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This note complemented my actual lecture with proper factors so that “~” becomes “=”, and some parts that perhaps I was not clear (from your faces) were more clearly and completely explained. Thank you for you the wonderful job. Yuichi

It was asked whether knowledge of the analytical solution to the harmonic oscillator potential would be needed for the test. It is possible that it will, but not likely due to time constraints. It would probably still be good to have some idea of how it works, even if you can't reproduce it entirely.

The only other question asked about the quiz was regarding how to determine the expectation values of the momentum and position for the simple harmonic oscillator by using the ladder operators. It is possible to write the position and momentum operators in terms of the ladder operators:

<p> = <math>i \sqrt{\frac{mw\hbar}{2}}(a_+ - a_-)</math>

<x> = <math>\sqrt{\frac{\hbar}{2m\omega}}(a_+ + a_-)</math>

Then just apply these operators as you would normally to the wavefunction and integrate.

For the free particle, we have the equation <math>\Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int\phi (k) e^{i(kx - \omega t)}dk</math>, where <math>\omega</math> is a function of k. But where does this come from?

For a bound state, such as a particle in the infinite square well, the particle is only allowed to have certain discrete quanta of energy. A particular stationary state in the infinite square well can be written as <math>\psi_n (x) = \sqrt{\frac{2}{L}}\sin{(\frac{n\pi}{L}x)}</math>. Here, <math>\frac{n\pi}{L}=k</math>. The general solution for the infinite square well can be written as a linear combination of these stationary states: <math>\Psi(x,0) = \sum_n c_n\psi_n(x)</math>. To get the full, time-dependent solution, we just add on the time component, <math>\phi(t)</math>, to get <math>\Psi(x,t) = \sum_n c_n\psi_n(x)e^{-i\omega t}</math>.

The difference for the free particle is that, since it is not bound in a potential, it has no restrictions on the energy it can have. The above sum goes to an integral and the coefficients <math>c_n</math> go to a continuous function of k. The dependence on n disappears completely, because there aren't discrete energy levels. What we end up with for the free particle is <math>\Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi(k)e^{i(kx - \omega t)}dk</math>. The <math>\frac{1}{\sqrt{2\pi}}</math> factor in front of the integral is somewhat arbitrary; remove it and your <math>\phi(k)</math> will end up differing by a constant factor from what it would have originally been. The final answer ends up being identical.

How do we determine <math>\phi(k)</math>?

For the wavefunctions of particles in bound potentials, we had <math>\Psi(x,0) = \sum_n c_n\psi_n(x) \Rightarrow c_n = \int \psi^*_n(x)\Psi(x,0)dx</math>, which we obtained by utilizing the orthogonality of the eigenfunctions <math>\psi_n(x)</math>.

For the free particle, we can use a Fourier transform to determine <math>\phi(k)</math>. For <math>\Psi(x,0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi(k)e^{ikx}dk</math>, we apply the transform to get <math>\phi(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}\Psi(x,0)e^{-ikx}dx</math>.

Yuichi suggested that we try Problem 2.20 in Griffiths to gain a better understanding.

The book defines the group velocity as <math>v_g=\frac{d\omega}{dk}</math>, but what does that mean? What is the group velocity? What makes it different from the phase velocity?

The real part of the wavefunction of a wave packet looks like a sinusoidal function bound within an envelope. The wavefunction can move faster or slower than the envelope. The speed of the envelope is the group velocity, and the speed of the waves inside the envelope is the phase velocity.

The group velocity represents the velocity that we would measure, <math>\frac{d<x>}{dt}</math>. One may wonder how this can be if <x> is at a place where the wavefunction plotted is 0; the answer is that the plotted part of the wavefunction does not include the imaginary component. The envelope represents the magnitude of the wavefunction (the real and imaginary parts added in quadrature, <math>\|\Psi\|=\sqrt{\Psi^{*}\Psi}=\sqrt{\(Re\(\Psi\)\)^2 + \(Im\(\Psi\)\)^2}</math>).

So why does <math>v_g = \frac{d\omega}{dt}</math>? Why would it be different in different systems? For a quantum mechanical particle, <math>\omega \propto E\propto \frac{p^2}{2m}\propto k^2</math>. For light, <math>\omega \propto k</math>. For water waves, <math>\omega \propto \sqrt{k}</math>.

Suppose that the wave packet has waves corresponding to <math>e^{ikx}</math> and <math>e^{i(k+\Delta k)x}</math>.

<math>\Psi (x, 0)=e^{ikx}+e^{i(k+\Delta k)x} = e^{i(\frac{k+k+\Delta k}{2})x}\(e^{-i\frac{\Delta k}{2} x}+e^{i\frac{\Delta k}{2} x}\)
= \[cos( \frac{(2k+\Delta k)}{2}x ) +i sin(\frac{(2k+\Delta k)}{2}x)\]\(2cos(\frac{\Delta k}{2} x)\)</math>

The second cosine in the last expression is the envelope, while the first term represents the phase inside the envelope.

We can also see that <math>\Psi (x, 0) = e^{ikx}+e^{i(k+\Delta k)x}=\(cos(kx)+cos((k+\Delta k)x)\)+i\(sin(kx) + sin((k+\Delta k)x)\)</math>. <math>\|\Psi\|</math> will then be at a maximum when the two waves are in phase, <math>kx_g=(k+\Delta k)x_g</math>, where <math>x_g</math> is the x corresponding to a maximum in the envelope. Taking time into consideration gives

<math>kx_g - \omega t = (k+\Delta k)x_g - (\omega + \Delta \omega)t</math>

At <math>t=0</math>, this gives <math>x_g = 0</math>, and at <math>t=\Delta t</math>, it gives <math>x_g=\frac{\Delta \omega}{\Delta k}\Delta t</math>. Then, we can see that

<math>v_g\approx \frac{\Delta x_g}{\Delta t}\approx \frac{d\omega}{dk}</math>

with the approximations becoming exact in the limit of small <math>\Delta</math>.

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