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--- classes:2009:fall:phys4101.001:lec_notes_1005 [2009/09/30 12:58] – prestegard
+++ classes:2009:fall:phys4101.001:lec_notes_1005 [2009/10/05 20:00] (current) – yk
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-===== Oct 05 (Mon) =====
+===== Oct 05 (Mon) delta-function potential well - Bound state =====
 ** Responsible party: Esquire, nikif002  **
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   * wonderful tricks which were used in the lecture.
-=== Main Points ===
+====nikif002====
+=== Quiz Comments ===
+Relying too much on tables and technology for integrals is counter-productive to developing an intuitive understanding.
+Example from quiz:
+<math>\int x^2 sin^2 x dx</math>
+Use the following trig identity: <math>sin^2 x = {1-cos2x}/2</math>, then integrate by parts. If you have a rough idea of what to do, ask the proctor and they may help you. For example, if someone knew that a trig identity needed to be used, but didn't know which, they could have asked the proctor on this last quiz.
+=== Delta-function potential well ===
+The delta-function potential is <math>V(x)=-\alpha \delta (x)</math>, with <math>\alpha >0</math>. It is zero everywhere except for x=0, and <math>-\infty</math> at x=0. While it may seem that the alpha makes no difference, <math>\delta (x)</math> is not simply defined as an infinitely sharp peak at 0. Another part of the delta-function definition is that <math>\int_{-\infty}^{\infty}{\delta (x)dx}=1</math>, so a constant multiplier does make a difference in V(x).
+Like the infinite square well, this is not a real potential, so why do physicists like it?
+  * It has few boundary conditions
+  * When studying crystals, putting a delta-function potential at each atomic site is a good approximation.
+== Bound states ==
+This lecture focused on bound states, where E < 0. Because the potential is 0 at <math>\pm\infty</math>, a particle in a state with E < 0 cannot go to <math>\pm\infty</math> - it cannot go arbitrarily far from the potential well.
+== Solving the Schrodinger equation ==
+For <math>x\neq 0, V=0</math>, so the Schrodinger equation becomes
+<math>-\frac{\hbar^2}{2m}\frac{\delta^2}{{\delta x}^2}\psi (x)=E\psi (x)</math>
+<math>\frac{\delta^2}{{\delta x}^2}\psi (x)=(-\frac{2mE}{\hbar^2})\psi (x)</math>
+Define <math>k^2=-\frac{2mE}{\hbar^2}</math>
+Since E is negative, the RHS is positive, so k is real. Assume k is positive - we can also assume it is negative, as long as we are consistent.
+Then, <math>\frac{\delta^2}{{\delta x}^2}\psi (x)=k^2 \psi(x)</math>
+We need a function that keeps its form when differentiated. This, of course, is the exponential function. So,
+<math>\psi (x)=Ae^{\pm kx}</math>
+Or, more accurately, we have a linear combination of the possible solutions:
+<math>\psi (x)=Ae^{kx}+Be^{-kx}</math>
+However, for <math>x\rightarrow \infty</math>, <math>Ae^{kx}\rightarrow \infty</math> and for <math>x\rightarrow -\infty</math>, <math>Be^{-kx}\rightarrow \infty</math>.
+Since we know that the wavefunction goes to 0 at <math>\pm\infty</math>, A=0 for x>0 and B=0 for x<0, so
+<math>\psi (x)=\psi_I (x)=Ae^{kx}</math> for x < 0
+and
+<math>\psi (x)=\psi_{II} (x)=Be^{-kx}</math> for x < 0
+We have 3 unknowns - //A//, //B//, //k//. Let's find 3 equations to solve for them.
+First, the wavefunction must be continuous. Since our definition changes at //x//=0, we need
+<math>\psi_I(0)=\psi_{II}(0)</math>
+This is our first equation. In a real case, the derivative of the wavefunction is also continuous. However, for some idealized potentials, it is not. For an example, think back to the ground state wavefunction of the infinite square well - outside the well it is flat, with a zero derivative, while inside it instantly changes to a finite slope.
+For the delta-function potential, the derivative is also discontinuous, but we can find the value of the discontinuity. It is similar to the electricity & magnetism example that follows.
+==EM example==
+Imagine a one-dimensional interface at x=0 between vacuum (x<0) and a dielectric material (x>0). There is an electric field <math>E_x</math> in the x direction and the electric field in the y and z directions is 0. The value of the electric field in the vacuum will be different from the electric field in the dielectric material. The negative charges in the dielectric material will concentrate at the interface, at x=0. What happens if we model this concentration as a delta-function since the thickness of the layer of charge is very thin, but when you integrate over this thickness, you get a finite amount of charge (still spread in the //y// and //z// directions)?
+Maxwell's equations state
+<math>\vec\bigtriangledown\vec E=\frac{\delta}{\delta x} E_x\sout{+\frac{\delta}{\delta y} E_y+\frac{\delta}{\delta y} E_y}=\frac{\rho}{\epsilon_0}=\frac{\sigma\delta(x)}{\epsilon_0}</math>
+Let <math>\epsilon</math> be any small number. Let's define a small neighborhood around the vacuum-dielectric interface: <math>[-\epsilon,\epsilon]</math> and integrate both sides of the above equation over this neighborhood.
+<math>\int_{-\epsilon}^\epsilon\frac{dE_x}{dx}dx=\frac{\sigma}{\epsilon_0}\int_{-\epsilon}^\epsilon\delta(x)dx</math>
+The integral of the delta function around any neighborhood of x=0 is 1, and integrating the left side simply reverses the derivative in the integrand, giving us the difference of the electric fields inside and outside the dielectric:
+<math>E_x(IN)-E_x(OUT)=\frac{\sigma}{\epsilon_0}</math>
+Since we can make <math>\epsilon</math> arbitrarily small, the electric field is discontinuous. This is, of course, an unrealistic artifact introduced by the approximation of the charge distribution by the delta function.  //(If we use more realistic finite - but thin - layer of charges, the electric field varies quickly across this layer, but it will not have a discontinuity.)//
+==Back to quantum==
+Similarly, there is a difference <math>\Delta=-\frac{2m\alpha}{\hbar^2}\psi_I(0)</math> between <math>\frac{\delta}{\delta x}\psi_I(0)</math> and <math>\frac{\delta}{\delta x}\psi_{II}(0)</math>
+By convention, it is defined the following way:
+<math>\frac{\delta}{\delta x}\psi_I(0)=\frac{\delta}{\delta x}\psi_{II}(0)-\Delta</math>
+Now, we have 2 equations to solve for our unknowns. The third equation is normalization. However, since we cannot measure the wavefunction in experiment, normalization is often uninformative. Since it turns out we can solve for //k// using only the first two equations, we will do just that.
+Equation I: <math>\psi_I(0)=\psi_{II}(0)</math>
+<math>\psi_I(0)=Ae^{0}=A</math>
+<math>\psi_{II}(0)=Be^{0}=B</math>
+<math>A=B</math>
+Equation II: <math>\frac{\delta}{\delta x}\psi_I(0)=\frac{\delta}{\delta x}\psi_{II}(0)-\Delta</math>
+<math>\frac{\delta}{\delta x}\psi_{II}(0)=-kBe^{-k*0}=-kB=-kA</math>
+<math>\frac{\delta}{\delta x}\psi_I(0)=kAe^{k*0}=kA=kA</math>
+<math>kA=-kA-\Delta</math>
+<math>2kA=-\Delta=\frac{2m\alpha}{\hbar^2}{A}</math>
+<math>k=\frac{m\alpha}{\hbar^2}</math>
+We now have the wavefunction up to a scaling constant:
+<math>\psi (x)=\psi_I (x)=Ae^{kx}</math> for x < 0
+and
+<math>\psi (x)=\psi_{II} (x)=Ae^{-kx}</math> for x < 0
+===Things that are not clear===
+I think I understood the lecture very well. However, I am not a physics major and have not had E&M in a very long time. I feel like I understood the idea behind the E&M example, but perhaps not as well as some of my classmates.  //I think that's OK, and hope you felt it's OK, too.  Yuichi//
 **To go back to the lecture note list, click [[lec_notes]]**\\

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