3D quantum harmonic oscillator: linear combination of states

[JulienB]

Homework Statement

Hi everybody! In my quantum mechanics introductory course we were given an exercise about the 3D quantum harmonic oscillator. We are supposed to write the state $l=2$, $m=2$ with energy $E=\frac{7}{2}\hbar \omega$ as a linear combination of Cartesian states $\psi_{n_x, n_y, n_z}$.

Homework Equations

$Y_{nlm} (r,\theta,\varphi)=R_{nl}(r) Y_l^m(\theta,\varphi)$

The Attempt at a Solution

This comes at the end of a series of exercises about the QHO, so I already know that for this energy I must have $n=2$ and $n_r=0$ since $n=2n_r+l$ and $E_n = (2n_r+l+\frac{3}{2})\hbar\omega$. I've also seen $k$ being used instead of $n_r$. Just to make sure what I mean, this is the degree of the polynomial in the radial equation for which any coefficient $a_{n_r+1}=0$ (it terminates the series). Please note that $n=0$ is my ground state energy (and not $n=1$ as for the hydrogen atom for example).

I've worked hard on the derivation of the radial equation, it was very long and tedious so I spare you that part. I found the following:

$R_{n l} (r) = \sum_{k=0}^{\infty} a_k \left( \frac{m\omega}{\hbar} r \right)^{l+2k} e^{-\frac{m\omega}{2\hbar} r^2}$

I believe this is correct (according to this source: http://quantummechanics.ucsd.edu/ph130a/130_notes/node244.html), and after normalization (not asked, but well) my whole wave function in spherical coordinates looks like:

$\psi_{220} (r,\theta) = \sqrt{\frac{16}{15}} \frac{1}{\sqrt{\pi}} \left( \frac{m\omega}{\hbar} \right)^{7/4} r^2 e^{-\frac{m\omega}{2\hbar} r^2} Y_2^0 (\theta)$

where $Y_2^0(\theta)=\sqrt{\frac{5}{16\pi}} (3 \cos^2 \theta - 1)$.

Now the part $r^2 (3 \cos^2 \theta - 1) = 2z^2-x^2-y^2$ and I rearranged the expressions until I got terms looking like

$\left(\frac{m\omega}{\pi \hbar} \right)^{1/4} \frac{1}{\sqrt{2^2 \cdot 2!}} H_2 \left( \sqrt{\frac{m\omega}{\hbar}} x \right) e^{-\frac{m\omega}{2\hbar} r^2}$

since that is the wave function for each 1D oscillator (of course you would have to replace $x$ with $y$ and $z$ for the other states). And by the way I chose the Hermite polynomial $H_2$ because my cartesian coordinates are squared and that's the only way I could fit them in the cartesian wave function.

So at the end I get a linear combination of cartesian states

$\psi(x,y,z)= \sqrt{\frac{m\omega}{\hbar}} \sqrt{\frac{2}{3}} \pi^{-1/4} \left[ \psi_2(z) - \frac{1}{2} \left( \psi_2(x)+\psi_2(y) \right) \right]$

I don't expect anyone to correct this, my problem is just that I don't understand at all the point of doing that. I don't understand why we would want a linear combination of states and not a product. Maybe I really misunderstood the question and we were really suppose to find a linear combination of 3D states (and not 1D...)? If so, I wouldn't know what to do with my $2z^2-x^2-y^2$...

And one last question: why am I not just retrieving one of the 6 possible ground states associated with energy $\frac{7}{2}\hbar \omega$? I guess that's similar to my previous question: I was expecting a product of states when transforming from spherical to cartesian coordinates, not a linear combination.

Thank you very much in advance for your answers and suggestions.Julien.

 

[TSny]

Hello.

The problem states that $m = 2$, but the spherical harmonic $Y^0_2$ implies $m = 0$.

As you note, $r^2Y^m_l$ yields a polynomial in $x, y, z$. But why do you say that this implies that the wavefunction $\psi_{n_x, n_y, n_z}(x,y,z)$ should be a sum of $\psi_{n_x}(x)$, $\psi_{n_y}(y)$, $\psi_{n_z}(z)$ rather than a linear combination of products of $\psi_{n_x}(x)$, $\psi_{n_y}(y)$, and $\psi_{n_z}(z)$?

 

[vela]

How did you manage to break up the exponential term since $e^{-\alpha r^2} = e^{-\alpha(x^2+y^2+z^2)} = e^{-\alpha x^2}e^{-\alpha y^2}e^{-\alpha z^2}$?

 

[TSny]

Maybe I really misunderstood the question and we were really suppose to find a linear combination of 3D states (and not 1D...)?

Yes, you want linear combinations of 3D states $\psi_{n_x, n_y, n_z}(x, y, z) = \psi_{n_x}(x) \psi_{n_y}(y) \psi_{n_z}(z)$

If so, I wouldn't know what to do with my $2z^2-x^2-y^2$...

You will get a different polynomial if you use $Y^2_2$ rather than $Y^0_2$. But, you should be able to express the polynomial as combinations of Hermite polynomials. This should help in finding the coefficients in the expansion $\psi(x, y, z) = \sum c_{n_x, n_y, n_z} \psi_{n_x, n_y, n_z}(x, y, z)$

 

[JulienB]

@TSny Oops sorry I made a mistake. $m=0$ in the problem. My bad.

@vela Good point, I didn't realize that...

Okay I guess I go back to the drawing board, I will post my progress as soon as I have something.

Thank you very much for your help.

Julien.

 

[JulienB]

Okay here is my new attempt:

$\psi_{220}(r,\theta)=\sqrt{\frac{16}{15}} \frac{1}{\sqrt{\pi}} \left( \frac{\hbar}{m\omega} \right)^{1/4} \left( \frac{m\omega}{\hbar} \right)^2 r^2 e^{-\frac{m\omega}{2\hbar} r^2} \sqrt{\frac{5}{16\pi}} (3 \cos^2\theta-1)$
$=\frac{1}{\sqrt{3}\pi} \left( \frac{m\omega}{\hbar} \right)^{7/4} r^2 (3\cos^2\theta - 1) e^{-\frac{m\omega}{2\hbar} r^2}$
$= \frac{1}{\sqrt{3} \pi} \left( \frac{m\omega}{\hbar} \right)^{7/4} (2z^2-x^2-y^2) e^{-\frac{m\omega}{2\hbar} (x^2+y^2+z^2)}$
$=\frac{1}{\sqrt{3} \pi}\left( \frac{m\omega}{\hbar} \right)^{7/4} \left[\frac{1}{2} H_2 \left(\sqrt{\frac{m\omega}{\hbar} z}\right) + 1 - \frac{1}{4} H_2\left(\sqrt{\frac{m\omega}{\hbar} x}\right) - \frac{1}{2} - \frac{1}{4} H_2\left(\sqrt{\frac{m\omega}{\hbar} y}\right) - \frac{1}{2} \right] e^{-\frac{m\omega}{2\hbar} (x^2+y^2+z^2)}$
$=\left( \frac{m\omega}{\pi \hbar} \right)^{3/4} \left( \frac{1}{\sqrt{2^2 \cdot 2!}} \right)^3 e^{-\frac{m\omega}{2\hbar} (x^2+y^2+z^2)} \left( \frac{m\omega}{\pi \hbar} \right) 8 \sqrt{\frac{2}{3}} \left( \frac{1}{\pi} \right)^{1/4} \left[ H_2\left(\sqrt{\frac{m\omega}{\hbar} z}\right) - \frac{1}{2} H_2\left(\sqrt{\frac{m\omega}{\hbar} x}\right) - \frac{1}{2} H_2\left(\sqrt{\frac{m\omega}{\hbar} y}\right) \right]$
$= \frac{8}{\pi^{1/4}} \sqrt{\frac{2}{3}} \frac{m\omega}{\pi \hbar} \left[ \psi_{002}(z) - \frac{1}{2} \left( \psi_{200}(x) + \psi_{020}(y) \right) \right]$

I wrote all my steps this time so that the train of thought is easier to follow. I took your remarks in consideration and tried to bring the spherical wave function in a linear combination of cartesian wave functions. Not quite sure I labeled the wave functions correctly, and I still don't quite get what this whole thing means.

Thank you in advance.Julien.

 

[TSny]

OK, that's starting to look good. But note that , for example, $\psi_{002}(z)$ is actually a function of $x$, $y$, and $z$:
$\psi_{002} = \psi_0(x) \psi_0(y)\psi_2(z)$,
where each of the $\psi$'s in the expression on the right is a 1D Harmonic oscillator function.

Also, I don't think your final expression is normalized (if that matters).

As far as what the whole thing means, you can think of the 3D Cartesian functions as forming a basis. You have expanded the given wavefunction (in spherical coordinates) in terms of the Cartesian basis.

 

[JulienB]

@TSny Ah yes, I oversaw the x,y,z dependence indeed. Your explanation makes sense about the basis. I think I can close this topic now, thank you very much for your input.

Julien.