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.

Exploring Quantum Physics - Final Part 1 Question 10

Question:

Considering the linear oscillator chain system, which was discussed in Lecture 8, what is the minimum energy required to excite an acoustic phonon?

Solution:

Any arbitrarily small energy can excite a phonon. I recalled this in the lectures.

This conclude the first part of the final exam.

Exploring Quantum Physics - Final Part 1 Question 9

Question:

What is the probability that the particle in the ground state of a 1 dimensional delta potential get to an excited state when the potential suddenly double its strength?

Solution:

I attempted the exam before I watch week 8 lecture 16, and the approach to this problem is discussed there. I am really thrilled because I used exactly the same procedure as the lecture told us to without watching it first!

The key idea is that the wave function is unchanged when the potential suddenly change, but then it suddenly becomes a linear combination of eigen functions in the doubled potential. To find out if it is excited, we compute the probability that it is still in the ground state using the inner product.

The probability that it is still in the ground state is $ (\langle \psi_{0, \alpha} | \psi_{0, 2\alpha} \rangle )^2 $.

We are given that
$ \psi_{0,\alpha}(x) = \frac{\sqrt{m\alpha}}{\hbar}\exp\left(-\frac{m\alpha|x|}{\hbar^2}\right) $

So the required integral can be computed as:

$ \begin{eqnarray*} & & \int\limits_{-\infty}^{\infty}{\frac{\sqrt{m\alpha}}{\hbar}\exp\left(-\frac{m\alpha|x|}{\hbar^2}\right)\frac{\sqrt{2m\alpha}}{\hbar}\exp\left(-\frac{2m\alpha|x|} {\hbar^2}\right)dx} \\ &=& \frac{\sqrt{2}m\alpha}{\hbar^2}\int\limits_{-\infty}^{\infty}{\exp\left(-\frac{3m\alpha|x|}{\hbar^2}\right)dx} \\ &=& \frac{\sqrt{2}m\alpha}{\hbar^2}\left(\int\limits_{-\infty}^{0}{\exp\left(-\frac{3m\alpha|x|}{\hbar^2}\right)dx} + \int\limits_{0}^{\infty}{\exp\left(-\frac{3m\alpha|x|}{\hbar^2}\right)dx}\right) \\ &=& \frac{\sqrt{2}m\alpha}{\hbar^2}\left(\int\limits_{-\infty}^{0}{\exp\left(\frac{3m\alpha x}{\hbar^2}\right)dx} + \int\limits_{0}^{\infty}{\exp\left(-\frac{3m\alpha x}{\hbar^2}\right)dx}\right) \\ &=& \frac{\sqrt{2}m\alpha}{\hbar^2}\left(\frac{\hbar^2}{3m\alpha}\exp\left(\frac{3m\alpha x}{\hbar^2}\right)|_{-\infty}^{0} + \frac{-\hbar^2}{3m\alpha}\exp\left(-\frac{3m\alpha x}{\hbar^2}\right)|_{0}^{\infty}\right) \\ &=& \frac{\sqrt{2}m\alpha}{\hbar^2}\frac{2\hbar^2}{3m\alpha} \\ &=& \frac{2\sqrt{2}}{3} \\ \end{eqnarray*} $

So the probability that the particle stay in ground state is $ \left(\frac{2\sqrt{2}}{3}\right)^2 = \frac{8}{9} $, and the probability that it get excited is $ \frac{1}{9} $.

Exploring Quantum Physics - Final Exam Part 1 Question 8

Question:

What are the eigen energies for this system?

$ \hat H = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + \frac{1}{2} m \omega^2 x^2 - q\mathcal{E}x $

Solution:

Using the given substitution $ z = x - \frac{q\mathcal{E}}{\omega^2m} $, we have

$ \begin{eqnarray*} \left(-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + \frac{1}{2} m \omega^2 x^2 - q\mathcal{E}x\right)\psi(x) &=& E\psi(x) \\ \left(-\frac{\hbar^2}{2m} \frac{d^2}{dz^2} + \frac{1}{2} m \omega^2 (z + \frac{q\mathcal{E}}{\omega^2m})^2 - q\mathcal{E}(z + \frac{q\mathcal{E}}{\omega^2m})\right)\psi(z + \frac{q\mathcal{E}}{\omega^2m}) &=& E\psi(z + \frac{q\mathcal{E}}{\omega^2m}) \\ \left(-\frac{\hbar^2}{2m} \frac{d^2}{dz^2} + \frac{1}{2} m \omega^2 (z + \frac{q\mathcal{E}}{\omega^2m})^2 - q\mathcal{E}z - \frac{q^2\mathcal{E}^2}{\omega^2m}\right)\psi(z + \frac{q\mathcal{E}}{\omega^2m}) &=& E\psi(z + \frac{q\mathcal{E}}{\omega^2m}) \\ \left(-\frac{\hbar^2}{2m} \frac{d^2}{dz^2} + \frac{1}{2} m \omega^2 (z^2 + 2z\frac{q\mathcal{E}}{\omega^2m}+(\frac{q\mathcal{E}}{\omega^2m})^2) - q\mathcal{E}z - \frac{q^2\mathcal{E}^2}{\omega^2m}\right)\psi(z + \frac{q\mathcal{E}}{\omega^2m}) &=& E\psi(z + \frac{q\mathcal{E}}{\omega^2m}) \\ \left(-\frac{\hbar^2}{2m} \frac{d^2}{dz^2} + \frac{1}{2} m \omega^2 z^2 + q\mathcal{E}z + \frac{q^2\mathcal{E}^2}{2\omega^2m} - q\mathcal{E}z - \frac{q^2\mathcal{E}^2}{\omega^2m}\right)\psi(z + \frac{q\mathcal{E}}{\omega^2m}) &=& E\psi(z + \frac{q\mathcal{E}}{\omega^2m}) \\ \left(-\frac{\hbar^2}{2m} \frac{d^2}{dz^2} + \frac{1}{2} m \omega^2 z^2 + \frac{q^2\mathcal{E}^2}{2\omega^2m} - \frac{q^2\mathcal{E}^2}{\omega^2m}\right)\psi(z + \frac{q\mathcal{E}}{\omega^2m}) &=& E\psi(z + \frac{q\mathcal{E}}{\omega^2m}) \\ \left(-\frac{\hbar^2}{2m} \frac{d^2}{dz^2} + \frac{1}{2} m \omega^2 z^2 - \frac{q^2\mathcal{E}^2}{2\omega^2m}\right)\psi(z + \frac{q\mathcal{E}}{\omega^2m}) &=& E\psi(z + \frac{q\mathcal{E}}{\omega^2m}) \\ \left(-\frac{\hbar^2}{2m} \frac{d^2}{dz^2} + \frac{1}{2} m \omega^2 z^2\right)\psi(z + \frac{q\mathcal{E}}{\omega^2m}) &=& (E + \frac{q^2\mathcal{E}^2}{2\omega^2m})\psi(z + \frac{q\mathcal{E}}{\omega^2m}) \\ \end{eqnarray*} $

Now we know $ E + \frac{q^2\mathcal{E}^2}{2\omega^2m} = \hbar\omega\left(n + \frac{1}{2}\right) $, in other words, $ E = \hbar\omega\left(n + \frac{1}{2}\right) - \frac{q^2\mathcal{E}^2}{2\omega^2m} $

Exploring Quantum Physics - Final Exam Part 1 Question 6/7

Question:

Given a system with non-positive energy and energy = 0 at infinity, do it have a bound state.

Solution:

For delta potential, we already know the answer, for 1 or 2 dimension we have bound state but for 3 we don't.

A reasonable conjecture, would then be, to conjecture we always have bound state for 1 or 2 dimension potential well, and may or may not have a bound state for 3 dimension.

It turns out to be true. See this as a theorem.

http://physics.stackexchange.com/questions/143630/why-the-statement-there-exist-at-least-one-bound-state-for-negative-potential

Exploring Quantum Physics - Final Exam Part 1 Question 5

Question:

Given a potential with a single discontinuity at $ x = 0 $, what is the right choice for the wave function?

Solution:

The basic principle is that the wave function must agree value at the boundary. The only choice that allow this is $ A\sin(kx)\theta(x) $. 

Exploring Quantum Physics - Final Exam Part 1 Question 4

Question:

How is the behavior of resitivity in weakly-disordered metals at low temperatures different in 3 dimensions versus one and two dimensions?

Solution:

The answer is:
In one and two dimensions weak scattering can lead to localization effects and an upturn in the resistivity at low temperatures.

I just vaguely recall this has to do with random walk. For 1 or 2 dimensional random walk they return to origin almost surely and therefore there are paths inference, but for 3 dimension that does not happen.

Exploring Quantum Physics - Final Exam Part 1 Question 3

Question:

How to reduce the Feynman Integral to classical action?

Solution:

Recall that the Feynman Integral reduce to classical action when we take the exponential term to minimum, as required for the Laplace method. So the answer is

$ \delta S[x(t)] = 0 $

Exploring Quantum Physics - Final Exam Part 1 Question 2

Question:

What are the following choices are valid solution for the time dependent Schrodinger equation?

Solution:

This is a trivial problem, just substitute the choices in the time dependent Schrodinger equation and verify.

Note that there cannot be a purely real solution - so two choices go away. Be careful with signs, that's all.

Exploring Quantum Physics - Final Exam Part 1 Question 1

Question:

What is the probability of finding the particle in the left hand side of the box under the particle in a box potential?

Solution:

$ \frac{1}{2} $. Left and right is symmetric.

Exploring Quantum Physics - Week 7 Question 10

Question:

Suppose we have a wave function of an electron under an electric field, what would the wave function be in the anti technique we just derived?

Solution:

We just simply do $ \alpha\beta $ to both side of the equation, that gives.

$ \begin{eqnarray*} i\hbar \frac{\partial \Psi_F \left(x,t\right)}{\partial t} &=& \left( c \hat{\alpha} \hat{p} - qF\hat{x} + mc^2 \hat{\beta}\right) \Psi_F \left(x,t\right) \\ -i\hbar \frac{\partial \hat{\alpha}\hat{\beta} \Psi_F \left(x,t\right)}{\partial t} &=& \left( c \hat{\alpha} \hat{p} - QF\hat{x} + Mc^2 \hat{\beta}\right) \hat{\alpha}\hat{\beta} \Psi_F \left(x,t\right) \\ -i\hbar \hat{\alpha}\hat{\beta}\frac{\partial \Psi_F \left(x,t\right)}{\partial t} &=& \left( c \hat{\alpha} \hat{p} - QF\hat{x} + Mc^2 \hat{\beta}\right) \hat{\alpha}\hat{\beta} \Psi_F \left(x,t\right) \\ -i\hbar \hat{\alpha}\hat{\beta}\frac{\partial \Psi_F \left(x,t\right)}{\partial t} &=& \hat{\alpha}\left( c \hat{\alpha} \hat{p} - QF\hat{x} - Mc^2 \hat{\beta}\right) \hat{\beta} \Psi_F \left(x,t\right) \\ -i\hbar \hat{\alpha}\hat{\beta}\frac{\partial \Psi_F \left(x,t\right)}{\partial t} &=& \hat{\alpha}\hat{\beta} \left( -c \hat{\alpha} \hat{p} - QF\hat{x} - Mc^2 \hat{\beta}\right) \Psi_F \left(x,t\right) \\ i\hbar \hat{\alpha}\hat{\beta}\frac{\partial \Psi_F \left(x,t\right)}{\partial t} &=& \hat{\alpha}\hat{\beta} \left( c \hat{\alpha} \hat{p} + QF\hat{x} + Mc^2 \hat{\beta}\right) \Psi_F \left(x,t\right) \\ \end{eqnarray*} $

So the new 'particle' has the charge flipped and the mass stay the same - that envisions positron!

Exploring Quantum Physics - Week 7 Question 9

Question:

How to construct another wave-function from one such that it overall gain a negative sign in the Dirac equation?

Solution:

The Dirac equation is:

$ i\hbar{\frac{\partial \Psi}{\partial t}} = (\hat{\alpha}\hat{p} + mc^2\hat{\beta})\Psi $.

We know $ \{\alpha, \beta\} = \alpha\beta + \beta\alpha = 0 $, so we have $\alpha\beta = -\beta\alpha $, that how a negative sign is introduced.

So all we need to do to make sure it gain an overall negative sign is simply make sure we have odd number of  $\alpha \beta $.

That explains the existence of anti matter!

Exploring Quantum Physics - Week 7 Question 8

Question:

What is the average velocity $ \langle \psi_0|v_1|\psi_0 \rangle $

Solution:

Without going through the calculation, there is no reason why there are any asymmetry. The solution should be 0.

Exploring Quantum Physics - Week 7 Question 7

Question:

Compute the root mean square of electron velocity in ground state quantized motion under the Earth's constant magnetic field.

Solution:

The problem seems daunting at first. But experience tell me once again this is most likely just number substituting exercise, so let's try expanding $ \hat{a}^\dagger \hat{a} $ and see what is going on there.

$ \begin{eqnarray*} & & \hat{a}^\dagger \hat{a} \\ &=& \frac{m}{2\hbar\omega}(\hat{v_1} - i\hat{v_2})(\hat{v_1} + i\hat{v_2}) \\ &=& \frac{m}{2\hbar\omega}(\hat{v_1}^2 + \hat{v_2}^2 + i[\hat{v_1}, \hat{v_2}]) \\ &=& \frac{m}{2\hbar\omega}(\hat{v_1}^2 + \hat{v_2}^2 + i(i\frac{\hbar\omega}{m})) \\ &=& \frac{m}{2\hbar\omega}(\hat{v_1}^2 + \hat{v_2}^2 - \frac{\hbar\omega}{m}) \\ \end{eqnarray*} $

So we can expand $ H_{2D} $ as follow:

$ \begin{eqnarray*} & & H_{2D} \\ &=& \hbar\omega(\hat{a}^\dagger \hat{a} + \frac{1}{2}) \\ &=& \hbar\omega(\frac{m}{2\hbar\omega}(\hat{v_1}^2 + \hat{v_2}^2 - \frac{\hbar\omega}{m}) + \frac{1}{2}) \\ &=& \frac{m}{2}(\hat{v_1}^2 + \hat{v_2}^2 - \frac{\hbar\omega}{m}) + \frac{\hbar\omega}{2} \\ &=& \frac{m}{2}(\hat{v_1}^2 + \hat{v_2}^2) - \frac{\hbar\omega}{2} + \frac{\hbar\omega}{2} \\ &=& \frac{m}{2}(\hat{v_1}^2 + \hat{v_2}^2) \\ \end{eqnarray*} $

Now we will equate the ground state energies

$ \begin{eqnarray*} \epsilon_0 | \psi_0 \rangle &=& H_{2D} | \psi_0 \rangle \\ \langle \psi_0 | \epsilon_0 | \psi_0 \rangle &=& \langle \psi_0 | H_{2D} | \psi_0 \rangle \\ \epsilon_0 \langle \psi_0 | \psi_0 \rangle &=& \langle \psi_0 | H_{2D} | \psi_0 \rangle \\ \epsilon_0 &=& \langle \psi_0 | H_{2D} | \psi_0 \rangle \\ \frac{h\omega}{2} &=& \langle \psi_0 | H_{2D} | \psi_0 \rangle \\ \frac{h\omega}{2} &=& \langle \psi_0 | \frac{m}{2}(\hat{v_1}^2 + \hat{v_2}^2) | \psi_0 \rangle \\ \frac{h\omega}{2} &=& \frac{m}{2}\langle \psi_0 | (\hat{v_1}^2 + \hat{v_2}^2) | \psi_0 \rangle \\ \frac{h\omega}{m} &=& \langle \psi_0 | (\hat{v_1}^2 + \hat{v_2}^2) | \psi_0 \rangle \\ \end{eqnarray*} $

Last but not least, the actual value root mean square value is 32.15.