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I understand that we can write any complex wavefunction on polar form $A\exp(iθ)$ with both $A,θ$ real. Following the logic of Griffiths on WKB (here, page 291):

  1. We write the energy wavefunction in the previous form.
  2. For $E>V$, we insert the previous form in S.E and demand $A^"=0$. The reason we can do this approximation is because we are lead to equation (8.6), which can give indeed real $A$ for $A^"=0$.

But what happens when $E<V $? In this case if we write again the wavefunction in the form $A\exp(iθ)$ with $A,θ$ real, then we cannot apply $A^"=0$, because equation (8.6) would not be able to give real $A$ (since $p^2$ will now be negative).

So what do we do to overcome this problem for $E<V $?

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  • $\begingroup$ Tunneling is explained in Griffiths section 8.2. $\endgroup$
    – Qmechanic
    Commented Jan 29, 2023 at 14:30
  • $\begingroup$ He doesn't explain it very well - he says that "only now p(x) is imaginary" but that means that θ is imaginary which is in contrast with his assumption that θ should be real. Also, if θ is imaginary he cannot break eq.(8.5) to two equations (8.6), (8.7). This is what is not clear to me. $\endgroup$
    – MTYS
    Commented Jan 29, 2023 at 14:36
  • $\begingroup$ In its current form your question is not very self-contained. Additionally, if you don’t understand Griffith’s derivation, what other derivations have you looked at? $\endgroup$ Commented Jan 29, 2023 at 15:18

1 Answer 1

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OP has a point: The polar form of the complex wavefunction $\psi$ is not useful in the classically forbidden region$E<V$ because the complex TISE then doesn't separate into 2 real equations.

Alternatively, consider the approach of Ref. 1 (which happens to be problem 8.2 in Ref. 2). Here the wavefunction is assumed to be on the semiclassical form

$$ \psi~=~\exp\left(\frac{i}{\hbar}\sigma\right), \tag{46.1} $$

where

$$ \sigma~=~\sum_{n=0}^{\infty}\left(\frac{\hbar}{i}\right)^n\sigma_n \tag{46.3}$$

is a complex power series in Planck's constant. The leading coefficient satisfies

$$ \sigma_0~=~\pm \int p \mathrm{d}x, \qquad p=\sqrt{2m(E-V)},\tag{46.5}$$

In eq. (46.5) the momentum $p$ is imaginary in the classically forbidden region $E<V$. The next-to-leading coefficient

$$ \sigma_1 ~=~-\frac{1}{2}{\rm Ln}(p) \tag{46.8}$$

is given in terms of a complex logarithm.

References:

  1. L.D. Landau & E.M. Lifshitz, QM, Vol. 3, 2nd & 3rd ed, 1981; $\S46$.

  2. D. Griffiths, Intro to QM, 1995; problem 8.2.

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  • $\begingroup$ thank you for your help! $\endgroup$
    – MTYS
    Commented Jan 30, 2023 at 20:23

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