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| classes:2009:fall:phys4101.001:lec_notes_0923 [2009/09/20 08:59] – yk | classes:2009:fall:phys4101.001:lec_notes_0923 [2009/09/29 19:12] (current) – johnson |
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| ===== Sept 23 (Wed) ===== | ===== Sept 23 (Wed) stationary state expansion of states ===== |
| ** Responsible party: joh04684, Aspirin ** | ** Responsible party: joh04684, Aspirin ** |
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| === Main Points === | === Main Points === |
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| | == 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. |
| | <del> * 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></del> |
| | == 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. |
| | * 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 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> <del>equation</del>//expression//, 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> |
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| | == 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> |
| | * Where <math>\sin{(\frac{\pi}{L}x)}</math> is proportional to <math>\psi_2</math> |
| | * //When you calculate// these coefficients <del>are highly dependent upon how you define the integral, as we're dealing with orthogonality</del> //one must pay close attention to the limits of the integral.// |
| | * If the limits are 0 and 2//L//, 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. |
| | <del> * 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.</del> |
| | * As we said before, doing the actual integrals is left to the students for an exercise. |
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| | == 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// |
| | * <math>E_n = (n + \frac{1}{2})\hbar\omega</math> for an oscillator |
| | * 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 |
| | * <math><V>_n = \frac{\hbar\omega}{2}(n + \frac{1}{2})</math> |
| | * Is this really the case? |
| | * <math>V = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2x^2</math> |
| | * <math><V> = \frac{1}{2}m\omega^2<x^2>_n</math> |
| | * To show that these two are equal, we need to calculate <math><x^2></math> |
| | * <math><x^2>_n = \int\psi^*_nx^2\psi_ndx</math> |
| | * Here, <math>\psi^*_n</math> are all the <math>\psi_0, \psi_1, \psi_2, etc</math> solutions, but it also might not be some closed-form expression for <math>\psi_n</math> |
| | * We can write p and x in terms of the raising and lowering operators: |
| | * <math>p = ( )(a_- - a_+)</math> |
| | * <math>x = ( )(a_- + a_+)</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) |
| | * k is related to energy, so it has to go to large distances to make <math>\xi >> k</math> simplifying things to: |
| | *<math>e^{-\frac{\xi^2}{2}}</math>, where we only take the -ve exponent, since as it approaches positive infinity, it is not normalizable. |
| | * Step three : Continued on Friday |
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