remains constant over time for a given quantum state. However, the angular momentum itself can change through quantum mechanical processes such as spin flips or changes in orbital angular momentum.

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