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I'm trying to learn how to apply WKB. I asked a similar question already, but that question was related to finding the energies. Here, I would like to understand how to find the wave functions using WKB.

An electron, say, in the nuclear potential $$U(r)=\begin{cases} & -U_{0} \;\;\;\;\;\;\text{ if } r < r_{0} \\ & k/r \;\;\;\;\;\;\;\;\text{ if } r > r_{0} \end{cases}$$ What is the wave function inside the barrier region ($r_{0} < r < k/E$)?

Shouldn't the wave function have the following form?

$$\psi(r)=\frac{A}{\sqrt{2m(E-U(r))}}e^{\phi(r)}+\frac{B}{\sqrt{2m(E-U(r))}}e^{-\phi(r)}$$ where $$\phi(r)=\frac{1}{\hbar}\int_{0}^{r} \sqrt{2m(E-U(r))} dr'$$

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  • $\begingroup$ What does your textbook say? What have you tried so far? $\endgroup$ Dec 15, 2013 at 17:28
  • $\begingroup$ My textbook (Griffiths) says that $\psi(x) = Ae^{\pm \kappa x}$, where $\kappa = \sqrt{2m(U-E)}/\hbar$ for $E<U$ and $U$ constant. But I'm not sure if $E<U$ here, so I'm not sure if the phase should be preceded by $i$. The textbook doesn't say anything about current density or escape rate to my knowledge. So far, I'm just trying to work through the explanation of WKB in the text. $\endgroup$
    – user27771
    Dec 15, 2013 at 17:52
  • $\begingroup$ From the wording of the question and your previous question, I would say that the electron has some energy $-U_0 \lt E \lt \frac{k}{r_0}$ $\endgroup$ Dec 15, 2013 at 18:17

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Inside vs outside, there is a sign change inside the square root, so that changes the nature of the "phase" $\phi(r)$.

Normally, when you match wave functions you require that $\psi_\mathrm{left}(x) = \psi_\mathrm{right}(x)$ (continuity) and that the derivative changes according to what you get when you integrate the Schrodinger equation: $\int_\mathrm{left}^\mathrm{right}\!dx\, \left( -\frac{\hbar^2}{2m} \frac{d ^2\psi(x)}{dx^2} + V(x) \psi(x) \right) = 0$ but in fact your WKB solutions are only approximations and they get worse as $V(x) \approx E$. So your textbook (Griffiths) devotes a few pages to deriving "matching conditions" instead. I suggest you start there and come back with specific questions if you are unsure about how to use those conditions.

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  • $\begingroup$ Thanks, but I'm still not sure how to even begin. Griffiths gives different formulas depending on whether $E<U(r)$ or $E>U(r)$, but you say that $-U_{0}<E<\frac{k}{r_{0}}$, so I don't know where to begin. Is the following the starting point for this problem: $\psi(r)=\frac{A}{\sqrt{|p(x)|}}e^{\frac{1}{\hbar}\int_{0}^{r}|p(r')|dr'}+\frac{B}{\sqrt{|p(x)|}}e^{\frac{-1}{\hbar}\int_{0}^{r}|p(r')|dr'}$? $\endgroup$
    – user27771
    Dec 15, 2013 at 23:01
  • $\begingroup$ Well, the potential is a piece-wise defined function, which you had better make a rough sketch of so you can visualize it. Inside the "well", $E \gt -U_0$, in other words, $E \gt U(r)$. And since $E \lt \frac{k}{r_0}$, what does that tell you about $E \,? \frac{k}{r}$ for $r \gt r_0$ ? $\endgroup$ Dec 15, 2013 at 23:59
  • $\begingroup$ Thanks for your help. I think I understand that inside the "barrier", meaning inside the "well", $E<U(r)$ so WKB gives the wave function of the form as I wrote in my comment above. Is that correct? Then, I would think that the exponent would be simply $\pm \frac{1}{\hbar}\int_{0}^{r}\sqrt{-U_{0}-E}dr'$ but Griffiths (Eq. 8.24, if you have a copy) shows in a similar example that the exponent is basically $\frac{1}{\hbar}\int_{r_{1}}^{r_{2}}\sqrt{\frac{k}{r'}-E}dr'$. So which one do you use? I don't get why he changes it, or if that's even relevant for my problem. $\endgroup$
    – user27771
    Dec 16, 2013 at 1:11
  • $\begingroup$ Oops, correction: I mean inside the well $E>U(r)$ and the exponents have an imaginary sign. Outside the barrier they don't. Is that correct? $\endgroup$
    – user27771
    Dec 16, 2013 at 1:18
  • $\begingroup$ yes. The form of the exponential depends on what the potential looks like, and since this potential is defined piecewise, it'll have different forms in different regions. $\endgroup$ Dec 16, 2013 at 19:34

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