Calculating Expected Values for 3D Harmonic Oscillator Wave Function

[Incand]

Homework Statement

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 $1/r$.

Homework Equations

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

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 $<1/r> = \frac{2}{r_0\sqrt{\pi}} \int_0^\infty \frac{e^{-r^2/r_0^2}}{r} dr$.

 

[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.

 

[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.

 

[PeroK]

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

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

 

[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!

 

[PeroK]

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!

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 $f$

 

[Incand]

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