31.12.2013 Views

Numerical Methods in Quantum Mechanics - Dipartimento di Fisica

Numerical Methods in Quantum Mechanics - Dipartimento di Fisica

Numerical Methods in Quantum Mechanics - Dipartimento di Fisica

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

whose goal is to give an estimate, to first order <strong>in</strong> perturbation theory, of the<br />

<strong>di</strong>fference δe between the current estimate of the eigenvalue and its f<strong>in</strong>al value.<br />

Rem<strong>in</strong>der: icl is the <strong>in</strong>dex correspond<strong>in</strong>g to the classical <strong>in</strong>version po<strong>in</strong>t.<br />

Integration is made with forward recursion up to this <strong>in</strong>dex, with backward<br />

recursion down to this <strong>in</strong>dex. icl is thus the <strong>in</strong>dex of the match<strong>in</strong>g po<strong>in</strong>t<br />

between the two functions. The function at the right is rescaled so that the<br />

total function is cont<strong>in</strong>uous, but the first derivative dy/dx will be <strong>in</strong> general<br />

<strong>di</strong>scont<strong>in</strong>uous, unless we have reached a good eigenvalue.<br />

In the section of the code shown above, y(icl) is the value given by Numerov’s<br />

method us<strong>in</strong>g either icl-1 or icl+1 as central po<strong>in</strong>t; ycusp is the value<br />

pre<strong>di</strong>cted by the Numerov’s method us<strong>in</strong>g icl as central po<strong>in</strong>t. The problem<br />

is that ycusp≠y(icl).<br />

What about if our function is the exact solution, but for a <strong>di</strong>fferent problem?<br />

It is easy to f<strong>in</strong>d what the <strong>di</strong>fferent problem could be: one <strong>in</strong> which a delta function,<br />

v 0 δ(x−x c ), is superimposed at x c ≡x(icl) to the potential. The presence<br />

of a delta function causes a <strong>di</strong>scont<strong>in</strong>uity (a ”cusp”) <strong>in</strong> the first derivative, as<br />

can be demonstrated by a limit procedure, and the size of the <strong>di</strong>scont<strong>in</strong>uity is<br />

related to the coefficient of the delta function. Once the latter is known, we<br />

can give an estimate, based on perturbation theory, of the <strong>di</strong>fference between<br />

the current eigenvalue (for the ”<strong>di</strong>fferent” potential) and the eigenvalue for the<br />

”true” potential.<br />

One may wonder how to deal with a delta function <strong>in</strong> numerical <strong>in</strong>tegration.<br />

In practice, we assume the delta to have a value only <strong>in</strong> the <strong>in</strong>terval ∆x centered<br />

on y(icl). The algorithm used to estimate its value is quite sophisticated. Let<br />

us look aga<strong>in</strong> at Numerov’s formula (1.32): note that the formula actually<br />

provides only the product y(icl)f(icl). From this we usually extract y(icl)<br />

s<strong>in</strong>ce f(icl) is assumed to be known. Now we suppose that f(icl) has a<br />

<strong>di</strong>fferent and unknown value fcusp, such that our function satisfies Numerov’s<br />

formula also <strong>in</strong> po<strong>in</strong>t icl. The follow<strong>in</strong>g must hold:<br />

fcusp*ycusp = f(icl)*y(icl)<br />

s<strong>in</strong>ce this product is provided by Numerov’s method (by <strong>in</strong>tegrat<strong>in</strong>g from icl-1<br />

to icl+1), and ycusp is that value that the function y must have <strong>in</strong> order to<br />

satisfy Numerov’s formula also <strong>in</strong> icl. As a consequence, the value of dfcusp<br />

calculated by the program is just fcusp-f(icl), or δf.<br />

The next step is to calculate the variation δV of the potential V (r) appear<strong>in</strong>g<br />

<strong>in</strong> Eq.(2.35) correspond<strong>in</strong>g to δf. By <strong>di</strong>fferentiat<strong>in</strong>g Eq.(2.36) one f<strong>in</strong>ds δg(x) =<br />

−(2m e /¯h 2 )r 2 δV . S<strong>in</strong>ce f(x) = 1 + g(x)(∆x) 2 /12, we have δg = 12/(∆x) 2 δf,<br />

and thus<br />

δV = − ¯h2 1 12<br />

2m e r 2 δf. (2.37)<br />

(∆x) 2<br />

First-order perturbation theory gives then the correspond<strong>in</strong>g variation of the<br />

eigenvalue:<br />

∫<br />

δe = 〈ψ|δV |ψ〉 = |y(x)| 2 r(x) 2 δV dx. (2.38)<br />

23

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!