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Thursday, April 30, 2015

Exploring Quantum Physics - Week 5 Problem 1

Question:

What value does Einstein quote for Lenard’s measurement of the “largest effective wavelength for air”? Express your answer as an integer in units of nanometers.

Solution:

Reading Einstein's original paper at wikisource, I found a line:

"According to Lenard's measurements for air the largest wavelength that has an effect is about 1.9·10-5 cm, so"

So the answer is $ 1.9 \times 10^{-5} \times 10^{-2} = 1.9 \times 10^{-7} = 190 \times 10^{-9} $ meters, or in other words, 190 nanometers.

Wednesday, April 29, 2015

Exploring Quantum Physics - Mystery about the resolution of the identity solved!

I was confused with the resolution of the identity, it states that if $ | \psi_i \rangle $ is a basis, then

$ \sum\limits_{i=0}^{\infty}{|\psi\rangle\langle\psi_i|} = 1 $.

After reading Quantum mechanics demystified, finally understood the meaning of the above.

As $ \psi $ is an abstract vector, we should NOT think of it as a wavefunction, or a tuple, or whatever, instead, we define $|a\rangle\langle b| $ as an operator to mean $ |a\rangle\langle b|( |c\rangle) = |a\rangle(\langle b|c\rangle) $,

Now it suddenly become clear! The 1 in the above identity is not the real/complex number 1, but the identity operator that takes a vector to itself!


Sunday, April 26, 2015

Exploring Quantum Physics - Week 4 Bonus Question 5

Question:

Investigate the value of $ \Psi(x, \frac{T}{2}) $.

Solution:

Just like homework 3 bonus question, we have the exponential term reverse the sign of all the odd terms and leave the even terms alone.

The odd terms are odd functions, so $ \psi_{2n+1}(-x) = -\psi_{2n+1}(x) $.
The even terms are even functions, os $ \psi_{2n}(x) = \psi_2n(-x) $.

So we see by simply substituting -x, we get what we want!

The last thing to notice is that we need to compensate for the extracted out phase term (which I forget), therefore the end result is $ \Psi(x, \frac{T}{2}) = -\Psi(-x, 0) $.

Exploring Quantum Physics - Week 4 Bonus Question 4

Question:

Given

$\phi\left(x, t \right) = \sum_{n=0}^{\infty}a_n \psi_n\left(x\right)\exp{\left(-i E_n t/\hbar \right)} = \exp{\left(-i \omega t/2 \right)} \Phi\left(x,t\right) $

What is true about $ \Phi\left(x,t\right) $?

Solution:

First, let's look at the energies. $ |\phi\left(x, t \right)|^2 = |\exp{\left(-i \omega t/2 \right)} \Phi\left(x,t\right)|^2 = |\exp{\left(-i \omega t/2 \right)}|^2 |\Phi\left(x,t\right)|^2 = |\Phi\left(x,t\right)|^2 $, so the energies are unchanged while pulling a phase term out.

Also, the Schrödinger equation guarantees that the wavefunction stay normalized, so if the wavefunction was normalized, it will still be at any time.

To look at the periodicity of the term, we simply need to follow what we have done in homework 3 bonus question. The exponential terms looks like $ e^{-\frac{i\hbar E_n}{\hbar}} $, and we also have $ E_n = \hbar\omega(n + \frac{1}{2}) $. With the phase term pulled out, we are left with $  e^{-i\omega n} $, that means we have the periods to be $ \frac{2\pi}{\omega n} $, with the largest one being $ \frac{2\pi}{\omega} $, that would be the same as question 1.

Exploring Quantum Physics - Week 3 Bonus Question 3

Question:

What is true about $ \hat{I}\psi_7(x) $?

Solution:

$ \psi_7(x) $ is an odd function of $ x $, so these choices are correct

$ \hat{I}\psi_7(x) = \psi_7(-x) $
$ \hat{I}\psi_7(x) = -\psi_7(x) $
$ \hat{I}\psi_7(x) = (-1)^7\psi_7(x) $

Exploring Quantum Physics - Week 4 Bonus Question 2

Question:

What is true about $ \hat{I} \psi_0(x) $? The operator is defined as $ \hat{I} \psi(x) = \psi(-x) $.

Solution:

$ \psi_0(x) $ is an even function, so these are correct choices:

$ \hat{I} \psi_0(x) = \psi_0(-x) $.
$ \hat{I} \psi_0(x) = \psi_0(x) $.

Exploring Quantum Physics - Week 4 Bonus Question 1

Question:

Given $x(t) = \cos {\left(\omega t\right)} x\left(0\right) + \frac{\sin {\left(\omega t\right)}}{\omega} \dot x\left(0\right) $, which of the choices are correct?

Solution:

$ \omega T = 2\pi $, the function is periodic. The period does not depend on $ x(0) $ or $ x'(0) $.

Saturday, April 25, 2015

Exploring Quantum Physics - Week 4 Question 6

Question:

Which of the following is most closely related to Goldstone's theorem?

Solution:

Without going through all the lectures in lecture 8, yet, I guess the answer is this one.

The spontaneously broken translational symmetry of the ground state of crystal implies the existence of gapless excitations, which turn out to be phonons.

We will see how that goes ... lecture time!

Exploring Quantum Physics - Week 4 Question 5

Question:

In the context of the model of a crystal we considered in the lectures, why does the ground state of a crystal break continuous translation symmetry?

Solution:

This is the closest choice I can think of given the lecture video, I still don't really understand what is going on there.

A crystal has a periodic structure, but it is not uniform in space. Translating by some distance which is not a multiple of the lattice spacing gives a state which is different from that of the original state.

Exploring Quantum Physics - Week 4 Question 4

Question:

What is the wavefunction of the first excited state of the quantum harmonic oscillator, $|1\rangle$?

Solution:

The answer can easily be read from the Quantum Harmonic Oscillator page from Wikipedia. But since this is a blog of doing exercise, what is the point of copying the answer? Let us derive it from the ground state using the creation operator.

First, we have the ground state wavefunction $ \psi_0(x) = \left(\frac{m\omega}{\pi\hbar}\right)^\frac{1}{4}e^{-\frac{m\omega x^2}{2\hbar}} $.

Now we apply the creation operator $ \hat{a}^\dagger = \sqrt{\frac{m\omega}{2\hbar}}(\hat{x} - \frac{i\hat{p}}{m\omega}) = \sqrt{\frac{m\omega}{2\hbar}}(x - \frac{\hbar}{m\omega}\frac{\partial}{\partial x}) $.

Let's first compute the derivative:

$ \frac{\partial}{\partial x} \psi_0(x) = \left(\frac{m\omega}{\pi\hbar}\right)^\frac{1}{4}e^{-\frac{m\omega x^2}{2\hbar}}\frac{-m\omega x}{\hbar} $

So we have the momentum term:

$ - \frac{\hbar}{m\omega}\frac{\partial}{\partial x}\psi_0(x) = \left(\frac{m\omega}{\pi\hbar}\right)^\frac{1}{4}e^{-\frac{m\omega x^2}{2\hbar}}x = x\psi_0(x) $

The position term is of course also $ x\psi_0(x) $. Combining using the creation operator we have

$ \hat{a}^\dagger \psi(x) = \sqrt{\frac{m\omega}{2\hbar}}(2x\psi_0(x)) = \frac{1}{\sqrt{2}}\psi_0(x)H_1(\frac{m\omega x}{\hbar}) $.

$ H_1 $ is the physicist Hermite polynomial of order 1 which is simply $ H_1(x) = 2x $.

Friday, April 24, 2015

Exploring Quantum Physics - Week 4 Question 3

Question:

Calculate the expectation value of the kinetic, $\langle \hat T \rangle = \langle \frac{\hat p^2}{2m}\rangle$, and potential, $\langle \hat V\rangle = \langle \frac{1}{2}m\omega^2 \hat x^2\rangle$, energy operators in the $n$-th excited state of the harmonic oscillator, $|n\rangle$

Solution:

The problem hint on not required to do integration, which is great. In fact, there is a easy way to just pick the right answer, the sum of the energies must be $ \hbar \omega\left(n + \frac{1}{2}\right) $ and there is only one answer that matches this, which is good for sanity check.

Now let's actually computes the kinetic energy, in the previous problem, we have the momentum operator, so we just have to apply that twice.

$ \begin{eqnarray*} \hat{p} \psi_n &=& i\sqrt{\frac{m\hbar\omega}{2}}(\hat{a}^\dagger - \hat{a})\psi_n \\ &=& i\sqrt{\frac{m\hbar\omega}{2}}(\hat{a}^\dagger\psi_n - \hat{a}\psi_n) \\ \hat{p}^2 \psi_n &=& i\sqrt{\frac{m\hbar\omega}{2}}(\hat{a}^\dagger - \hat{a})(i\sqrt{\frac{m\hbar\omega}{2}}(\hat{a}^\dagger\psi_n - \hat{a}\psi_n)) \\ &=& \frac{m\hbar\omega}{2}(\hat{a} - \hat{a}^\dagger)(\hat{a}^\dagger\psi_n - \hat{a}\psi_n) \\ &=& \frac{m\hbar\omega}{2}(\hat{a}\hat{a}^\dagger\psi_n - \hat{a}\hat{a}\psi_n - \hat{a}^\dagger\hat{a}^\dagger\psi_n + \hat{a}^\dagger\hat{a}\psi_n) \\ &=& \frac{m\hbar\omega}{2}((1+\hat{a}^\dagger\hat{a})\psi_n - \hat{a}\hat{a}\psi_n - \hat{a}^\dagger\hat{a}^\dagger\psi_n + \hat{a}^\dagger\hat{a}\psi_n) \\ &=& \frac{m\hbar\omega}{2}(\psi_n - \hat{a}\hat{a}\psi_n - \hat{a}^\dagger\hat{a}^\dagger\psi_n + 2\hat{a}^\dagger\hat{a}\psi_n) \\ &=& \frac{m\hbar\omega}{2}(\psi_n - \hat{a}\hat{a}\psi_n - \hat{a}^\dagger\hat{a}^\dagger\psi_n + 2n\psi_n) \\ &=& \frac{m\hbar\omega}{2}((2n + 1)\psi_n - \hat{a}\hat{a}\psi_n - \hat{a}^\dagger\hat{a}^\dagger\psi_n) \\ &=& \frac{m\hbar\omega}{2}((2n + 1)\psi_n - C_1\psi_{n-2} - C_2\psi_{n+2}) \\ \frac{\hat{p}^2}{2m} \psi_n &=& i\sqrt{\frac{\hbar\omega}{4}}(\hat{a}^\dagger - \hat{a})(i\sqrt{\frac{m\hbar\omega}{2}}(\hat{a}^\dagger\psi_n - \hat{a}\psi_n)) \\ \end{eqnarray*} $

$ C_1 $ and $ C_2 $ are simply constant that I don't bother computing. Now we know the wavefunctions are orthonormal, so the final answer for the kinetic energy is $ \frac{\hbar\omega}{4}(2n+1) $.

Exploring Quantum Physics - Week 4 Question 2

Question:

Recalling our definition of the creation, $\hat a^\dagger$, and annihilation, $\hat a$, operators, how can we express the momentum operator in terms of these?

Solution:

Let's actually recall the operators as follow:

$ \hat{a} = \sqrt{\frac{m\omega}{2\hbar}} \hat{x} + \frac{i\hat{p}}{\sqrt{2m\hbar \omega}} $
$ \hat{a}^\dagger = \sqrt{\frac{m\omega}{2\hbar}} \hat{x} - \frac{i\hat{p}}{\sqrt{2m\hbar \omega}} $

The key idea is to take the difference of the two operators above, we have

$ \begin{eqnarray*} \hat{a} - \hat{a}^\dagger &=& \frac{2i\hat{p}}{\sqrt{2m\hbar \omega}} \\ \frac{-i}{2}\frac{2i\hat{p}}{\sqrt{2m\hbar \omega}} &=& \frac{-i}{2}(\hat{a} - \hat{a}^\dagger) \\ \frac{\hat{p}}{\sqrt{2m\hbar \omega}} &=& \frac{i}{2}(\hat{a}^\dagger - \hat{a}) \\ \sqrt{2}\frac{\hat{p}}{\sqrt{2m\hbar \omega}} &=& \sqrt{2}\frac{i}{2}(\hat{a}^\dagger - \hat{a}) \\ \frac{\hat{p}}{\sqrt{m\hbar \omega}} &=& \frac{i}{\sqrt{2}}(\hat{a}^\dagger - \hat{a}) \\ \hat{p} &=& i\frac{\sqrt{m\hbar \omega}}{\sqrt{2}}(\hat{a}^\dagger - \hat{a}) \\ &=& i\sqrt{\frac{m\hbar \omega}{2}}(\hat{a}^\dagger - \hat{a}) \\ \end{eqnarray*} $


Exploring Quantum Physics - Week 4 Question 1

Question:

Suppose we have a potential $V(x) = ax^4+bx^2+c$, with $a,b,c$ positive, and imagine that we are examining a low energy bound state of this system. Expressing the low energy approximation of this potential as a harmonic oscillator, what is its characteristic frequency, $\omega$?

Solution:

That would be a local Taylor approximation at the lowest energy.

First, let's get a feeling on how $ V(x) $ look like by computing some derivatives. We have:

$ \frac{dV}{dx} = 4ax^3 + 2bx = x(4ax^2 + 2b) $

$ \frac{d^2V}{dx^2} = 12ax^2 + 2b > 0 $.

Therefore, the function is convex, with the minimum achieved at 0, $ \sqrt{\frac{-2b}{4a}} $. So the only real minimum is at 0.

Approximating the potential there, we have $ V(x) \approx V(0) + \frac{V'(0)(x - 0)}{1!} + \frac{V''(0)(x-0)^2}{2!} = c + bx^2 $.

As with the typical prescription, we should have the equation

$ \hat{H} = \frac{\hat{p}^2}{2m} + b\hat{x}^2 + c $

Ignoring the constant part $ c $ and do some matching we have $ \frac{1}{2}m\omega^2x^2 = bx^2 $, that makes $ \omega = \sqrt{\frac{2b}{m}} $. 

Thursday, April 23, 2015

Exploring Quantum Physics - Week 3 Extra Credits Question 10

Question:

... snip ...

Given $\phi(x,T/4)= \exp{(-i\theta)}\left(\cos{(\theta)}\phi(x,0) + i \, \sin{(\theta)} \phi(x,T/2)\right)$ where $ \theta = \pi/k$ Find $ k $.

Solution:

Using the same approach we did in question 9, we found out that the exponential term becomes $ \exp(-in^2\frac{\pi}{2}) $ when $ t = \frac{T}{4} $.

Inspecting the values, we see what the term really do is that it brings odd terms to the imaginary part and even terms to the real parts.

Next, we look at what $ \theta $ can do for us on the left hand side.

For simplicity, let $ A = \phi(x,0) $ and $ B = \phi(x,\frac{T}{2}) $, the expression is now simplified to
$ \exp{(-i\theta)}(\cos{(\theta)}A + i \, \sin{(\theta)} B) = (\cos(\theta) - i\sin(\theta))(\cos{(\theta)}A + i \, \sin{(\theta)} B) = cos^2(\theta)A + \sin^2(\theta)B + i(\cos(\theta)\sin(\theta)B - \cos(\theta)\sin(\theta)A) $.

We already knew from question 9 that applying $ t = \frac{T}{2} $ inverting the odd terms and leaving the even terms alone.

Here is the key intuition, to get rid of odd terms in the real part, we make $ \cos^2(\theta) = \sin^2(\theta) = \frac{1}{2} $. That makes the real part simply $ \frac{1}{2}(A + B) $, which eliminates the odd terms and leave the even terms alone. So $ \theta = \frac{\pi}{4} $. The same $ \theta $ works for the imaginary part as well and therefore it is the answer!


Wednesday, April 22, 2015

Exploring Quantum Physics - Week 3 Extra Credit Question 9

Question

... snip ... 
Suppose $ \psi(x, \frac{T}{2}) = \alpha\psi(\beta x + \gamma L, 0) $, we are the values of $ \alpha $, $ \beta $ and $ \gamma $?

Solution:

We have $\phi(x, t) = \sum_{n=1}^{\infty} a_n \psi_n(x) \exp \left(-iE_n t/\hbar \right) $. Notice that the value of $ t $ only impacts the exponential term, so let's focus there.

Substitute in $ E_n = \frac{\pi^2 \hbar^2 n^2}{2mL^2} $ and $ T = \frac{8mL^2}{h} $, we have the exponential term at $ t = \frac{T}{2} $ equals to $ \exp \left(-i\frac{\pi^2 \hbar^2 n^2}{2mL^2} \frac{8mL^2}{h}/\hbar \right) =  \exp \left(-in^2\pi \right) = (-1)^n $.

The last equality comes from nowhere, really, it is just an observation on those values, now we know that by shifting time by $ \frac{T}{2} $, we are inverting the odd terms and keeping the even terms untouched.

Next we wanted to derive the values for $ \alpha $, $ \beta $ and $ \gamma $ to achieve the effect of inverting the odd terms and leaving the even terms untouched. The identity $ \sin(x + n\pi) = (-1)^n \sin(x) $ comes in really handy. It is really telling us that set $ \gamma $ to 1.

Thanks for the discussion forum's reminder, I should make sure the range I passed into $ \psi $ lies between $ 0 $ and $ L $, so it leaves me no choice but set $ \beta $ to -1 to maintain that constraint.

Last but not least, setting $ \beta $ to -1 inverted the terms unconditionally by 1, so we need to set $ \alpha $ to -1 to flip it back, so the answer is -1,-1,1

Exploring Quantum Physics - Week 3 Extra Credit Question 8

Question: 

... snip ...

What is the value of T for an electron in GaAs?

Solution:

It is a value hunting exercise. By one paper, we found out $ L = 175 \unicode{x212B} $. An $ \unicode{x212B} $ is a Angstrom, which is basically $ 10^{-10} $ meters.

On the effective mass of an electron in GaAs, another paper tell us the answer is $ 0.067 m_0 $, but what is $ m_0 $? Turn out $ m_0 $ is the electron mass and we can find it in the NIST values ,which is $ 9.10938291 \times 10^{-31} $ kg

To compute T, we will also need the Planck's constant, from the NIST website, it is $ 6.62606957 \times 10^{-34} $ J s.

So we compute $ T = \frac{8mL^2}{h} =2.2567 \times 10^{-13}$ s.
The last piece of the puzzle is a femtosecond, a femtosecond is basically $ 10^{-15} $ second, so the final answer is 225.67 femtoseconds.

Monday, April 20, 2015

Exploring Quantum Physics - Week 3 Extra Credit Question 7

Question:

... snip ...
What is the fundamental period of the wavefunctions?

Solution:

Notice that we have $\phi(x, t) = \sum_{n=1}^{\infty} a_n \psi_n(x) \exp \left(-iE_n t/\hbar \right)$, it is just a sum of complex exponentials, so let's take a look at their periods.

The periods are just $ \frac{2\pi}{E_n/\hbar} = \frac{2\pi\hbar}{E_n}$ and the energies are $ \frac{\pi^2\hbar^2n^2}{2mL^2} $, so can compute the periods as $ \frac{2\pi\hbar}{\frac{\pi^2\hbar^2n^2}{2mL^2}} = \frac{4mL^2}{\pi\hbar n^2} = \frac{8mL^2}{h n^2}$

Now we see the largest period happens when $ n = 1 $ and it is an integral multiple of all other periods, so $ \frac{8mL^2}{h} $ is the answer.

Exploring Quantum Physics - Week 3 Extra Credit Question 6

Question:

One of the correct choices in question 5 is associated with which of the following principles?

Solution:

I am uncertain about this one, but I would pick conservation of energy, and the average energy level does not change with time.

Exploring Quantum Physics - Week 3 Extra Credit Question 5 - Final Part 5

Reaching the end of Question 5 - we will conclude with the answer.

These are the correct choices:

  • $X(t) \leq L$
  • $P(t) = 0$
  • $\frac{\partial P^2(t)}{\partial t} = 0$
  • $\frac{\partial P(t)}{\partial t} = 0$
  • $\frac{\partial X(t)}{\partial t} = 0$

And this is not

  • $X(t) = 0$

Looking in retrospect, I don't have to compute all those moments. All of these can be argued without derivation at all. First of all, of course it makes sense the particle stay within the well. Second option is not obvious, but if the particle take an average drift, the probability density would not be stationary. The rest are just the fact that the probability density is stationary and therefore have no relation to $ t $. 

The wrong one is simply ridiculous. How could the particle has an average position of 0?




Sunday, April 19, 2015

Exploring Quantum Physics - Week 3 Extra Credit Question 5 - Part 4

Question 5 is a hard question, so we will deal with it in parts. In the fourth part we computes the expectation of momentum squared, it is similar to the previous calculation by not quite.

$ \begin{eqnarray*} \langle p^2 \rangle &=& \int\limits_{0}^{L}{\psi(x, t)^* i\hbar\frac{\partial}{\partial x}(i\hbar\frac{\partial}{\partial x} \psi(x, t)) dx} \\ &=& -\hbar^2\int\limits_{0}^{L}{\psi(x, t)^* \frac{\partial^2}{\partial x^2} \psi(x, t)) dx} \\ &=& -\hbar^2\int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}})^* \frac{\partial^2}{\partial x^2} (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}}) dx} \\ &=& -\hbar^2\int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{\frac{i E_n t}{\hbar}}) \frac{\partial^2}{\partial x^2} (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}}) dx} \\ &=& -\hbar^2\int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) \frac{\partial^2}{\partial x^2} (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) dx} \\ &=& -\hbar^2\int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) (-(\frac{n\pi}{L})^2) (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) dx} \\ &=& \frac{2n^2\hbar^2\pi^2}{L^3}\int\limits_{0}^{L}{\sin^2(\frac{n\pi x}{L}) dx} \\ \end{eqnarray*} $

We let $ y = \frac{2 n \pi x}{L} $, when $ x = 0 $, we have $ y = 0 $, when $ x = L $, we have $ y = 2 n \pi $, lastly $ dy = \frac{2 n \pi}{L} dx $.

$ \begin{eqnarray*} \langle p^2 \rangle &=& \frac{2n^2\hbar^2\pi^2}{L^3}\int\limits_{0}^{L}{\sin^2(\frac{n\pi x}{L}) dx} \\ &=& \frac{2n^2\hbar^2\pi^2}{L^3}\int\limits_{0}^{2n\pi}{\sin^2(y) \frac{L}{2n\pi}dy} \\ &=& \frac{n\hbar^2\pi}{L^2}\int\limits_{0}^{2n\pi}{\sin^2(y) dy} \\ &=& \frac{n\hbar^2\pi}{L^2}\int\limits_{0}^{2n\pi}{\frac{1 - \cos(2y)}{2} dy} \\ &=& \frac{n\hbar^2\pi}{2L^2}\int\limits_{0}^{2n\pi}{(1 - \cos(2y)) dy} \\ &=& \frac{n\hbar^2\pi}{2L^2}(y - \sin(2y)|_{0}^{2n\pi} \\ &=& \frac{n^2\hbar^2\pi^2}{L^2} \end{eqnarray*} $

Therefore the expectation of the momentum squared operator is $ \frac{n^2\hbar^2\pi^2}{L^2} $

Exploring Quantum Physics - Week 3 Extra Credit Question 5 - Part 3

Question 5 is a hard question, so we will deal with it in parts. In the third part we computes the expectation of momentum, it is similar to the previous calculation by not quite.

$ \begin{eqnarray*} \langle p \rangle &=& \int\limits_{0}^{L}{\psi(x, t)^* i\hbar\frac{\partial}{\partial x} \psi(x, t) dx} \\ &=& \int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}})^* i\hbar\frac{\partial}{\partial x} (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}}) dx} \\ &=& \int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{\frac{i E_n t}{\hbar}}) i\hbar\frac{\partial}{\partial x} (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}}) dx} \\ &=& \int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) i\hbar\frac{\partial}{\partial x} (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) dx} \\ &=& \int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) i\hbar\frac{n\pi}{L} (\sqrt{\frac{2}{L}}\cos(\frac{n\pi x}{L})) dx} \\ &=& \frac{2i\hbar n\pi}{L^2}\int\limits_{0}^{L}{\sin(\frac{n\pi x}{L})\cos(\frac{n\pi x}{L}) dx} \\ &=& \frac{2i\hbar n\pi}{L^2}\int\limits_{0}^{L}{\frac{1}{2}\sin(\frac{2n\pi x}{L}) dx} \\ &=& \frac{i\hbar n\pi}{L^2}\int\limits_{0}^{L}{\sin(\frac{2n\pi x}{L}) dx} \\ &=& \frac{i\hbar n\pi}{L^2}(\frac{L}{2n\pi})\cos(\frac{2n\pi x}{L})|_0^{L} \\ &=& 0 \end{eqnarray*} $

Therefore the expectation of the momentum operator is $ 0 $, that of course doesn't mean the particle is not moving, it merely means the average velocity is 0, that is, it is not drifting left or right.

Exploring Quantum Physics - Week 3 Extra Credit Question 5 - Part 2

Question 5 is a hard question, so we will deal with it in parts. In the second part we computes the expectation of position squared. This is almost identical to the previous post except a single difference in the power of $ x $.

$ \begin{eqnarray*} \langle x^2 \rangle &=& \int\limits_{0}^{L}{\psi(x, t)^* x^2 \psi(x, t) dx} \\ &=& \int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}})^* x^2 (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}}) dx} \\ &=& \int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{\frac{i E_n t}{\hbar}}) x^2 (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}}) dx} \\ &=& \int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) x^2 (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) dx} \\ &=& \frac{2}{L}\int\limits_{0}^{L}{x^2 \sin^2(\frac{n\pi x}{L}) dx} \\ &=& \frac{2}{L}\int\limits_{0}^{L}{x^2 \frac{1 - \cos(\frac{2n\pi x}{L})}{2} dx} \\ &=& \frac{1}{L}\int\limits_{0}^{L}{x^2 (1 - \cos(\frac{2n\pi x}{L})) dx} \\ &=& \frac{1}{L}\int\limits_{0}^{L}{x^2dx} - \frac{1}{L}\int\limits_{0}^{L}{x^2\cos(\frac{2n\pi x}{L})) dx} \\ &=& \frac{1}{L}(\frac{x^3}{3})|_{0}^{L} - \frac{1}{L}\int\limits_{0}^{L}{x^2\cos(\frac{2n\pi x}{L})) dx} \\ &=& \frac{L^2}{3} - \frac{1}{L}\int\limits_{0}^{L}{x^2\cos(\frac{2n\pi x}{L})) dx} \\ \end{eqnarray*} $

The remaining integral seems complicated, but we have done with the preparation in the previous post, to use that, we let $ y = \frac{2 n \pi x}{L} $, when $ x = 0 $, we have $ y = 0 $, when $ x = L $, we have $ y = 2 n \pi $, lastly $ dy = \frac{2 n \pi}{L} dx $.

$ \begin{eqnarray*} \langle x^2 \rangle &=& \frac{L^2}{3} - \frac{1}{L}\int\limits_{0}^{L}{x^2\cos(\frac{2n\pi x}{L})) dx} \\ &=& \frac{L^2}{3} - \frac{1}{L}\int\limits_{0}^{2n\pi}{(\frac{L y}{2 n \pi})^2\cos(y) \frac{L}{2 n \pi} dy} \\ &=& \frac{L^2}{3} - \frac{1}{L}(\frac{L}{2 n \pi})^3\int\limits_{0}^{2n\pi}{y^2 \cos(y) dy} \\ &=& \frac{L^2}{3} - \frac{1}{L}(\frac{L}{2 n \pi})^3(t^2 \sin t + 2t\cos t - 2 \sin t)|_0^{2n\pi} \\ &=& \frac{L^2}{3} - \frac{1}{L}(\frac{L}{2 n \pi})^3(4n\pi) \\ &=& L^2(\frac{1}{3} - \frac{2}{(2 n \pi)^2}) \end{eqnarray*} $

Therefore the expectation of the position squared operator is $ L^2(\frac{1}{3} - \frac{2}{(2 n \pi)^2}) $

Exploring Quantum Physics - Week 3 Extra Credits Question 5 - Part 1

Question 5 is a hard question, so we will deal with it in parts. Let's first compute the expectation of the position operator on the wavefunctions.

$ \begin{eqnarray*} \langle x \rangle &=& \int\limits_{0}^{L}{\psi(x, t)^* x \psi(x, t) dx} \\ &=& \int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}})^* x (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}}) dx} \\ &=& \int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{\frac{i E_n t}{\hbar}}) x (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}}) dx} \\ &=& \int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) x (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) dx} \\ &=& \frac{2}{L}\int\limits_{0}^{L}{x \sin^2(\frac{n\pi x}{L}) dx} \\ &=& \frac{2}{L}\int\limits_{0}^{L}{x \frac{1 - \cos(\frac{2n\pi x}{L})}{2} dx} \\ &=& \frac{1}{L}\int\limits_{0}^{L}{x (1 - \cos(\frac{2n\pi x}{L})) dx} \\ &=& \frac{1}{L}\int\limits_{0}^{L}{xdx} - \frac{1}{L}\int\limits_{0}^{L}{x\cos(\frac{2n\pi x}{L})) dx} \\ &=& \frac{1}{L}(\frac{x^2}{2})|_{0}^{L} - \frac{1}{L}\int\limits_{0}^{L}{x\cos(\frac{2n\pi x}{L})) dx} \\ &=& \frac{L}{2} - \frac{1}{L}\int\limits_{0}^{L}{x\cos(\frac{2n\pi x}{L})) dx} \\ &=& \frac{L}{2} - \frac{1}{L}\int\limits_{0}^{L}{x\cos(\frac{2n\pi x}{L})) dx} \\ \end{eqnarray*} $

The remaining integral seems complicated, but we have done with the preparation in the previous post, to use that, we let $ y = \frac{2 n \pi x}{L} $, when $ x = 0 $, we have $ y = 0 $, when $ x = L $, we have $ y = 2 n \pi $, lastly $ dy = \frac{2 n \pi}{L} dx $.

$ \begin{eqnarray*} \langle x \rangle &=& \frac{L}{2} - \frac{1}{L}\int\limits_{0}^{L}{x\cos(\frac{2n\pi x}{L})) dx} \\ &=& \frac{L}{2} - \frac{1}{L}\int\limits_{0}^{2n\pi}{\frac{L y}{2 n \pi}\cos(y) \frac{L}{2 n \pi} dy} \\ &=& \frac{L}{2} - \frac{1}{L}(\frac{L}{2 n \pi})^2\int\limits_{0}^{2n\pi}{y \cos(y) dy} \\ &=& \frac{L}{2} - \frac{1}{L}(\frac{L}{2 n \pi})^2(y\sin y - \cos y)|_0^{2n\pi} \\ &=& \frac{L}{2} \end{eqnarray*} $

Therefore the expectation of the position operator is $ \frac{L}{2} $, not a surprising result.

Differential Geometry - rational parameterization of the Cissoid of Diocles (1)

Question:

Find a rational parameterization of the Cissoid of Diocles.

The curve is defined as follow.

Draw a circle of radius 0.5 centered at (0, 0.5).
Draw a line y = 1
Shoot a ray from the origin, measure the distance between the intersection point with the circle and the intersection point with the line.
Compute the point with that length from the origin lying of the same ray.
The locus is the Cissoid of Diocles.

Solution:

We know a rational parameterization of the circle through the t-substitution.

$ \begin{eqnarray*} \tan^2 \theta &=& \frac{\sin^2 \theta}{\cos^2 \theta} \\ 1 + \tan^2 \theta &=& 1 + \frac{\sin^2 \theta}{\cos^2 \theta} \\ &=& \frac{\cos^2 \theta}{\cos^2 \theta} + \frac{\sin^2 \theta}{\cos^2 \theta} \\ &=& \frac{1}{\cos^2 \theta} \\ \cos^2 \theta &=& \frac{1}{1 + \tan^2\theta} \\ 2\cos^2 \theta - 1 &=& 2 \frac{1}{1 + \tan^2\theta} - 1 \\ \cos 2\theta &=& 2 \frac{1}{1 + \tan^2\theta} - \frac{1 + \tan^2\theta}{1 + \tan^2\theta} \\ &=& \frac{1 - \tan^2 \theta}{1 + \tan^2\theta} \\ \sqrt{1 - \cos^2 2\theta} &=& \sqrt{1 - (\frac{1 - \tan^2 \theta}{1 + \tan^2\theta})^2} \\ \sin 2\theta &=& \sqrt{(\frac{1 + \tan^2 \theta}{1 + \tan^2\theta})^2 - (\frac{1 - \tan^2 \theta}{1 + \tan^2\theta})^2} \\ &=& \sqrt{(\frac{2 \tan \theta}{1 + \tan^2\theta})^2} \\ &=& \frac{2 \tan \theta}{1 + \tan^2\theta} \\ \end{eqnarray*} $

The key above is that we have a square root of a perfect square above, so we can get to the rational parameterization of the circle. With careful scaling and shifting, we have $ c_x $ and $ c_y $ in the below code represents the circle. To compute the intersection of the ray with the line, we compute the slope $ m $ of the ray, and then since the line has $ y $ coordinate 1, it must have $ x $ coordinate $ l_x = \frac{1}{m} $. Finally we construct $ d_x = l_x - c_x $, that will create the $ x $ coordinate of the point on the Cissoid of Diocles, $ d_y $ is simply $ m d_x $.

Last but not least, the key is to simplify it not manually, but automatically using MATLAB, here is the code for the purpose.

syms t;
cx = 0.5*(1 - t*t)/(1 + t*t);
cy = t/(1 + t*t) + 0.5;
m  = cy/cx;
lx = 1/m;
dx = lx - cx;
dy = m * dx;

Now it gives our answer: $ x = \frac{(1-t)^3}{2(1+t^2)(1+t)} $ and $ y = \frac{(1-t)^2}{2(1+t^2)} $

Exploring Quantum Physics - computing a few trigonometric integrals

As part of Exploring Quantum Physics - I need to evaluate this integral. Turn out an exercise I did sometimes ago give this as a special case.

An intermediate result of that exercise is this:

$ \begin{eqnarray*} \int{t^ne^{-at}dt} &=& (\frac{1}{a})^n n! \frac{e^{-at}}{-a} -\frac{1}{a}e^{-at}\sum\limits_{k=0}^{n-1} \frac{n!t^{n-k}}{a^k(n-k)!} \end{eqnarray*} $

Given this result, we substitute $ a = -i $ , we can simplify this to:

$ \begin{eqnarray*} \int{t^ne^{it}dt} &=& (\frac{1}{-i})^n n! \frac{e^{it}}{i} -\frac{1}{-i}e^{it}\sum\limits_{k=0}^{n-1} \frac{n!t^{n-k}}{(-i)^k(n-k)!} \\ &=& i^{n-1} n! e^{it} + e^{it}\sum\limits_{k=0}^{n-1} i^{k+3}\frac{n!t^{n-k}}{(n-k)!} \end{eqnarray*} $

In particular, we are interested in $ n = 1 $ and $ n = 2 $.

$ \begin{eqnarray*} \int{t^1e^{it}dt} &=& i^{1-1} 1! e^{it} + e^{it}\sum\limits_{k=0}^{1-1} i^{k+3}\frac{1!t^{1-k}}{(1-k)!} \\ &=& e^{it} + e^{it} i^{0+3}\frac{1!t^{1-0}}{(1-0)!} \\ &=& e^{it} - ite^{it} \\ &=& e^{it}(-it + 1) \\ \int{t^2e^{it}dt} &=& i^{n-1} n! e^{it} + e^{it}\sum\limits_{k=0}^{n-1} i^{k+3}\frac{n!t^{n-k}}{(n-k)!} \\ &=& i^{2-1} 2! e^{it} + e^{it}\sum\limits_{k=0}^{2-1} i^{k+3}\frac{2!t^{2-k}}{(2-k)!} \\ &=& 2 i e^{it} + e^{it}(i^{0+3}\frac{2!t^{2-0}}{(2-0)!} + i^{1+3}\frac{2!t^{2-1}}{(2-1)!}) \\ &=& 2 i e^{it} + e^{it}(-it^2 + 2t) \\ &=& e^{it}(-it^2 + 2t + 2i) \end{eqnarray*} $

Using Euler's identity we can now write:

$ \begin{eqnarray*} \int{te^{it}dt} &=& e^{it}(-it + it) \\ \int{t(\cos t + i\sin t)dt} &=& (\cos t + i\sin t)(-it + 1) \\ &=& (t\sin t + \cos t) + i(-t\cos t + \sin t) \\ \int{t^2e^{it}dt} &=& e^{it}(-it^2 + 2t + 2i) \\ \int{t^2(\cos t + i\sin t)dt} &=& (\cos t + i\sin t)(-it^2 + 2t + 2i) \\ &=& (t^2 \sin t + 2t \cos t - 2 \sin t) + i(-t^2 \cos t + 2t \sin t + 2 \cos t) \\ \end{eqnarray*} $

Equating the real parts and imaginary parts we get these equations:

  • $ \int{t \cos t dt} = t\sin t + \cos t $
  • $ \int{t \sin t dt} = -t \cos t + \sin t $
  • $ \int{t^2 \cos t dt} = t^2 \sin t + 2t \cos t - 2 \sin t $
  • $ \int{t^2 \sin t dt} = -t^2 \cos t + 2t \sin t + 2 \cos t $


Friday, April 17, 2015

Exploring Quantum Physics - Week 3 Extra Credit Question 4

Question:

The correct answer to Question 3 reflects which of the following principles?

Solution:

Conservation of probability i.e. $ \int\limits_{0}^{L}{ \phi^*(x,t)\phi(x,t) dt} = 1 $ for all times $ t $.
In fact, we derived the result based on that.

Exploring Quantum Physics - Week 3 Extra Credit Question 3

Question:

... snip ...

Given $ \phi(x, t) = \sum_{n=1}^{\infty} a_n \psi_n(x) \exp \left(-iE_n t/\hbar \right) $, what must be true?

Solution:

This is the Parseval's theorem. Since $ \phi(x, t)^* \phi(x, t) $ represents a probability, we have $ \int\limits_{-\infty}{^\infty}{\phi(x, t)^* \phi(x, t) dx} = 1 $. Substitute in the definition, we have:

$ \begin{eqnarray*} 1 &=& \int\limits_{-\infty}^{\infty}{\phi(x, t)^* \phi(x, t) dx} \\ &=& \int\limits_{-\infty}^{\infty}{(\sum_{n=1}^{\infty} a_n^* \psi_n(x) \exp \left(iE_n t/\hbar \right))(\sum_{n=1}^{\infty} a_n \psi_n(x) \exp \left(-iE_n t/\hbar \right)) dx} \\ &=& \int\limits_{-\infty}^{\infty}{\sum_{p=1}^{\infty}{\sum_{q=1}^{\infty}{a_p^* a_q \exp \left(iE_p t/\hbar \right) \exp \left(-iE_q t/\hbar \right) \psi_p(x) \psi_q(x)}} dx} \\ &=& \sum_{p=1}^{\infty}{\sum_{q=1}^{\infty}{a_p^* a_q \exp \left(iE_p t/\hbar \right) \exp \left(-iE_q t/\hbar \right) \int\limits_{-\infty}^{\infty}{\psi_p(x) \psi_q(x) dx}}} \\ &=& \sum_{p=1}^{\infty}{\sum_{q=1}^{\infty}{a_p^* a_q \exp \left(iE_p t/\hbar \right) \exp \left(-iE_q t/\hbar \right) \delta_{pq}}} \\ &=& \sum_{p=1}^{\infty}{|a_p|^2 \exp \left(iE_p t/\hbar \right) \exp \left(-iE_p t/\hbar \right)} \\ &=& \sum_{p=1}^{\infty}{|a_p|^2} \\ \end{eqnarray*} $

Therefore the answer is $ \sum_{p=1}^{\infty}{|a_p|^2} = 1 $.

Exploring Quantum Physics - Week 3 Extra Credit Question 2

Question:

Which of these is the correct expression for the nth eigenenergy $ E_n $ that appears in the figure above, where n=1,2,3,... is an index that labels the energies in increasing order?

Solution

I struggled on this quite a bit because the wave-function is not dependent on time. This is strange to me at first and I finally figured out that we are really abusing the term wavefunction here. Sometimes the position component of the full wavefunction is also called a wavefunction and leading to confusion. The full wave function also contains a term that with exponential dependence to time. But for that part, the energy is in the exponential term and eventually you will only get $ E = E $ as a trivial and useless result. The key to this is solving the problem based on the momentum operator instead.

$ \begin{eqnarray*} \hat{E} &=& \frac{\hat{p}^2}{2m} \\ &=& \frac{(-i\hbar\frac{\partial}{\partial x})^2}{2m} \\ &=& \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \\ \hat{E} \psi(x) &=& \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}(\sqrt{\frac{2}{L}})\sin(\frac{2 n \pi x}{L}) \\ &=& \frac{\hbar^2}{2m}(\frac{2n\pi}{L})^2(\sqrt{\frac{2}{L}})(-\sin(\frac{2 n \pi x}{L})) \\ \end{eqnarray*} $

So the energies are given by: $ \frac{\hbar^2}{2m}(\frac{2n\pi}{L})^2 = \frac{\pi^2\hbar^2n^2}{2mL^2} $.

Exploring Quantum Physics - Week 3 Extra Credit Question 1

Question:

... Snip ...

What is the coefficient $ N $ for the equation $ \psi_n(x) = N\sin(\frac{n \pi x}{L}) $.

Solution:

We will do that by normalizing the wavefunction.

$ \begin{eqnarray*} 1 &=& \int\limits_{0}^{L}{(N\sin(\frac{n \pi x}{L}))^2dx} \\ &=& N^2\int\limits_{0}^{L}{\sin^2(\frac{n \pi x}{L})dx} \\ &=& N^2\int\limits_{0}^{L}{\frac{1 - \cos(\frac{2 n \pi x}{L})}{2}dx} \\ &=& N^2\frac{L}{2} \\ N &=& \sqrt{\frac{2}{L}} \end{eqnarray*} $

So the answer is $ \sqrt{\frac{2}{L}} $.

Exploring Quantum Physics - Week 3 Bonus Question 2

Question:

Prove that the only eigenvalues of the parity operator, $ \hat{I} $, are 1 and −1.

Solution:

Recall that $ \hat{I}(f(x)) = f(-x) $, and let $ \lambda $ be its eigenvalue, so we have:

$ \begin{eqnarray*} f(x) &=& \hat{I}(f(-x)) \\ &=& \lambda f(-x) \\ &=& \lambda (\hat{I}(f(x))) \\ &=& \lambda (\lambda f(x)) \\ 1 &=& \lambda^2 \end{eqnarray*} $

So the only eigenvalues are 1 and -1.

Exploring Quantum Physics - Week 3 Bonus Question 1

Question:

Recall that in Lecture 5, Part II, we derived an energy quantization condition for the even parity bound states of a finite potential well of width a and depth $U_0 $. What is the corresponding condition for the odd parity states in terms of the dimensionless variables $ x=\frac{ka}{2} $ and $ \xi^2 = \frac{ma^2U_0}{2\hbar^2} $?

Solution:

The derivation of the odd parity solution is almost exactly the same as the even version, so here it is:

$ \begin{eqnarray*} \tilde{C}\sin(\frac{ka}{2}) &=& Ae^{-\gamma x} \\ \tilde{C}k\cos(\frac{ka}{2}) &=& -\gamma Ae^{-\gamma x} \\ k\cot(\frac{ka}{2}) &=& -\gamma \\ \cot(\frac{ka}{2}) &=& -\frac{\gamma}{k} \\ \cot(\frac{ka}{2}) &=& -\sqrt{\frac{\gamma^2}{k^2}} \\ &=& -\sqrt{\frac{\frac{2m}{\hbar^2}(U_0 - E)}{\frac{2m}{\hbar^2}E}} \\ &=& -\sqrt{\frac{U_0 - E}{E}} \\ &=& -\sqrt{\frac{U_0}{E} - 1} \\ &=& -\sqrt{(\frac{\xi}{x})^2 - 1} \\ \end{eqnarray*} $

So this is the answer!

Thursday, April 16, 2015

Exploring Quantum Physics - Week 3 Question 8

Question:

In lecture 6, part III, we showed that the two-particle wave function for particles with center of mass position $\vec{R}$, relative separation $\vec{r}$, and central potential $V(\vec{r})$, in a state of energy $E$, can be written as $\Psi(\vec R, \vec r)=e^{\frac{i}{\hbar}\vec{P}\cdot \vec{R}} \psi(\vec r)$ where $\psi(\vec{r})$ obeys the single particle Schrödinger equation:
$\left(-\frac{\hbar^2 \nabla^2}{2\mu} + V(\vec{r})\right)\psi(\vec r) = E' \psi(\vec r)$
with $\mu = \frac{m_1 m_2}{m_1+m_2}$ being the reduced mass, and $E'=E-\frac{\vec P^2}{2(m_1+m_2)}$.
Suppose we have two identical, free ($V(\vec r) = 0$), fermions of spin $\frac{1}{2}$ (just for concreteness) and mass $m$. Which of the following is a valid two-particle wavefunction $\Psi(\vec R, \vec r)$ for a state of energy $E$, where $\vec p$ is a vector of magnitude $|\vec p| = \sqrt{m E'}$?
Assume that the spin part of the wavefunction is symmetric or more simply that the scalar wave function $\psi$ determines the symmetry properties under exchange.

Solution:

It appears the problem is quite complicated but it really isn't. The method in lecture already tell us the form of the solution, and the fact that the particles are fermions means they are bounded by Pauli exclusion principle that the probability of finding them with together at the same place is 0, so the only plausible solution is $\Psi(\vec R, \vec r) = \exp\left(\frac{i}{\hbar} \vec P \cdot \vec R\right) \sin(\frac{1}{\hbar} \vec p \cdot \vec r)$

Exploring Quantum Physics - Week 3 Question 7

Question:

Why might one expect that superconductivity could be achieved by pairing electrons in a metal?

Solution:

Pairing the electrons can makeup a total spin of 1, which is the spin for a Boson. Boson is not restricted by the Pauli exclusion principle and therefore might be able to achieve super conductivity.

We know that bosons can condense to their ground state and form a superfluid which allows for dissipation-less flow. One can pair electrons to form bosons.

Truth to be told - I am not sure how the fact that electron cannot occupy the same state has anything to do with the friction blocking an electron to move or the chance of colliding an atom nucleus. sounds like a good question to ask. 

Exploring Quantum Physics - Week 3 Question 6

Question:

Which of the following is responsible for the existence of a Fermi surface in metals?

Answer:

First, electron, like any particle, like low energy state, and Pauli exclusion principle dictates that each energy level (i.e. state) can have only one electron. So we should expect to find electrons to clutter at the lowest possible level.

There the answer is Pauli exclusion principle.

Exploring Quantum Physics - Week 3 Question 5

Question:

Recall that in the lecture 5 we found the Fourier transform of the wavefunction for a bound state of the delta function potential, $ V(x)=−\alpha \delta(x) $, in 1 dimension. Our result was $ \tilde{\phi}_k=\frac{\alpha\psi(0)}{\frac{\hbar^2 k^2}{2m}−E} $, where in the bound state, $ E = −\frac{m\alpha^2}{2\hbar^2} $. What is the wavefunction of the bound state in real space, $ \psi(x) $? (Don't forget to normalize the wavefunction. You may take $ \psi(0) $ to be real and positive.)

Solution:

It is pretty obvious that the question is asking for the inverse Fourier transform of the function. Working backwards from the solution, I need to prove two lemmas about Fourier transform for myself, first, let look at this Fourier transform pair with $ a < 0 $.

$ \begin{eqnarray*} \mathcal{F}(e^{a|t|}) &=& F(\omega) \\ &=& \int\limits_{-\infty}^{\infty}{e^{a|t|}e^{-i\omega t}dt} \\ &=& \int\limits_{-\infty}^{0}{e^{-at}e^{-i\omega t}dt} + \int\limits_{0}^{\infty}{e^{at}e^{-i\omega t}dt} \\ &=& \int\limits_{-\infty}^{0}{e^{(-a-i\omega) t}dt} + \int\limits_{0}^{\infty}{e^{(a-i\omega) t}dt} \\ &=& \frac{e^{(-a-i\omega)t}}{-a-i\omega}|_{-\infty}^{0} + \frac{e^{(a-i\omega) t}}{a-i\omega}|_{0}^{+\infty} \\ &=& \frac{1}{-a-i\omega} + \frac{-1}{a-i\omega} \\ &=& \frac{-1}{a+i\omega} + \frac{-1}{a-i\omega} \\ &=& \frac{-1(a-i\omega)}{(a+i\omega)(a-i\omega)} + \frac{-1(a+i\omega)}{(a+i\omega)(a-i\omega)} \\ &=& \frac{-2a}{a^2+\omega^2} \end{eqnarray*} $

It does look like our form, but well, our $ k^2 $ comes with a coefficient, so how do we fix this? Let consider frequency scaling as follow:

$ \begin{eqnarray*} F(\omega) &=& \mathcal{F}(f(t)) \\ &=& \int\limits_{-\infty}^{\infty}{f(t)e^{-i\omega t}dt} \\ F(a\omega) &=& \int\limits_{-\infty}^{\infty}{f(t)e^{-i a \omega t} dt} \\ \end{eqnarray*} $

If $ a > 0 $, we can do a substitution $ T = at $, we have $ dT = a dt $, and so we have

$ \begin{eqnarray*} F(a\omega) &=& \int\limits_{-\infty}^{\infty}{f(t)e^{-i a \omega t} dt} \\ &=& \int\limits_{-\infty}^{\infty}{f(\frac{T}{a})e^{-i \omega T} \frac{1}{a}dT} \\ &=& \mathcal{F}(\frac{1}{a}f(\frac{T}{a})) \end{eqnarray*} $

Otherwise, we can still do the substitution, but the limit changes sign

$ \begin{eqnarray*} F(a\omega) &=& \int\limits_{-\infty}^{\infty}{f(t)e^{-i a \omega t} dt} \\ &=& \int\limits_{\infty}^{-\infty}{f(\frac{T}{a})e^{-i \omega T} \frac{1}{a}dT} \\ &=& -\mathcal{F}(\frac{1}{a}f(\frac{T}{a})) \end{eqnarray*} $

So we can compactly represent this as

$ \begin{eqnarray*} F(a\omega) &=& \mathcal{F}(\frac{1}{|a|}f(\frac{T}{a})) \end{eqnarray*} $

With the two lemmas, we can now proceed to the problem as follow:

$ \begin{eqnarray*} F(e^{a|t|}) &=& \frac{-2a}{a^2+\omega^2} \\ F(\frac{1}{|b|}e^{a|\frac{t}{b}|}) &=& \frac{-2a}{a^2+(b\omega)^2} \\ \end{eqnarray*} $

Since we will need to normalize anyway, we will simply ignore any constant factor.
By just pattern matching, we have got $ b = \frac{\hbar}{\sqrt{2m}} > 0 $, $ a = -\sqrt{-E} $. Plugging back in we can get:

$ \begin{eqnarray*} F(e^{a|\frac{t}{b}|}) & \propto & \frac{-2a}{a^2+(b\omega)^2} \\ F(e^{-\sqrt{-E}\frac{|t|}{\frac{\hbar}{\sqrt{2m}}}}) & \propto & \frac{a\psi(0)}{E+\frac{\hbar^2\omega^2}{2m}} \end{eqnarray*} $

Let's see what is $ -\sqrt{-E}\frac{1}{\frac{\hbar}{\sqrt{2m}}} $, putting back the solution $ -E = \frac{m\alpha^2}{2\hbar^2} $, we get

$ \begin{eqnarray*} -\sqrt{-E}\frac{1}{\frac{\hbar}{\sqrt{2m}}} &=& \sqrt{\frac{m\alpha^2}{2\hbar^2}}\frac{1}{\frac{\hbar}{\sqrt{2m}}} \\ &=& -\frac{m\alpha}{\hbar^2} \\ \end{eqnarray*} $

We are almost there, now we know $ \psi(x) \propto e^{\frac{-m\alpha |x|}{\hbar^2}} $, it remains to normalize it

$ \begin{eqnarray*} \int\limits_{-\infty}^{\infty}{(e^{\frac{-m\alpha |x|}{\hbar^2}})^2 dt} &=& 2 \int\limits_{0}^{\infty}{(e^{\frac{-m\alpha x}{\hbar^2}})^2 dt} \\ &=& 2 \int\limits_{0}^{\infty}{e^{\frac{-2m\alpha x}{\hbar^2}} dt} \\ &=& 2 \frac{e^{\frac{-2m\alpha x}{\hbar^2}}}{\frac{-2m\alpha}{\hbar^2}}|_0^{\infty} \\ &=& \frac{\hbar^2}{m\alpha} \end{eqnarray*} $

Now it is obvious the solution is $ \frac{\sqrt{m\alpha}}{\hbar}e^{\frac{-m\alpha |x|}{\hbar^2}} $.
Phew - now that's a lot of calculation - I have to confess, that I engineered this presentation towards the answer (because there are only 4 options) I knew it is right. I'd have made tons of mistakes without that (well you may want to take into account it is 1:48 am my time)

Exploring Quantum Physics - Week 3 Question 4

Question 4:

Which of the following are properties of superconductors? (Check all that apply)

Solution:

This is really easy, even if I don't know about quantum physics at all, I would simply guess

  • Expulsion of magnetic field lines, and
  • Zero electrical resistance.

Exploring Quantum Physics - Week 3 Question 3

Question:

Recall the definition of the parity operator $ \hat{I} $ from lecture 5. What is $ \hat{I}(xe^{-\frac{x^2}{2l^2}}) $?

Answer:

This one is really simple, the operator does nothing but just substitute $ x $ by $ -x $, so we have the answer is $ (-x)e^{-\frac{(-x)^2}{2l^2}} = -xe^{-\frac{x^2}{2l^2}} $

Exploring Quantum Physics - Week 3 Question 2

Question:

Recall how in Lecture 5, we used graphical methods to obtain the energy levels of a finite potential well. Suppose we are using the same methods in the following graphs. Which graph corresponds to the deepest potential well? (Assume all the wells are of the same width.)

Solution:

We should pick the one such that the curve representing $ \sqrt{(\frac{\xi}{x})^2 - 1} $ has a largest zero crossing, that is because when the curves crosses 0, it correspond to the point when $ x = \xi $, so the curve with the largest zero crossing has the largest $ \xi $ value.

Exploring Quantum Physics - Week 3 Question 1

Question:

Why was the isotope effect a hint towards the mechanism of superconductivity?

Solution:

I didn't attempt this question before I watch the last lecture, but looking backward, I could have. Isotopes are identical material with only differences in mass, the charges stay the same, which means the major player in superconductivity has to do with electron's interaction with the lattice, so the answer is this:

Different isotopes have the same chemical properties, meaning that their electrons behave more or less the same. Thus, the isotope effect suggested that the lattice ions were important for superconductivity.

Wednesday, April 15, 2015

Laplace Transform of a polynomial

Now we are getting to the fun, here is a key result that we will need to know about differentiating a power multiplied with an exponential.

$ \begin{eqnarray*} \frac{d}{dt}t^n e^{-at} &=& nt^{n-1}e^{-at} - at^ne^{-at} \end{eqnarray*} $

The fun begins with integrating both sides:

$ \begin{eqnarray*} \int{\frac{d}{dt}t^n e^{-at}dt} &=& \int{(nt^{n-1}e^{-at} - at^ne^{-at})dt} \end{eqnarray*} $

To make life simple we denote $ I(n) = \int{t^ne^{-at}dt} $, so we have:

$ \begin{eqnarray*} t^n e^{at} &=& nI(n-1) - aI(n) \\ I(n) &=& \frac{n}{a}I(n-1) -\frac{1}{a}t^n e^{-at} \end{eqnarray*} $

We could have just expanded this, but the notation is so cubersome, let consider a simpler problem below like this:

$ \begin{eqnarray*} A(n) &=& npA(n-1) + qt^n \\ &=& \frac{n!p}{(n-1)!} A(n-1) + q\sum\limits_{k=0}^{0} \frac{n!p^k t^{n-k}}{(n-k)!} \\ &=& \frac{n!p}{(n-1)!} ((n-1)pA(n-2) + qt^{n-1}) + q\sum\limits_{k=0}^{0} \frac{n!p^k t^{n-k}}{(n-k)!} \\ &=& \frac{n!p}{(n-1)!} (n-1)pA(n-2) + q\frac{n!p t^{n-1}}{(n-1)!} + q\sum\limits_{k=0}^{0} \frac{n!p^k t^{n-k}}{(n-k)!} \\ &=& \frac{n!p^2}{(n-2)!} A(n-2) + q\sum\limits_{k=0}^{1} \frac{n!p^k t^{n-k}}{(n-k)!} \\ &=& ... \\ &=& \frac{n!p^n }{(n-n)!} A(n-n) + q\sum\limits_{k=0}^{n-1} \frac{n!p^k t^{n-k}}{(n-k)!} \\ &=& p^n n! A(0) + q\sum\limits_{k=0}^{n-1} \frac{n!p^k t^{n-k}}{(n-k)!} \end{eqnarray*} $

Imagine how much more complicated if we have not used the substitution. Now substituting back we have this:

$ \begin{eqnarray*} I(n) &=& \frac{n}{a}I(n-1) -\frac{1}{a}t^n e^{-at} \\ &=& (\frac{1}{a})^n n! I(0) -\frac{1}{a}e^{-at}\sum\limits_{k=0}^{n-1} \frac{n!(\frac{1}{a})^kt^{n-k}}{(n-k)!} \\ &=& (\frac{1}{a})^n n! \frac{e^{-at}}{-a} -\frac{1}{a}e^{-at}\sum\limits_{k=0}^{n-1} \frac{n!t^{n-k}}{a^k(n-k)!} \\ \end{eqnarray*} $

I know, the last summation term is ugly, bear with me, we will kill it, by taking $ t \to \infty $ and $ t \to 0 $. Using l'hopital's rule, we can write

$ \begin{eqnarray*} \lim_{t \to \infty}{t^n e^{-at}} &=& \lim_{t \to \infty}{\frac{t^n}{e^{at}}} \\ &=& \lim_{t \to \infty}{\frac{nt^{n-1}}{ae^{at}}} \\ &=& ... \\ &=& \lim_{t \to \infty}{\frac{n!}{a^ne^{at}}} \\ &=& 0 \end{eqnarray*} $

So as promised, the ugly terms goes away, and therefore:

$ \begin{eqnarray*} \mathcal{L}(t^n) &=& \int\limits_{0}^{\infty}{t^ne^{-st}dt} \\ &=& I(n)|_{0}^{\infty} \\ &=& \frac{n!}{s^{n+1}} \end{eqnarray*} $

QED

Laplace Transform of complex exponential and its consequences

There is nothing blocking us from making the $ a $ in the previous post complex, so we get this Laplace transform pair for free!


$ \begin{eqnarray*} \int\limits_{0}^{\infty}{e^{at}e^{-st}dt} &=& \frac{1}{s-a} \end{eqnarray*} $

This, of course, imply $ |a - s| < 0 $. Interesting things happen when we use the Euler's formula. Let $ a = r + i \theta $, we have $ e^{at} = e^{rt}(\cos \theta t + i \sin \theta t) $. Plugging this into the formula above we have

$ \begin{eqnarray*} \mathcal{L}(e^{rt}(\cos \theta t + i \sin \theta t)) &=& \frac{1}{s-a} \\ \mathcal{L}(e^{rt}(\cos \theta t)) &=& \operatorname{Re}(\frac{1}{s-a}) \\ &=& \operatorname{Re}(\frac{1}{s-(r + i \theta))}) \\ &=& \operatorname{Re}(\frac{1}{s - r - i \theta}) \\ &=& \operatorname{Re}(\frac{s - r + i \theta}{(s - r - i \theta)(s - r + i \theta)}) \\ &=& \frac{s-r}{(s-r)^2 + \theta^2} \\ \mathcal{L}(e^{rt}(\sin \theta t)) &=& \operatorname{Im}(\frac{1}{s-a}) \\ &=& \operatorname{Im}(\frac{1}{s-(r + i \theta))}) \\ &=& \operatorname{Im}(\frac{1}{s - r - i \theta}) \\ &=& \operatorname{Im}(\frac{s - r + i \theta}{(s - r - i \theta)(s - r + i \theta)}) \\ &=& \frac{\theta}{(s-r)^2 + \theta^2} \\ \end{eqnarray*} $

Of course we can further specialize so that $ r = 0 $, and in that case we will have

$ \begin{eqnarray*} \mathcal{L}(\cos \theta t) &=& \frac{s}{s^2 + \theta^2} \\ \mathcal{L}(\sin \theta t) &=& \frac{\theta}{s^2 + \theta^2} \end{eqnarray*} $

See now little we need to come up with so many results!

Laplace Transform of exponential

Now let's go slightly more complicated:


$ \begin{eqnarray*} \int\limits_{0}^{\infty}{e^{at}e^{-st}dt} &=& \int\limits_{0}^{\infty}{e^{(a-s)t}dt} \\ &=& \frac{{e^{(a-s)t}dt}}{(a-s)}|_{0}^{\infty} \\ &=& \frac{1}{s-a} \end{eqnarray*} $

Note that in the third equal sign above, we choose $ s \lt a $ so that we $ t \to \infty $, the exponential converge to $ 0 $.

Laplace Transform of 1

This is the easiest Laplace Transform I can think of

$ \begin{eqnarray*} \int\limits_{0}^{\infty}{e^{-st}dt} &=& \frac{e^{-st}}{-(s)}|_{0}^{\infty} \\ &=& \frac{1}{s} \end{eqnarray*} $

Introduction to Laplace Transform

I have been working on a few problems around Laplace Transform on Yahoo Answer (in fact, the one in Hong Kong/Taiwan). It would be a waste for those content to only post there, so I am posting them here. Feel free to checkout my profile there if you are interested

This post is about Laplace Transform, the Laplace Transform of a function can be simply written as $ Y(s) = \mathcal{L}(y) = \int\limits_{0}^{\infty}{y(t)e^{-st}dt} $.

There are a few things to note:


  1. $ s $ is in general a complex number, $ t $ is usually real.
  2. The integral is improper, in particular, by writing $\int\limits_{0}^{\infty}{y(t)e^{-st}dt} $, we really mean $ \lim_{T \to \infty}{\int\limits_{0}^{\infty}{y(t)e^{-st}dt}} $.
  3. When necessary, we pick ranges of $ s $ such that the limit converges and define $ Y(s) $ there only.
That's it for this post. Next we will talk about some common Laplace Transform pair and properties.

Friday, April 10, 2015

Exploring Quantum Physics - Week 2 Bonus Question 1

Question:

Estimate the value of $ \int\limits_{-\infty}^{\infty}{e^{\frac{−1}{\epsilon}f(x)}dx} $ using the saddle point approximation, with $ \epsilon = 0.05 $ and $ f (x) = \frac{1}{2}x^4 + x + 1 $.

Solution:

Let's take a look at the derivatives of $ f(x) $, we have

$ \begin{eqnarray*} \frac{d}{dx} f(x) &=& 2x^3 + 1 \\ \frac{d^2}{dx^2} f(x) &=& 6x^2 \\ \end{eqnarray*} $

Now we see the function is convex, we also see $ m = -\frac{1}{\sqrt[3]{2}} $ is the global minimum of the function. The power in the exponential is negative, which means for large numbers, the value of the exponential is essentially 0, and therefore we can concentrate near the global minimum. Let's expand $ f(x) $ as a Taylor series around the global minimum, we have:

$ \begin{eqnarray*} f(x) &=& f(m) + \frac{f''(m)(x-m)^2}{2} + ... \end{eqnarray*} $

Ignoring the other coefficients we have got a Gaussian integral, that can be computed as:

$ \begin{eqnarray*} \int\limits_{-\infty}^{\infty}{e^{\frac{−1}{\epsilon}f(x)}dx} & \approx & \int\limits_{-\infty}^{\infty}{e^{\frac{−1}{\epsilon}(f(m) + \frac{f''(m)(x-m)^2}{2})}dx} \\ & = & \frac{\sqrt{\pi}e^{-\frac{1}{\epsilon} f(m)}}{\sqrt{\frac{1}{2\epsilon} f''(m)}} \\ \end{eqnarray*} $

The numeric value of the above expression is $ 8.799 \times 10^-5 $, which matches quite well with a numerical integration, so that's the answer! To confess - I made a mistake once forgetting the 2 comes from the Taylor series, but thanks Coursera for allowing a second attempt.

Exploring Quantum Physics - Week 2 Question 10

Question:

Say we have the action $ S = \int{dt[\frac{1}{2}mv^2 + qEx]} $. What equation of motion does the principle of least action give?

Solution:

To minimize the classical action functional - we will use the Euler-Lagrange equations as follow:

$ \begin{eqnarray*} \frac{\partial}{\partial x} L - \frac{d}{dx}\frac{\partial}{\partial v} L & = & 0 \\ qE - \frac{d}{dx} mv & = & 0 \\ ma & = & qE \\ a & = & \frac{qE}{m} \end{eqnarray*} $

There you go!

Exploring Quantum Physics - Week 2 Question 9

Question:

Suppose we have the Hamiltonian $ H=\frac{p^2}{2m}+\frac{1}{2}m\omega^2 x^2 $. What is the corresponding action?

Solution:

This problem is nothing but just substituting in the definitions. We have $ E = \frac{p^2}{2m} $ and therefore $ V = \frac{1}{2}m\omega^2 x^2 $. Therefore the action would be $ \int{E - V dt} = \int{ \frac{p^2}{2m} - \frac{1}{2}m\omega^2 x^2 dt} $. Now note that $ \frac{p^2}{2m} = \frac{(mv)^2}{2m} = \frac{mv^2}{2} $, so the answer is $ \int{ \frac{mv^2}{2} - \frac{1}{2}m\omega^2 x^2 dt} $.

Exploring Quantum Physics - Week 2 Question 8

Question:

Why are interference effects for particles less important as we go to the classical regime?

Solution:

The lecture video isn't really explicit about this - I think it is intended for us to think. In the last video of the 4th lecture, it is mentioned that the interference is really the cosine terms, and as phases changes rapidly the cosine terms sums to zero on average. So I think the answer is this one:

As $ \hbar \to 0 $ the phases of paths, given by $ e^{\frac{iS}{\hbar}} $, oscillate rapidly, so only paths that interfere constructively (classical paths) contribute.

Thursday, April 9, 2015

Exploring Quantum Physics - Week 2 Question 7

Question: 

Which of the following is most suitable for a saddle point approximation (Laplace's method)?
The problem comes with a set of graphs which I omitted here.

Solution:

The only one with distinct global maximum, this feature make the function particularly suitable for Laplace's method, as the rest of the function will fade away easily.

Exploring Quantum Physics - Week 2 Question 6

Problem:

Which of the following satisfy the diffusion equation in one dimension (for $ t > 0 $) (including, if applicable, unbounded solutions which however satisfy the equation at any finite $ x $ and $ t $)? (Check all that apply)


  • $ e^{\frac{−x^2}{4 D t}} $
  • $ \frac{1}{\sqrt{2\pi D t}}e^{\frac{−x^2}{4 D t}} $
  • $ A⋅(t+\frac{1}{2D}x^2) $
  • $ e^{i D ( p x − E t )} $ where $ p $ and $ E $ are non-zero real numbers.


Solution:

From the lecture, we see the diffusion equation is $ \frac{\partial \rho}{\partial t} = D \nabla^2 \rho $, reducing to one dimension, we have $ \frac{\partial \rho}{\partial t} = D \frac{\partial^2}{\partial x^2} \rho $, so we need to differentiate the equation once by $ t $ and twice by $ x $, then we can know the answer by simply checking if they are equals (of course, with the multiplicative constant).

Differentiating the options, we have



$ \begin{eqnarray*} \frac{\partial}{\partial t} e^{\frac{−x^2}{4 D t}} & = & \frac{x^2}{4 D t^2} e^{\frac{−x^2}{4 D t}} \\ \frac{\partial^2}{\partial x^2} e^{\frac{−x^2}{4 D t}} & = & \frac{\partial}{\partial x} \frac{-x}{2 D t} e^{\frac{−x^2}{4 D t}} \\ & = & \frac{-1}{2 D t} e^{\frac{−x^2}{4 D t}} + \frac{-x}{2 D t} \frac{-x}{2 D t} e^{\frac{−x^2}{4 D t}} \\ & = & \frac{-1}{2 D t} e^{\frac{−x^2}{4 D t}} + \frac{x^2}{4 D^2 t^2} e^{\frac{−x^2}{4 D t}} \\ & = & e^{\frac{−x^2}{4 D t}} (\frac{x^2}{4 D^2 t^2} - \frac{1}{2 D t}) \end{eqnarray*} $

So the first options is incorrect, but it does help the computation of the below, see

$ \begin{eqnarray*} \frac{\partial}{\partial t} \frac{1}{\sqrt{2\pi D t}} e^{\frac{−x^2}{4 D t}} & = & \frac{1}{\sqrt{2\pi D t}} \frac{x^2}{4 D t^2} e^{\frac{−x^2}{4 D t}} + \frac{-1}{2t\sqrt{2\pi D t}} e^{\frac{−x^2}{4 D t}} \\ & = & \frac{1}{\sqrt{2\pi D t}} e^{\frac{−x^2}{4 D t}} (\frac{x^2}{4 D t^2} - \frac{1}{2t}) \\ \frac{\partial^2}{\partial x^2} \frac{1}{\sqrt{2\pi D t}} e^{\frac{−x^2}{4 D t}} & = & \frac{1}{\sqrt{2\pi D t}} \frac{\partial^2}{\partial x^2} e^{\frac{−x^2}{4 D t}} \\ & = & \frac{1}{\sqrt{2\pi D t}} e^{\frac{−x^2}{4 D t}} (\frac{x^2}{4 D^2 t^2} - \frac{1}{2 D t}) \end{eqnarray*} $

So the second option is right - with the complex expressions behind us, the next two is less complex.

$ \begin{eqnarray*} \frac{\partial}{\partial t} A (t+\frac{1}{2D}x^2) & = & A \\ \frac{\partial^2}{\partial x^2} A (t+\frac{1}{2D}x^2) & = & \frac{\partial}{\partial x} \frac{A}{D}x \\ & = & \frac{A}{D} \end{eqnarray*} $

So the third option is also right! Last but not least for the fourth option, we have

$ \begin{eqnarray*} \frac{\partial}{\partial t} e^{i D ( p x − E t )} & = & -i D E e^{i D ( p x − E t )} \\ \frac{\partial^2}{\partial x^2} e^{i D ( p x − E t )} & = & \frac{\partial}{\partial x} i D p e^{i D ( p x − E t )} \\ & = & (i D p )(i D p )e^{i D ( p x − E t )} \\ & = & - D^2 p^2 e^{i D ( p x − E t )} \end{eqnarray*} $

So the fourth option is wrong, as if there are equal, as $ D $, $ E $ and $ p $ cannot be real number at the same time.

Phew, that's a lot of calculation!

Exploring Quantum Physics - Week 2 Question 5

Question:

For the same class of weakly disordered systems as problem 4, in which of the following dimensions might we expect to find localization effects causing an increase in resistance at low temperatures?

Solution:

Again, this problem is trivial after watching the video, only if dimension < 3 will work because random walk with less than 3 dimension is going to be go back to where it was.

Exploring Quantum Physics - Week 4 Question 4

Question:

Recall that in lecture 4, parts II and III, we considered the effects of localization in weakly disordered systems. In this class of systems, why are closed loops the dominant contribution to localization?

Solution:

This question is trivial after watching the video. The reason why closed loop contributes to localization is because a closed loop can be traversed in two ways so that they don't cancel each other.

Wednesday, April 8, 2015

Exploring Quantum Physics - Week 2 Question 3

Question:

Consider the time evolution operator $ \hat{U}(\tau) = e^{\frac{−i}{\hbar} \hat{H} \tau} $. Given an eigenstate of the Hamiltonian, $ \hat{H} |E\rangle = E | E\rangle $, at $ t = 0 $, what is $ \hat{U}(\tau) | E \rangle $, where $ \tau $ is some time, $ \tau > 0 $.

Solution:

To solve this problem, we expand the exponential operator using its Taylor series definition as follows:

$ \begin{eqnarray*} \hat{U}(\tau) |E\rangle &=& e^{\frac{−i}{\hbar} \hat{H} \tau} |E\rangle \\ &=& (\sum\limits_{k = 0}^{\infty}{\frac{(\frac{−i}{\hbar} \hat{H} \tau)^k}{k!}}) |E\rangle \\ &=& \sum\limits_{k = 0}^{\infty}{\frac{(\frac{−i}{\hbar} \hat{H} \tau)^k |E\rangle}{k!}} \\ &=& \sum\limits_{k = 0}^{\infty}{\frac{(\frac{−i}{\hbar} \tau)^k \hat{H}^k |E\rangle}{k!}} \\ &=& \sum\limits_{k = 0}^{\infty}{\frac{(\frac{−i}{\hbar} \tau)^k E^k |E\rangle}{k!}} \\ &=& (\sum\limits_{k = 0}^{\infty}{\frac{(\frac{−i}{\hbar} \tau E)^k }{k!}}) |E\rangle \\ &=& e^{\frac{−i}{\hbar} \tau E} |E\rangle \end{eqnarray*} $

There you go!

Exploring Quantum Physics - Week 2 Question 2

Question:

The probability for particle to be at $ \vec{r_2} $ at $ t_2 $ if it was at $ \vec{r_1} $ at $ t_1 $ is given by $ | G(\vec{r_2}, t_2; \vec{r_1}, t_1) |^2 $ . Let us consider a propagator between points $ \vec{r_1} $  and $ \vec{r_2} $  at time t_1 and t_2 which has the form $ G ( \vec{r_2} ,t2; \vec{r_1} , t_1) = a + b $, where $ a $ and $ b $ are complex and non-zero. This can, in some sense, be considered the case where there are two (and only two) possible ways for particle to travel from $ ( \vec{r_1} , t_1 ) $ to $ (\vec{r_2}, t_2 ) $. What is the probability for the particle to go from $ ( \vec{r_1} , t_1 ) $ to $ (\vec{r_2} , t_2 ) $ ?

Recall that the complex conjugate of a complex number $ z = x + i y $ is $ z^∗ = x − i y $, where $ x $ and $ y $ are real numbers.

Solution:

This is a really complicated way to simply asking what is $ | a + b |^2 $, We can simply compute that as $ | a + b |^2 = ( a + b)(a + b)^* = ( a + b)(a^* + b^*) = aa^* + ab^* + ba^* + bb^* = |a|^2 + ab^* + ba^* + |b|^2 $. 

Exploring Quantum Physics - Week 2 Question 1

Question:

Suppose we have a state $ | \psi \rangle $ that is a linear combination of two other states $ | \psi \rangle = a | 1 \rangle + b | 2 \rangle $, where $ a $ and $ b $ are non-zero, but otherwise unspecified, and $ | 1 \rangle $ and $ | 2 \rangle $ are orthonormal ($ \langle 1 | 2 \rangle = \langle 2 | 1 \rangle = 0 $, $ \langle 1 | 1 \rangle = \langle 2 | 2 \rangle = 1 $). What is $ \langle 1 | \psi \rangle $ ?

Solution:

This is in fact easy, we have:

$ \begin{eqnarray*} \langle 1 | \psi \rangle &=& \langle 1 | (a | 1 \rangle + b | 2 \rangle) \rangle \\ &=& \langle 1 | a | 1 \rangle + \langle 1 | b | 2 \rangle \\ &=& a \langle 1 | 1 \rangle + b \langle 1 | 2 \rangle \\ &=& a \end{eqnarray*} $

So that is it - consider this is a warm up for week 2.

Tuesday, April 7, 2015

Exploring Quantum Physics - Week 1 Bonus Question 1

Question:

Which is the position operator in momentum space?

Solution:

By watching the lecture video alone, I have no idea how to approach this. Fortunately, I have got from the library a copy of "Quantum Mechanics Demystified" which give me an idea what is the momentum space.

So basically, we have $ \psi(x) $ in position space, then we have $ \phi(p) $ in momentum space as Fourier transform pair

$ \begin{eqnarray*} \psi(x) &=& \frac{1}{\sqrt{2\pi\hbar}}\int\limits_{-\infty}^{\infty}{\phi(p) e^{\frac{ i p x}{\hbar}} dp} \\ \phi(p) &=& \frac{1}{\sqrt{2\pi\hbar}}\int\limits_{-\infty}^{\infty}{\psi(x) e^{\frac{-i p x}{\hbar}} dx} \end{eqnarray*} $

So for the position operator, in position space, it should have representation $ x \psi(x) $. Now in momentum space, it should have representation $ \frac{1}{\sqrt{2\pi\hbar}}\int\limits_{-\infty}^{\infty}{\psi(x) x e^{\frac{-i p x}{\hbar}}} dx $. Note that we already have $ \phi(p) = \frac{1}{\sqrt{2\pi\hbar}}\int\limits_{-\infty}^{\infty}{\psi(x) e^{\frac{-i p x}{\hbar}} dx} $, we just need to get one more x in the integrand, and here is the Eureka moment! Differentiate the integrand by $ p $, that will yield:

$ \begin{eqnarray*} \frac{\partial}{\partial p}\phi(p) &=& \frac{\partial}{\partial p}\frac{1}{\sqrt{2\pi\hbar}}\int\limits_{-\infty}^{\infty}{\psi(x) e^{\frac{-i p x}{\hbar}} dx} \\ &=& \frac{1}{\sqrt{2\pi\hbar}}\int\limits_{-\infty}^{\infty}{\psi(x) \frac{\partial}{\partial p}e^{\frac{-i p x}{\hbar}} dx} \\ &=& \frac{1}{\sqrt{2\pi\hbar}}\int\limits_{-\infty}^{\infty}{\psi(x) \frac{-i x}{\hbar} e^{\frac{-i p x}{\hbar}} dx} \\ i\hbar \frac{\partial}{\partial p}\phi(p) &=& \frac{1}{\sqrt{2\pi\hbar}}\int\limits_{-\infty}^{\infty}{\psi(x) x e^{\frac{-i p x}{\hbar}} dx} \end{eqnarray*} $

So $ i\hbar \frac{\partial}{\partial p}\phi(p) $ is exactly what we will need :) Sometimes you need another book, and sometimes, you need an Eureka moment!

Sunday, April 5, 2015

Exploring Quantum Physics - Week 1 Question 9

Question:

Suppose that $ \psi_1 $ and $ \psi_2 $ satisfy the time independent Schrödinger equation for some potential. Which of the following necessarily also satisfies the Schrödinger equation?

Solution:

The correct choice is $ c_1 \psi_1 + c_2 \psi_2 $. This is because the Schrödinger equation is linear and homogeneous.

Exploring Quantum Physics - Week 1 Question 8

Question:

The variance $ \sigma^2 = \langle(\hat{X} - \langle \hat{X} \rangle)^2 \rangle $ of an operator, $ \hat{X} $, is a measure of how large a range its possible values are spread over (the standard deviation is given by $ \sigma = \sqrt{\sigma^2} $). Suppose that $ | X \rangle $ is an eigenstate of some operator $ \hat{X} $, what is the variance of $ \hat{X} $ in this state? You may assume that $ |X\rangle $ is normalized ($ \langle X | X\rangle = 1 $).

Solution:

The real difficult part of this is notation and definitions. Let's start with "eigenstate". The statement $ |X \rangle $ is an eigenstate of $ \hat{X} $ means $ \hat{X}| X \rangle = \lambda | X \rangle $ for some eigenvalue $ \lambda $.

Next, we computes the innermost parenthesis, $ \langle \hat{X} \rangle $. Recall that surrounding an operator using a pair of angle brackets means expectation, which means $ \langle \hat{X} \rangle = \int\limits_{-\infty}^{\infty}{|X\rangle^* \hat{X} | X \rangle dx} = \int\limits_{-\infty}^{\infty}{|X\rangle^* \lambda | X \rangle dx} = \lambda $. The last equality follows because it is given that the eigenstate $ | X \rangle $ is normalized.

Then we proceed to the variance. Again, the angle brackets means expectation. So the variance is given by

$ \begin{eqnarray*} \sigma^2 &=& \langle(\hat{X} - \langle \hat{X} \rangle)^2 \rangle \\ &=& \int\limits_{-\infty}^{\infty}{| X \rangle^* (\hat{X} - \lambda)^2 | X \rangle dx} \\ &=& \int\limits_{-\infty}^{\infty}{| X \rangle^* (\hat{X}^2 - 2\lambda\hat{X} +\lambda^2) | X \rangle dx} \\ &=& \int\limits_{-\infty}^{\infty}{| X \rangle^* (\lambda^2| X \rangle - 2\lambda^2| X \rangle +\lambda^2| X \rangle) dx} \\ &=& 0 \end{eqnarray*} $

The last equality follows as the terms in the bracket cancel each other. I am surprised with this result. Does it mean if the state get into an eigenstate, a certain observable will not have any variances, it is hard to believe both questions leads to 0.

Exploring Quantum Physics - Week 1 Question 7

Question:

Suppose a particle has wavefunction $ \psi(x, t = 0) = Ae^{-\frac{x^2}{2l^2}} $. What is the average value (expectation value) of $ \hat{p} $, $ \langle \hat{p} \rangle $, for this state at $ t = 0$ ?

Solution:

Recall the definition of expectation of an operator is given by $ \langle \hat{p} \rangle = \int\limits_{-\infty}^{\infty}{\psi^{*}(x)\hat{p}\psi(x)dx} $, so again, the problem becomes evaluating this integral, so let's do it

$ \begin{eqnarray*} \hat{p} &=& -i\hbar \frac{\partial}{\partial x} \\ \hat{p}\psi &=& -i\hbar \frac{\partial}{\partial x} A e^{-\frac{x^2}{2l^2}} \\ &=& -i\hbar A e^{-\frac{x^2}{2l^2}} \frac{-2x}{2l^2} \\ &=& \frac{i\hbar x}{l^2} A e^{-\frac{x^2}{2l^2}} \end{eqnarray*} $

With that, we can move on to evaluate the integral

$ \begin{eqnarray*} \langle \hat{p} \rangle &=& \int\limits_{-\infty}^{\infty}{\psi^{*}(x)\hat{p}\psi(x)dx} \\ &=& \int\limits_{-\infty}^{\infty}{(Ae^{-\frac{x^2}{2l^2}})(\frac{i\hbar x}{l^2} A e^{-\frac{x^2}{2l^2}}) dx} \\ &=& \frac{A^2 \hbar i}{l^2}\int\limits_{-\infty}^{\infty}{(e^{-\frac{x^2}{2l^2}})(x e^{-\frac{x^2}{2l^2}}) dx} \\ &=& \frac{A^2 \hbar i}{l^2}\int\limits_{-\infty}^{\infty}{x e^{-\frac{x^2}{l^2}} dx} \\ &=& 0 \end{eqnarray*} $

Surprise! The last line comes from the fact that the integrand is an odd function. One could have think of that as the scaled mean of some Gaussian, too!

Exploring Quantum Physics - Week 1 Question 6

Question:

What is the probability current density of a particle with wavefunction $\psi(x, t) = e^{\frac{i}{\hbar}(p x - E t)} $.

Recall that the probability current density can be computed as: $ j = \frac{\hbar}{2mi}[\psi^{*}\frac{\partial}{\partial x}\psi - \psi \frac{\partial}{\partial x}\psi^{*}] $

Solution:

The problem is in some sense easy, because it really just ask for computing an expression. All I have to do is simply do it.


$ \begin{eqnarray*} \frac{\partial}{\partial x} \psi &=& \frac{\partial}{\partial x} e^{\frac{i}{\hbar}(p x - E t)} \\ &=& \frac{\partial}{\partial x} e^{\frac{i}{\hbar}p x} e^{\frac{-i}{\hbar} E t} \\ &=& e^{\frac{-i}{\hbar} E t} \frac{\partial}{\partial x} e^{\frac{i}{\hbar}p x} \\ &=& e^{\frac{-i}{\hbar} E t} \frac{ip}{\hbar} e^{\frac{i}{\hbar}p x} \\ &=& \frac{ip}{\hbar} e^{\frac{i}{\hbar}p x} e^{\frac{-i}{\hbar} E t} \\ &=& \frac{ip}{\hbar} e^{\frac{i}{\hbar}(p x - E t)} \\ &=& \frac{ip}{\hbar} \psi \end{eqnarray*} $

Similarly, we have


$ \begin{eqnarray*} \frac{\partial}{\partial x} \psi^{*} &=& \frac{\partial}{\partial x} e^{\frac{-i}{\hbar}(p x - E t)} \\ &=& \frac{\partial}{\partial x} e^{\frac{-i}{\hbar}p x} e^{\frac{i}{\hbar} E t} \\ &=& e^{\frac{i}{\hbar} E t} \frac{\partial}{\partial x} e^{\frac{-i}{\hbar}p x} \\ &=& e^{\frac{i}{\hbar} E t} \frac{-ip}{\hbar} e^{\frac{-i}{\hbar}p x} \\ &=& \frac{-ip}{\hbar} e^{\frac{-i}{\hbar}p x} e^{\frac{i}{\hbar} E t} \\ &=& \frac{-ip}{\hbar} e^{\frac{-i}{\hbar}(p x - E t)} \\ &=& \frac{-ip}{\hbar} \psi^{*} \end{eqnarray*} $

With all these preparation, we can get the final answer as follow:

$ \begin{eqnarray*} j &=& \frac{\hbar}{2mi}[\psi^{*}\frac{\partial}{\partial x}\psi - \psi \frac{\partial}{\partial x}\psi^{*}] \\ &=& \frac{\hbar}{2mi}[\psi^{*}\frac{ip}{\hbar}\psi - \psi \frac{-ip}{\hbar}\psi^{*}] \\ &=& \frac{p}{2m}[\psi^{*}\psi + \psi \psi^{*}] \\ &=& \frac{p}{m}[\psi \psi^{*}] \\ &=& \frac{p}{m} \end{eqnarray*} $

The last equality follow from the fact the $ \psi $ is just a complex exponential, which means its norm is 1.

Exploring Quantum Physics - Basic Wave Equation

In this post I would like to summarize what I understand about the wave equation.

The wave equation look like this $ u(x, t) = A \sin ( k x - \omega t + \phi) $.

The number $ k $ is the wave vector. This confused me because it looks like a scalar to me.

Suppose time freezes, if we advance $ x $ by $ \frac{2\pi}{k} $, we get $ u(x + \frac{2\pi}{k}, t) = A \sin ( k (x + \frac{2\pi}{k}) - \omega t + \phi) = A \sin ( k x + 2\pi - \omega t + \phi) = A \sin ( k x - \omega t + \phi) $. Therefore, $ \frac{2\pi}{k} $ is the wave length.

Switching to the time scale, if position freeze and we advance time by $ \frac{2\pi}{\omega} $. It is apparent that using the same trick we get the same value. This means $ \frac{2\pi}{\omega} $ is the period, $ \frac{\omega}{2\pi} $ is the frequency, and $ \omega $ is the angular frequency, nothing special about this.

To advance one wavelength, one need one period of time. Therefore the speed of the wave is $ v = \frac{\lambda}{T} = f\lambda = \frac{\omega}{2\pi}\frac{2\pi}{k} = \frac{\omega}{k} $. 

Saturday, April 4, 2015

Exploring Quantum Physics - Week 1 Question 5

Question:

Suppose we have a particle in 1-dimension, with wavefunction $ Ae^{-\frac{|x|}{2d}} $. What is the probability to find the particle in the interval $ [0, d] $?

Solution:

The born's interpretation tell us the probability is:

$ \begin{eqnarray*} P &=& \int\limits_{0}^{d}{\psi(x, t)^2dx} \\ &=& \int\limits_{0}^{d}{(Ae^{-\frac{|x|}{2d}})^2dx} \\ &=& A^2\int\limits_{0}^{d}{(e^{-\frac{x}{d}})dx} \\ &=& A^2(-de^{-\frac{x}{d}})|_0^d \\ &=& A^2(-de^{-1}-(-d)) \\ &=& A^2d(1-e^{-1}) \\ \end{eqnarray*} $

If we wish, we can substitute the $ A = \frac{1}{\sqrt{2d}} $ in it.

Exploring Quantum Physics - Week 1 Question 4

Problem:

The ground state wavefunction for a particle in a shallow potential well in 1 dimension is of the form: $ A e^{-\frac{|x|}{2d}} $. Given that the particle must be found somewhere in the range $ x \in (-\infty, \infty) $, the born rule then places a constraint on the modulus of A. Assuming that A is real and positive, what is the value of A?

Solution:

The born rule requires the squared wavefunction is the probability density function of finding the particle, which means it is simply an integration problem.

$ \begin{eqnarray*} \int\limits_{-\infty}^{\infty}{\psi^2(x)dx} &=& 1 \\ \int\limits_{-\infty}^{\infty}{(Ae^{-\frac{|x|}{2d}})^2dx} &=& 1 \\ A^2\int\limits_{-\infty}^{\infty}{e^{-\frac{|x|}{d}}dx} &=& 1 \\ 2A^2\int\limits_{0}^{\infty}{e^{-\frac{x}{d}}dx} &=& 1 \\ 2A^2(-de^{-\frac{x}{d}})|_{0}^{\infty} &=& 1 \\ 2A^2(0 - (-d)) &=& 1 \\ A^2 &=& \frac{1}{2d} \\ A &=& \frac{1}{\sqrt{2d}} \end{eqnarray*} $

That is the answer we wanted.

Exploring Quantum Physics - Week 1 Question 3

Problem:

What is the time-dependent Schrödinger equation for a particle in a potential $ V = \frac{1}{2}m\omega^2x^2 $ ?

Solution:

The Schrödinger equation with potential energy is summarized as:

$ i \hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2 \nabla^2}{2m}\psi + \frac{1}{2}m\omega^2x^2\psi $.

Note the potential energy term also have to multiply the wave function.

Exploring Quantum Physics - What the operator?

In the last post, I mentioned about operator. Here is what I think what they should mean.

An operator acts on a function, gives another function, very much like a 'system' in a standard signal processing class do.

E is an operator, in particular, any scalar value is an operator, it acts on the function by multiplying with in and gives another function.

Partial differentiation is an operator, of course, by partial differentiating a function, we get another function as output.

Operators can be composed, for example, multiplying and then differentiating is another operator.

In the last post, all the equal signs, except the last one, equates operator to operator.

Why are we allowed to equates operator to operator? I think this is nothing but just a shorthand to skip writing the wave function on every line.

That's how I interpreted the video.

Exploring Quantum Physics - Week 1 Question 2

Question:

Recall how the Schrödinger equation was motivated by the non-relativistic dispersion relation $ E = \frac{p^2}{2m} $. If we follow the same procedure for the case of a relativistic dispersion relation $ E^2 = p^2 c^2 + m^2 c^4 $), what equation do we arrive at? (For simplicity consider the one-dimensional case)

Solution:

Here comes the math. This problem scare me at first, but once I go through the lecture on deriving the Schrödinger equation it wasn't so hard to copy cat the approach here.

First of all, the operator $ i \hbar \frac{\partial}{\partial t} = E $, and $ -i \hbar \frac{\partial}{\partial x} = p $, we have:

$ \begin{eqnarray*} E^2 &=& p^2 c^2 + m^2 c^4 \\ (i \hbar \frac{\partial}{\partial t})^2 &=& (-i \hbar \frac{\partial}{\partial x})^2 c^2 + m^2 c^4 \\ - \hbar^2 \frac{\partial^2}{\partial t^2} &=& - \hbar^2 c^2 \frac{\partial^2}{\partial x^2} + m^2 c^4 \\ \hbar^2 \frac{\partial^2}{\partial t^2} - \hbar^2 c^2 \frac{\partial^2}{\partial x^2} + m^2 c^4 &=& 0 \\ \frac{1}{c^2} \frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2} + \frac{m^2 c^2}{\hbar^2} &=& 0 \\ (\frac{1}{c^2} \frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2} + \frac{m^2 c^2}{\hbar^2}) \phi &=& 0 \end{eqnarray*} $

To be honest, I don't really sure what I am doing, I am just copying what the lecture did. I will explain what these means in the next post.

Exploring Quantum Physics - Week 1 Question 1

Question: 

For what was Albert Einstein awarded the Nobel prize?

Solution:

Photoelectric effect.

Consider this problem a warm up to this crazy hard course.