Exploring Quantum Physics - Final Exam Part 2 Question 4

Question:

What is the Gaussian variational estimate of the ground state energy for the following potential:

$\hat{H}\Psi\left(\vec{r}\right) = -\frac {\hbar^2} {2 \mu} \nabla^2 \Psi\left(\vec{r}\right) + kr \Psi\left(\vec{r}\right) = E \Psi\left(\vec{r}\right)$

Solution:

I see this as the climax of the whole exam. The original problem statement has a lot of other hints, but I still get this wrong due to some inaccuracy.

First, we substitute the standard Gaussian to the kinetic energy term. I've got inaccuracy here so the whole problem is wrong. So let's do that here again.


$\begin{eqnarray*} & & -\frac {\hbar^2} {2 \mu} \nabla^2 \Psi\left(\vec{r}\right) \\ &=& -\frac {\hbar^2} {2 \mu} \nabla^2 \frac{1}{\pi^{3/4} d^{3/2}} \exp\left(-\frac{r^2}{2 d^2}\right) \\ &=& -\frac {\hbar^2} {2 \mu} \nabla^2 \frac{1}{\pi^{3/4} d^{3/2}} \exp\left(-\frac{r_1^2 + r_2^2 + r_3^2}{2 d^2}\right) \\ &=& -\frac {\hbar^2} {2 \mu \pi^{3/4}d^{3/2}} \nabla^2 \exp\left(-\frac{r_1^2 + r_2^2 + r_3^2}{2 d^2}\right) \\ &=& -\frac {\hbar^2} {2 \mu \pi^{3/4}d^{3/2}} \nabla \cdot \left(\frac{-r_1}{d^2}, \frac{-r_2}{d^2}, \frac{-r_3}{d^2}\right) \exp\left(-\frac{r_1^2 + r_2^2 + r_3^2}{2 d^2}\right) \\ &=& -\frac {\hbar^2} {2 \mu \pi^{3/4}d^{3/2}} \left(\frac{r_1^2}{d^4} - \frac{1}{d^2} + \frac{r_2^2}{d^4} - \frac{1}{d^2} + \frac{r_3^2}{d^4} - \frac{1}{d^2}\right) \exp\left(-\frac{r_1^2 + r_2^2 + r_3^2}{2 d^2}\right) \\ &=& -\frac {\hbar^2} {2 \mu \pi^{3/4}d^{3/2}} \left(\frac{r_1^2}{d^4} + \frac{r_2^2}{d^4} + \frac{r_3^2}{d^4} - \frac{3}{d^2}\right) \exp\left(-\frac{r_1^2 + r_2^2 + r_3^2}{2 d^2}\right) \\ \end{eqnarray*}$

To find the energy, we do an integration

$\begin{eqnarray*} & & \langle T \rangle \\ &=& \int\limits_{-\infty}^{\infty}{\frac{1}{\pi^{3/4} d^{3/2}} \exp\left(-\frac{r^2}{2 d^2}\right)\left(-\frac {\hbar^2} {2 \mu \pi^{3/4}d^{3/2}} \left(\frac{r_1^2}{d^4} + \frac{r_2^2}{d^4} + \frac{r_3^2}{d^4} - \frac{3}{d^2}\right) \exp\left(-\frac{r_1^2 + r_2^2 + r_3^2}{2 d^2}\right)\right)dV} \\ &=& \int\limits_{-\infty}^{\infty}{\frac{1}{\pi^{3/4} d^{3/2}} \exp\left(-\frac{r_1^2 + r_2^2 + r_3^2}{2 d^2}\right)\left(-\frac {\hbar^2} {2 \mu \pi^{3/4}d^{3/2}} \left(\frac{r_1^2}{d^4} + \frac{r_2^2}{d^4} + \frac{r_3^2}{d^4} - \frac{3}{d^2}\right) \exp\left(-\frac{r_1^2 + r_2^2 + r_3^2}{2 d^2}\right)\right)dV} \\ &=& -\frac{1}{\pi^{3/4} d^{3/2}}\frac {\hbar^2} {2 \mu \pi^{3/4}d^{3/2}} \int\limits_{-\infty}^{\infty}{ \exp\left(-\frac{r_1^2 + r_2^2 + r_3^2}{2 d^2}\right)\left( \left(\frac{r_1^2}{d^4} + \frac{r_2^2}{d^4} + \frac{r_3^2}{d^4} - \frac{3}{d^2}\right) \exp\left(-\frac{r_1^2 + r_2^2 + r_3^2}{2 d^2}\right)\right)dV} \\ &=& -\frac {\hbar^2} {2 \mu \pi^{3/2}d^3} \int\limits_{-\infty}^{\infty}{ \left( \left(\frac{r_1^2}{d^4} + \frac{r_2^2}{d^4} + \frac{r_3^2}{d^4} - \frac{3}{d^2}\right) \exp\left(-\frac{r_1^2 + r_2^2 + r_3^2}{d^2}\right)\right)dV} \\ &=& -\frac {\hbar^2} {2 \mu \pi^{3/2}d^5} \int\limits_{-\infty}^{\infty}{ \left( \left(\frac{r_1^2}{d^2} + \frac{r_2^2}{d^2} + \frac{r_3^2}{d^2} - 3\right) \exp\left(-\frac{r_1^2 + r_2^2 + r_3^2}{d^2}\right)\right)dV} \\ &=& -\frac {\hbar^2} {2 \mu \pi^{3/2}d^5} \int\limits_{-\infty}^{\infty}{ \left( \left(\frac{r_1^2}{d^2} + \frac{r_2^2}{d^2} + \frac{r_3^2}{d^2} \right) \exp\left(-\frac{r_1^2 + r_2^2 + r_3^2}{d^2}\right)\right)dV} + \frac {\hbar^2} {2 \mu \pi^{3/2}d^5} \int\limits_{-\infty}^{\infty}{ \left( 3 \exp\left(-\frac{r_1^2 + r_2^2 + r_3^2}{d^2}\right)\right)dV} \\ &=& -\frac {\hbar^2} {2 \mu \pi^{3/2}d^7} \int\limits_{-\infty}^{\infty}{ \left( \left(r_1^2 + r_2^2 + r_3^2 \right) \exp\left(-\frac{r_1^2 + r_2^2 + r_3^2}{d^2}\right)\right)dV} + \frac {\hbar^2} {2 \mu \pi^{3/2}d^5} \int\limits_{-\infty}^{\infty}{ \left( 3 \exp\left(-\frac{r_1^2 + r_2^2 + r_3^2}{d^2}\right)\right)dV} \end{eqnarray*}$

In this form, we can see this is a Gaussian variance integral. We know that $\int\limits_{-\infty}^{\infty}{\frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{x^2}{2\sigma^2}}dx} = 1$ and $\int\limits_{-\infty}^{\infty}{x^2\frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{x^2}{2\sigma^2}}dx} = \sigma^2$, so we can compute the basic integrals as follow:

$\begin{eqnarray*} & & \int\limits_{-\infty}^{\infty}{\exp\left(-\frac{x^2}{d^2}\right)dx} \\ &=& \int\limits_{-\infty}^{\infty}{\exp\left(-\frac{x^2}{2\left(\frac{1}{2}\right)d^2}\right)dx} \\ &=& \int\limits_{-\infty}^{\infty}{\exp\left(-\frac{x^2}{2\left(\frac{1}{\sqrt{2}}\right)^2d^2}\right)dx} \\ &=& \int\limits_{-\infty}^{\infty}{\exp\left(-\frac{x^2}{2\left(\frac{d}{\sqrt{2}}\right)^2}\right)dx} \\ &=& \sqrt{2\pi\left(\frac{d}{\sqrt{2}}\right)^2}\frac{1}{\sqrt{2\pi\left(\frac{d}{\sqrt{2}}\right)^2}}\int\limits_{-\infty}^{\infty}{\exp\left(-\frac{x^2}{2\left(\frac{d}{\sqrt{2}}\right)^2}\right)dx} \\ &=& \sqrt{2\pi\left(\frac{d}{\sqrt{2}}\right)^2} \\ &=& \sqrt{\pi}d \end{eqnarray*}$

$\begin{eqnarray*} & & \int\limits_{-\infty}^{\infty}{x^2\exp\left(-\frac{x^2}{d^2}\right)dx} \\ &=& \int\limits_{-\infty}^{\infty}{x^2\exp\left(-\frac{x^2}{2\left(\frac{1}{2}\right)d^2}\right)dx} \\ &=& \int\limits_{-\infty}^{\infty}{x^2\exp\left(-\frac{x^2}{2\left(\frac{1}{\sqrt{2}}\right)^2d^2}\right)dx} \\ &=& \int\limits_{-\infty}^{\infty}{x^2\exp\left(-\frac{x^2}{2\left(\frac{d}{\sqrt{2}}\right)^2}\right)dx} \\ &=& \sqrt{2\pi\left(\frac{d}{\sqrt{2}}\right)^2}\frac{1}{\sqrt{2\pi\left(\frac{d}{\sqrt{2}}\right)^2}}\int\limits_{-\infty}^{\infty}{x^2\exp\left(-\frac{x^2}{2\left(\frac{d}{\sqrt{2}}\right)^2}\right)dx} \\ &=& \sqrt{2\pi\left(\frac{d}{\sqrt{2}}\right)^2} \left(\frac{d}{\sqrt{2}}\right)^2 \\ &=& \frac{\sqrt{\pi}d^3}{2} \\ \end{eqnarray*}$

With these basic integral - computing $\langle T \rangle$ becomes a substitution exercise. The above are triple integrals and therefore we turn it into iterated integral and compute them.

$\begin{eqnarray*} & & \langle T \rangle \\ &=& -\frac {\hbar^2} {2 \mu \pi^{3/2}d^7} \int\limits_{-\infty}^{\infty}{ \left( \left(r_1^2 + r_2^2 + r_3^2 \right) \exp\left(-\frac{r_1^2 + r_2^2 + r_3^2}{d^2}\right)\right)dV} + \frac {\hbar^2} {2 \mu \pi^{3/2}d^5} \int\limits_{-\infty}^{\infty}{ \left( 3 \exp\left(-\frac{r_1^2 + r_2^2 + r_3^2}{d^2}\right)\right)dV} \\ &=& -\frac {\hbar^2} {2 \mu \pi^{3/2}d^7} \left(3\frac{\sqrt{\pi}d^3}{2}\sqrt{\pi}d\sqrt{\pi}d\right) + \frac {\hbar^2} {2 \mu \pi^{3/2}d^5} \left(3\sqrt{\pi}d\sqrt{\pi}d\sqrt{\pi}d\right) \\ &=& -\frac {3\hbar^2} {4 \mu d^2} + \frac {3\hbar^2} {2 \mu d^2} \\ &=& \frac {3\hbar^2} {4 \mu d^2} \end{eqnarray*}$

Next we move on an compute $\langle V \rangle$, another integral

$\begin{eqnarray*} & & \langle V \rangle \\ &=& \int\limits_{-\infty}^{\infty}{\Psi^*kr\Psi dV} \\ &=& \int\limits_{-\infty}^{\infty}{\frac{1}{\pi^{3/4} d^{3/2}} \exp\left(-\frac{r^2}{2 d^2}\right)kr\frac{1}{\pi^{3/4} d^{3/2}} \exp\left(-\frac{r^2}{2 d^2}\right)dV} \\ &=& \frac{1}{\pi^{3/2} d^3} \int\limits_{-\infty}^{\infty}{\exp\left(-\frac{r^2}{2 d^2}\right)kr\exp\left(-\frac{r^2}{2 d^2}\right)dV} \\ &=& \frac{1}{\pi^{3/2} d^3} \int\limits_{-\infty}^{\infty}{kr\exp\left(-\frac{r^2}{d^2}\right)dV} \\ \end{eqnarray*}$

Despite the deceptive simplicity in the formula, it is a triple integral with $r$ being the length of the vector. Now it make sense to convert this to spherical coordinates.

$\begin{eqnarray*} &=& \frac{1}{\pi^{3/2} d^3} 4\pi \int\limits_{-\infty}^{\infty}{kr^3\exp\left(-\frac{r^2}{d^2}\right)dr} \\ \end{eqnarray*}$

Let we let $s = \frac{r^2}{d^2}$, $ds = \frac{2r}{d^2}dr$, so we further simplify this to

$\begin{eqnarray*} &=& \frac{1}{\pi^{3/2} d^3} 4\pi \int\limits_{-\infty}^{\infty}{kr^3\exp\left(-s\right)dr} \\ &=& \frac{1}{\pi^{3/2} d^3} 4\pi \int\limits_{-\infty}^{\infty}{kr^3\frac{d^2}{2r}\exp\left(-s\right)ds} \\ &=& \frac{1}{\pi^{3/2} d^3} 4\pi \int\limits_{-\infty}^{\infty}{k\frac{r^2d^2}{2}\exp\left(-s\right)ds} \\ &=& \frac{1}{\pi^{3/2} d^3} 4\pi \int\limits_{-\infty}^{\infty}{k\frac{r^2d^4}{2d^2}\exp\left(-s\right)ds} \\ &=& \frac{1}{\pi^{3/2} d^3} 4\pi \frac{kd^4}{2} \int\limits_{-\infty}^{\infty}{s\exp\left(-s\right)ds} \\ &=& \frac{2 kd}{\pi^{1/2}} \int\limits_{-\infty}^{\infty}{s\exp\left(-s\right)ds} \\ &=& \frac{2 kd}{\pi^{1/2}} \Gamma(2) \\ &=& \frac{2 kd}{\pi^{1/2}} \\ \end{eqnarray*}$

So the total energy is $\langle T \rangle + \langle V \rangle = \frac {3\hbar^2} {4 \mu d^2} + \frac{2 kd}{\pi^{1/2}}$. To estimate the ground state we would want to minimize it.

$\begin{eqnarray*} 0 &=& \frac{d}{dd} \left(\langle T \rangle + \langle V \rangle \right) \\ &=& \frac{d}{dd} \left(\frac {3\hbar^2} {4 \mu d^2} + \frac{2 kd}{\pi^{1/2}}\right) \\ &=& \frac {3\left(-2\right)\hbar^2} {4 \mu d^3} + \frac{2 k}{\pi^{1/2}} \\ \frac {3\left(2\right)\hbar^2} {4 \mu d^3} &=& \frac{2 k}{\pi^{1/2}} \\ d^3 &=& \frac {3\left(2\right)\hbar^2\pi^{1/2}} {4 \left(2k\right )\mu}\\ &=& \frac {3\hbar^2\pi^{1/2}} {4k\mu} \\ \end{eqnarray*}$

Finally, we substitute this back into the energy expression

$\begin{eqnarray*} \langle T \rangle + \langle V \rangle &=& \frac {3\hbar^2} {4 \mu d^2} + \frac{2 kd}{\pi^{1/2}} \\ &=& \frac {3\hbar^2d} {4 \mu d^3} + \frac{2 kd}{\pi^{1/2}} \\ &=& \frac {3\hbar^2d} {4 \mu \left(\frac {3\hbar^2\pi^{1/2}} {4k\mu}\right)} + \frac{2 kd}{\pi^{1/2}} \\ &=& \frac {kd} {\pi^{1/2}} + \frac{2 kd}{\pi^{1/2}} \\ &=& \frac{3 kd}{\pi^{1/2}} \\ &=& \frac{3 k}{\pi^{1/2}} \left(\frac {3\hbar^2\pi^{1/2}} {4k\mu} \right)^{1/3} \\ &=& \left(\frac {81k^3\hbar^2\pi^{1/2}} {4k\mu\pi^{3/2}}\right)^{1/3} \\ &=& \left(\frac {81k^2\hbar^2} {4\pi\mu} \right)^{1/3} \\ &=& \left(\frac {81} {4\pi} \right)^{1/3}\left(\frac {\hbar^2k^2}{\mu}\right)^{1/3} \\ \end{eqnarray*}$

Phew - that's it! What a climax.