# 3d Harmonic oscillator

1. Feb 21, 2016

### Incand

1. The problem statement, all variables and given/known data
The wave function for the three dimensional oscillator can be written
$\Psi(\mathbf r) = Ce^{-\frac{1}{2}(r/r_0)^2}$
where $C$ and $r_0$ are constants and $r$ the distance from the origen.
Calculate
a) The most probably value for $r$
b) The expected value of $r$
c) The expected value of $1/r$.

2. Relevant equations
Expected value of function with a normed wave function
$<f(r)> = \int dr \Psi(r)^*f(r) \Psi$.

3. The attempt at a solution
a) It is my understanding that the most probably value is the maximum value of $\Psi^* \Psi$ or equivalently in this case the maximum of $\Psi$ which is at $r=0$.
The answer to the exercise however disagrees and says it's at $r_0$.
If the oscillator is centered at $r_0$ this makes physical sense with the most probably value of course being the centre but it doesn't seem to agree with maximizing the function.

b) Norming the wave function
$1 = C^2 \int_0^\infty e^{-r^2/r_0^2}dr = C^2 \int_0^\infty r_0 e^{-s^2}ds = C^2\frac{r_0 \sqrt{\pi}}{2}$ and hence that $C^2 = \frac{2}{r_0 \sqrt{\pi}}$.
The expected value of $r$ is then
$<r> = \frac{2}{r_0\sqrt{\pi}} \int_0^\infty re^{-r^2/r_0^2} dr = \frac{r_0}{\sqrt{\pi}}$.
The answer however says $\frac{2r_0}{\sqrt{\pi}}$.

c) For this one, I get a divergent integral $<1/r> = \frac{2}{r_0\sqrt{\pi}} \int_0^\infty \frac{e^{-r^2/r_0^2}}{r} dr$.

2. Feb 21, 2016

### PeroK

You're very modest to post QM under "introductory" homework. In any case, your problems stem from not thinking enough about integrating in spherical coordinates.

3. Feb 21, 2016

### Incand

Well it's a first course in QM for second year university students (at an European university) so I felt it being more introductory. The advanced forum I thought were mainly for graduate students.

Thanks! I forgot I have to integrate over the volume in this case.
Norming the function
$1 = C^2 \int_0^\infty \int_0^\pi \int_0^{2\pi} r^2\sin \theta e^{-r^2/r_0^2}d\phi d\theta dr = C^2 4\pi \frac{\sqrt{\pi}r_0^3}{4} = C^2 \pi \sqrt{\pi} r_0^3 \Longrightarrow C^2 = \frac{1}{\pi \sqrt{\pi }r_0^3}$.

The expected value is then
$<r> = \frac{1}{\pi \sqrt{\pi} r_0^3} 4\pi \int_0^\infty r^3e^{-r^2/r_0^2}dr = \frac{1}{\pi \sqrt{\pi} r_0^3} 4\pi \frac{r_0^4}{2} = \frac{2r_0}{\sqrt{\pi}}$ as expected!

And for $1/r$
$<1/r> = \frac{4}{\sqrt{\pi} r_0^3} \int_0^\infty re^{-r^2/r_0^2}dr = \frac{2}{\sqrt{\pi} r_0}$.

However as for the a) question I'm still not sure? that doesn't have anything with integrating to do.

4. Feb 21, 2016

### PeroK

Why not calculate the probability that the particle is found near a given distance from the origin: between $R$ and $R+ \epsilon$ say?

5. Feb 21, 2016

### Incand

After thinking about this for a while it seems my error here just as before is that I need the volume element.

As far as I understand it, the probability distribution is
$P(\mathbf r) = |\Psi(\mathbf r)|^2 = \Psi^*(\mathbf r) \Psi(\mathbf r).$
And the probability of finding the particle within a certain volume is
$\int_V P(\mathbf r) dV$.
In our case that is
$I = \frac{4}{\sqrt{\pi}r_0^3} \int_R^{R+\epsilon} r^2e^{-r^2/r_0^2}dr$
So we want to find the $R$ that maximizes $\int_R^{R+\epsilon} r^2e^{-r^2/r_0^2}dr$ when $\epsilon \to 0$.
However since this seem complicated we can instead maximize the integrand $r^2e^{-r^2/r_0^2}$ which has a maximum for $r=r_0$.
So I need to account for the scale factors when I'm not working in Cartesian coordinates. Thanks for helping!

6. Feb 21, 2016

### PeroK

Yes. The thing you're missing is that the limit of an integral such as the one you have is simply the function value at the point:

$\lim_{\epsilon \rightarrow 0} \int_{R}^{R+\epsilon} f(r) dr = f(R)$

So, maximising that integral is equivalent to maximising $f$

7. Feb 21, 2016

### Incand

Thanks for pointing that out. I had forgotten about the mean value theorem!