Sept 23 (Wed) stationary state expansion of states
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.
main points which are not clear. - describe what you have understood and what the remain questions surrounding the point(s).
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.
Main Points
Setup for Discussion Problem
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}{a}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, (as well as x > 2L, and x < 0)
We want to find <math>\Psi(x, t)</math>
It's tempting to say, <math>\Psi(x, 0) </math> is a stationary state (first excited state) of the infinite square well after the wall has moved since the new set of (after the wall's move) stationary state wave functions are, <math>\psi_n(x) = \sqrt{\frac{2}{2L}}\sin{(\frac{n\pi}{2L}x)}</math>, and for <math>n=2</math>, it's <math>\psi_2(x) = \sqrt{\frac{2}{2L}}\sin{(\frac{2\pi}{2L}x)}=\sqrt{\frac{1}{L}}\sin{(\frac{\pi}{L}x)}</math>, which is very similar to <math>\Psi(x, 0) </math>.
If this is the case, we can say that <math>\Psi(x, t)=\Psi(x, 0)e^{-iE_2t/\hbar}</math>, where <math>E_2</math> is the energy of the 1st excited state, and we are done. But this is not correct.
There is important difference between <math>\Psi(x, 0) </math> and the 2nd new stationary state, <math>\psi_2(x)</math>. The normalization factors are different: <math>\sqrt{\frac{1}{L}}</math> v.s. <math>\sqrt{\frac{2}{L}}</math> may not look so significant, but the major difference is that for <math>L<x<2L</math>, <math>\Psi(x, 0) </math> is zero, while the stationary state wave function is still non-zero at <math>\psi_2(x) =\sqrt{\frac{1}{L}}\sin{(\frac{\pi}{L}x)}</math>. These differences lead to the following.
When you expand <math>\Psi(x, 0) </math> in terms of the linear combination of the (new) stationary state wave functions (Fourier series in this case - infinite square well stationary state wave functions), <math>\sum_n^\infty c_n \psi_n(x)</math>, <math>c_2</math> is large but not 1 as you may think. Actually it will be <math>\sqrt{\frac{1}{2}}</math>. The remaining components are spread among all stationary state wave functions with odd indeces.
The energy associated with <math>\Psi(x, 0) </math> is not uniquely that of <math>\psi_2(x)</math> (only half of the time) but energies associated with many other states are involved.
i.e. <math>\Psi(x, 0) = \sqrt{\frac{1}{2}}\psi_2(x)e^{\frac{-iE_2t}{\hbar}}+\sum_n^{\rm odd}c_n\psi_n(x)e^{\frac{-iE_nt}{\hbar}}</math>. The calculations of <math>c_n</math>'s are left to the students.
* When you expand
* 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>
Energy
What then is our <math>E_n</math>? You don't have to memorize this for infinite square well. You can always use the kind of thinking shown below.
The momentum of a deBroglie wavelength has the form:
<math>p = \frac{h}{\lambda}</math>
Wave number k is related to the wavelength if you remember their definitions: <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_n(x)</math> equationexpression, 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>
Coefficients
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>
When you calculate these coefficients are highly dependent upon how you define the integral, as we're dealing with orthogonality one must pay close attention to the limits of the integral.
If the limits are 0 and 2L, the integral above seems to involve functions in an orthonormal set, so <math>c_m</math>'s appear to be zero for <math>m\not = 2</math>, and 1 for m=2. But since the integral upper limit is L, these conclusions drawn from the orthonormal assumption turn out to be incorrect.
* 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.
Simple Harmonic Oscillator
For a SHO, <math><p^2> = \frac{\hbar\omega m}{2}(2n+1)</math>
Where the kinetic energy is <math><K> = \frac{1}{2m}<p^2></math>
Or <math><K> = \frac{\hbar\omega}{4}(2n+1)</math>, where this is the energy an oscillator has in any state n
Here we can see that <math>E_n = \frac{1}{2}K</math> (half of the kinetic energy) so the potential must be the other half, such that
Is this really the case?
To show that these two are equal, we need to calculate <math><x^2></math>
Simple Harmonic Oscillator, Analytical Solution
The first step to solving the SHO analytically is to remove dimensions and x and all its constants in terms of one variable:
<math>\xi = ( )x</math>, where the parenthesis contain a combination of <math>m, \hbar, \omega</math>
Next, we need to normalize the energy with hbar
<math>(2n + 1) = K = \frac{2E}{\hbar\omega}</math> or <math> (n + \frac{1}{2}) = k = \frac{E}{\hbar\omega}</math>
Doing this makes the differential equation more simple to work with
<math>\frac{d^2\psi}{d\xi^2} = (\xi^2 - k)\psi</math>
Step two is to consider our variable as it approaches plus or minus infinity (such that x is also approaching these limits)
Step three : Continued on Friday
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