# 3D harmonic oscillator orbital angular momentum

1. May 20, 2013

### ProPatto16

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

i need to calculate the orbital angular momentum for 3D isotropic harmonic oscillator is the first excited state

3. The attempt at a solution

for the first excited state:

$$\psi_{100}=\left(\frac{4m^3\omega^3}{\pi\hbar^3}\right)^{3/4}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}r sin\theta cos\phi e^{-m\omega r^2/2\hbar}$$

now as far as i can work out to find the angular momentum i need to apply this function to the spherical harmonics $Y_{l,m_l}(\theta,\phi)$

but im not sure what to try?

ive been floating around the web and cant find a single example of a solution to orbital angular moment of this type, all i can find are countless derivations.

what baffles me the most is how to get an eigenvalue for momentum of a multiple of $\hbar$ when multiplying those functions as theres no derivation and the only occurence of $\hbar$ is in the exponent.

in general to find the value of an observable i.e. momentum then i need to apply the wave function to the eigenfunctions of the observable i.e. momentum and solve for the eigenvalue, yes?

not quite sure how to approach this problem..

normally i would apply the wavefunction to the orbital angular momentum operators, but ive been told to apply it to the spherical harmonics.

thanks

2. May 20, 2013

### vela

Staff Emeritus
What do you mean by "apply this function to the spherical harmonics"?

3. May 20, 2013

### Simon Bridge

Have you tried the angular momentum operator?

No. To find the value of an observable, you need to apply the operator for the observable to the wavefunction. If the wavefunction is an eigenvector of the operator (the observable) then you will have a single eigenvalue - which is the value you are looking for. As you observe below:
... you did know to use the operator.

Compare your wavefunction above with the general solution.
$$\psi_{klm}(r,\theta,\phi) = R_k(r)Y_{lm}(\theta,\phi)$$

... is there any other part of the wavefunction that the angular momentum operator would apply to? Can you express the operator on spherical coordinates?
http://quantummechanics.ucsd.edu/ph130a/130_notes/node216.html

4. May 20, 2013

### ProPatto16

By "apply" I mean multiply the wave function and the spherical harmonics together and simplify to the wave function multiplied by a constant.

However I would personally multiply by momentum operators in each direction not spherical harmonics. However to do that I'd need to separate the variables first but that's easy to do by expanding the exponent.

But I've been told to "apply" the wave function to the spherical harmonics. But I've never seen these practically applied so I'm not sure how this goes.

Also, the question specifically says calculate the orbital angular momentum... But if I was to use the spherical harmonics wouldn't I have a momentum eigenvalue for each harmonic?

5. May 20, 2013

### ProPatto16

im beginning to question my understanding of how to use the spherical harmonics...

i always thought to find the observable i take the wave function, multiply it with the operators, in this case each component of the wave function with the appropriate directional momentum operator. and in reducing it to the original wave function times a constant, the constant is the value of the observable, i.e. momentum.

so why would i be told to, or need to, include the spherical harmonics.

im not sure whether this is right or not and it would be extensive to try it... but..

To find the momentum

If i take the wave function, multiply it by the operator, reduce it down, but instead of a single eigenfunction, reduce it to the original wave function multiplied by a spherical harmonic, and an eigenvalue... and the eigenvalue is then the momentum?

im not 100% sure how the spherical harmonics fit into the big picture, what they physically represent etc. i understand the derivation, but i cant rely on my intuition.

and by the way, i have the operators in spherical co-ordinates already. and initially i would just directly multiply the necessary parts of the wave function to each operator but because of the harmonics, im not so sure

6. May 20, 2013

### Simon Bridge

You are not asked to find the momentum - you are asked to find angular momentum.
What is the angular momentum operator in spherical coordinates?

You do not apply spherical harmonics to the wavefunction - they are the wavefunction: the angular part of it.
Have you followed the suggestions in post #3 yet?

7. May 20, 2013

### ProPatto16

Assume i mean angular momentum unless i say otherwise but ill specify from now on..

$$L_{xop}=\frac{\hbar}{i}\left(-sin\phi \frac{∂}{∂\theta} - cot\theta cos\phi \frac{∂}{∂\phi}\right)$$

$$L_{yop}=\frac{\hbar}{i}\left(cos\phi \frac{∂}{∂\theta} - cot\theta sin\phi \frac{∂}{∂\phi}\right)$$

$$L_{zop}=\frac{\hbar}{i} \frac{∂}{∂\phi}$$

Thats what i was a bit cloudy about, but that clears that up

I tried separating the function into each variable and applying the operators individually, but it turned a bit messy.

However given your general solution, when separating the variables i did it in cartesian and converted to polar afterwards which leaves an r in the exponent but i need to get that out?

should i begin with the Laplacian and the SE and then separate in polar coordinates?

8. May 20, 2013

### vela

Staff Emeritus
You really don't want to conflate making a measurement with applying the operator to a state.

When you make a measurement of some quantity, like energy, there's an associated Hermitian operator, called an observable. In the case of energy, it would be the Hamiltonian. For the z-component of angular momentum, it would be $\hat{L}_z$, and so on.

The possible results of a measurement correspond to the eigenvalues of the associated observable. For each eigenvalue $\lambda_n$, there's an associated eigenvector/eigenfunction $\phi_n$. These eigenvectors form a basis, so given a state $\psi$, you can express it as a linear combination of the eigenvectors/functions:
$$\psi = c_1 \phi_1 + c_2 \phi_2 + \cdots$$
The probability amplitude an of obtaining $\lambda_n$ when you make the measurement on a system in state $\psi$ is given by
$$a_n = \int \phi_n^*(x) \psi(x)\,dx,$$ which is generally a complex quantity. The corresponding probability pn is given by $p_n=|a_n|^2$. You should be able to convince yourself that $a_n = c_n$.

Note that in all of this, the operator was never applied to $\psi$.

Last edited: May 20, 2013
9. May 20, 2013

### Simon Bridge

Perhaps if we put it in more familiar terms:
You are given $\psi$ and asked to find the energy of the state.
Is this something you've done before?

if $\hat{H}\psi = E\psi$ ? Then you'd say the energy of the state is the eigenvalue E, and that $\psi$ is an energy eigenstate?

But what if $\psi$ is not in an energy eigenstate?
What is the value of the energy then? How would you answer that question?

What if you know it's an eigenstate, but when you apply the operator the eigenvalue is not easy to separate out? Is there something additional you can try?

Back to your questions:
Notice how the angular momentum operators have nothing to do with the radial component, only the angular component? i.e. the spherical harmonic part?

Is the spherical harmonic an eigenstate of the angular momentum operator(s)?
I suspect that the point of the exercise is to get you thinking about the properties of the spherical harmonics.

Note: you shouldn't need to convert to Cartesian coords.
Notes: may help understand spherical harmonics better in relation to angular momentum:
http://www.phy.duke.edu/~rgb/Class/phy319/phy319/node100.html
http://quantummechanics.ucsd.edu/ph130a/130_notes/node210.html

10. May 20, 2013

### ProPatto16

Okay, so from a combination of both of your replies...

i need to manipulate $\psi_{100}$ such that it can be expressed like this

$$\psi_{100}=c_1 Y_{l,m_l}+c_2 Y_{l,m_l}+c_3 Y_{l,m_l} \dots$$

so lets just scale down to ground state for a second...

$$\psi_{000}=\left(\frac{m\omega}{\pi \hbar}\right)^{3/4} e^{-m\omega r^2 /2 \hbar }$$

so looking now to where the spherical harmonics can be applied... since they are in terms of $\theta$ and $\phi$ theres nowhere. however since any function of $\theta$ and $\phi$ can be expressed in terms of the harmonics...

$$\psi_{000}=\left(\frac{m\omega}{\hbar}\right)^{3/4}\left( \frac{1}{\pi}\right)^{3/4}e^{-m\omega r^2 /2 \hbar }$$

and so

$$\psi_{000}=\left(\frac{m\omega}{\hbar}\right)^{3/4}\left( \frac{2}{\pi^{1/4}}\right)\left(\sqrt{\frac{1}{4\pi}}\right)e^{-m\omega r^2 /2 \hbar }$$

so we have

$$c_1 Y_{0,0} \dots with \dots c_1=\left( \frac{2}{\pi^{1/4}}\right)$$

or have i gone bush?

11. May 21, 2013

### Simon Bridge

I can't tell if you have followed any of the suggestions.

Your wavefunction has form: $\psi = Are^{-\nu r^2}\sin\theta\cos\phi$ ... where $A$ and $\nu$ are constants.

Is this oscillator supposed to be the isotropic harmonic oscillator?

12. May 21, 2013

### ProPatto16

yes it is meant to be the isotopic harmonic oscillator... in 3D and its first excited state.

how does my wave function have that form?

more specifically how is there an r in front of the e? assuming thats the r the spherical coordinate r, given that only one of the coordinates are in the excited state.

13. May 21, 2013

### ProPatto16

So the solution is not to expand the wave fiction into a sum of spherical harmonics?

Although the wave function is a function of r,theta,phi but I need a function just in terms of theta phi? Which brings me back to that general solution that was posted..

14. May 21, 2013

### Simon Bridge

From what you wrote in post #1 - I just combined all the constants so your wavefunction can be more easily compared with the general form:

$$\psi_{100} = N_{10}e^{-\nu r^2}L_1^{(1/2)}(2\nu r^2)Y_{00}(\theta ,\phi)$$
$$N_{10} = \left ( \frac{2\nu^3}{\pi} \right )^{\frac{1}{4}}$$
$$L_{1}^{(1/2)} = 3\nu r^2 - 4\nu^2 r^4$$

$$Y_{00} =\frac{1}{\sqrt{4\pi}}$$ ... i.e, there is no angular dependence to $\psi_{100}$ for an isotropic 3D HO, but you have an angular dependence in yours. Where did yours come from?

15. May 21, 2013

### Simon Bridge

Oh no - expanding in another basis and applying operators are all part of the general process of predicting the outcome of a measurement. But you will not know how to apply them until you understand the wavefunction in front of you.

The responses you have been getting are aimed at helping you to this understanding.
Unfortunately you don't seem to want to follow suggestions, read the links, or answer questions.
This means that I cannot help you. Sorry.

16. May 21, 2013

### ProPatto16

$$\psi_{100}(x,y,z)=\psi_1(x) \psi_0(y) \psi_0(z)$$

im given

$$\psi_1(x)=\left(\frac{4m^3\omega^3}{\pi \hbar^3}\right)^{3/4} xe^{-m\omega x^2/2 \hbar}$$

and from my text

$$\psi_0(x)=\left(\frac{m\omega}{\pi \hbar}\right)^{1/4}e^{-m\omega x^2/2 \hbar}$$

so hence

$$\psi_{100}(x,y,z)=\left(\frac{4m^3\omega^3}{\pi \hbar^3}\right)^{3/4}\left(\frac{m\omega}{\pi \hbar}\right)^{1/4}\left(\frac{m\omega}{\pi \hbar}\right)^{1/4} xe^{-m\omega x^2/2 \hbar} e^{-m\omega y^2/2 \hbar} e^{-m\omega z^2/2 \hbar}$$

and so

$$\psi_{100}(x,y,z)=\left(\frac{4m^3\omega^3}{\pi \hbar^3}\right)^{3/4}\left(\frac{m\omega}{\pi \hbar}\right)^{1/4}\left(\frac{m\omega}{\pi \hbar}\right)^{1/4} xe^{-m\omega(x^2+y^2+z^2)/2 \hbar}$$

then since $r^2=x^2+y^2+z^2$ and $x=rsin\theta cos\phi$

then thats how i got my wave function.

You are given $\Psi$ and asked to find the energy of the state.
Is this something you've done before?

yes

if $\hat H \Psi=E\Psi$ ? Then you'd say the energy of the state is the eigenvalue E, and that $\Psi$ is an energy eigenstate?

yes

But what if $\Psi$ is not in an energy eigenstate?
What is the value of the energy then? How would you answer that question?

If the particle is not in an energy eigenstate then it does not have a definitive energy. however the wave function can still be expressed in terms of a sum of energy eigenstates and expansion co-efficients.

For energy it would be something like this i believe

$$\psi=\sum c_n \sqrt{2/L} sin\left(\frac{n \pi x}{L}\right)$$

where $c_n$ are the expansion co-efficients.

and then each eigenfunction in the state has a definitive energy?

What if you know it's an eigenstate, but when you apply the operator the eigenvalue is not easy to separate out? Is there something additional you can try?

17. May 21, 2013

### Simon Bridge

Ahhhh ... that is $\psi_{n_xn_yn_z}$ rather than the $\psi_{nlm}$ that is the notation used in all the references.

... and so
$$\psi = \left ( \frac{32^3\nu^5}{\pi^5} \right )^{1/4} r e^{-\nu r^2}\sin\theta\cos\phi \; : \nu=\frac{m\omega}{2\hbar}$$ ... gotcha.

note: the new variable $\nu$ is conventional and, also, we probably want to be able to use "m" as the spin quantum number that will appear in the spherical harmonics. I dropped the index to avoid confusion later on.

You would basically proceed in the way you would have for the version of the question for energy.
i.e. Is $\psi$ an eigenfunction of angular momentum?
Of $L_z$ or $L^2$?

You also need to decide what is meant by "find the angular momentum of the state".
It could be that you are expected to produce the distribution of possible angular momenta, or the expectation value or something else. It's not clear from where I am sitting so I cannot advise you - which was the point of asking you about that earlier.

Anyway - the expansion in spherical harmonics would get you to (something like):

$\psi = \left ( \frac{32^3\nu^5}{\pi^5} \right )^{1/4} r e^{-\nu r^2}\sin\theta\cos\phi = R(r) \sum_{l,m} c_{lm}Y_{lm}$

The reason you want to do that is because of the special relationship between spherical harmonics and angular momentum - which is why I gave you those links earlier.

Here is a list of spherical harmonics:
http://en.wikipedia.org/wiki/Table_of_spherical_harmonics
... notice that not all of them have a $\sin\theta$ in them, and that $2\cos\phi = e^{i\phi}+e^{-i\phi}$

Last edited: May 21, 2013
18. May 21, 2013

### Simon Bridge

Aside:
You can use: $$\int \psi_k\hat{H}\psi_n dx = E_n\delta_{nk}$$

Last edited: May 22, 2013
19. May 22, 2013

### ProPatto16

alright i think i got it... thanks for all your help.. and patience.. and tolerance

Last edited: May 22, 2013
20. May 22, 2013

### Simon Bridge

Ah excellent - so what did you find? (in case it helps someone else...)