1
$\begingroup$

The derivation is obtained from Introduction to Quantum Mechanics by Griffiths

Let assume the turning point occurs at $x=0$, then the WKB solutions right and left to the turning point are:

$$ \psi=\dfrac{C}{\sqrt{|p|}}\exp{\left(-\dfrac{1}{\hbar}\left|\int_0^x pdx\right|\right)} \quad \text{for} \quad x>0, \tag{8.31a} $$

$$ \psi=\dfrac{C_1}{\sqrt{p}}\exp{\left(\dfrac{i}{\hbar}\int_0^x pdx\right)}+\dfrac{C_2}{\sqrt{p}}\exp{\left(-\dfrac{i}{\hbar}\int_0^x p dx\right)}\quad \text{for} \quad x<0.\tag{8.31b} $$

By assuming the potential near $x=0$ to be $$ V(x) \cong E + V'(0)x \tag{8.32} $$

and solving the Schrödinger equation, he takes the patching wave function as $$ \psi_p(x) = aAi(\alpha x) + bBi(\alpha x).\tag{8.37} $$

Then for $x > 0$, he uses the asymptotic forms of the Airy functions to obtain $$ \psi_p(x) \cong \dfrac{a}{2\sqrt{\pi}(\alpha x)^{1/4}}\exp(-\dfrac{2}{3}(\alpha x)^{3/2}) + \dfrac{b}{\sqrt{\pi}(\alpha x)^{1/4}}\exp(\dfrac{2}{3}(\alpha x)^{3/2}),\tag{8.40} $$

and use this to obtain the constant $C$,

My question is, why can we assume $x \gg 0$? Shouldn't this wave function stay in the neighborhood of the origin?

$\endgroup$
1
$\begingroup$

Actually, Griffiths does not assume that the position coordinate $x\to \infty$ is unbounded, cf. footnote 8 on p. 288. To the contrary, he assumes that $x$ does not leave the patching region so that the linear approximation (8.32) to the potential is valid, cf. Fig. 8.9. Instead he assumes that the argument (to the Airy function) $$ z~\equiv~\alpha x~\gg~1 \tag{8.35}$$ is large, where $$ \alpha~~\equiv~\left[\frac{2m}{\hbar^2}V^{\prime}(0)\right]^{1/3}. \tag{8.34}$$ Semiclassically, $\alpha\to \infty$ for $\hbar\to 0$. Griffiths makes the assumption $z\gg 1$, since he is only after the leading semiclassical behavior.

For alternative derivations of the WKB connection formulas, see e.g. this related Phys.SE post.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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