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.

Exploring Quantum Physics - Week 7 Question 6

Question:

What is the exact energy?

Solution:

To solve the equation, we use a substitution $z = \zeta - z_0$.

$\begin{eqnarray*} -\frac{\hbar^2}{2m}\frac{d^2\Psi}{dz^2} + \frac{m\omega^2}{2}z^2\Psi + mgz\Psi = E_{exact} \Psi \\ -\frac{\hbar^2}{2m}\frac{d^2\Psi}{d\zeta^2} + \frac{m\omega^2}{2}(\zeta - z_0)^2\Psi + mg(\zeta - z_0)\Psi = E_{exact} \Psi \\ -\frac{\hbar^2}{2m}\frac{d^2\Psi}{d\zeta^2} + \frac{m\omega^2}{2}(\zeta^2 - 2\zeta z_0 + z_0^2)\Psi + mg(\zeta - z_0)\Psi = E_{exact} \Psi \\ \end{eqnarray*}$

Now we see a convenient choice for $z_0$ would be $\frac{g}{\omega^2}$, because that would give

$\begin{eqnarray*} -\frac{\hbar^2}{2m}\frac{d^2\Psi}{d\zeta^2} + \frac{m\omega^2}{2}(\zeta^2 - 2\zeta z_0 + z_0^2)\Psi + mg(\zeta - z_0)\Psi = E_{exact} \Psi \\ -\frac{\hbar^2}{2m}\frac{d^2\Psi}{d\zeta^2} + \frac{m\omega^2}{2}(\zeta^2 - 2\zeta \frac{g}{\omega^2} + z_0^2)\Psi + mg(\zeta - z_0)\Psi = E_{exact} \Psi \\ -\frac{\hbar^2}{2m}\frac{d^2\Psi}{d\zeta^2} + \frac{m\omega^2}{2}(\zeta^2 + z_0^2)\Psi - mgz_0\Psi = E_{exact} \Psi \\ -\frac{\hbar^2}{2m}\frac{d^2\Psi}{d\zeta^2} + \frac{m\omega^2}{2}\zeta^2\Psi = (E_{exact} + mgz_0 - \frac{m\omega^2}{2}z_0^2) \Psi \\ \end{eqnarray*}$

Therefore, for ground state, we have $,$.

$\begin{eqnarray*} \frac{h\omega}{2} &=& E_{exact} + mgz_0 - \frac{m\omega^2}{2}z_0^2 \\ E_{exact} &=& \frac{h\omega}{2} - mgz_0 + \frac{m\omega^2}{2}z_0^2 \\ &=& \frac{h\omega}{2} - mg\frac{g}{\omega^2} + \frac{m\omega^2}{2}(\frac{g}{\omega^2})^2 \\ &=& \frac{h\omega}{2} - \frac{mg^2}{\omega^2} + \frac{m\omega^2g^2}{2\omega^4} \\ &=& \frac{h\omega}{2} - \frac{mg^2}{\omega^2} + \frac{mg^2}{2\omega^2} \\ &=& \frac{h\omega}{2} - \frac{mg^2}{2\omega^2} \end{eqnarray*}$

Exploring Quantum Physics - Week 7 Question 5

Question:

What happen to $E_2$ in classical limit $\hbar \to 0$?

Solution:

$E_2$ is independent of $\hbar$.

Exploring Quantum Physics - Week 7 Question 4

Question:

Compute the second order energy.

Solution:

Now we pick the terms for second order.

$H_0\psi_2 + V\psi_1 = E_0\psi_2 + E_1\psi_1 + E_2\psi_0$.

To find $E_2$, we take inner product with $\phi_0$. That gives

$\langle \phi_0 | H_0 |\psi_2 \rangle + \langle \phi_0 | V | \psi_1 \rangle = \langle \phi_0 | E_0 | \psi_2 \rangle + \langle \phi_0 | E_1 | \psi_1 \rangle + \langle \phi_0 | E_2 | \psi_0 \rangle$.

As in lecture, the terms cancels, and we already establish $E_1 = 0$, so we have

$\langle \phi_0 | V | \psi_1 \rangle =  E_2$.

But we still don't know what is $\psi_1$, what can we do now?

Note that $V = mgz$, so we can relate $\langle \phi_0 | V$ with $\phi_1$ as follow.

$\begin{eqnarray*} \langle \phi_1 | &=& \frac{1}{\sqrt{2}}\phi_0(2\sqrt{\frac{m\omega}{\hbar}}z) \\ &=& 2\sqrt{\frac{m\omega}{2\hbar}}\frac{1}{mg}\phi_0 mgz \\ &=& 2\sqrt{\frac{m\omega}{2\hbar}}\frac{1}{mg}\langle \phi_0 | V \\ \langle \phi_1 | \psi_1 \rangle &=& 2\sqrt{\frac{m\omega}{2\hbar}}\frac{1}{mg}\langle \phi_0 | V | \psi_1 \rangle \\ -\frac{mg}{\hbar\omega}\sqrt{\frac{\hbar}{2m\omega}} &=& 2\sqrt{\frac{m\omega}{2\hbar}}\frac{1}{mg}\langle \phi_0 | V | \psi_1 \rangle \\ \frac{mg^2}{2\omega^2} &=& \langle \phi_0 | V | \psi_1 \rangle \\ \end{eqnarray*}$

The last equal sign involve messy cancelation, but at the end that's the perfect result I needed!

Exploring Quantum Physics - Week 7 Question 3

Question:

What are the conditions that is true about $\psi_1$.

Solution:

We had the equation $H_0 \psi_1 + V \psi_0 = E_0 \psi_1 + E_1 \psi_0$.

With the last solution we can drop the last term.

Now consider the inner product of this equation with the eigenfunctions of the Quantum Harmonic Oscillator.

$\langle \phi_k | H_0 | \psi_1 \rangle + \langle \phi_k | V | \psi_0 \rangle =  E_0 \langle \phi_k | \psi_1 \rangle$

Being eigenfunction of the $H_0$, we can simplify the first term, and substitute in the eigen energies.

$\hbar\omega(k+\frac{1}{2})\langle \phi_k | \psi_1 \rangle + \langle \phi_k | V | \psi_0 \rangle =  \hbar\omega(\frac{1}{2}) \langle \phi_k | \psi_1 \rangle$

So it obviously simplify to

$\hbar\omega k\langle \phi_k | \psi_1 \rangle + \langle \phi_k | V | \psi_0 \rangle =  0$

From this form, it is clear that we need to evaluate $\langle \phi_k | V | \psi_0 \rangle$. By symmetry, $\langle \phi_0 | V | \psi_0 \rangle$ and $\langle \phi_2 | V | \psi_0 \rangle$ are 0, so the key term to evaluate is $\langle \phi_1 | V | \psi_0 \rangle$

$\begin{eqnarray*} & & \langle \phi_1 | V | \psi_0 \rangle \\ &=& \int\limits_{\infty}^{\infty}{dz(\frac{1}{\sqrt{2}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp(-\frac{m\omega z^2}{2\hbar})2\left(\frac{m\omega}{\hbar}\right)^{1/2}z)mgz(\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp(-\frac{m\omega z^2}{2\hbar})) } \\ &=& 2mg\frac{1}{\sqrt{2}}\left(\frac{m\omega}{\hbar}\right)^{1/2}\left(\frac{m\omega}{\pi\hbar}\right)^{1/2}\int\limits_{\infty}^{\infty}{dz z^2(\exp(-\frac{m\omega z^2}{\hbar})) } \\ &=& 2mg\left(\frac{m\omega}{\hbar}\right)\frac{1}{\sqrt{2\pi}}\int\limits_{\infty}^{\infty}{dz z^2(\exp(-\frac{m\omega z^2}{\hbar})) } \\ \end{eqnarray*}$

The integral can be seen as the variance of a normal distribution with variance = $\frac{\hbar}{2m\omega}$, so we will do the integral as follow:

$\begin{eqnarray*} & & 2mg\left(\frac{m\omega}{\hbar}\right)\frac{1}{\sqrt{2\pi}}\int\limits_{\infty}^{\infty}{dz z^2(\exp(-\frac{m\omega z^2}{\hbar})) } \\ &=& 2mg\left(\frac{m\omega}{\hbar}\right)\sqrt{\frac{\hbar}{2m\omega}}\frac{1}{\sqrt{2\pi}\sqrt{\frac{\hbar}{2m\omega}}}\int\limits_{\infty}^{\infty}{dz z^2(\exp(-\frac{m\omega z^2}{\hbar})) } \\ &=& 2mg\left(\frac{m\omega}{\hbar}\right)\sqrt{\frac{\hbar}{2m\omega}}(\frac{\hbar}{2m\omega}) \\ &=& mg\sqrt{\frac{\hbar}{2m\omega}} \end{eqnarray*}$

To complete this, we substitute it back to get the result we need.

$\begin{eqnarray*} \hbar\omega k \langle \phi_k | \psi_1 \rangle + \langle \phi_k | V | \psi_0 \rangle &=& 0 \\ \hbar\omega \langle \phi_1 | \psi_1 \rangle + \langle \phi_1 | V | \psi_0 \rangle &=& 0 \\ \hbar\omega \langle \phi_1 | \psi_1 \rangle + mg\sqrt{\frac{\hbar}{2m\omega}} &=& 0 \\ \hbar\omega \langle \phi_1 | \psi_1 \rangle &=& -mg\sqrt{\frac{\hbar}{2m\omega}} \\ \langle \phi_1 | \psi_1 \rangle &=& -\frac{mg}{\hbar\omega}\sqrt{\frac{\hbar}{2m\omega}} \\ \end{eqnarray*}$

Lot of work just to show it is non-zero, but we will need it for the next problem.

Exploring Quantum Physics - Week 7 Question 2

Question:

What is the first order energy for the Quantum Harmonic Oscillator under gravity?

Solution:

We will use perturbation theory.

First, as usual, we let $E = \sum\limits_{k = 0}^{\infty}{E_k\lambda^k}$, $\psi = \sum\limits_{k = 0}^{\infty}{\psi_k \lambda^k}$.

Substitute these back to the Schrodinger equation. We get this by picking the terms with $\lambda$ to the first order.

$\lambda(H_0\psi_1 + V\psi_0) = \lambda(E_0\psi_1 + E_1\psi_0)$

That allow us to show, $E_1 = \langle \psi_0 | V | \psi_0 \rangle$.

We know from the Quantum Harmonic Oscillator that $\psi_0 = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \exp(-\frac{m\omega z^2}{2\hbar})$.

So the rest is just an integration.

$\begin{eqnarray*} & & \int\limits_{-\infty}^{\infty}{dx (\left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \exp(-\frac{m\omega z^2}{2\hbar})) mgz (\left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \exp(-\frac{m\omega z^2}{2\hbar}))} \\ &=& \left(\frac{m\omega}{\pi\hbar}\right)^{1/2} mg \int\limits_{-\infty}^{\infty}{dx z (\exp(-\frac{m\omega z^2}{\hbar}))} \\ \end{eqnarray*}$

Lot of typesetting, but it is now obvious the first order energy is 0 because of the symmetry.

Exploring Quantum Physics - Week 7 Question 1

Question:

What are the parameters in the exponential function of the base state of the Quantum Harmonic Oscillator.

Solution:

$m, \omega, \hbar$.

Exploring Quantum Physics - Week 6 Bonus Question

Question:

Prove that $-l \le m \le l$ for $|lm\rangle$ is both an eigenvector of $L^2$ and $L_3$.

Solution:

Thanks Gang Xu for the ladder operator hint.

$\begin{eqnarray*} & & L_3 L_\pm|lm\rangle \\ &=& ( L_\pm L_3 - L_\pm L_3 + L_3 L_\pm)|lm\rangle \\ &=& ( L_\pm L_3 + L_3 L_\pm - L_\pm L_3)|lm\rangle \\ &=& ( L_\pm L_3 + [L_3, L_\pm])|lm\rangle \\ &=& ( L_\pm L_3 + \pm\hbar L_\pm)|lm\rangle \\ &=& ( L_\pm \hbar m + \pm\hbar L_\pm)|lm\rangle \\ &=& \hbar (m \pm 1) L_\pm |lm\rangle \\ \end{eqnarray*}$

To proceed, we need to know one more identity

$\begin{eqnarray*} & & L_- L_+ \\ &=& (L_1 - i L_2)(L_1 + i L_2) \\ &=& (L_1^2 - i L_2 L_1 + i L_1 L_2 + L_2^2) \\ &=& (L_1^2 + L_2^2) + i L_1 L_2 - i L_2 L_1 \\ &=& (L_1^2 + L_2^2) + i [L_1, L_2] \\ &=& (L_1^2 + L_2^2) + i (i\hbar L_3) \\ &=& (L_1^2 + L_2^2) - \hbar L_3 \\ &=& L^2 - L_3^2 - \hbar L_3 \\ \end{eqnarray*}$

So we can combine the two identities into a single one $L_\pm L_\mp = L^2 - L_3^2 \pm \hbar L_3$. Now we can efficiently do both at the same time

$\begin{eqnarray*} & & L^2 L_\pm | lm\rangle \\ &=& (L_\pm L_\mp + L_3^2 \mp \hbar L_3) L_\pm | lm\rangle \\ &=& (L_\pm L_\mp L_\pm + L_3^2 L_\pm \mp \hbar L_3 L_\pm) | lm\rangle \\ &=& (L_\pm (L^2 - L_3^2 \mp \hbar L_3) + L_3^2 L_\pm \mp \hbar L_3 L_\pm) | lm\rangle \\ &=& (L_\pm (L^2 - L_3^2 \mp \hbar L_3) + L_3^2 L_\pm \mp \hbar^2 (m \pm 1) L_\pm) | lm\rangle \\ &=& (L_\pm (L^2 - L_3^2 \mp \hbar L_3) + L_3 \hbar (m \pm 1) L_\pm \mp \hbar^2 (m \pm 1) L_\pm) | lm\rangle \\ &=& (L_\pm (L^2 - L_3^2 \mp \hbar L_3) + \hbar (m \pm 1) L_3 L_\pm \mp \hbar^2 (m \pm 1) L_\pm) | lm\rangle \\ &=& (L_\pm (L^2 - L_3^2 \mp \hbar L_3) + \hbar^2 (m \pm 1)^2 L_\pm \mp \hbar^2 (m \pm 1) L_\pm) | lm\rangle \\ &=& (L_\pm (L^2 - L_3^2 \mp \hbar^2 m) + \hbar^2 (m \pm 1)^2 L_\pm \mp \hbar^2 (m \pm 1) L_\pm) | lm\rangle \\ &=& (L_\pm (L^2 - L_3 \hbar m \mp \hbar^2 m) + \hbar^2 (m \pm 1)^2 L_\pm \mp \hbar^2 (m \pm 1) L_\pm) | lm\rangle \\ &=& (L_\pm (L^2 - \hbar m L_3 \mp \hbar^2 m) + \hbar^2 (m \pm 1)^2 L_\pm \mp \hbar^2 (m \pm 1) L_\pm) | lm\rangle \\ &=& (L_\pm (L^2 - \hbar^2 m^2 \mp \hbar^2 m) + \hbar^2 (m \pm 1)^2 L_\pm \mp \hbar^2 (m \pm 1) L_\pm) | lm\rangle \\ &=& (L_\pm (\hbar^2 l (l+1) - \hbar^2 m^2 \mp \hbar^2 m) + \hbar^2 (m \pm 1)^2 L_\pm \mp \hbar^2 (m \pm 1) L_\pm) | lm\rangle \\ &=& ((\hbar^2 l (l+1) - \hbar^2 m^2 \mp \hbar^2 m) L_\pm + \hbar^2 (m \pm 1)^2 L_\pm \mp \hbar^2 (m \pm 1) L_\pm) | lm\rangle \\ &=& (\hbar^2 l (l+1) - \hbar^2 m^2 \mp \hbar^2 m + \hbar^2 (m \pm 1)^2 \mp \hbar^2 (m \pm 1)) L_\pm | lm\rangle \\ &=& \hbar^2 (l (l+1) - m^2 \mp m + (m \pm 1)^2 \mp (m \pm 1)) L_\pm | lm\rangle \\ &=& \hbar^2 (l (l+1) - m^2 \mp m + m^2 \pm 2m + 1 \mp (m \pm 1)) L_\pm | lm\rangle \\ &=& \hbar^2 (l (l+1) \mp m \pm 2m + 1 \mp (m \pm 1)) L_\pm | lm\rangle \\ &=& \hbar^2 (l (l+1)) L_\pm | lm\rangle \\ \end{eqnarray*}$

The last equality comes from just plus or minus sign case study - they will all cancel in either case. In retrospect - this result is not surprising, as $L_2$ is supposed to commute with $L_1$ and $L_2$. That is so far I can go.

Exploring Quantum Physics - More angular momentum commutators

The goal of this post is to compute $[L_a, L_d]$.

We'll prove something more basic first, let's start with $[x_a p_b, x_c]$. This one is basic, we have

$[x_a p_b, x_c] = x_a p_b x_c - x_c x_a p_b = x_a p_b x_c - x_a x_c p_b = x_a ( p_b x_c - x_c p_b ) = x_a [p_b, x_c] = -i\hbar \delta_{bc} x_a$.

Similarly we can also do $[x_a p_b, p_c]$, using essentially the same trick

$[x_a p_b, p_c] =  x_a p_b p_c - p_c x_a p_b = x_a p_c p_b - p_c x_a p_b = (x_a p_c - p_c x_a) p_b = [x_a p_c] p_b = i\hbar \delta_{ac} p_b$.

I thought I will use these relations to prove the $[L_a, L_d]$ relation, but I was wrong. We should always use closest established result, which is $[L_a, x_b]$ and $[L_a. p_b]$.

$\begin{eqnarray*} [L_a, L_d] &=& [L_a, \epsilon_{bcd}x_b p_c] \\ &=& \epsilon_{bcd}[L_a, x_b p_c] \\ &=& \epsilon_{bcd}([L_a, x_b] p_c + x_b [L_a, p_c]) \\ &=& \epsilon_{bcd}(i\hbar\epsilon_{abe} x_e p_c + i\hbar\epsilon_{acf} x_b p_f ) \\ \end{eqnarray*}$

The expression seems complicated but it could be simplified. Let focus on one term at a time

$\begin{eqnarray*} & & i\hbar\epsilon_{bcd}\epsilon_{abe} x_e p_c \\ &=& i\hbar\epsilon_{bcd}\epsilon_{bea} x_e p_c \\ &=& i\hbar(\delta_{ad}\delta_{ce} - \delta_{ac}\delta_{de}) x_e p_c \end{eqnarray*}$

Similarly, we can get the right hand side as

$\begin{eqnarray*} & & i\hbar\epsilon_{bcd}\epsilon_{acf} x_b p_f \\ & & i\hbar\epsilon_{cdb}\epsilon_{cfa} x_b p_f \\ &=& i\hbar(\delta_{ab}\delta_{df} - \delta_{ad}\delta_{bf}) x_b p_f \end{eqnarray*}$

The key idea to simplification is that $b$ and $f$ are just running index in the second term, we can rename it as we wish. Apparently, we want $b \to e$ and $f \to c$, so we can write the second term as

$\begin{eqnarray*} & & i\hbar(\delta_{ab}\delta_{df} - \delta_{ad}\delta_{bf}) x_b p_f \\ &=& i\hbar(\delta_{ae}\delta_{cd} - \delta_{ad}\delta_{ce}) x_e p_c \end{eqnarray*}$

So if we put them back together, we get:

$\begin{eqnarray*} & & [L_a, L_d] \\ &=& i\hbar(\delta_{ad}\delta_{ce} - \delta_{ac}\delta_{de}) x_e p_c + (\delta_{ae}\delta_{cd} - \delta_{ad}\delta_{ce}) x_e p_c \\ &=& i\hbar(\delta_{ae}\delta_{cd} - \delta_{ac}\delta_{de}) x_e p_c \\ &=& i\hbar(x_a p_d - x_d p_a) \\ \end{eqnarray*}$

To complete the circle, let's prove the answer is $i\hbar\epsilon_{adb}L_b$

$\begin{eqnarray*} & & i\hbar\epsilon_{adb}L_b \\ &=& i\hbar\epsilon_{adb}\epsilon_{ceb}x_c p_e \\ &=& i\hbar\epsilon_{bad}\epsilon_{bce}x_c p_e \\ &=& i\hbar(\delta_{ac}\delta_{de} - \delta_{ae}\delta_{cd})x_c p_e \\ &=& i\hbar(x_a p_d - x_d p_a) \end{eqnarray*}$

No wonder professor said this is complicated.

Exploring Quantum Physics - More about Hermitian Operator

It always confuse me when we have equation like this

$\langle \psi|\hat{A}$

What on earth does that mean? Why do we put an operator to the left of the function being applied?
I decided to dig deeper.

According to Wikipedia, a Hermitian Operator in a Hilbert space has this property.

$\langle \hat{A} x, y \rangle = \langle x, \hat{A} y \rangle$. The angle brackets are NOT bra-ket notation, they represents inner product.

So we can write $\langle \hat{A} x| \hat{B} y \rangle = \langle x | \hat{A} \hat{B} y \rangle$. This time we indeed mean the bra-ket notation.

Exploring Quantum Physics - Week 6 Question 11

Question:

What is the ground state energy of a bouncing neutron under the simplified, constant gravitational field?

Solution:

Just substitute the values into the equations in the lectures to give the answer:

$\alpha = (\frac{\hbar^2 M g^2}{2})^{1/3} = 9.64 \times 10^{-32}$
$E = -\rho \alpha = 2.25 \times 10^{-31}$
$T = \frac{E}{k} = 1.63 \times 10^{-8}$

So the answer is 16 nano-kelvin.

Exploring Quantum Physics - Week 6 Question 10

Question:

Do all states of this new Hamiltonian that have the same energy also have the same parity?

Solution:

No, it is possible to have $|1\rangle|1\rangle|2\rangle$ to have energy $6\hbar\omega$ and $|1\rangle|3\rangle|1\rangle$ to have energy $6\hbar\omega$. But they have different parity. The former is even and the latter is odd.

Exploring Quantum Physics - Week 6 Question 9

Question:

Does the angular momentum operator commute with the Hamilionian in this asymmetric potential?

Solution:

Yes. The same reasoning applies just as question 7 does.

Turn out I am wrong - it commutes only with $L_3$.
The expression I used in question 7 assumes all coefficients are the same, but this one isn't. Some terms may not cancel.

Exploring Quantum Physics - Week 6 Question 8

Question:

What is the ground state energy of the potential $V\left(x_1,x_2,x_3\right)= \frac{M\omega^2}{2}\left(x_1^2+x_2^2+4x_3^2\right)$

Solution:

Following the same mindset in Question 4. The ground state energy is really just the sum of ground state energies, so that would be

$\hbar\omega\frac{1}{2} + \hbar\omega\frac{1}{2} + \hbar 2\omega\frac{1}{2} = 2\hbar\omega$.

Exploring Quantum Physics - Week 6 Question 7

Question:

Does the angular momentum operator commutes with the Hamilitonian in the isotropic potential?

Solution:

The Hamilitonian operator is $\hat{H} = \frac{\hat{p}^2}{2m} + \hat{V}$.

$\begin{eqnarray*} [L_k, H] &=& [L_k, \frac{\hat{p}^2}{2m} + \hat{V}] \\ &=& \frac{1}{2m}[L_k, \hat{p}^2] + [L_k, \hat{V}] \\ \end{eqnarray*}$

Again, in order not to clutter the derivation, let's solve a subproblem first.

$\begin{eqnarray*} & & [L_a, \hat{p}^2] \\ &=& [L_a, p_b p_b] \\ &=& [L_a, p_b] p_b + p_b [L_a, p_b] \\ &=& \epsilon_{abc} p_c p_b + p_b \epsilon_{abc} p_c \\ &=& \epsilon_{abc} p_c p_b + p_b \epsilon_{abc} p_c \\ &=& 2\epsilon_{abc} p_b p_c \\ &=& 0 \end{eqnarray*}$

The last equality comes from the fact that the sum cancel each other! Suppose $a = 1$, then the non-vanishing terms of the sum is just $p_2 p_3 - p_3 p_2$, which is simply 0. Looking backward, if I go just a little further with my question 3, I would have reach the same result. Now let's solve the potential side

$\begin{eqnarray*} & & [L_a, x_b x_b] \\ &=& [L_a, x_b] x_b + x_b [L_a, x_b] \\ &=& \epsilon_{abc} x_c x_b + x_b \epsilon_{abc} x_c \\ &=& \epsilon_{abc} x_c x_b + x_b \epsilon_{abc} x_c \\ &=& 2\epsilon_{abc} x_b x_c \\ &=& 0 \end{eqnarray*}$

The reasoning is identical, so the commutator is just 0.

Exploring Quantum Physics - Week 6 Question 5, 6

Question:

A state $\Psi\left(x_1,x_2,x_3\right)$ is said to have even parity if $\Psi\left(x_1,x_2,x_3\right) = \Psi\left(-x_1,-x_2,-x_3\right)$ and odd parity if $\Psi\left(x_1,x_2,x_3\right) = -\Psi\left(-x_1,-x_2,-x_3\right)$.

5) What is the parity of the wave function $|n_1\rangle|n_2\rangle|n_3\rangle$

6) Is it true that same energy implies same parity.

Solution:

This one is simple. For the basic quantum harmonic oscillator solution, even energy is even function, odd energy is odd function.

Together with the basic facts

Product of two even functions is a even function.
Product of two odd functions is a even function.
Product of an even function with an odd function is an odd function.

We deduce the the wavefunction is even parity if $n_1 + n_2 + n_3$ is even, and it is odd parity if $n_1 + n_2 + n_3$ is odd.

Now question 6 is obvious - by question 4 energy depends on sum of indices - and by question 5 parity depends on sum of indices. So if we have the same energy, we must also have the same parity!

Exploring Quantum Physics - Week 6 Question 4

Question:

Given an isotropic Quantum Harmonic Oscillator, what is the ground state energy.

$\hat{H} = -\frac{\hbar^2}{2M} \nabla^2 + V\left(x_1,x_2,x_3\right)= -\frac{\hbar^2}{2M}\nabla^2 + \frac{M\omega^2}{2}\left(x_1^2+x_2^2+x_3^2\right).$

Solution:

The Hamilitonian can be written as a sum
$\begin{eqnarray*} \hat{H} &=& -\frac{\hbar^2}{2M}\nabla^2 + \frac{M\omega^2}{2}\left(x_1^2+x_2^2+x_3^2\right) \\ &=& -\frac{\hbar^2}{2M}(\frac{\partial^2}{\partial x_1^2} + \frac{\partial^2}{\partial x_2^2} + \frac{\partial^2}{\partial x_3^2}) + \frac{M\omega^2}{2}\left(x_1^2+x_2^2+x_3^2\right) \\ &=& \sum\limits_{k=1}^{3}(-\frac{\hbar^2}{2M}\frac{\partial^2}{\partial x_k^2} + \frac{M\omega^2x_k^2}{2}) \end{eqnarray*}$

In this form, we can easily see the energy is just the sum of three harmonic oscillator, so the answer is $\hbar\omega(n_1+n_2+n_3+\frac{3}{2})$.
Phew - finally an easier question - blocked on question 3 for so long.

Exploring Quantum Physics - Week 6 Question 3

Question:

What is $[L \cdot L, \hat{r} \cdot \hat{p} ]$?

Solution:

At this point I still don't know the solution. But I think it might be beneficial at this point to try to typeset whatever I have right now. Maybe that's the key to completion of the complicated piece.

This is just really complicated. Anyway, we need to start somewhere.

$\begin{eqnarray*} & & [L \cdot L, \hat{r} \cdot \hat{p} ] \\ &=& [L_a L_a, x_b p_b ] \\ &=& [L_a L_a, x_b] p_b + x_b[L_a L_a, p_b] \\ \end{eqnarray*}$

To avoid getting too complicated - now we focus on the first term

$\begin{eqnarray*} & & [L_a L_a, x_b] \\ &=& [L_a, x_b] L_a + L_a[L_a, x_b] \\ &=& (\epsilon_{abc}i\hbar x_c) L_a + L_a(\epsilon_{abc}i\hbar x_c) \\ &=& \epsilon_{abc}i\hbar(x_c L_a + L_a x_c) \\ &=& \epsilon_{abc}i\hbar(x_c (\epsilon_{ade} x_d p_e) + (\epsilon_{ade} x_d p_e) x_c) \\ &=& \epsilon_{abc}\epsilon_{ade}i\hbar(x_c x_d p_e + x_d p_e x_c) \end{eqnarray*}$

Now we work on the second term

$\begin{eqnarray*} & & [L_a L_a, p_b] \\ &=& [L_a, p_b] L_a + L_a[L_a, p_b] \\ &=& (\epsilon_{abc}i\hbar p_c) L_a + L_a(\epsilon_{abc}i\hbar p_c) \\ &=& \epsilon_{abc}i\hbar(p_c L_a + L_a p_c) \\ &=& \epsilon_{abc}i\hbar(p_c (\epsilon_{ade} x_d p_e) + (\epsilon_{ade} x_d p_e) p_c) \\ &=& \epsilon_{abc}\epsilon_{ade}i\hbar(p_c x_d p_e + x_d p_e p_c) \end{eqnarray*}$

And then we substitute them back

$\begin{eqnarray*} & & [L \cdot L, \hat{r} \cdot \hat{p} ] \\ &=& [L_a L_a, x_b] p_b + x_b[L_a L_a, p_b] \\ &=& \epsilon_{abc}\epsilon_{ade}i\hbar(x_c x_d p_e + x_d p_e x_c) p_b + x_b\epsilon_{abc}\epsilon_{ade}i\hbar(p_c x_d p_e + x_d p_e p_c) \\ &=& \epsilon_{abc}\epsilon_{ade}i\hbar(x_c x_d p_e p_b + x_d p_e x_c p_b + x_b p_c x_d p_e + x_b x_d p_e p_c) \\ &=& (\delta_{bd}\delta_{ce} - \delta_{be}\delta_{cd})i\hbar(x_c x_d p_e p_b + x_d p_e x_c p_b + x_b p_c x_d p_e + x_b x_d p_e p_c) \\ &=& i\hbar((x_c x_b p_c p_b + x_b p_c x_c p_b + x_b p_c x_b p_c + x_b x_b p_c p_c)-(x_c x_c p_b p_b + x_c p_b x_c p_b + x_b p_c x_c p_b + x_b x_c p_b p_c)) \\ &=& i\hbar((x_b x_c p_b p_c + x_b p_c x_c p_b + x_b p_c x_b p_c + x_b x_b p_c p_c)-(x_c x_c p_b p_b + x_c p_b x_c p_b + x_b p_c x_c p_b + x_b x_c p_b p_c)) \\ &=& i\hbar((x_b p_c x_b p_c + x_b x_b p_c p_c)-(x_c x_c p_b p_b + x_c p_b x_c p_b)) \end{eqnarray*}$

When $b$ and $c$ are equal, the expression above is 0, so we can focus on the case when $b \ne c$, in that case the operator commutes!

$\begin{eqnarray*} & & [L \cdot L, \hat{r} \cdot \hat{p} ] \\ &=& i\hbar((x_b p_c x_b p_c + x_b x_b p_c p_c)-(x_c x_c p_b p_b + x_c p_b x_c p_b)) \\ &=& i\hbar((x_b x_b p_c p_c + x_b x_b p_c p_c)-(x_c x_c p_b p_b + x_c x_c p_b p_b)) \\ &=& i\hbar(2 x_b x_b p_c p_c - 2 x_c x_c p_b p_b) \\ &=& 2i\hbar(x_b x_b p_c p_c - x_c x_c p_b p_b) \end{eqnarray*}$

This is the end - I have no trick to further simplify this. It appears to me no choice matches this.

Now we try something else - discussion forum suggest I can try expanding the operators the other way round.

$\begin{eqnarray*} & & [L \cdot L, \hat{r} \cdot \hat{p} ] \\ &=& [L_a L_a, x_b p_b ] \\ &=& [L_a L_a, x_b p_b ] \\ &=& [L_a, x_b p_b] L_a + L_a [L_a, x_b p_b] \end{eqnarray*}$

Oh - leveraging the previous question - this is just zero?

Exploring Quantum Physics - Question 2

Question:

What is $[\hat{L_1}, \hat{r} \cdot \hat{p}]$

Solution:

With the identity and the experience deriving those, it is not complicated anymore.

$$\begin{eqnarray*} [\hat{L_1}, \hat{r} \cdot \hat{p}] &=& [\hat{L_1}, \hat{x_a}\hat{p_a}] \\ &=& [\hat{L_1}, \hat{x_a}]\hat{p_a} + \hat{x_a}[\hat{L_1}, \hat{p_a}] \\ &=& (\epsilon_{1ab}\hat{x_b})\hat{p_a} + \hat{x_a}(\epsilon_{1ab}\hat{p_b}) \\ &=& \epsilon_{1ab}(\hat{x_b}\hat{p_a} + \hat{x_a}\hat{p_b}) \\ &=& \hat{x_3}\hat{p_2} + \hat{x_2}\hat{p_3} - \hat{x_2}\hat{p_3} - \hat{x_3}\hat{p_2} \\ &=& 0 \end{eqnarray*}$$

Exploring Quantum Physics - More commutators

In this post we will talk about the angular momentum operator and their commutators. The angular momentum operator is $\hat{L} = \hat{r} \times \hat{p}$, so we can write $\hat{L_c} = \epsilon_{abc}\hat{x_a}\hat{p_b}$, and we wanted to compute $[\hat{L_a}, \hat{x_d}]$ and $[\hat{L_a}, \hat{p_b}]$.

Given the principle we learnt in the last post. I will try to work in abstract as much as I can. It works great this way.

$\begin{eqnarray*} [\hat{L_c}, \hat{x_d}] &=& [\epsilon_{abc}\hat{x_a}\hat{p_b}, \hat{x_d}] \\ &=& \epsilon_{abc}[\hat{x_a}\hat{p_b}, \hat{x_d}] \\ &=& \epsilon_{abc}([\hat{x_a}, \hat{x_d}]\hat{p_b} + \hat{x_a}[\hat{p_b}, \hat{x_d}]) \\ &=& \epsilon_{abc}\hat{x_a}[\hat{p_b}, \hat{x_d}] \\ &=& -\epsilon_{abc}\hat{x_a}[\hat{x_b}, \hat{p_d}] \\ &=& -\epsilon_{abc}\hat{x_a}(\delta_{bd}i\hbar) \\ &=& -i\hbar\epsilon_{adc}\hat{x_a} \\ &=& i\hbar\epsilon_{cda}\hat{x_a} \\ \end{eqnarray*}$

Let's also solve the momentum problem:

$\begin{eqnarray*} [\hat{L_c}, \hat{p_d}] &=& [\epsilon_{abc}\hat{x_a}\hat{p_b}, \hat{p_d}] \\ &=& \epsilon_{abc}[\hat{x_a}\hat{p_b}, \hat{p_d}] \\ &=& \epsilon_{abc}([\hat{x_a}, \hat{p_d}]\hat{p_b} + \hat{x_a}[\hat{p_b}, \hat{p_d}]) \\ &=& \epsilon_{abc}[\hat{x_a}, \hat{p_d}]\hat{p_b} \\ &=& \epsilon_{abc}(\delta_{ad}i\hbar)\hat{p_b} \\ &=& i\hbar\epsilon_{dbc}\hat{p_b} \\ &=& i\hbar\epsilon_{cdb}\hat{p_b} \end{eqnarray*}$

These match perfectly with the standard result. Imagine how complicated and convoluted it would look like without all these simplifying identity and symbols.

Exploring Quantum Physics - Generic Operator Calculus

Operators, symbols, are no good if one need to keep thinking about the underlying representation in order to use it. To this end, we need to be able to reason about operators without going back to its underlying representation. I call it operator calculus.

The simplest rule of operator is linearity.

$(\hat{A + B})\hat{C} = \hat{A}\hat{C} + \hat{B}\hat{C}$.

In general, operators do not commute, i.e. $\hat{A} \hat{B} \ne \hat{B}\hat{A}$. But they can be compensated by commutator $[\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A}$. This definition automatically implies $[\hat{A}, \hat{B}] = -[\hat{B}, \hat{A}]$.

Commutators are also linear $[\hat{A} + \hat{B}, \hat{C}] = (\hat{A} + \hat{B})\hat{C} - \hat{C}(\hat{A} + \hat{B}) = \hat{A}\hat{C} + \hat{B}\hat{C} - \hat{C}\hat{A} + \hat{C}\hat{B} = [\hat{A},\hat{C}] + [\hat{B}, \hat{C}]$.

This identity is useful to decompose product of operators inside a commutator.

$[\hat{A}, \hat{B}\hat{C}] = [\hat{A}, \hat{B}]\hat{C}+\hat{B}[\hat{A}, \hat{C}]$.

To show that, we expands the commutators on the right.

$[\hat{A}, \hat{B}]\hat{C} + \hat{B}[\hat{A}, \hat{C}] = (\hat{A}\hat{B}-\hat{B}\hat{A})\hat{C} + \hat{B}(\hat{A}\hat{C}-\hat{C}\hat{A}) = \hat{A}\hat{B}\hat{C} - \hat{B}\hat{A}\hat{C} + \hat{B}\hat{A}\hat{C} -\hat{B}\hat{C}\hat{A} = \hat{A}\hat{B}\hat{C} - \hat{B}\hat{C}\hat{A} = [\hat{A},\hat{B}\hat{C}]$

Same principle applies to commutator, once we prove a certain identity in commutator, then let's not expand the operator. That's how one maybe able to handle the complexity.

To apply that, let's see what can we do to deal with an operator product on the left, we have

$[\hat{A}\hat{B}, \hat{C}] = -[\hat{C}, \hat{A}\hat{B}] = -([\hat{C}, \hat{A}]\hat{B} + \hat{A}[\hat{C}, \hat{B}]) = [\hat{A}, \hat{C}]\hat{B} + \hat{A}[\hat{B}, \hat{C}]$

Exploring Quantum Physics - Basic commutators

The goal of this post is to show the basic commutator relationships. Recall that the position operator is simply multiplying with the position, and the momentum operator is $-i\hbar\frac{\partial}{\partial x}$. Now we work in 3 dimensional space.

Also recall $[\hat{A}, \hat{B}] =  \hat{A}\hat{B} - \hat{B}\hat{A}$.

Now we wanted to know the commutation relationship between position and momentum operator, we have

$\begin{eqnarray*} [\hat{x_m}, \hat{p_n}] &=& \hat{x_m}\hat{p_n} - \hat{p_n}\hat{x_m} \\ [\hat{x_m}, \hat{p_n}] \psi &=& x_m(-i\hbar)\frac{\partial}{\partial x_n}\psi - (-i\hbar)\frac{\partial}{\partial x_n}x_m\psi \\ &=& (-i\hbar)(x_m\frac{\partial}{\partial x_n}\psi - \frac{\partial}{\partial x_n}x_m\psi) \\ \end{eqnarray*}$

In this form, it is obvious that if $m \ne n$, then we can simply pull $x_m$ out from the partial derivative at the latter term and the whole thing cancel out. So let focus on the case when $m = n$, the expression becomes:

$\begin{eqnarray*} [\hat{x_m}, \hat{p_m}] \psi &=& (-i\hbar)(x_m\frac{\partial}{\partial x_m}\psi - \frac{\partial}{\partial x_m}x_m\psi) \\ &=& (-i\hbar)(x_m\frac{\partial}{\partial x_m}\psi - (\psi + x_m\frac{\partial}{\partial x_m}\psi)) \\ &=& (-i\hbar)(-\psi) \\ [\hat{x_m}, \hat{p_m}] &=& i\hbar \\ \end{eqnarray*}$

Exploring Quantum Physics - Week 6 Question 1

Question:

What is $\epsilon_{abc}\epsilon_{cde}$.

Solution:

The contraction identity we just shown in the previous post is applied.

$\epsilon_{abc}\epsilon_{cde} = \epsilon_{cab}\epsilon_{cde} = \delta_{ad}\delta_{be}-\delta_{ae}\delta_{bd}$.

Exploring Quantum Physics - The contraction identity

The proof of that identity is a straightforward verification, so here is the code that does the job for us.

Note in particular that the Einstein summation convention in action in the left hand side as $i$ is a repeated index.

namespace ManagedWorkspace
{
    using System;
    using System.Collections.Generic;
    using System.Linq;

    internal static class Program
    {
        static int levi(int i, int j, int k)
        {
            int c = i * 100 + j * 10 + k;
            switch (c)
            {
                case 123: return 1;
                case 231: return 1;
                case 312: return 1;
                case 132: return -1;
                case 213: return -1;
                case 321: return -1;
            }
            return 0;
        }

        static int delta(int i, int j)
        {
            return i == j ? 1 : 0;
        }

        static void Main(string[] args)
        {
                for (int j = 1; j <= 3; j++)
                {
                    for (int k = 1; k <= 3; k++)
                    {
                        for (int s = 1; s <= 3; s++)
                        {
                            for (int t = 1; t <= 3; t++)
                            {
                                int lhs = 0;
                                for (int i = 1; i <= 3; i++)
                                {
                                    lhs += levi(i, j, k) * levi(i, s, t);
                                }
                                int rhs = delta(j, s) * delta(k, t) - delta(j, t) * delta(k, s);
                                if (lhs != rhs)
                                {
                                    Console.WriteLine("Error!");
                                }
                            }
                        }
                    }
                }
            }
        }
    }

Exploring Quantum Mechanics - even more about cross product.

Believe it or not, there are even more property in the cross product. This time our goal is the BAC-CAB identity

$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b})$.

To prove this, we need this result first, the so-called contraction identity.

$\epsilon_{ijk}\epsilon_{ist} = \delta_{js}\delta_{kt} - \delta_{jt}\delta_{ks}$.

To be honest, I don't really understand how does one come up with the identity, but it is relatively easy to verify that it is correct, and under what condition (that $i$ does not equal any of $j, k, s, t$

With that, we express the right hand side as follow:

$\begin{eqnarray*} \vec{a} \times (\vec{b} \times \vec{c}) &=& \vec{a} \times (\epsilon_{pqr}b_pc_qe_r) \\ &=& \epsilon_{srt}a_s\epsilon_{pqr}b_pc_qe_t \\ &=& \epsilon_{srt}\epsilon_{pqr}a_sb_pc_qe_t \\ &=& \epsilon_{rts}\epsilon_{rpq}a_sb_pc_qe_t \\ &=& (\delta_{tp}\delta_{sq} - \delta_{tq}\delta_{sp})a_sb_pc_qe_t \\ &=& \delta_{tp}\delta_{sq}a_sb_pc_qe_t - \delta_{tq}\delta_{sp}a_sb_pc_qe_t \\ &=& a_sb_tc_se_t - a_sb_sc_te_t \\ &=& \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b}) \end{eqnarray*}$

Exploring Quantum Physics - More about cross product

In this post, I want to summarize more about the cross product. We will first prove a simple fact about dot product.

It is well known that $\vec{a} \cdot \vec{b} = |a| |b| \cos \theta$ where $\theta$ is the angle between the two vectors, but why?

To see that, note that the dot product can be written as a matrix product $x^Ty$ where vectors are written as column. Now we insert $R^TR = I$ into the product with $R$ being a rotation (i.e. orthogonal), we have $x^Ty = x^TR^TRy = (Rx)^TRy$, that mean dot product is rotational invariant.

Now we scale all vectors to have unit length and rotate them such that they both lie on the same XY plane with $\vec{x}$ being parallel to the $x$ axis, it is obvious that it is possible, then the rest is obvious. The dot product is simply the length of the x projection of the other vector, which is exactly $\cos \theta$! Mapping it back to the original problem, multiplying back the scale, we get the formula we want.

The next interesting thing is to prove the also familar identity $\vec{a} \times \vec{b} = |a||b|\sin \theta$. To that end, it is easier to prove $|\vec{a} \times \vec{b}|^2 = |a||b| - (\vec{a} \cdot \vec{b})^2$, as this will automatically implies the result. To prove this one, however, is easy too as one only need to multiply out all the terms, detail skipped.

Exploring Quantum Mechanics - Cross Product

We are going to look into angular momentum soon. Before that, let's review the vector cross product.

The definition of the cross product is as follow

$\begin{eqnarray*} \vec{x} \times \vec{y} &=& \left|\begin{array}{ccc}i & j & k\\x_1 & x_2 & x_3\\y_1 & y_2 & y_3\end{array}\right| \\ &=& (x_2 y_3 - x_3 y_2) e_1 + (x_3 y_1 - x_1 y_3 ) e_2 + (x_1 y_2 - x_2 y_1) e_3 \end{eqnarray*}$

In this form, we can easily seen we can write it as a summation of these indices

$\begin{eqnarray*} \vec{x} \times \vec{y} &=& \sum\limits_{i,j,k = 0}^{3}{f(i, j, k)x_i y_j e_k} \\ \end{eqnarray*}$

i	j	k	f(i,j,k)
1	2	3	1
1	3	2	-1
2	1	3	-1
2	3	1	1
3	1	2	1
3	2	1	-1
otherwise			0

It is easy to verify that this is true, therefore we have $\vec{x} \times \vec{y} = \sum\limits_{i,j,k = 0}^{3}{\epsilon_{ijk}x_iy_je_k}$, where $\epsilon_{ijk}$ is the Levi-Civita symbol. We can even simplify it to just $\epsilon_{ijk}x_iy_je_k$, using the Einstein's summation notation.

As a simple sum, differentiate it will be easy. Suppose we have the vectors as a function of time, then we have

$\begin{eqnarray*} \frac{d}{dt} \vec{x} \times \vec{y} &=& \frac{d}{dt}(\epsilon_{ijk}x_i(t)y_j(t)e_k) \\ &=& \epsilon_{ijk}(x_i'(t)y_j(t) + x_i(t)y_j'(t))e_k \\ &=& \epsilon_{ijk}x_i'(t)y_j(t)e_k + \epsilon_{ijk}x_i(t)y_j'(t)e_k \\ &=& \vec{x'} \times \vec{y} + \vec{x} \times \vec{y'} \end{eqnarray*}$

Exploring Quantum Physics - Solving the Quantum Harmonic Oscillator using the series method

Disclaimer - this is just an attempt - it never finishes. Blogging this so that I don't forget where I were:
Recall the Quantum Harmonic Oscillator problem as follow:

$\begin{eqnarray*} \hat{H}\psi &=& E\psi \\ (\frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2x^2)\psi &=& E\psi \\ (\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + \frac{1}{2}m\omega^2x^2)\psi &=& E\psi \\ \end{eqnarray*}$

The so called series method is basically assume $\psi$ as an analytic function, as such, it can be (locally) represented as a power series.

$\begin{eqnarray*} \psi(x) &=& \sum\limits_{k=0}^{\infty}{a_kx^k} \\ \end{eqnarray*}$

Substituting it back to the problem, we have

$\begin{eqnarray*} E\psi &=& (\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + \frac{1}{2}m\omega^2x^2)\psi \\ E(\sum\limits_{k=0}^{\infty}{a_kx^k}) &=& (\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + \frac{1}{2}m\omega^2x^2)(\sum\limits_{k=0}^{\infty}{a_kx^k}) \\ \end{eqnarray*}$

So we simply match the coefficients.
$\begin{eqnarray*} Ea_0 &=& \frac{-\hbar^2}{2m}(2)a_2 \\ Ea_1 &=& \frac{-\hbar^2}{2m}(3)(2)a_3 \\ Ea_k &=& \frac{-\hbar^2}{2m}(k+1)(k+2)a_{k+2} + \frac{1}{2}m\omega^2x^2(a_{k-2}) \\ \end{eqnarray*}$

This apparently lead to the recurrence that we can use to solve for $a_k$, but I am stuck on the next step that I can foresee, how do I know I have found the Gaussian?

Exploring Quantum Physics - Week 5 Bonus Question

Question:

Prove the Ehrenfest theorem.

Solution:

I could have write cool equations like this

$\frac{d}{dt}\int{\psi^* \hat{A} \psi} = \int{\frac{d}{dt}\psi^* \hat{A} \psi + \psi^* \frac{d\hat{A}}{dt} \psi + \psi^* \hat{A} \frac{d}{dt}\psi }$ and then use the relation $\hat{H} = ih\frac{d}{dt}$ to present a 'proof', just like the wikipedia page does.

The key difficulty is defining the derivative of an operator, otherwise the right hand side of the theorem just cannot be understood.

There is nothing magic in it, except this. The operator itself is a function of $t$ in order for the differentiation to make sense

So we go back to the first principle and define the derivative of an operator to be a limit.

$\frac{d\hat{A(t)}}{dt}\psi = \lim\limits_{\Delta t \to 0}{\frac{\hat{A(t + \Delta t)}\psi - \hat{A(t)}\psi}{\Delta t}}$

Then we can proof the product rule on operator differentiation just like one would prove it in analysis.

$\begin{eqnarray*} \frac{d}{dt}{\hat{A(t)}\psi(t)} &=& \lim\limits_{\Delta t \to 0}{\frac{\hat{A(t + \Delta t)}\psi(t + \Delta t) - \hat{A(t)}\psi(t)}{\Delta t}} \\ &=& \lim\limits_{\Delta t \to 0}{\frac{\hat{A(t + \Delta t)}\psi(t + \Delta t) - \hat{A(t + \Delta t)}\psi(t) + \hat{A(t + \Delta t)}\psi(t) - \hat{A(t)}\psi(t)}{\Delta t}} \\ &=& \hat{A(t)}\frac{d\psi(t)}{dt} + \frac{d\hat{A(t)}}{dt}\psi(t) \end{eqnarray*}$

The rest should follows ...

Exploring Quantum Physics - Week 5 Question 7, 8, 9

Question:

... snip ...
Compute the energies of a particle under a field of constant force towards the center using Bohr's model.

Solution:

The particle is going in circular motion, so it is running on a particular plane, choose coordinate system so that the plane is the xy plane (i.e. z-coordinate is always zero), the trajectory of the particle can then be simplified to $(r \cos \omega t, r \sin \omega t)$. The velocity is the first derivative $(-r \omega \sin \omega t, r \omega\cos \omega t)$, and the acceleration would be the second derivative $(-r \omega^2 \cos \omega t, -r \omega^2 \sin \omega t)$

Now let's work on the magnitudes, $|v| = r \omega$, $|a| = r\omega^2$.

The total energy is given by

$\begin{eqnarray*} E & = & \frac{p^2}{2m} + Fr \\ & = & \frac{(mv)^2}{2m} + mar \\ & = & \frac{mv^2}{2} + mar \\ & = & \frac{m(r\omega)^2}{2} + m(r\omega^2)r \\ & = & \frac{m(r\omega)^2}{2} + m(r\omega)^2 \\ & = & \frac{3m(r\omega)^2}{2} \end{eqnarray*}$

So we need to find $r \omega$. The Bohr's model give us $mr^2\omega = m(r\omega)r = mvr = n\hbar$, and the Newton's second law gives us $F = ma = mr\omega^2$. A little arithmetic trick gives

$\begin{eqnarray*} (r\omega)^3 &=& (r^2\omega)(r\omega^2) \\ &=& \frac{mr^2\omega}{m}\frac{mr\omega^2}{m} \\ &=& \frac{n\hbar}{m}\frac{F}{m} \\ &=& \frac{n\hbar F}{m^2} \\ (r\omega) &=& (\frac{n\hbar F}{m^2})^{1/3} \end{eqnarray*}$

Substitute this back to the equation we get the answer

$\begin{eqnarray*} E & = & \frac{3m(r\omega)^2}{2} \\ & = & \frac{3m((\frac{n\hbar F}{m^2})^{1/3})^2}{2} \\ & = & \frac{3}{2}m(\frac{n\hbar F}{m^2})^{2/3} \\ & = & \frac{3}{2}(m^{3/2}\frac{n\hbar F}{m^2})^{2/3} \\ & = & \frac{3}{2}(\frac{n\hbar F}{\sqrt{m}})^{2/3} \\ & = & \frac{3}{2}n^{2/3} (\frac{\hbar F}{\sqrt{m}})^{2/3} \end{eqnarray*}$

This formula solved question 7, 8, 9 as follow:

Question 7 requires the quantity that is a factor that is independent of $n$, that would be $(\frac{\hbar F}{\sqrt{m}})^{2/3}$.

Question 8 ask for the number part of the ground state, that would be $\frac{3}{2}1^{2/3} = 1.5$.

Question 9 ask for the exponent, so it is $\frac{2}{3}$.

Exploring Quantum Physics - Week 5 Question 6

Question:

Computing the Gaussian Variational Estimate for the ground state of this potential.

$V(x) = - V_0 a \delta(x)$

Solution:

First, let's compute the expected energy for the Gaussian 'solution'.

$\begin{eqnarray*} \psi_d(x) &=& \frac{1}{\pi^{1/4}\sqrt{d}} \exp\left[-\frac{x^2}{2d^2}\right] \\ E_d &=& \langle \psi_d(x) | \hat{H}|\psi_d(x)\rangle \\ &=& \int\limits_{-\infty}^{\infty}{\psi_d^*(x)(\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi_d(x) + V(x) \psi_d(x))dx} \\ &=& \int\limits_{-\infty}^{\infty}{\psi_d(x)(\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi_d(x) + V(x) \psi_d(x))dx} \\ &=& \int\limits_{-\infty}^{\infty}{\psi_d(x)(\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi_d(x))dx} + \int\limits_{-\infty}^{\infty}{\psi_d(x)(V(x) \psi_d(x))dx} \\ &=& \frac{-\hbar^2}{2m}\int\limits_{-\infty}^{\infty}{\psi_d(x)\frac{\partial^2}{\partial x^2}\psi_d(x)dx} + \int\limits_{-\infty}^{\infty}{(V(x) \psi_d^2(x))dx} \\ \end{eqnarray*}$

To compute that, let's start with the second term.

$\begin{eqnarray*} \int\limits_{-\infty}^{\infty}{(V(x) \psi_d^2(x))dx} &=& \int\limits_{-\infty}^{\infty}{(-V_0a\delta(x) \psi_d^2(x))dx} \\ &=& -V_0a \psi_d^2(0) \\ &=& -V_0a (\frac{1}{\pi^{1/4}\sqrt{d}})^2 \\ &=& \frac{-V_0a}{\pi^{1/2}d} \end{eqnarray*}$

Next, we deal with the second derivative

$\begin{eqnarray*} \frac{\partial^2}{\partial x^2}\psi_d(x) &=& \frac{\partial^2}{\partial x^2} \frac{1}{\pi^{1/4}\sqrt{d}} \exp\left[-\frac{x^2}{2d^2}\right] \\ &=& \frac{1}{\pi^{1/4}\sqrt{d}} \frac{\partial^2}{\partial x^2} \exp\left[-\frac{x^2}{2d^2}\right] \\ &=& \frac{1}{\pi^{1/4}\sqrt{d}} \frac{\partial}{\partial x} \frac{-x}{d^2} \exp\left[-\frac{x^2}{2d^2}\right] \\ &=& \frac{-1}{\pi^{1/4}d^{5/2}} \frac{\partial}{\partial x} x \exp\left[-\frac{x^2}{2d^2}\right] \\ &=& \frac{-1}{\pi^{1/4}d^{5/2}} (\exp\left[-\frac{x^2}{2d^2}\right] + \frac{-x^2}{d^2} \exp\left[-\frac{x^2}{2d^2}\right]) \\ &=& \frac{-1}{\pi^{1/4}d^{5/2}} \exp\left[-\frac{x^2}{2d^2}\right](1 - \frac{x^2}{d^2}) \\ \end{eqnarray*}$

Now we substitute this back into the first term and see how it works out:

$\begin{eqnarray*} & & \frac{-\hbar^2}{2m}\int\limits_{-\infty}^{\infty}{\psi_d(x)\frac{\partial^2}{\partial x^2}\psi_d(x)dx} \\ &=& \frac{-\hbar^2}{2m}\int\limits_{-\infty}^{\infty}{(\frac{1}{\pi^{1/4}\sqrt{d}} \exp\left[-\frac{x^2}{2d^2}\right])\frac{-1}{\pi^{1/4}d^{5/2}} \exp\left[-\frac{x^2}{2d^2}\right](1 - \frac{x^2}{d^2})dx} \\ &=& \frac{\hbar^2}{2\pi^{1/2}md^{3}} \int\limits_{-\infty}^{\infty}{(\exp\left[-\frac{x^2}{2d^2}\right]) \exp\left[-\frac{x^2}{2d^2}\right](1 - \frac{x^2}{d^2})dx} \\ &=& \frac{\hbar^2}{2\pi^{1/2}md^{3}} \int\limits_{-\infty}^{\infty}{\exp\left[-\frac{x^2}{d^2}\right](1 - \frac{x^2}{d^2})dx} \\ \end{eqnarray*}$

Now it looks simpler, let's further simplify it further by letting $y = \frac{x}{d}$, so $d dy = dx$

$\begin{eqnarray*} & & \frac{\hbar^2}{2\pi^{1/2}md^{3}} \int\limits_{-\infty}^{\infty}{\exp\left[-\frac{x^2}{d^2}\right](1 - \frac{x^2}{d^2})dx} \\ &=& \frac{\hbar^2}{2\pi^{1/2}md^{3}} \int\limits_{-\infty}^{\infty}{\exp\left[-y^2\right](1 - y^2)ddy} \\ &=& \frac{\hbar^2}{2\pi^{1/2}md^{2}} \int\limits_{-\infty}^{\infty}{\exp\left[-y^2\right](1 - y^2)dy} \\ \end{eqnarray*}$

The integral value is $\frac{\sqrt{\pi}}{2}$. it can be seen as computing the total probability minus the variance of a standard Gaussian distribution with variance $\sqrt{\frac{1}{2}}$. So the final value of the first term is

$\begin{eqnarray*} & & \frac{\hbar^2}{2\pi^{1/2}md^{2}} \int\limits_{-\infty}^{\infty}{\exp\left[-y^2\right](1 - y^2)dy} \\ &=& \frac{\hbar^2}{2\pi^{1/2}md^{2}} \frac{\sqrt{\pi}}{2} \\ &=& \frac{\hbar^2}{4md^{2}} \end{eqnarray*}$

Now we combine them back together, the expected energy for the Gaussian 'solution' would be $\frac{\hbar^2}{4md^{2}} - \frac{V_0a}{\pi^{1/2}d}$. To minimize the quantity, we compute the derivative.

$\begin{eqnarray*} 0 &=& \frac{\partial}{\partial d} \frac{\hbar^2}{4md^{2}} - \frac{V_0a}{\pi^{1/2}d} \\ &=& \frac{-2\hbar^2}{4md^{3}} + \frac{V_0a}{\pi^{1/2}d^2} \\ &=& \frac{-2\hbar^2\pi^{1/2}}{4\pi^{1/2}md^{3}} + \frac{4mV_0ad}{4\pi^{1/2}md^3} \\ 4mV_0ad -2\hbar^2\pi^{1/2} &=& 0 \\ d &=& \frac{2\hbar^2\pi^{1/2}}{4mV_0a} \\ &=& \frac{\hbar^2\pi^{1/2}}{2mV_0a} \\ \end{eqnarray*}$

Last but not least, we just substitute this back to the energy to get the minimum possible energy for the 'solution'

$\begin{eqnarray*} & & \frac{\hbar^2}{4md^{2}} - \frac{V_0a}{\pi^{1/2}d} \\ &=& \frac{\hbar^2}{4m(\frac{\hbar^2\pi^{1/2}}{2mV_0a})^2} - \frac{V_0a}{\pi^{1/2}(\frac{\hbar^2\pi^{1/2}}{2mV_0a})} \\ &=& \frac{\hbar^2(2mV_0a)^2}{4m(\hbar^2\pi^{1/2})^2} - \frac{V_0a(2mV_0a)}{\pi^{1/2}(\hbar^2\pi^{1/2})} \\ &=& \frac{4m^2\hbar^2V_0^2a^2}{4m\hbar^4\pi} - \frac{2mV_0^2a^2}{\hbar^2\pi} \\ &=& \frac{mV_0^2a^2}{\hbar^2\pi} - \frac{2mV_0^2a^2}{\hbar^2\pi} \\ &=& - \frac{mV_0^2a^2}{\hbar^2\pi} \end{eqnarray*}$

Can you believe it all simplify to this? I am shocked now when I get this result! The final answer is to compare this with $E_0$ and get a numerical answer, which is $\frac{2}{\pi}$.

Exploring Quantum Physics - Week 5 Question 5

Question:

How does he state the dependence of $M_0$ upon $E$?

Solution:

In Bohr's paper, $M_0 = \frac{h}{2\pi}$, it has nothing to do with electron charge.

Exploring Quantum Physics - Week 5 Question 4

Question:

On p. 15 of Bohr’s paper there is a discussion of the quantization of the angular momentum, $M$, of the electron orbiting an atomic nucleus. Bohr states that $M$ must be an integral multiple of a fundamental quantity, $M_0$. Bohr states a numerical value of $M_0$ in an equation on that page. How
much does his number differ from the presently accepted value of that quantity?

Note that Bohr does not explicitly state on that page what system of units he is using!

Solution:

It is quite obvious that on page 15 Bohr mentioned the quantity is $1.04 \times 10^{-27}$. The challenge, really, is to determine the unit of that value Bohr meant.

Today's value should be $1.05 \times 10^{-34}$. There is a 10 million fold changes, so the changes should be accounted for by the unit of measurement.

After messing up around online searching, I found the Gaussian units (or cgs units). If we apply the units to the planck's constant, we will find agreement.

Energy = Force $\times$ distance = Mass $\times$ acceleration $\times$ distance = $(kgms^{-2})m = kgm^2s^{-2}$.

Therefore, the Planck constant has unit energy $\times$ time = $(kgm^2s^{-2})s  = kgm^2s^{-1}$. Now we substitute in the units using gram and cm instead, we will get a 1000 fold from the kg to gram conversion, and 10,000 fold from the $m^2$ to $cm^2$.

Wow - that explain the 10,000,000 fold of the value difference, now we see the values is surprisingly close, the relative error is less than 1%!

Exploring Quantum Physics - Week 5 Question 3

Question:

How many photons is required to excite a hydrogen atom for the following lasers?

Red R: 650 nm
Green G: 532 nm
Violet V: 405 nm

Solution:

We need to have $0.99R_\infty hc = E \le \frac{nhc}{\lambda}$, therefore $n \ge  \frac{0.99R_\infty hc \lambda}{hc} = 0.99R_\infty \lambda$.

Plugging the values in, we have minimum n = 8, 6, 5 respectively.

Exploring Quantum Physics - Week 5 Question 2

Question:

What is the maximum value of the wavelength of light that will ionize the hydrogen atom?
Take the energy of the ground state of the hydrogen atom to be $-0.99946650834 R_\infty h c \,$, where $R_\infty$ is the Rydberg constant and $c$ is the speed of light.

Solution:

Recall the potential energy of an item in a gravitational field is define to have 0 potential energy if and only if it is at infinitely far. I guess the same idea apply here when we say the energy of the ground state of the hydrogen atom is a negative number.

So we need a photon with energy just enough of make the overall energy 0.

A photon has energy $E = hf = \frac{hc}{\lambda}$. Therefore we have $\lambda = \frac{hc}{0.99R_\infty hc} = 9.11 \times 10^-{8}$, or 91 nanometers.