D.46 De­riva­tion of the WKB ap­prox­i­ma­tion

The pur­pose in this note is to de­rive an ap­prox­i­mate so­lu­tion to the Hamil­ton­ian eigen­value prob­lem

$$\frac{{\rm d}^2\psi}{{\rm d}x^2} = - \frac{p_{\rm {c}}^2}{\hbar^2}\psi$$

where the clas­si­cal mo­men­tum $p_{\rm {c}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sqrt{2m(E-V)}$ is a known func­tion for given en­ergy. The ap­prox­i­ma­tion is to be valid when the val­ues of $p_{\rm {c}}$$\raisebox{.5pt}{$/$}$​$\hbar$ are large. In quan­tum terms, you can think of that as due to an en­ergy that is macro­scop­i­cally large. But to do the math­e­mat­ics, it is eas­ier to take a macro­scopic point of view; in macro­scopic terms, $p_{\rm {c}}$$\raisebox{.5pt}{$/$}$​$\hbar$ is large be­cause Planck’s con­stant $\hbar$ is so small.

Since ei­ther way $p_{\rm {c}}$$\raisebox{.5pt}{$/$}$​$\hbar$ is a large quan­tity, for the left hand side of the Hamil­ton­ian eigen­value prob­lem above to bal­ance the right hand side, the wave func­tion must vary rapidly with po­si­tion. Some­thing that varies rapidly and non­triv­ially with po­si­tion tends to be hard to an­a­lyze, so it turns out to be a good idea to write the wave func­tion as an ex­po­nen­tial,

$$\psi = e^{{\rm i}\tilde\theta}$$

and then ap­prox­i­mate the ar­gu­ment $\tilde\theta$ of that ex­po­nen­tial.

To do so, first the equa­tion for $\tilde\theta$ will be needed. Tak­ing de­riv­a­tives of $\psi$ us­ing the chain rule gives in terms of $\tilde\theta$

$$\frac{{\rm d}\psi}{{\rm d}x} = e^{{\rm i}\tilde\theta} {\r... ...\tilde\theta} {\rm i}\frac{{\rm d}^2\tilde\theta}{{\rm d}x^2}$$

Then plug­ging $\psi$ and its sec­ond de­riv­a­tive above into the Hamil­ton­ian eigen­value prob­lem and clean­ing up gives:

$$\left(\frac{{\rm d}\tilde\theta}{{\rm d}x}\right)^2 = \fra... ...{\hbar^2} + {\rm i}\frac{{\rm d}^2\tilde\theta}{{\rm d}x^2} %$$ (D.30)

For a given en­ergy, $\tilde\theta$ will de­pend on both what $x$ is and what $\hbar$ is. Now, since $\hbar$ is small, math­e­mat­i­cally it sim­pli­fies things if you ex­pand $\tilde\theta$ in a power se­ries with re­spect to $\hbar$:

$$\tilde\theta = \frac{1}{\hbar}\left(f_0 + \hbar f_1 + {\textstyle\frac{1}{2}} \hbar^2 f_2 + \ldots \right)$$

You can think of this as writ­ing $\hbar\theta$ as a Tay­lor se­ries in $\hbar$. The co­ef­fi­cients $f_0,f_1,f_2,\ldots$ will de­pend on $x$. Since $\hbar$ is small, the con­tri­bu­tion of $f_2$ and fur­ther terms to $\psi$ is small and can be ig­nored; only $f_0$ and $f_1$ will need to be fig­ured out.

Plug­ging the power se­ries into the equa­tion for $\tilde\theta$ pro­duces

$$\frac{1}{\hbar^2}f_0'^2 + \frac{1}{\hbar} 2 f_0' f_1' + \ld... ...{\hbar^2}p_{\rm {c}}^2 + \frac{1}{\hbar} {\rm i}f_0'' + \ldots$$

where primes de­note $x$-​de­riv­a­tives and the dots stand for pow­ers of $\hbar$ greater than $\hbar^{-1}$ that will not be needed. Now for two power se­ries to be equal, the co­ef­fi­cients of each in­di­vid­ual power must be equal. In par­tic­u­lar, the co­ef­fi­cients of 1$\raisebox{.5pt}{$/$}$​$\hbar^2$ must be equal, $f_0'^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $p_{\rm {c}}^2$, so there are two pos­si­ble so­lu­tions

$$f_0' = \pm p_{\rm {c}}$$

For the co­ef­fi­cients of 1$\raisebox{.5pt}{$/$}$​$\hbar$ to be equal, $2f_0'f_1'$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${{\rm i}}f_0''$, or plug­ging in the so­lu­tion for $f_0'$,

$$f_1'= {\rm i}\frac{p_{\rm {c}}'}{2p_{\rm {c}}}$$

It fol­lows that the $x$-​de­riv­a­tive of $\tilde\theta$ is given by

$$\tilde\theta' = \frac{1}{\hbar} \left( \pm p_{\rm {c}} + \hbar {\rm i}\frac{p_{\rm {c}}'}{2p_{\rm {c}}} + \ldots \right)$$

and in­te­grat­ing gives $\tilde\theta$ as

$$\tilde\theta = \pm \frac{1}{\hbar} \int p_{\rm {c}}{ \rm... ...rm i}{\textstyle\frac{1}{2}} \ln p_{\rm {c}} + \tilde C \ldots$$

where $\tilde{C}$ is an in­te­gra­tion con­stant. Fi­nally, $e^{{\rm i}\tilde\theta}$ now gives the two terms in the WKB so­lu­tion, one for each pos­si­ble sign, with $e^{{\rm i}\tilde{C}}$ equal to the con­stant $C_{\rm {f}}$ or $C_{\rm {b}}$.