Responsible party: liux0756, Dagny

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Main class wiki page: home

Please try to include the following

• main points understood, and expand them - what is your understanding of what the points were.
  • expand these points by including many of the details the class discussed.
• main points which are not clear. - describe what you have understood and what the remain questions surrounding the point(s).
  • Other classmates can step in and clarify the points, and expand them.
• How the main points fit with the big picture of QM. Or what is not clear about how today's points fit in in a big picture.
• wonderful tricks which were used in the lecture.
  

Today we focus on the following two points.

• rest of Legendre polynomial
• Radial part of Schrodinger equation

In the last class the power series method is used:

<math>U(\xi)=\sum_{n=0}^\infty a_n \xi^2</math>

Substitute it into differential equation derives the recursive relation:

<math>a_k=\frac{(\nu+1)(\nu+2)…(\nu+k)(-\nu)(-\nu+1)…(-\nu+k-1)}{(k!)^2} a_0</math>

<math>a_0</math> is determined by normalization.

Now consider the convergence requirement. Just as the simple harmonic oscillator, for normalizable solutions the power series must terminate. And if <math>\nu</math> is an integer, <math>a_k</math> will be zero if k is large enough. If <math>\nu</math> is negative, there exists a corresponding positive <math>\nu</math> that leads to the same recursive relation. So <math>\nu</math> can be limited to non-negative numbers:

<math>\nu=0,1,2,…</math>

<math>\alpha=\nu(\nu+1)=0,2,6,12,…</math>

Generally <math>P_\nu (z) </math> is called Legendre function, <math>\nu</math> is any real number. If convergence is important, <math>\nu=l</math> is integer, we deal with the Legendre polynomial <math>P_l (z) </math>

The equation is written as:

<math>\frac{d}{dr} (r^2 \frac{dR}{dr})- \frac{2mr^2}{\hbar^2}[V®-E]R=l(l+1)R</math>

<math>l(l+1)</math> is related with the square of angular momentum:

<math>L^2=l(l+1)\hbar^2</math>

Define <math> R®=\frac{u®}{r}</math>, then the differential equation is transformed from <math>R®</math> to <math>u®</math>:

<math> -\frac{\hbar^2}{2m} \frac{d^2u}{dr^2}+[V+\frac{\hbar^2}{2m} \frac{l(l+1)}{r^2}]u=Eu </math>

This equation is similar to 1D Schrodinger equations discussed in Chapter 2.

The equation above cannot be solved further before one knows the potential distribution in the system.

Now consider a 3D infinite square well.

When <math>r>a</math>, the potential goes to infinity.

When <math>r<a</math>, the potential is 0.

The radial differential equation is:

<math>\frac{d^2u}{dr^2}-[\frac{l(l+1)}{r^2}-k^2]u=0</math>

where <math>k^2=\frac{2mE}{\hbar^2} </math>

The solution is Bessel function <math>j_l(kr)</math>

Now take a look at the 0th order solution.

When <math>l=0</math>, the differential equation is <math>\frac{d^2u}{dr^2}=-k^2u</math>

The solution is: <math>u®=Ae^{ikr}+Be^{-ikr}=A'sinkr+B'coskr</math>

The boundary condition is <math>u(a)=0</math>

At r=0, if u is not 0, then <math>R=\frac{u}{r}</math> will go to infinity. So <math>u(0)=0</math> should be satisfied. (However, this is not quite correct because even if the value of wave function is infinite, the wave function may still be able to be normalized: <math>\int_0^a \frac{1}{r^2}dV = \int_0^a \frac{1}{r^2} r^2 sin\theta dr d\theta d\phi=\int_0^a sin\theta dr d\theta d\phi</math> is normalizable)

So the <math>B'coskr</math> term should be 0. (In fact this term is allowed)

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