# How to use the WKB approximation to find wave functions?

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 $$\phi(r)=\frac{1}{\hbar}\int_{0}^{r} \sqrt{2m(E-U(r))} dr'$$

• What does your textbook say? What have you tried so far? Dec 15, 2013 at 17:28
• 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.
– user27771
Dec 15, 2013 at 17:52
• From the wording of the question and your previous question, I would say that the electron has some energy $-U_0 \lt E \lt \frac{k}{r_0}$ Dec 15, 2013 at 18:17

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.

• 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: $\psi(r)=\frac{A}{\sqrt{|p(x)|}}e^{\frac{1}{\hbar}\int_{0}^{r}|p(r')|dr'}+\frac{B}{\sqrt{|p(x)|}}e^{\frac{-1}{\hbar}\int_{0}^{r}|p(r')|dr'}$?
– user27771
Dec 15, 2013 at 23:01
• 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 $r \gt r_0$ ? Dec 15, 2013 at 23:59
• 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.
– user27771
Dec 16, 2013 at 1:11
• 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?
– user27771
Dec 16, 2013 at 1:18
• 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. Dec 16, 2013 at 19:34