Exploring Quantum Physics - Good bye

With the last final exam solution posted - this post mark the end to this series of Quantum physics posts. 

Looking backward in this couple month, it seems so amazing that I could accomplish this many problems - with around 10 problems every week for 8 weeks, that amounts to about 80 problems.

Let me end this post with my special thanks to the professors who gave us the course, this journey is simply amazing.

Exploring Quantum Physics - Final Exam Part 2 Question 6

Question:

What is the required frequency stir an atom is a trap of 20 micron radius to put the an atom with in $ L_3 = 1 \times \hbar $?

Solution:

Disclaimer - I have got this problem wrong - and there is no official solution yet, so this is at best a sharing of my idea.

The Bohr model gives $ mvr = n\hbar $. The problem requires $ n = 1 $, so we got almost everything in this equation, except $ v $ .

If a stirrer is operating at a certain frequency $ f $, then a particle being stirred will have travelled a circle in one period of time, in other words, $ v = \frac{d}{T} = \frac{2\pi r}{T} = 2\pi r f $.

Substitute this back to the equation we get

$ m (2 \pi r f) r = \hbar $. So $ f = \frac{\hbar}{2 \pi r^2 m} = \frac{h}{ r^2 m} = 43 Hz $

I didn't know what I pick $ 10^6 Hz $ at that point - perhaps I was just too nervous. I am not sure if the current answer is correct either. But that's the idea - the problem isn't too hard. It is just that the climax was just too high :p

Exploring Quantum Physics - Final Exam Part 2 Question 5

Question:

What is the exact energy for the Hamiltonian of the previous problem.

Solution:

So we get started with the original equation again

$ \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) $

We were given a wonderful substitution$\Psi(r)= \frac{\psi(r)}{r} $ so that $\nabla^2 \Psi(r) = \frac{1}{r} \frac{\partial^2 \psi(r)}{\partial r^2} $. Using it, we get

$ -\frac {\hbar^2} {2 \mu}\frac{1}{r} \frac{\partial^2 \psi(r)}{\partial r^2}+ kr\frac{\psi(r)}{r}= E\frac{\psi(r)}{r}$

Multiply by $ r $ on both side, we get the standard form

$ -\frac {\hbar^2} {2 \mu} \frac{\partial^2 \psi(r)}{\partial r^2}+ kr\psi(r)= E\psi(r) $

This is simply the bouncing ball potential if we set $ \mu = M $ and $ k = Mg $, therefore the ground state energy is simply

$ \begin{eqnarray*} E_0 &=& - \alpha \rho_0 \\ &=& \left(\frac{\hbar^2 M g^2}{2}\right)^{1/3} \rho_0 \\ &=& \left(\frac{\hbar^2 M \left(\frac{k}{M}\right)^2}{2}\right)^{1/3} \rho_0 \\ &=& \left(\frac{\hbar^2 \frac{k^2}{M}}{2}\right)^{1/3} \rho_0 \\ &=& \left(\frac{\hbar^2 k^2}{2M}\right)^{1/3} \rho_0 \\ &=& \left(\frac{\hbar^2 k^2}{2\mu}\right)^{1/3} \rho_0 \\ &=& \left(\frac{\hbar^2 k^2}{\mu}\right)^{1/3}\left(\frac{1}{2}\right)^{1/3} \rho_0 \\ \end{eqnarray*} $

So that's the answer - I have got this wrong just because I am not using the numerically accurate enough $ \rho_0 $ - Ooops - too bad.

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.

Exploring Quantum Physics - Final Exam Part 2 Question 3

Question:

Despite the very long description - question 3 by itself is trivial. In some sense, it is just a hint for the upcoming questions. To a bare minimum, the question listed a bunch of quantity and asked which one has to unit of energy. So all we have to do is to match dimensions.

Solution:

Energy unit
= Force times Distance
= Mass times Acceleration times Distance
= M(LT^-2)(L)
= ML²T^-2

kr = Energy
Therefore k has a unit of Force MLT^-2

Planck's constant has unit of Joule Second = Energy Second = ML²T^-1

Finally $ \mu $ has unit of mass.

So $ \frac{\hbar^2 k^2}{\mu} $ has unit $ \frac{(ml^2t^{-1})^2(mlt^{-2})^2}{m} = m^3l^6t^{-6} $

Therefore we see $ \left(\frac{\hbar^2 k^2}{\mu}\right)^{1/3} $ has the unit of energy.

Exploring Quantum Physics - Final Exam Part 2 Question 2

Question:

What is the value of $ \langle j | x | k \rangle $ where $ | j \rangle $ and $ | k \rangle $ are the j and k eigen function of the Quantum Harmonic Oscillator?

Solution:

As a disclaimer, I didn't quite solve the problem completely in the exam. But I get the correct answer.

The key annoying piece is the $ x $ inside the sandwich. We just break it down into ladder operators.

$ \hat{x} = \sqrt{\frac{2\hbar}{2m\omega}}(\hat{a}^{\dagger} + \hat{a}) $.

Substitute this back into $ \langle j | x | k \rangle $, we have got


$ \begin{eqnarray*} & & \langle j | x | k \rangle \\ &=& \langle j | \sqrt{\frac{2\hbar}{2m\omega}}(\hat{a}^{\dagger} + \hat{a}) | k \rangle \\ &=& \sqrt{\frac{2\hbar}{2m\omega}} \langle j | (\hat{a}^{\dagger} + \hat{a}) | k \rangle \\ &=& \sqrt{\frac{2\hbar}{2m\omega}} (\langle j | \hat{a}^{\dagger} | k \rangle + \langle j | \hat{a} | k \rangle)\\ &=& \sqrt{\frac{2\hbar}{2m\omega}} (\sqrt{k+1}\langle j | k + 1 \rangle + \sqrt{k} \langle j | k - 1 \rangle)\\ &=& \sqrt{\frac{2\hbar}{2m\omega}} (\sqrt{k+1}\delta_{j, k + 1} + \sqrt{k} \delta_{j, k - 1})\\ \end{eqnarray*} $

Exploring Quantum Physics - Final Exam Part 2 Question 1

Question:

Given:

$ s_3 | \uparrow \rangle = \frac{\hbar}{2} | \uparrow \rangle $ and $ s_3 | \downarrow \rangle = -\frac{\hbar}{2} | \downarrow \rangle $

Find

$ s_3 \frac{1}{\sqrt{2}}(| \uparrow \downarrow \rangle - | \downarrow \uparrow \rangle) $

Solution:

Disclaimer: My answer is wrong.

It seems to be a really simple question, just apply the definitions.

$ \begin{eqnarray*} & & s_3 \frac{1}{\sqrt{2}}(| \uparrow \downarrow \rangle - | \downarrow \uparrow \rangle) \\ &=& \frac{1}{\sqrt{2}} s_3 [| \uparrow \rangle - | \downarrow \rangle , | \downarrow \rangle - | \uparrow \rangle] \\ &=& \frac{1}{\sqrt{2}} [| \frac{\hbar}{2}\uparrow \rangle + \frac{\hbar}{2}| \downarrow \rangle , -\frac{\hbar}{2}| \downarrow \rangle - \frac{\hbar}{2}| \uparrow \rangle] \\ &=& \frac{\hbar}{2}\frac{1}{\sqrt{2}} [| \uparrow \rangle + | \downarrow \rangle , -| \downarrow \rangle - | \uparrow \rangle] \\ \end{eqnarray*} $

Now I am stuck, it appears the vector I get is not in the choices. For the exam, I guessed an answer, and got the wrong answer.