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Connection Formulas in the WKB approximation

  1. Apr 4, 2009 #1
    Hello friends,

    I've been reading Schiff's book on QM (3rd Edition), esp the section on the WKB approximation. (This isn't homework.)

    I have a few questions:

    What is the physical significance of the arrow on the connection formulas, like

    [tex]\frac{1}{2}\frac{1}{\sqrt{\kappa}} e^{-\zeta_{2}} \longrightarrow \frac{1}{\sqrt{k}}\cos\left(\zeta_{1}-\frac{\pi}{4}\right)[/tex]


    [tex]\frac{e^{-\zeta_{2}}\sin\eta}{\sqrt{\kappa}} \longleftarrow \frac{1}{\sqrt{k}}\cos\left(\zeta_{1}-\frac{\pi}{4}+\eta\right)[/tex]


    [tex]k = \sqrt{\frac{2m(E-V(x))}{\hbar^2}}[/tex]
    [tex]\kappa = \sqrt{\frac{2m(V(x)-E)}{\hbar^2}}[/tex]
    [tex]\zeta_{1} = \int_{0}^{x}k(x')dx'[/tex]
    [tex]\zeta_{2} = \int_{x}^{0}\kappa(x')dx'[/tex]

    According to the book, these formulas indicate that the wavefunction on the left (in the first formula) goes on to the right with this form, but the converse is not true as the cosine picks up a phase of [itex]\pi/4[/itex]. While I've been able to "prove" both these formulas mathematically, I do not have a good feel of what they're saying exactly.

    First of all, when I write an arbitrary linear combination of the "exact" solutions to Schrodinger's equation near the turning point (where I have expanded the potential V(x) as a linear function), how do I know what the correct direction of the arrow in the connection formula is?

    In other words, after I form the linear combination, I can use the asymptotic forms of the Bessel Function (or the Airy Function) for [itex]x\rightarrow\infty[/itex] and [itex]x\rightarrow -\infty[/itex] thereby obtaining two asymptotic forms of the linear combination. How do I determine whether the arrow points from region 2 (the left of the turning point at x = 0) to region 1 or vice versa?

    Thanks for your help.
  2. jcsd
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