1

In the derivation, we assume the eigenfunctions of $H$ to have the form $$\psi (x)=e^{i \phi(x)/\hbar}$$ where $\phi (x)$ is allowed to take any complex value.

But then suddenly we assume this expansion of $\phi (x)$ in terms of $\hbar$:

$$\phi (x)=\phi _0 (x)+ \phi _1 (x)\hbar + \phi _2 (x) \hbar^2+....$$

We assume the higher powers of $\hbar$ are negligible (in our approximation) and then feed this into the Schrodinger equation.

My question is about the expansion. What on earth is this expansion? It's not a Taylor series or anything (where we can neglect higher order terms). How is this expansion of $\phi (x)$ even unique? What determines $\phi _0 (x)$, $\phi_1 (x)$, etc in this expansion? Can't we make this expansion whatever we want?

Qmechanic
  • 201,751
Ryder Rude
  • 6,312

1 Answers1

2

That expansion is a Taylor expansion (actually an asymptotic expansion), of the phase of the wavefunction. And the point of the WKB approximation is that you will eventually truncate the series.

The way the different terms at each order in $h$ are fixed are following an iterative procedure. At order $h^n$, you plug the ansatz into the Schrodinger equation and keep all terms at the same order in $h$ (note that the momentum-squared operator $p^2/2m$ is intrinsically order $h^2$, since the Laplacian is multiplied by $\hbar^2$; see wikipedia). This will give you a differential equation that lets you solve for $\phi_n$ in terms of lower order approximations to the phase, as well as any source terms (zeroth-order terms in $\phi$).

There's an explicit example of how this works on wikipedia.

Andrew
  • 48,573