10
$\begingroup$

In the first term the energies are given by the Wentzel–Kramers–Brillouin (WKB) formula

$$ \oint p dq = 2\pi \left( n+\frac{1}{2} \right) $$

However, can this formula be improved to include further corrections? For example the wave function in the WKB approach can be evaluated to several orders of magnitude in $ \hbar $

$\endgroup$
1

1 Answer 1

5
$\begingroup$

In the following, I'll describe to you how in principle one can compute higher order corrections to the Bohr-Sommerfeld condition. In order to find higher order corrections to the Bohr-Sommerfeld formula, we need to include higher order corrections of the wavefunction of the form: $$ \Psi(x) = \sum_n \hbar^n a_n(x) \exp\left(\frac{i}{\hbar} \int^x \sqrt{2m(E-V(y)) }dy\right)\,. $$ Inserting in the Schrodinger equation and requiring fullfilment to each order of $\hbar$, we obtain: $$ a_0 = \frac{const.}{(2m(E-V(y)))^{\frac{1}{4}} } $$ and $$ a_k \Delta S + 2 \nabla a_k\cdot \nabla S = i \Delta a_{k-1}\,, $$ where $$ S= \int^x \sqrt{2m(E-V(y)) } dy\,. $$ In one dimension these equations can solved in closed form.

The quantization condition is achieved through the requirement that the wave function phase does not change over a closed loop: $$ \oint^x \frac{1}{\Psi} \frac{d\Psi}{dx} dx = 2\pi i n\hbar\,. $$ Please see, the following article by A. Voros where this procedure was applied to the anharmonic oscillator. Equation (4.4), describes an explicit solution of the wave function to the first higher order.

Remark: Voros uses the Bargmann representation of the phase space instead of the usual coordinate-momentum representation.

$\endgroup$
2
  • $\begingroup$ have you got a good, accessible reference for the Bargmann representation? I remember having trouble finding one last time round. $\endgroup$ Commented Dec 10, 2012 at 19:00
  • $\begingroup$ @Emilio You may find these lecture note by B.C. Hall useful: math.cinvestav.mx/~NorteSur/Taller10/notas/Hall.pdf $\endgroup$ Commented Dec 11, 2012 at 13:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.