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| classes:2009:fall:phys4101.001:lec_notes_0928 [2009/09/30 00:29] – yk | classes:2009:fall:phys4101.001:lec_notes_0928 [2010/11/01 16:02] (current) – external edit 127.0.0.1 |
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| //Now, let's step back, and look at the bigger picture to see how the wave function will look like. We focus on the sign of the 2nd derivative of the wave function. Knowing that when the 2nd derivative is zero (at some value of //<math>\xi</math>//), the wave function has an inflection point there, while if it is positive, the wave function is concave up, and if it is negative, it is concave down.// | //Now, let's step back, and look at the bigger picture to see how the wave function will look like. We focus on the sign of the 2nd derivative of the wave function. Knowing that when the 2nd derivative is zero (at some value of //<math>\xi</math>//), the wave function has an inflection point there, while if it is positive, the wave function is concave up, and if it is negative, it is concave down.// |
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| {{https://zzz.physics.umn.edu/_media/doctor_physics/sho1.jpg}} | {{https://zzz.physics.umn.edu/_media/dgs_advice/sho1.jpg}} |
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| //Suppose the K takes the smallest value, 1. Then when //<math>0<\xi<1</math>//, the term //<math>\xi^2-K</math>// is negative. So the sign of //<math>\psi</math>// and //<math>\frac{d^2\psi}{d\xi^2}</math>// is opposite. For simplicity, let's assume that //<math>\psi(0)>0</math>//. Then //<math>\frac{d^2\psi}{d\xi^2}</math>// will be negative (meaning that the wave function is concave down) until either //<math>\psi</math>// reaches zero and going negative or //<math>\xi=1</math>// is reached. It turns out that since //<math>\xi^2-K</math>// is not a big negative (K is small in the current example) so that the wave function never reaches zero before //<math>\xi=1</math>// causes the inflection, where the concave up region starts. When <math>K</math> is just right (at 1), the downward slope of the wave function at //<math>\xi=1</math>// is just right so that the wave function will slowly approach zero while the slope also approach zero. If <math>K</math> is slightly less than 1, the curvature of the concave is looser, and the slope of the wave function is less negative. With the concave up region above //<math>\xi=1</math>// will gradually increase the wave function's slope to a positive value before //<math>\xi=\infty</math>//, and the wave function diverge to the //<math>\infty</math>// from there. On the other hand, if <math>K</math> is slightly more than 1, the curvature of the concave is tighter, and the slope of the wave function is more negative. With the concave up region above //<math>\xi=1</math>// will gradually increase the wave function's slope to a less negative value but before //<math>\xi=\infty</math>//, the wave function will reach zero. At the point, the sign of the 2nd derivative changes its sign! and the slope starts to become more negative, and the wave function will diverge to the //<math>-\infty</math>// from there. **End of Yuichi's clarification**// | //Suppose the K takes the smallest value, 1. Then when //<math>0<\xi<1</math>//, the term //<math>\xi^2-K</math>// is negative. So the sign of //<math>\psi</math>// and //<math>\frac{d^2\psi}{d\xi^2}</math>// is opposite. For simplicity, let's assume that //<math>\psi(0)>0</math>//. Then //<math>\frac{d^2\psi}{d\xi^2}</math>// will be negative (meaning that the wave function is concave down) until either //<math>\psi</math>// reaches zero and going negative or //<math>\xi=1</math>// is reached. It turns out that since //<math>\xi^2-K</math>// is not a big negative (K is small in the current example) so that the wave function never reaches zero before //<math>\xi=1</math>// causes the inflection, where the concave up region starts. When <math>K</math> is just right (at 1), the downward slope of the wave function at //<math>\xi=1</math>// is just right so that the wave function will slowly approach zero while the slope also approach zero. If <math>K</math> is slightly less than 1, the curvature of the concave is looser, and the slope of the wave function is less negative. With the concave up region above //<math>\xi=1</math>// will gradually increase the wave function's slope to a positive value before //<math>\xi=\infty</math>//, and the wave function diverge to the //<math>\infty</math>// from there. On the other hand, if <math>K</math> is slightly more than 1, the curvature of the concave is tighter, and the slope of the wave function is more negative. With the concave up region above //<math>\xi=1</math>// will gradually increase the wave function's slope to a less negative value but before //<math>\xi=\infty</math>//, the wave function will reach zero. At the point, the sign of the 2nd derivative changes its sign! and the slope starts to become more negative, and the wave function will diverge to the //<math>-\infty</math>// from there. **End of Yuichi's clarification**// |
| <math>K = 2n + 1 </math> | <math>K = 2n + 1 </math> |
| thus for n = 0,1,2... , //K// is 1, 3, 5.... | thus for n = 0,1,2... , //K// is 1, 3, 5.... |
| {{https://zzz.physics.umn.edu/_media/doctor_physics/sho2.jpg}} | {{https://zzz.physics.umn.edu/_media/dgs_advice/sho2.jpg}} |
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| For the SHO, as //K// becomes larger, the inflection points will be spread out (<math>\xi=1, \sqrt{2}, \sqrt{3}</math>, ... for the first, second and third energy states. See the graph above.) The wiggly region of the wave function slowly grows longer). The final inflection point is where | For the SHO, as //K// becomes larger, the inflection points will be spread out (<math>\xi=1, \sqrt{2}, \sqrt{3}</math>, ... for the first, second and third energy states. See the graph above.) The wiggly region of the wave function slowly grows longer). The final inflection point is where |
| * This is a no calculator quiz, as well. No extravagant calculations should be necessary. | * This is a no calculator quiz, as well. No extravagant calculations should be necessary. |
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| ====also ==== | ==== Free Particle (Begun) ==== |
| There are obviously some formatting errors I've not been able to finish yet. Anyone is welcome to help me out! (or I'll be trying to proof read it later. Yay for class!) | |
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| I didn't finish the notes for the start of the "Free Particle" either... | |
| | For the Free Particle, we have the equation |
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| | <math>\psi(x) = e^{ikx} </math> |
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| | (Which can also be written cos (kx) + isin(kx) ) |
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| | With the time dependance it is written |
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| | <math>\Psi(x,t) = e^{i(kx-\omega t)}</math> |
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| | where omega is |
| | <math>\omega(k) = \frac{E}{\hbar} = \frac{k^2 h^2}{\hbar2m}</math> |
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| | (Note that these are not normalized waves.) |
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| | The wave packet can be written as |
| | * a wave packet |
| | *linear combination of plane waves |
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| | Which means that if we can understand the second representation, we can understand the first better. (It is much easier to compute the second). |
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| | From 2601 last year we remember representing a particle wave hitting a finite potential. Part of the wave bounces back, part and part of it goes through. In this way we can calculate the wave forms from the boundary conditions. |
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| | <math> \left|\frac{B}{A}\right |^2 = Reflection </math> |
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| | A is the particles going in, and B is the particles going out. |
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| | <math>\left | \frac{C}{A} \right| ^2 \frac{K'}{K} = Transmission </math> |
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| | A is the number of particles / unit length, and C is number of particles transmitted right. |
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| | Thus if the potential goes up, the speed of particles going through is slower. |
| | <math> Transmission = \frac{# Particles/second}{# Particles / Second}</math> |
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| | and |
| | <math>\frac{K'}{K} <math> |
| | is the conversion of particles/meter to particles to second. |
| | (these are the velocity ratios of the two areas). |
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| | Now we've done this in terms of the plane wave. Now we'll do it in terms of the wave packet. The reason we do it for the plane wave is because it's much easier than the wave packet. |
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