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.

Since we cannot go much further without specifying potential energy, 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 - finite)

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

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