Responsible party: prest121, Anaximenes

To go back to the lecture note list, click lec_notes
previous lecture note: lec_notes_0928
next lecture note: lec_notes_1005

Main class wiki page: home

Please try to include the following

• main points understood, and expand them - what is your understanding of what the points were.
  • expand these points by including many of the details the class discussed.
• main points which are not clear. - describe what you have understood and what the remain questions surrounding the point(s).
  • Other classmates can step in and clarify the points, and expand them.
• How the main points fit with the big picture of QM. Or what is not clear about how today's points fit in in a big picture.
• wonderful tricks which were used in the lecture.

The only 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.

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.

To go back to the lecture note list, click lec_notes
previous lecture note: lec_notes_0928
next lecture note: lec_notes_1002