Responsible party: joh04684, Aspirin

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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 particle is confined in its ground state in an infinite square well of width L.
• Right at time t = 0, the infinite square well is expanded to width 2L.
• The time-independent form of the particle at t = 0 before expansion is <math>\psi(x) = \sqrt{\frac{2}{L}}\sin{(\frac{\pi}{2}x)}</math>
• Once the width of the well has expanded to 2L, the initial conditions for our function function becomes:
  • <math>\Psi(x, 0) = \sqrt{\frac{2}{L}}\sin{(\frac{\pi}{L}x)}</math> for 0 < x < L
  • <math>\Psi(x, 0) = 0</math> for L < x < 2L, x > 2L, and x < 0
• We want to find <math>\Psi(x, t)</math>
• The general form is <math>\psi(x)\phi(t)</math>, where <math>\psi_n(x) = A\sin{Bx}</math>, where A and B are some constants
• Thus, our general form for the particle from 0 < x < L before widening the well is:
  • <math>\psi_n(x) = \sqrt{\frac{2}{L}}\sin{(\frac{n\pi}{2L}x)}</math>
• And after widening the well:
  • <math>\psi_n(x) = \sqrt{\frac{2}{2L}}\sin{(\frac{n\pi}{2L}x)}</math>
  • <math>\phi_n(t) = e^{\frac{-iE_nt}{\hbar}}</math>

• What then is our <math>E_n</math>?
  • Well, potential is zero inside the well, so all of the energy must be Kinetic Energy
• The momentum of a deBroglie wavelength has the form:
  • <math>p = \frac{h}{\lambda}</math>
  • Wave number k is then <math>k = \frac{2\pi}{\lambda}</math>
  • This gives us a simpler form: <math>p = k\hbar</math>
• In a general wave equation, we have the form <math>\sin{kx}</math>
• So from our general <math>\Psi(x, t)</math> equation, we can say that <math>k_n = \frac{n\pi}{2L}</math>
• Substituting back in, we now have <math>p_n = \frac{n\pi}{2L}\hbar</math>, giving us <math>E_n = \frac{p_n^2}{2m} = \frac{1}{2m}\frac{n^2\pi^2\hbar^2}{4L^2}</math>
  • Finally: <math>E_n = \frac{n^2\pi^2\hbar^2}{8mL}</math>

• We want to express <math>\psi</math> in terms of <math>\Psi (x, 0) = … </math>
  • <math>\Psi (x, t) = \Psi(x, 0)e^{\frac{-iEt}{\hbar}}</math>
  • However, <math>\Psi(x, 0)</math> no longer has unique energies, so we can't do it in this form.
• We can write:
  • <math>c_m = \int_{-\infty}^{\infty}\psi_m(x)\Psi(x, 0) dx</math>
  • But everywhere but 0 < x < 2L is 0, so we can really integrate from 0 to L (<math>\Psi(x,0)</math> is already 0 between L and 2L)
  • More simply: <math>c_m = \int_{0}^{L}\psi_m(x)\Psi(x, 0) dx</math>
• So which form of <math>\psi_n</math> are we talking about? The one where the width is L or 2L?
  • <math>\psi_m = \sqrt{\frac{1}{L}}\sin{(\frac{n\pi}{2L}x)}</math>
  • And: <math>\Psi(x,0) = \sqrt{\frac{2}{L}}\sin{(\frac{\pi}{L}x)}</math>
• Substituting these into our equation, we get:
  • <math>c_m = \int_0^L\frac{1}{\sqrt{L}}\sin{(\frac{m\pi}{2L}x)}\sqrt{\frac{2}{L}}\sin{(\frac{\pi}{L}x)}dx</math>
  • <math>c_m = \frac{\sqrt{2}}{L}\int_0^L\sin{(\frac{m\pi}{2L}x)}\sin{(\frac{\pi}{L}x)}dx</math>
    • Where <math>\sin{(\frac{\pi}{L}x)}</math> is proportional to <math>\psi_2</math>
• These coefficients are highly dependent upon how you define the integral, as we're dealing with orthogonality
• Here, we are only integrating from 0 to L, not 0 to 2L
  • We may need to do this integral not just for the general case where m does not equal 2, but also for where m = 2.

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