# Dealing with addition of cosntant to wave equation? Spherical Harmonics

1. Dec 11, 2009

### Deathfrost

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

I am trying to calculate the angular momenta for

$\psi(x,y,z) = A(ar^2 + bz^2)$

A is given as a constant.

2. Relevant equations

3. The attempt at a solution

I know that $z=r\sqrt{4\pi/3} * Y_0^1$

What I have so far is:-
$\psi(x,y,z) = r^2Aa + r^2b\sqrt{4\pi/3} * Y_0^1)$

and thus one of the possible values of momentum is hbar since l is one. but how do I deal with the constant part added to the equation, the A*a*r^2 when I try to calculate the probability?

A problem like $\psi(x,y,z) = A(x+y+2z)*exp^{-ar}$ I can deal with, since everything is multiplied across by the same constant, that just differ by some known factor. When I try to find the probability, and divide each angular part, the constants cancel out.

Thanks for any help

2. Dec 11, 2009

### jdwood983

Are you sure?? (Maybe it's me, but also, shouldn't it be $Y_1^0=Y_\ell^m$? Never seen it done $Y_m^\ell$ )

My suggestion would be to use the fact that $r^2=x^2+y^2+z^2$ and use the relations between Cartesian coordinates and spherical harmonics. That might make it a little more clear what you need to do.

3. Dec 12, 2009

### Deathfrost

Re: dealing with addition of constant to wave equation? Spherical Harmonics

Ah I guess I just reversed the notations for l and m, anyway I was trying to say l=1, and m=0.

Ok thanks for the suggestion, I see what I can do to proceed from there, but as a follow up question,

If the wave equation is composed of a radial and a angular part,

$$\psi_n_l_m(r,\theta ,\phi) = R_n_l(r)Y_l_m(\theta,\phi)$$

then cant I just ignore the r's when Im looking for angular momentum?

Any basic information provided is appreciated :) I have no Physics, or Math background beyond basic calculus, and just crazily signed up for a quantum class.

Thanks.

4. Dec 12, 2009

### gabbagabbahey

Re: dealing with addition of constant to wave equation? Spherical Harmonics

That's a bit of an odd looking wave function....I assume that this is only the wavefunction in some finite region, and the wavefunction is zero elsewhere?....If so, you can get rid of $A$ by normalizing the wave function.

Why is there no $A$ in the second term? And didn't you have $z$ squared in your equation...$z^2=r^2\cos^2\theta$, so you should have $Y_2{}^0$ and $Y_0{}^0$ in there instead of $Y_1{}^0$

Yes, there is no need to write $r$ in terms of Cartesian coordinates....However, $\psi_n_l_m(r,\theta ,\phi) = R_n_l(r)Y_l{}^m(\theta,\phi) [/tex] are the spherical harmonics...your given wave function need not be in one of these harmonics (In fact, it isn't!) but instead is more generally going to be in a superposition of these harmonics: $$\psi(x)=A(ar^2+bz^2)=\sum_{n,l,m}C_{n,l,m}R_n_l(r)Y_l{}^m(\theta,\phi)$$ What you need to do is find all the nonzero [itex]C_{n,l,m}$

5. Dec 13, 2009

### Deathfrost

Re: dealing with addition of constant to wave equation? Spherical Harmonics

Tons of clumsy mistakes on my part.

For z^2 :-

Why not just $z^2=r^2 * 4\pi/3 * Y^0_1 * Y^0_1?$

Though Ive never seen any wave-function with 2 angular parts multiplied together, so I don't know how to interpret that.

6. Dec 13, 2009

### gabbagabbahey

Re: dealing with addition of constant to wave equation? Spherical Harmonics

Which is exactly why doing that is useless to you....write $z^2$ in terms of $Y_0{}^0$ and $Y_2{}^0$ instead...this gives you a linear sum (superposition) of harmonics instead of a product of them.

7. Dec 13, 2009

### Cryxic

Everyone is making this too complicated. The allowed angular momentum is just 0. There is no phi dependence in that wavefunction! You just have r and z. You're always going to get 0 with 100% chance.

8. Dec 13, 2009

### jdwood983

You seem to be forgetting that $z$ is a Cartesian coordinate and $r$ is a spherical coordinate. Since $z$ can be expressed in terms of $r$ and $\theta$, it would appear that there is an angular dependence and thus $\mathbf{L}\neq0$.

Gabba is right, I don't know why I first thought to convert $r^2=x^2+y^2+z^2$ because you'd get $r^2$ out of the conversion from Cartesian coordinates to spherical harmonics anyway. You should have something along the lines of

$$z^2\propto r^2\left(AY_2^0(\theta,\phi)+BY_0^0(\theta,\phi)\right)$$

which given the normal definition of $z$ in terms of spherical coordinates should reduce to $z^2=r^2\cos^2\theta$

9. Dec 13, 2009

### Cryxic

No the question is about the z part of the angular momentum (implicitly), otherwise you'd be right. But in this case, no phi dependence, and so measured L(z) will always be 0.

10. Dec 13, 2009

### Cryxic

But even if it wants the azimuthal quantum numbers (little L).....that's easy too: just 0 and 2 (given it's a 2nd degree equation)....

11. Dec 13, 2009

### jdwood983

Right, this is what he's trying to find.

12. Dec 13, 2009

### Cryxic

Shankar's Principle of Quantum Mechanics has a similar question, and that question asks for the possible L(z) angular momenta and their probabilities. That's what I assumed this person was trying to find, although that's not stated anywhere. Either way, it's all simple. The r^2 part will just be a constant (4pi in this case) and the z^2 part will be the Y(20) spherical harmonic. That should be enough to find the allowed values and the probabilities.

13. Dec 13, 2009

### Deathfrost

Yeah Ive worked through the problem in shankar, the last equation in my original post is in fact the Shankar problem. I had no trouble with that one, as I was easily able to put x , y and z in Spherical coordinates. Then calculating the probabilities was no problem, and finding the possible values of the angular momentum lz was also trivial.

My problem with the current problem is just the algebra, as my math is really weak right now. And we just covered Spherical harmonics last week.

I appreciate all the help though.

Ive been trying to find a source that covers spherical harmonics somewhere online, somethign that would have a bunch of generic functions and examples of how to convert them would be helpful. I would never have figured out that
$$z^2\propto r^2\left(AY_2^0(\theta,\phi)+BY_0^0(\theta,\phi)\right )$$

14. Dec 13, 2009

### jdwood983

http://en.wikipedia.org/wiki/Table_of_spherical_harmonics" [Broken] has a table of the first 11 spherical harmonics ($\ell=0\rightarrow10$). I would suggest, since I haven't found a good source for what you are looking for, that you make your own. Take a rainy afternoon (or anytime you are bored) and start cranking out simple functions (linear relations is always a safe start) and then build it up to more complex functions.

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