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| classes:2009:fall:phys4101.001:q_a_1009 [2009/10/11 11:33] – yk | classes:2009:fall:phys4101.001:q_a_1009 [2009/10/22 12:21] (current) – czhang |
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| **If you want to see lecture notes, click [[lec_notes]]** | **If you want to see lecture notes, click [[lec_notes]]** |
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| **Main class wiki page: ** [[home]] | **Main class wiki page: ** [[home]] |
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| ==== Spherical Chicken 10/7 12:38 ==== | ==== Spherical Chicken 10/7 12:38 ==== |
| == liux0756 3:30pm 10/09 == | == liux0756 3:30pm 10/09 == |
| Yes because V=0 when x is not 0, E>0 is for scattering state and E<0 is for bound state. | Yes because V=0 when x is not 0, E>0 is for scattering state and E<0 is for bound state. |
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| | == Blackbox 8:00pm 10/11 == |
| | If I add a little more on the above two opinions,,, |
| | There are two different states which are bound state and scattering state as you know. |
| | In the case of E>0, Energy is always larger than potential energy in the entire location. |
| | Differently with the bound state, a particle can transmit through the position of x=0 due to the potential has negative infinite value at x=0. I guess that's the reason why the scattering state focuses only on when E>0. |
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| ====Hydra 10/9 21:30==== | ====Hydra 10/9 21:30==== |
| Ok, I just need some verification on the delta-function..... It is not so much a "function" but instead a distribution. And we know that a distribution represents a probability..... But since the delta-function is spiked at one point only....does this mean there is only one probable outcome? I apologize for my fragmented question full of dotted pauses..... but I think it effectively reflects my confusion & frustration with this seemingly hand-waving method. It works, but why? | Ok, I just need some verification on the delta-function..... It is not so much a "function" but instead a distribution. And we know that a distribution represents a probability..... But since the delta-function is spiked at one point only....does this mean there is only one probable outcome? I apologize for my fragmented question full of dotted pauses..... but I think it effectively reflects my confusion & frustration with this seemingly hand-waving method. It works, but why? |
| ===David Hilbert's Hat 10/10 9:20am=== | ===David Hilbert's Hat 10/10 9:20am=== |
| I think Griffiths makes a reference to the delta function when it's used as a distribution of a point particle's mass/charge. Everywhere that's not located at the particle it's zero, but at the particle it's infinity; when you integrate over all space it comes out to be 1, because you have one particle. That seems to be the easiest way to see the delta function - it is very peculiar because it is zero everywhere except at one point it's infinity, but it integrates to 1, like a point particle's distribution (which seems a lot more familiar). | I think Griffiths makes a reference to the delta function when it's used as a distribution of a point particle's mass/charge. Everywhere that's not located at the particle it's zero, but at the particle it's infinity; when you integrate over all space it comes out to be 1, because you have one particle. That seems to be the easiest way to see the delta function - it is very peculiar because it is zero everywhere except at one point it's infinity, but it integrates to 1, like a point particle's distribution (which seems a lot more familiar). |
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| ==== Anaximenes - 22:30 - 10/09/09 ==== | ==== Anaximenes - 22:30 - 10/09/09 ==== |
| This question is about problem 2.34. The problem asks us in part c to show that <math>T=\sqrt{\frac{E-V_0}{E}}\frac{|F|^2}{|A|^2}</math>. However, Eq. 2.139 on page 75 says <math>T\equiv \frac{|F|^2}{|A|^2}</math>. That's with three lines, as in identically equal, as in any statement that they're not equal (such as that in the prompt in 2.34c) is incongruent. Now, I remember the professor said in class that <math>T=\frac{|F|^2}{|A|^2} \frac{k_2}{k_1}</math>, but how do we //show// that? We have no definition of T other than the (patently false) one in 2.139. For shame, Griffiths. For shame. | This question is about problem 2.34. The problem asks us in part c to show that <math>T=\sqrt{\frac{E-V_0}{E}}\frac{|F|^2}{|A|^2}</math>. However, Eq. 2.139 on page 75 says <math>T\equiv \frac{|F|^2}{|A|^2}</math>. That's with three lines, as in identically equal, as in any statement that they're not equal (such as that in the prompt in 2.34c) is incongruent. Now, I remember the professor said in class that <math>T=\frac{|F|^2}{|A|^2} \frac{k_2}{k_1}</math>, but how do we //show// that? We have no definition of T other than the (patently false) one in 2.139. For shame, Griffiths. For shame. |
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| | === Yuichi 11:44 10/11 === |
| | OK. Let's start from a simple case when the momentum before and after the obstacle represented by the potential is the same to show that 2.139 is sensible (if you accept the idea that |A|^2 can be interpreted as the probability for the incoming particle, which should be somehow normalized to 1. Even though you cannot really normalize the plain wave, close your eyes to such a detail.*) If |A|^2 represents that the total probability for the incoming particle (integrated over the entire space!) is one, |B|^2 must be related to the similar probability for the reflected wave, and |C|^2 is for the transmitted wave. Is this acceptable? Or more quantitatively, |B/A|^2 is the probability that the particle is in the "reflected" state, and |C/A|^2 is the probability that the particle would be found in the transmitted state. |
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| | Now in a more general case when the momentum changes after the collision with the potential well (or bump). Paying a bit more attention to what |A|^2, |B|^2 and |C|^2 are related to. They are related to the probability DENSITies that the particle is found __in a unit length in the incident, reflected and transmitted states.__ But if you think about how the reflection and transmission is measured, it's not the spacial densities of the particle in the three waves (incident, transmitted and reflected), but temporal densities which is more directly relevant. //i.e.// how many particle (fractional!) is arriving as an incident wave, going out as reflected and transmitted waves __per second.__ When you do this conversion of probability density per length to per second, you will need to consider |A|^2*k_1, etc. Hence the equation described in problem 2.34. |
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| | If you doing like to ignore the "detail" at *, we have to consider wave packets of incident, reflected and transmitted waves. Following the lecture note for 10/9, |
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| | <math>\psi(x) = e^{-ikx}</math> alone for x<0 region and <math>\Psi(x) = 0</math> will not satisfy the Schrodinger equation and boundary conditions. However, |
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| | <math>\psi_I(x) = e^{ikx}+\frac{B}{A}e^{-ikx}</math> for x<0 |
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| | and |
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| | <math>\psi_{II}(x) = \frac{C}{A}e^{ikx}</math> for x>0 |
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| | will for any value of //k//'s. By taking a linear combination of these for different values of //k//'s, the result should still satisfy the Schrodinger Equation. Actually, at this point, we are really thinking about time dependent Schrodinger equation so the time dependence should also be included in the wave function. So we are thinking about |
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| | <math>\Psi_I(x,t) = e^{i(kx-\omega t)}+\frac{B}{A}e^{-i(kx-\omega t)}</math> for x<0 |
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| | and |
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| | <math>\Psi_{II}(x,t) = \frac{C}{A}e^{i(kx-\omega t)}</math> for x>0 |
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| | before taking a linear combination (<math>\omega=\omega(k)</math> is a function of //k// and <math>\omega(k)\sim k^2</math>), and afterward |
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| | <math>\Psi_I(x) = \int \phi(k) [e^{i(kx-\omega t)}+\frac{B}{A}e^{-i(kx+\omega t)}] dk</math> for x<0 |
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| | and |
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| | <math>\Psi_{II}(x) = \int \phi(k) [\frac{C}{A}e^{i(kx-\omega t)}] dk</math> for x>0. |
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| | <math>\phi(k)</math> is such that <math>\Psi_I(x) = \int \phi(k) e^{i(kx-\omega t)} dk</math> describes the wave packet for the incident particle at t<<0 (well before the collision) situated in where? (x << 0). It will pass through the potential regions of x ~ 0 at t ~ 0, and moves on the x >> 0 at t >> 0 like as if this is a free particle. |
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| | You can show (for you to do if you are looking for a challenge) that given this situation described in the above paragraph, that the terms containing "B" is a wave packet located at x >> 0 at t << 0, and x << 0 at t >> 0, while the "C" term corresponds to a wave packet located at x << 0 for t << 0 and x >> 0 for t >> 0 (not dis-similar to the incident wave packet). Note that for t << 0, only the incident wave packet is in the x region which is appropriate for it (//i.e.// x < 0) and the other two wave packets are ghosts existing only in equations, but not in real life (not dis-similar to image charges in E&M, or if you remember something from waves on a string where waves are reflected, where you were probably told that reflected wave come from the point beyond the fix point of the string where waves cannot physically exist). For t >> 0, the wave packet for the incident particle becomes a ghost sitting in the x >> 0 region, while the other two wave packets are real. |
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| | Here are some more challenges for you, but one can show that |
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| | - the total probability for the incident particle <math>\int_{-\infty}^\infty |\Psi_{\rm inc}(x,t)|^2 dx</math> where <math>\Psi_{\rm inc}(x) = \int \phi(k) e^{i(kx-\omega t)} dk</math> remains unity for any time, //t//, once <math>\psi(k)</math> is determined appropriately (so that the total probability is unity at some time t << 0). |
| | - the total probability for the particle associated with the reflected wave <math>\int_{-\infty}^\infty |\Psi_{\rm ref}(x,t)|^2 dx</math> where <math>\Psi_{\rm ref}(x) = \int \phi(k) \frac{B}{A}e^{-i(kx+\omega t)} dk</math> will be |B/A|^2 and remains the same for any time, //t//. For R, the reflection coefficient, one is interested in the total probability in the reflected wave packet for t >> 0 over the incident wave packet for t << 0, which must be 1 if <math>\psi(k)</math> is correctly determined. This is |B/A|^2.\\ \\ To figure out the total probability associated with this wave packet, you need to modify this expression so that you can use the fact that the probability for the incident wave packet integrate to 1 |
| | - Very similar claim can be made for the transmitted part of the waves. Note that in case that the momentum of the transmitted wave is different, "k" in the plane wave expression has to be changed to <math>e^{i(k'x-\omega t)}</math> and the wave packet will look like,\\ \\ <math>\Psi_{\rm trans}(x) = \int \phi(k) [\frac{C}{A}e^{i(k'x-\omega t)}] dk</math>. Note that only "k" in the plane wave is changed to "k'", but not the other. (think about why if you don't want to just accept it from me.)\\ \\ To figure out the total probability for the transmitted wave packet, make a reasonable assumption that <math>\phi(k)</math> has significant non-zero values only for a small //k'// range around it peak at <math>k_0'</math>, which corresponds to <math>k_0</math> for the incident wave. Then one can approximate that \\ \\ <math>k' = k_0'+\frac{k_0}{k_0'}(k-k_0)</math>. I will leave the rest for you to prove. |
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| | The following may be useful for some of the calculations needed to prove some of the things above. When the incident wave packet, <math>\Psi_{\rm inc}(x) = \int \phi(k) e^{i(kx-\omega t)} dk</math>, is properly normalized, <math>\int_{-\infty}^\infty [\Psi_{\rm inc}(x)]^*\Psi_{\rm inc}(x) dx = 1</math>. By substituting the former into the latter, and change one occurrence of //k// to //k_2// to remember that the two integrals involve different variable of integrals, |
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| | <math>\int_{-\infty}^\infty [\int \phi^*(k_2) e^{i(k_2x-\omega_2 t)} dk_2]^*\int \phi(k) e^{i(kx-\omega t)} dk dx = 1</math>, where <math>\omega_2 \equiv \omega(k_2)</math>. |
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| | Since we are sloppy physicists, we freely change the order of the integrals. So rather than doing //k// and //k_2// integrals, we decide to do the //x// integral first. This will result in |
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| | <math>\int \phi^*(k_2)dk_2 \int \phi(k) dk \int_{-\infty}^\infty dx [e^{-i(k_2x-\omega_2 t)} e^{i(kx-\omega t)}] = 1</math>. |
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| | <math>\int \phi^*(k_2)dk_2 \int \phi(k) dk [e^{+i\omega_2 t} e^{-i\omega t}]\int_{-\infty}^\infty dx [e^{-ik_2x} e^{ikx}] = 1</math>. |
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| | Since the most inside integral is one way to express the delta-function (times 2pi) (see Griffiths' hand-wave argument for this), //k//_2 integral becomes easy: just replace //k//_2 with //k//. |
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| | <math> 2\pi \int \phi^*(k)\phi(k) dk [e^{+i\omega t} e^{-i\omega t}] = 2\pi \int |\phi(k)|^2 dk = 1</math> |
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| | This is the normalization equation for <math>\phi(k)</math> similar to what we used to have for "A" in some practice problems, or <math>\sum_n |c_n|^2 = 1 </math>. |
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| | I may have made some mistakes like where 2pi goes, so please carefully check if you are interested in this issue. |
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| | === Can 10/21 10:52am === |
| | For Yuichi, instead of saying A^2 is probability, I think it would make much more sense if we think of A^2 as incident intensity, just my personal opinion not sure if it is rigorous. Recall that the square of the amplitude of for E field is proportional to the intensity , E^2 is proportional to I, same analogy can be made here, then the probability can be interpreted as transmitted intensity versus incident intensity. <math>T=\frac{F^2}{A^2}</math>. At least, I think this might be easier to understand transmission intuitively, maybe not mathematically, one still has to work out all the math. |
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| | ====Liam Devlin 10/12, 10:45am==== |
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| | Why is k a known for a scattering problem? |
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| | ===Can 10/22 12:15pm === |
| | Since for scattering problem <math>E=\frac{h^2k^2}{2m}</math> which means <math>k=\frac{\sqrt{2mE}}{h}</math>, and E is the incident energy of the particle, which can be manually tuned. E is known , so k is known. |
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| | ====Aspirin 10/12, 2:30pm==== |
| | Why does the coefficient, D, of the genernal solution which is <math> \psi_I_I = C e^{ikx} + D e^{-ikx} </math> cancel out? |
| | Yuichi mentioned the reason, but I didn't fully get it. |
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| | ===Blackbox 10/12, 3:20pm=== |
| | I'm not quite sure I understood that correctly, but if I say, when a particle moves from the left to the right direction, it will transmitt through the Delta function well. In other words, there is no physical values from the right to the left direction. If we think of the opposite situation, a particle moves from the right to the left, in the first region, <math> \psi_I = Ae^{ikx} + Be^{-ikx} </math>, A would be removed because nothing can go back to the right direction. |
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| | ====Blackbox 10/12, 6:00pm==== |
| | I might write wrong in my note though. On last Wednesday, Yuichi mentioned the unit of α is Energy/length. Isn't it Energy*length? |
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