Background of the question (see pp. 161, section 47 in Landau & Lifshitz's quantum mechanics textbook Vol3, 2nd Ed. Pergamon Press). We have the following potential well $$U(x)\leq E \quad\text{for} \quad x \leq a ,$$ $$U(x)>E \quad\text{for} \quad x>a .\tag{47.0}$$

The WKB solutions right and left to the turning point are

$$\begin{align}\psi=&\dfrac{C}{2\sqrt{p}}\exp{\left(-\dfrac{1}{\hbar}\left|\int_a^x pdx\right|\right)} \cr &\quad \text{for} \quad x>a, \end{align} \tag{47.1}$$

$$\begin{align} \psi= & \dfrac{C_1}{\sqrt{p}}\exp{\left(\dfrac{i}{\hbar}\int_a^x pdx\right)}\cr +&\dfrac{C_2}{\sqrt{p}}\exp{\left(-\dfrac{i}{\hbar}\int_a^x p dx\right)}\cr &\quad \text{for} \quad x<a,\end{align} \tag{47.2}$$

respectively. Most quantum mechanics textbooks determine the relation between $C$ and $C_i$'s by finding the exact solution near the turning point. And then let the exact solution match with the WKB solutions.

However, in Landau & Lifshitz's quantum mechanics textbooks (vol3, section 47) they let $x$ vary in the complex plane and pass around the turning point $a$ from right to left through a large semicircle in the upper complex plane. Landau claims starting from +$\infty$, when arriving at $-\infty$ (left to $a$), there is phase gain $\pi$ in the denominator of prefactor in Eq.(2). From this we can determine $$C_2=\frac{C}{2}\exp\left(i\frac{\pi}{4}\right).\tag{47.4a}$$

They also claim the first term will exponentially decay along the semicircle in the upper half plane. Question is why? Can we show $$\Im{\left(\int_a^x pdx\right)},$$ where $\Im$ stands for the imaginary part, is positive?


1 Answer 1


I) L&L are referring to a linearized model where the TISE

$$\begin{align} \hbar^2\psi^{\prime\prime}(x) +P^2(x) \psi(x)~=~&0,\cr P(x)~:=~ \sqrt{2m (E-V(x))}~=~& |P(x)| e^{i\theta(x)}, \end{align} \tag{1}$$

becomes the Airy differential equation. Near a turning point $x_0$, we can approximate the square momentum

$$\begin{align} P^2(x)~\approx~& \alpha (x\!-\!x_0), \cr (P^2)^{\prime}(x)~\approx~&\alpha~=~2mF_0 ~\neq~0,\end{align} \tag{2} $$

with an affine function, where $\alpha\in \mathbb{R}\backslash\{0\}$ is a non-zero real constant. (In L&L's example $\alpha < 0$ is negative).

II) Let us first reproduce L&L's argument. In the classically allowed region $x<x_0$, L&L choose the momentum $P^{>}>0$ positive. (The superscripts $>$ and $<$ refer to the classically allowed and forbidden regions, respectively.) Hence the WKB wave function (47.2) becomes

$$\begin{align} \psi^{>}~\stackrel{(47.2)}{=}&~\frac{1}{\sqrt{P^{>}}}\sum_{\pm}C_{\pm} \exp\left(\pm \frac{i}{\hbar}\int_{x_0}^x \! P^{>}~\mathrm{d}x\right) \cr ~=~&\frac{1}{\sqrt{|P|}} \sum_{\pm}C_{\pm} \exp\left( \mp \frac{2i\sqrt{|\alpha|}}{3\hbar}\left(x_0-x \right)^{\frac{3}{2}}\right)\cr & \qquad\text{for}\qquad x<x_0.\end{align}\tag{3}$$

III) L&L is next analytically continuing the WKB wave function (47.1) (which is valid in the classically forbidden region) along an (upper/lower) semicircle in the complex $x$-plane

$$ x-x_0~=~\rho e^{\pm i\phi} , \qquad \phi~\in~[0,\pi], \tag{47.4a}$$

in order to derive the WKB connection formulas. L&L choose the momentum variable

$$\begin{align} P^{<}~=~&\sqrt{|\alpha|(x-x_0)}\cr ~=~&\sqrt{|\alpha|}\rho^{\frac{1}{2}} e^{\pm\frac{ i\phi}{2}}, \qquad \rho~>~0,\end{align}\tag{4} $$

such that it starts out positive

$$ \left. P^{<}\right|_{\phi=0} ~=~|P|~>~0. \tag{5} $$

The analytical continuation of the WKB wave function (47.1) then becomes

$$\begin{align} \psi^{<}~\stackrel{(47.1)}{=}&~\frac{C}{2\sqrt{P^{<}}} \exp\left( -\frac{1}{\hbar}\int_{x_0}^x \! P^{<}~\mathrm{d}x\right)\cr ~=~&\frac{C}{2\sqrt{|P|}} \exp\left( -\frac{2\sqrt{|\alpha|}\rho^{\frac{3}{2}}e^{\pm\frac{ i3\phi}{2}} }{3\hbar} \mp i\frac{\phi}{4}\right). \end{align}\tag{6} $$

By setting $\phi=\pi$ in eq. (6), and comparing with the WKB wave function (3), one deduces from the analytical continuation along the (upper/lower) semicircle that

$$\begin{align} C_2~\equiv~& C_-~=~\frac{C}{2}e^{-\frac{i\pi}{4}}, \tag{47.4b}\cr C_1~\equiv~& C_+~=~\frac{C}{2}e^{\frac{i\pi}{4}} , \tag{47.4c}\end{align} $$

respectively. This is L&L's main argument. (We will return to the question of what happens to the other branch in Section VII below.)

IV) Alternatively, since the Airy functions have Fourier integral representations (cf. e.g. my Math.SE answer here), one may instead use the method of steepest descent to derive matching asymptotic expansions on each side of the turning point. It is interesting to compare this method with L&L's above argument.

One may show that the Airy-like wave function

$$ \psi(x) ~=~ \sqrt{\frac{|\alpha|}{\pi\hbar}}\int_{\mathbb{R}-i~{\rm sgn}(\alpha)~\varepsilon} \! \mathrm{d}p~ \exp\left(\frac{i}{\hbar}S(p,x)\right),\tag{7}$$ $$\begin{align} S(p,x)~:=~& px - \tilde{S}(p), \cr \tilde{S}(p)~:=~& p\left(x_0+\frac{p^2}{3\alpha}\right),\end{align} \tag{8}$$ $$\begin{align} S^{\prime}(p,x)~=~&x-\tilde{S}^{\prime}(p)\cr ~=~& x\!-\!x_0 - \frac{p^2}{\alpha}\cr ~\stackrel{(2)}{\approx}~& \frac{P^2(x)-p^2}{\alpha},\end{align} \tag{9} $$ $$-S^{\prime\prime}(p,x)~=~\tilde{S}^{\prime\prime}(p)~=~ \frac{2p}{\alpha},\tag{10} $$

satisfies the TISE (1):

$$\begin{align} \hbar^2& \psi^{\prime\prime}(x) +P^2(x) \psi(x)\cr ~\stackrel{(7)}{=}~& \sqrt{\frac{|\alpha|}{\pi\hbar}} \int_{\mathbb{R}-i~{\rm sgn}(\alpha)~\varepsilon} \! \mathrm{d}p~\left(P^2(x)- p^2\right) \exp\left(\frac{i}{\hbar}S(p,x)\right)\cr ~\stackrel{(9)}{\approx}~& -i \hbar \alpha\sqrt{\frac{|\alpha|}{\pi\hbar}} \int_{\mathbb{R}-i~{\rm sgn}(\alpha)~\varepsilon} \! \mathrm{d}p~\frac{d}{dp} \exp\left(\frac{i}{\hbar}S(p,x)\right)\cr ~=~&0 \end{align}\tag{11}$$

for the chosen contour in the complex $p$-plane.

V) There are 2 critical points:

$$\begin{align} p_{\sigma} ~=~& \sigma~ {\rm sgn}(\alpha)~P , \cr S_{\sigma}~=~& \frac{2\sigma P^3 }{3|\alpha|} , \cr S^{\prime\prime}_{\sigma}~=~& -2\sigma |\alpha| P , \cr \sigma ~\in~&\{\pm 1\}.\end{align} \tag{12}$$

The steepest descent contribution from each critical point comes from a Gaussian integral:

$$\begin{align} I_{\sigma}~=~& \sqrt{\frac{|\alpha|}{\pi\hbar}} \sqrt{\frac{2\pi i\hbar}{S^{\prime\prime}_{\sigma} } } \exp\left(\frac{i S_{\sigma}}{\hbar}\right)\cr ~=~&\frac{1}{\sqrt{P}} \exp\left(\frac{2i\sigma P^3}{3\hbar |\alpha|} -\frac{i\sigma \pi}{4}\right)\cr ~=~& \frac{1}{\sqrt{|P|}} \exp\left(\frac{2i\sigma |P|^3e^{3i\theta}}{3\hbar |\alpha|} -i\left(\frac{\sigma \pi}{4}+\frac{\theta}{2}\right)\right) \end{align} \tag{13} $$

with angular steepest descent direction $-\frac{\sigma \pi}{4}-\frac{\theta}{2}$.

$\uparrow$ Fig. 1. The complex $p$-plane with 3 possible integration contours ${\cal C}_1$, ${\cal C}_2$, ${\cal C}_3$ for $\alpha<0$. The shaded regions denote exponentially decaying sectors. (If $\alpha>0$, it is opposite.) Since the integrand is an entire function in $p$, the $p$-integral only depend on the monodromy of the contour. (Figure taken from Ref. [W].)

VI) Now let us return to L&L's example with $\alpha<0$.

Classically forbidden region $x>x_0$: Then $P=\pm i|P|$, with $\theta=\pm \frac{\pi}{2}$, where $\pm$ corresponds to two different possible sign choices. The contour along the real $p$-axis is reproduced by the steepest descent contour corresponding to the critical point $p_{\mp 1}$:

$$\begin{align} \psi_{<}(x)~\sim~& I_{\mp 1}\cr ~=~& \frac{1}{\sqrt{|P|}} \exp\left(-\frac{2|P|^3}{3\hbar |\alpha|} \right). \end{align} \tag{14} $$

Classically allowed region $x<x_0$: Then $P=|P|>0$ with $\theta=0$. The contour along the real $p$-axis is reproduced by the sum of both steepest descent contours:

$$\begin{align} \psi_{>}(x)~\sim~& \sum_{\sigma\in\{\pm 1\}}I_{\sigma}\cr ~=~& \frac{1}{\sqrt{|P|}}\sum_{\sigma\in\{\pm 1\}} \exp\left(i\sigma\left(\frac{2|P|^3}{3\hbar |\alpha|} -\frac{ \pi}{4}\right)\right)\cr ~=~&\frac{2}{\sqrt{|P|}} \cos\left(\frac{2|P|^3}{3\hbar |\alpha|} -\frac{\pi}{4} \right)\cr ~=~& \frac{2}{\sqrt{|P|}} \sin\left(\frac{2|P|^3}{3\hbar |\alpha|} +\frac{\pi}{4} \right).\end{align} \tag{15} $$

By comparing eqs. (14) & (15), we have derived a connection formula. In detail, by comparing the contributions to the critical points $p_{\mp 1}$, we deduce L&L's eqs. (47.4c) and (47.4b), respectively. So the two methods agree.

VII) Exponentially growing solutions are physically possible if the classically forbidden region has finite length, e.g. in quantum tunneling. However, in L&L's example, the classically forbidden region is non-compact, so exponentially growing solutions must be discarded.

It remains to explain why only one of the 2 oscillatory branches in the classically allowed side gets analytically continued to the classically forbidden side.

On one hand, L&L argue that the exponentially growing WKB branch cannot be trusted during the $\frac{3\pi}{2}$ phase shift in the Boltzmann factor because it becomes momentarily exponentially suppressed. This highlights the fact that the connection formulas are one-directional.

On the other hand, from the perspective of the method of steepest descent in Sections IV-VI, one crosses a Stokes line in the complex $x$-plane, so that the steepest descent contours pick up monodromy in the complex $p$-plane, cf. Fig. 1.

For more information, see also e.g. my related Phys.SE answer here.

VIII) Finally, we should mention that the $\frac{\pi}{2}$ phase shift between the incoming and outgoing wave at the turning point in eq. (15) corresponds to a Maslov index $|\mu|=1$, cf. e.g. this Phys.SE post.


  • [LL] L.D. Landau & E.M. Lifshitz, QM, Vol. 3, 2nd & 3rd ed, 1981; $\S47$ & $\S b$.

  • [W] E. Witten, Analytic Continuation Of Chern-Simons Theory, arXiv:1001.2933; p. 23-29, 48-49. A related 2015 KITP lecture by Witten, A New Look At The Path Integral Of Quantum Mechanics, can be found on YouTube.

  • $\begingroup$ L&L propose to go backwards from $\pi$ to $0$ in the plane $y>0$. Then $\frac{c_2}{\sqrt{p}}e^{-i\int p(x)dx}$ becomes small. To make the term $\frac{c_1}{\sqrt{p}}e^{+i\int p(x)dx}$, small one has to do the same for $y<0$. Could you please comment how one could possibly deduce this by looking at $\frac{c_1}{\sqrt{p}}e^{+i\int p(x)dx}+\frac{c_2}{\sqrt{p}}e^{-i\int p(x)dx}$. I don't see how the formula $x-a=\rho e^{i\phi}$ is applied here, if applied at all. I don't really get it from your answer. thank you in advance $\endgroup$ Commented Apr 25, 2018 at 18:23
  • $\begingroup$ Thanks for the feedback. I plan to review this in the near future. $\endgroup$
    – Qmechanic
    Commented May 1, 2018 at 19:28
  • $\begingroup$ I gave it a little taught. I think that this is quite simple, as Landau himself writes. If one would continue $\frac{c_1}{\sqrt{p(x)}}e^{i\int p(x)dx}$ in the complex plane via $x-a=\rho e^{i\phi}$ one gets $e^{-\sin{\frac{3\phi}{2}}+i \cos{\frac{3\phi}{2}}}$. The same can be done for $\frac{c_1}{\sqrt{p(x)}}e^{-i\int p(x)dx}=e^{+\sin{\frac{3\phi}{2}}-i \cos{\frac{3\phi}{2}}}$. The sum of these two terms would give the desired result if one would go from $0,2\pi$ or backwards $2\pi,0$. $\endgroup$ Commented May 2, 2018 at 9:17

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