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classes:2009:fall:phys4101.001:lec_notes_1005 [2009/10/05 16:21] – x500_nikif002classes: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  **  ** Responsible party: Esquire, nikif002  ** 
  
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 <math>\psi (x)=Ae^{kx}+Be^{-kx}</math> <math>\psi (x)=Ae^{kx}+Be^{-kx}</math>
  
-However, for <math>x\rightarrow \inf</math>, <math>Ae^{kx}\rightarrow \inf</math> and for <math>x\rightarrow -\infty</math>, <math>Be^{-kx}\rightarrow \inf</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\inf</math>, A=0 for x>0 and B=0 for x<0, so +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 <math>\psi (x)=\psi_I (x)=Ae^{kx}</math> for x < 0
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 <math>\psi (x)=\psi_{II} (x)=Be^{-kx}</math> for x < 0 <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.+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+First, the wavefunction must be continuous. Since our definition changes at //x//=0, we need
  
 <math>\psi_I(0)=\psi_{II}(0)</math> <math>\psi_I(0)=\psi_{II}(0)</math>
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 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. 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 that follows.+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== ==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?+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 Maxwell's equations state
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 <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> <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.+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> <math>\int_{-\epsilon}^\epsilon\frac{dE_x}{dx}dx=\frac{\sigma}{\epsilon_0}\int_{-\epsilon}^\epsilon\delta(x)dx</math>
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 <math>E_x(IN)-E_x(OUT)=\frac{\sigma}{\epsilon_0}</math> <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.+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== ==Back to quantum==
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 <math>\frac{\delta}{\delta x}\psi_I(0)=\frac{\delta}{\delta x}\psi_{II}(0)-\Delta</math> <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.+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>  Equation I: <math>\psi_I(0)=\psi_{II}(0)</math> 
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 ===Things that are not clear=== ===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 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]]**\\ **To go back to the lecture note list, click [[lec_notes]]**\\
classes/2009/fall/phys4101.001/lec_notes_1005.1254777666.txt.gz · Last modified: 2009/10/05 16:21 by x500_nikif002

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