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Posts Tagged ‘WKB method’

Review of Quantum mechanics approximation results.

Posted by peeterjoot on November 10, 2011

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Motivation.

Here I’ll summarize what I’d put on a cheat sheet for the tests or exam, if one would be allowed. While I can derive these results, memorization unfortunately appears required for good test performance in this class, and this will give me a good reference of what to memorize.

This set of review notes covers all the approximation methods we covered except for Fermi’s golden rule.

Variational method

We can find an estimate of our ground state energy using

\begin{aligned}\boxed{\frac{{\left\langle {\Psi} \right\rvert} H {\left\lvert {\Psi} \right\rangle}}{\left\langle{{\Psi}} \vert {{\Psi}}\right\rangle}\ge E_0}\end{aligned} \hspace{\stretch{1}}(2.1)

Time independent perturbation

Given a perturbed Hamiltonian and an associated solution for the unperturbed state

\begin{aligned}\boxed{\begin{aligned}H &= H_0 + \lambda H', \qquad \lambda \in [0,1] \\ H_0 {\left\lvert {{\psi_{m\alpha}}^{(0)}} \right\rangle} &= {E_m}^{(0)} {\left\lvert {{\psi_{m\alpha}}^{(0)}} \right\rangle},\end{aligned}}\end{aligned} \hspace{\stretch{1}}(3.2)

we assume a power series solution for the energy

\begin{aligned}E_m = {E_m}^{(0)} + \lambda {E_m}^{(1)} + \lambda^2 {E_m}^{(2)} + \cdots\end{aligned} \hspace{\stretch{1}}(3.3)

For a non-degenerate state ${\left\lvert {\psi_m} \right\rangle} = {\left\lvert {\psi_{m1}} \right\rangle}$, with an unperturbed value of ${\left\lvert {\psi_{m}^{(0)}} \right\rangle} = {\left\lvert {\psi_{m1}^{(0)}} \right\rangle}$, we seek a power series expansion of this ket in the perturbed system

\begin{aligned}\begin{aligned}{\left\lvert {\psi_m} \right\rangle} &= \sum_{n,\alpha} {c_{n\alpha;m}}^{(0)} {\left\lvert {{\psi_{n\alpha}}^{(0)}} \right\rangle} +\lambda\sum_{n,\alpha} {c_{n\alpha;m}}^{(1)} {\left\lvert {{\psi_{n\alpha}}^{(0)}} \right\rangle} + \lambda^2\sum_{n,\alpha} {c_{n\alpha;m}}^{(2)} {\left\lvert {{\psi_{n\alpha}}^{(0)}} \right\rangle} + \cdots \\ &\propto{\left\lvert {{\psi_m}^{(0)}} \right\rangle} + \lambda\sum_{n \ne m, \alpha} {\bar{c}_{n\alpha;m}}^{(1)} {\left\lvert {{\psi_{n\alpha}}^{(0)}} \right\rangle} +\lambda^2\sum_{n \ne m, \alpha} {\bar{c}_{n\alpha;m}}^{(2)} {\left\lvert {{\psi_{n\alpha}}^{(0)}} \right\rangle} + \cdots\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.4)

Any states $n \ne m$ are allowed to have degeneracy. For this case, we found to second order in energy and first order in the kets

\begin{aligned}\boxed{\begin{aligned}E_m &= E_m^{(0)} + \lambda {H_{m1;m1}}' + \lambda^2 \sum_{n \ne m, \alpha} \frac{{\left\lvert{{H_{n\alpha;m1}}'}\right\rvert}^2 }{ E_m^{(0)} - E_n^{(0)} } + \cdots\\ {\left\lvert {\psi_m} \right\rangle} &\propto {\left\lvert {{\psi_m}^{(0)}} \right\rangle} + \lambda\sum_{n \ne m, \alpha} \frac{{H_{n\alpha;m1}}'}{ E_m^{(0)} - E_n^{(0)} } {\left\lvert {{\psi_{n\alpha}}^{(0)}} \right\rangle}+ \cdots \\ H_{n\alpha;s\beta}' &={\left\langle {{\psi_{n\alpha}}^{(0)}} \right\rvert}H'{\left\lvert {{\psi_{s\beta}}^{(0)}} \right\rangle}.\end{aligned}}\end{aligned} \hspace{\stretch{1}}(3.5)

Degeneracy.

When the initial energy eigenvalue $E_m$ has a degeneracy $\gamma_m > 1$ we use a different approach to compute the perturbed energy eigenkets and perturbed energy eigenvalues. Writing the kets as ${\left\lvert {m\alpha} \right\rangle}$, then we assume that the perturbed ket is a superposition of the kets in the degenerate energy level

\begin{aligned}{\left\lvert {m \alpha} \right\rangle}' = \sum_i c_i {\left\lvert {m i} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(4.6)

We find that we must have

\begin{aligned}\left( (E^0 - E)I + \lambda \begin{bmatrix} H_{mi;mj}' \end{bmatrix} \right)\begin{bmatrix}c_1 \\ c_2 \\ \dot{v}s \\ c_{\gamma_m}\end{bmatrix}= 0.\end{aligned} \hspace{\stretch{1}}(4.7)

Diagonalizing this matrix $\begin{bmatrix} H_{mi;mj}' \end{bmatrix}$ (a subset of the complete $H'$ matrix element)

\begin{aligned}\begin{bmatrix}{\left\langle {m i} \right\rvert} H' {\left\lvert {m j} \right\rangle}\end{bmatrix}= U_m \begin{bmatrix}\delta_{ij} \mathcal{H}_{m,i}'\end{bmatrix} U_m^\dagger,\end{aligned} \hspace{\stretch{1}}(4.8)

we find, by taking the determinant, that the perturbed energy eigenvalues are in the set

\begin{aligned}\boxed{E = E_m^0 + \lambda \mathcal{H}_{m,i}', \quad i \in [1, \gamma_m]}\end{aligned} \hspace{\stretch{1}}(4.9)

To compute the perturbed kets we must work in a basis for which the block diagonal matrix elements are diagonal for all $m$, as in

\begin{aligned}\begin{bmatrix}{\left\langle {m i} \right\rvert} H' {\left\lvert {m j} \right\rangle}\end{bmatrix}= \begin{bmatrix}\delta_{ij} \mathcal{H}_{m,i}'\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(4.10)

If that is not the case, then the unitary matrices of 4.8 can be computed, and the matrix

\begin{aligned}U = \begin{bmatrix}U_1 & & & \\ & U_2 & & \\ & & \ddots & \\ & & & U_N \\ \end{bmatrix},\end{aligned} \hspace{\stretch{1}}(4.11)

can be formed. The kets

\begin{aligned}{\left\lvert {\overline{m \alpha}} \right\rangle} = U^\dagger {\left\lvert {m \alpha} \right\rangle},\end{aligned} \hspace{\stretch{1}}(4.12)

will still be energy eigenkets of the unperturbed Hamiltonian

\begin{aligned}H_0 {\left\lvert {\overline{m \alpha}} \right\rangle} = E_m^0 {\left\lvert {\overline{m \alpha}} \right\rangle},\end{aligned} \hspace{\stretch{1}}(4.13)

but also ensure that the partial diagonalization condition of 4.8 is satisfied. In this basis, dropping overbars, the first order perturbation results found previously for perturbation about a non-degenerate state also hold, allowing us to write

\begin{aligned}\boxed{{\left\lvert {s \alpha} \right\rangle}' = {\left\lvert {s \alpha} \right\rangle} + \lambda \sum_{m \ne s, \beta} \frac{{H'}_{m \beta ; s \alpha}}{ E_s^{(0)} - E_m^{(0)} } {\left\lvert {m \beta} \right\rangle}+ \cdots}\end{aligned} \hspace{\stretch{1}}(4.14)

Interaction picture.

We split of the Hamiltonian into time independent and time dependent parts, and also factorize the time evolution operator

\begin{aligned}\boxed{\begin{aligned}H &= H_0 + H_I(t) \\ {\left\lvert {\alpha_S} \right\rangle} &= e^{-i H_0 t/\hbar } {\left\lvert {\alpha_I(t)} \right\rangle} = e^{-i H_0 t/\hbar } U_I(t) {\left\lvert {\alpha_I(0)} \right\rangle} .\end{aligned}}\end{aligned} \hspace{\stretch{1}}(5.15)

Plugging into Schr\”{o}dinger’s equation we find

\begin{aligned}\boxed{\begin{aligned}i \hbar \frac{d{{}}}{dt} {\left\lvert {\alpha_I(t)} \right\rangle} &= H_I(t) {\left\lvert {\alpha_I(t)} \right\rangle} \\ i \hbar \frac{d{{U_I}}}{dt} &= H_I' U_I \\ H_I'(t) &= e^{i H_0 t/\hbar } H_I(t) e^{-i H_0 t/\hbar } \end{aligned}}\end{aligned} \hspace{\stretch{1}}(5.16)

Time dependent perturbation.

We moved on to time dependent perturbations of the form

\begin{aligned}\boxed{\begin{aligned}H(t) &= H_0 + H'(t) \\ H_0 {\left\lvert {\psi_n^{(0)} } \right\rangle} &= \hbar \omega_n {\left\lvert {\psi_n^{(0)} } \right\rangle}.\end{aligned}}\end{aligned} \hspace{\stretch{1}}(6.17)

where $\hbar \omega_n$ are the energy eigenvalues, and ${\left\lvert {\psi_n^{(0)} } \right\rangle}$ the energy eigenstates of the unperturbed Hamiltonian.

Use of the interaction picture led quickly to the problem of seeking the coefficients describing the perturbed state

\begin{aligned}{\left\lvert {\psi(t)} \right\rangle} = \sum_n c_n(t) e^{-i \omega_n t} {\left\lvert {\psi_n^{(0)} } \right\rangle},\end{aligned} \hspace{\stretch{1}}(6.18)

and plugging in we found

\begin{aligned}\boxed{\begin{aligned}i \hbar \cdot_s &= \sum_n H_{sn}'(t) e^{i \omega_{sn} t} c_n(t) \\ \omega_{sn} &= \omega_s - \omega_n \\ H_{sn}'(t) &= {\left\langle {\psi_s^{(0)}} \right\rvert} H'(t) {\left\lvert {\psi_n^{(0)} } \right\rangle},\end{aligned}}\end{aligned} \hspace{\stretch{1}}(6.19)

Perturbation expansion in series.

Introducing a $\lambda$ parametrized dependence in the perturbation above, and assuming a power series expansion of our coefficients

\begin{aligned}\boxed{\begin{aligned}H'(t) &\rightarrow \lambda H'(t) \\ c_s(t) &= c_s^{(0)}(t) + \lambda c_s^{(1)}(t) + \lambda^2 c_s^{(2)}(t) + \cdots\end{aligned}}\end{aligned} \hspace{\stretch{1}}(6.20)

we found, after equating powers of $\lambda$ a set of coupled differential equations

\begin{aligned}\begin{aligned}i \hbar \cdot_s^{(0)}(t) &= 0 \\ i \hbar \cdot_s^{(1)}(t) &= \sum_{n} H_{sn}'(t) e^{i \omega_{sn} t} c_n^{(0)}(t) \\ i \hbar \cdot_s^{(2)}(t) &= \sum_{n} H_{sn}'(t) e^{i \omega_{sn} t} c_n^{(1)}(t) \\ &\dot{v}s\end{aligned}\end{aligned} \hspace{\stretch{1}}(6.21)

Of particular value was the expansion, assuming that we started with an initial state in energy level $m$ before the perturbation was “turned on” (ie: $\lambda = 0$).

\begin{aligned}{\left\lvert {\psi(t)} \right\rangle} = e^{-i \omega_m t} {\left\lvert {\psi_m^{(0)} } \right\rangle}\end{aligned} \hspace{\stretch{1}}(6.22)

So that $c_n^{(0)}(t) = \delta_{nm}$. We then found a first order approximation for the transition probability coefficient of

\begin{aligned}\boxed{i \hbar \cdot_m^{(1)} = H_{ms}'(t) e^{i \omega_{ms} t}}\end{aligned} \hspace{\stretch{1}}(6.23)

Sudden perturbations.

The idea here is that we integrate Schr\”{o}dinger’s equation over the small interval containing the changing Hamiltonian

\begin{aligned}{\left\lvert {\psi(t)} \right\rangle} = {\left\lvert {\psi(t_0)} \right\rangle} + \frac{1}{{i\hbar}} \int_{t_0}^t H(t') {\left\lvert {\psi(t')} \right\rangle} dt'\end{aligned} \hspace{\stretch{1}}(7.24)

and find

\begin{aligned}\boxed{{\left\lvert {\psi_\text{after}} \right\rangle} = {\left\lvert {\psi_\text{before}} \right\rangle}.}\end{aligned} \hspace{\stretch{1}}(7.25)

An implication is that, say, we start with a system measured in a given energy, that same system after the change to the Hamiltonian will then be in a state that is now a superposition of eigenkets from the new Hamiltonian.

Given a Hamiltonian that turns on slowly at $t=0$, a set of instantaneous eigenkets for the duration of the time dependent interval, and a representation in terms of the instantaneous eigenkets

\begin{aligned}\boxed{\begin{aligned}H(t) &= H_0, \qquad t \le 0 \\ H(t) {\left\lvert {\hat{\psi}_n(t)} \right\rangle} &= E_n(t) {\left\lvert {\hat{\psi}_n(t)} \right\rangle} \\ {\left\lvert {\psi} \right\rangle} &= \sum_n \bar{b}_n(t) e^{-i\alpha_n + i \beta_n} {\left\lvert {\hat{\psi}_n} \right\rangle} \\ \alpha_n(t) &= \frac{1}{{\hbar}} \int_0^t dt' E_n(t'),\end{aligned}}\end{aligned} \hspace{\stretch{1}}(8.26)

plugging into Schr\”{o}dinger’s equation we find

\begin{aligned}\boxed{\begin{aligned}\frac{d{{\bar{b}_m}}}{dt} &= - \sum_{n \ne m} \bar{b}_n e^{-i \gamma_{nm} } {\left\langle {\hat{\psi}_m(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} \\ \gamma_{nm}(t) &= \alpha_n(t) - \alpha_m(t) - (\beta_n(t) - \beta_m(t)) \\ \beta_n(t) &= \int_0^t dt' \Gamma_n(t') \\ \Gamma_n(t) &= i {\left\langle {\hat{\psi}_n(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} \\ \end{aligned}}\end{aligned} \hspace{\stretch{1}}(8.27)

Evolution of a given state.

Given a system initially measured with energy $E_m(0)$ before the time dependence is “turned on”

\begin{aligned}\boxed{{\left\lvert {\psi(0)} \right\rangle} = {\left\lvert {\hat{\psi}_m(0)} \right\rangle},}\end{aligned} \hspace{\stretch{1}}(8.28)

we find that the first order Taylor series expansion for the transition probability coefficients are

\begin{aligned}\boxed{\bar{b}_s(t) = \delta_{sm} - t (1 - \delta_{sm}) {\left\langle {\hat{\psi}_s(0)} \right\rvert} {\left.{{\frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_m(t)} \right\rangle}}}\right\vert}_{{t=0}}.}\end{aligned} \hspace{\stretch{1}}(8.29)

If we introduce a $\lambda$ perturbation, separating all the (slowly changing) time dependent part of the Hamiltonian $H'$ from the non time dependent parts $H_0$ as in

\begin{aligned}H(t) = H_0 + \lambda H'(t)\end{aligned} \hspace{\stretch{1}}(8.30)

then we find our perturbed coefficients are

\begin{aligned}\boxed{\bar{b}_s(t) =\delta_{ms}(1 + \lambda \text{constant})- (1-\delta_{ms}) \lambda\int_0^t dt'e^{i \gamma_{sm}(t') } {\left\langle {\hat{\psi}_s(t')} \right\rvert} \frac{d}{dt'} {\left\lvert {\hat{\psi}_m(t')} \right\rangle} }\end{aligned} \hspace{\stretch{1}}(8.31)

WKB.

We write Schr\”{o}dinger’s equation as

\begin{aligned}\boxed{\begin{aligned}0 &= \frac{d^2 U}{dx^2} + k^2 U \\ k^2 &= -\kappa^2 = \frac{2m (E - V)}{\hbar}.\end{aligned}}\end{aligned} \hspace{\stretch{1}}(9.32)

and seek solutions of the form $U \propto e^{i\phi}$. Schr\”{o}dinger’s equation takes the form

\begin{aligned}- (\phi'(x))^2 + i \phi''(x) + k^2(x) = 0.\end{aligned} \hspace{\stretch{1}}(9.33)

Initially setting $\phi'' = 0$ we refine our approximation to find

\begin{aligned}\phi'(x) = k(x) \sqrt{ 1 + i \frac{k'(x)}{k^2(x)} } .\end{aligned} \hspace{\stretch{1}}(9.34)

To first order, this gives us

\begin{aligned}\boxed{U(x) \propto \frac{1}{{\sqrt{k(x)}}} e^{\pm i \int dx k(x)} }\end{aligned} \hspace{\stretch{1}}(9.35)

What we didn’t cover in class, but required in the problems was the Bohr-Sommerfeld condition described in section 24.1.2 of the text [1].

\begin{aligned}\boxed{\int_{x_1}^{x_2} dx \sqrt{ 2m (E - V(x))} = \left( n + \frac{1}{{2}} \right) \pi.}\end{aligned} \hspace{\stretch{1}}(9.36)

This was found from the WKB connection formulas, themselves found my some Bessel function arguments that I have to admit that I didn’t understand.

References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

PHY456H1F: Quantum Mechanics II. Lecture 12 (Taught by Mr. Federico Duque Gomez). WKB Method

Posted by peeterjoot on October 21, 2011

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

WKB (Wentzel-Kramers-Brillouin) Method.

This is covered in section 24 in the text [1]. Also section 8 of [2].

\begin{aligned}-\frac{\hbar^2}{2m} \frac{d^2 U}{dx^2} + V(x) U(x) = E U(x)\end{aligned} \hspace{\stretch{1}}(2.1)

which we can write as

\begin{aligned}\frac{d^2 U}{dx^2} + \frac{2m}{\hbar^2} (E - V(x)) U(x) = 0\end{aligned} \hspace{\stretch{1}}(2.2)

Consider a finite well potential as in figure (\ref{fig:qmTwoL13:qmTwoL12fig1})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL12fig1}
\caption{Finite well potential}
\end{figure}

With

\begin{aligned}k &= \frac{2m (E - V)}{\hbar},\qquad E > V \\ \kappa &= \frac{2m (V - E)}{\hbar}, \qquad V > E,\end{aligned} \hspace{\stretch{1}}(2.3)

we have for a bound state within the well

\begin{aligned}U \propto e^{\pm i k x}\end{aligned} \hspace{\stretch{1}}(2.5)

and for that state outside the well

\begin{aligned}U \propto e^{\pm \kappa x}\end{aligned} \hspace{\stretch{1}}(2.6)

In general we can hope for something similar. Let’s look for that something, but allow the constants $k$ and $\kappa$ to be functions of position

\begin{aligned}k^2(x) &= \frac{2m (E - V(x))}{\hbar},\qquad E > V \\ \kappa^2(x) &= \frac{2m (V(x) - E)}{\hbar}, \qquad V > E.\end{aligned} \hspace{\stretch{1}}(2.7)

In terms of $k$ Schr\”{o}dinger’s equation is just

\begin{aligned}\frac{d^2 U(x)}{dx^2} + k^2(x) U(x) = 0.\end{aligned} \hspace{\stretch{1}}(2.9)

We use the trial solution

\begin{aligned}U(x) = A e^{i \phi(x)},\end{aligned} \hspace{\stretch{1}}(2.10)

allowing $\phi(x)$ to be complex

\begin{aligned}\phi(x) = \phi_R(x) + i \phi_I(x).\end{aligned} \hspace{\stretch{1}}(2.11)

We need second derivatives

\begin{aligned}(e^{i \phi})'' &=(i \phi' e^{i \phi})' \\ &=(i \phi')^2 e^{i \phi} + i \phi'' e^{i \phi},\end{aligned}

and plug back into our Schr\”{o}dinger equation to obtain

\begin{aligned}- (\phi'(x))^2 + i \phi''(x) + k^2(x) = 0.\end{aligned} \hspace{\stretch{1}}(2.12)

For the first round of approximation we assume

\begin{aligned}\phi''(x) \approx 0,\end{aligned} \hspace{\stretch{1}}(2.13)

and obtain

\begin{aligned}(\phi'(x))^2 = k^2(x),\end{aligned} \hspace{\stretch{1}}(2.14)

or

\begin{aligned}\phi'(x) = \pm k(x).\end{aligned} \hspace{\stretch{1}}(2.15)

A second round of approximation we use 2.15 and obtain

\begin{aligned}\phi''(x) = \pm k'(x)\end{aligned} \hspace{\stretch{1}}(2.16)

Plugging back into 2.12 we have

\begin{aligned}-(\phi'(x))^2 \pm i k'(x) + k^2(x) = 0,\end{aligned} \hspace{\stretch{1}}(2.17)

or

\begin{aligned}\begin{aligned}\phi'(x) &= \pm \sqrt{ \pm i k'(x) + k^2(x) } \\ &= \pm k(x) \sqrt{ 1 \pm i \frac{k'(x)}{k^2(x)} } .\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.18)

If $k'$ is small compared to $k^2$

\begin{aligned}\frac{k'(x)}{k^2(x)} \ll 1, \end{aligned} \hspace{\stretch{1}}(2.19)

we have

\begin{aligned}\phi'(x) = \pm k(x) \pm i \frac{k'(x)}{2 k(x)} \end{aligned} \hspace{\stretch{1}}(2.20)

Integrating

\begin{aligned}\phi(x) &= \pm \int dx k(x) \pm i \int dx \frac{k'(x)}{2 k(x)} + \text{const} \\ &= \pm \int dx k(x) \pm i \frac{1}{{2}} \ln k(x) + \text{const} \end{aligned}

Going back to our wavefunction, if $E > V(x)$ we have

\begin{aligned}U(x) &\sim A e^{i \phi(x)} \\ &= \exp \left(i\left( \pm \int dx k(x) \pm i \frac{1}{{2}} \ln k(x) + \text{const} \right)\right) \\ &\sim \exp \left(i\left( \pm \int dx k(x) \pm i \frac{1}{{2}} \ln k(x) \right)\right) \\ &= e^{\pm i \int dx k(x)} e^{\mp \frac{1}{{2}} \ln k(x)} \\ \end{aligned}

or

\begin{aligned}U(x) \propto \frac{1}{{\sqrt{k(x)}}} e^{\pm i \int dx k(x)} \end{aligned} \hspace{\stretch{1}}(2.21)

FIXME: Question: the $\pm$ on the real exponential got absorbed here, but would not $U(x) \propto \sqrt{k(x)} e^{\pm i \int dx k(x)}$ also be a solution? If so, why is that one excluded?

Similarly for the $E < V(x)$ case we can find

\begin{aligned}U(x) \propto \frac{1}{{\sqrt{\kappa(x)}}} e^{\pm i \int dx \kappa(x)}.\end{aligned} \hspace{\stretch{1}}(2.22)

Validity
\begin{enumerate}
\item V(x) changes very slowly $\implies k'(x)$ small, and $k(x) = \sqrt{2 m (E - V(x))}/\hbar$.
\item $E$ very far away from the potential ${\left\lvert{(E - V(x))/V(x)}\right\rvert} \gg 1$.
\end{enumerate}

Examples

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL12fig2}
\caption{Example of a general potential}
\end{figure}

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL12fig3}
\caption{Turning points where WKB won’t work}
\end{figure}

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL12fig4}
\caption{Diagram for patching method discussion}
\end{figure}

WKB won’t work at the turning points in this figure since our main assumption was that

\begin{aligned}{\left\lvert{\frac{k'(x)}{k^2(x)}}\right\rvert} \ll 1\end{aligned} \hspace{\stretch{1}}(3.23)

so we get into trouble where $k(x) \sim 0$. There are some methods for dealing with this. Our text as well as Griffiths give some examples, but they require Bessel functions and more complex mathematics.

The idea is that one finds the WKB solution in the regions of validity, and then looks for a polynomial solution in the patching region where we are closer to the turning point, probably requiring lookup of various special functions.

This power series method is also outlined in [3], where solutions to connect the regions are expressed in terms of Airy functions.

References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

[2] D.J. Griffiths. Introduction to quantum mechanics, volume 1. Pearson Prentice Hall, 2005.

[3] Wikipedia. Wkb approximation — wikipedia, the free encyclopedia, 2011. [Online; accessed 19-October-2011]. http://en.wikipedia.org/w/index.php?title=WKB_approximation&oldid=453833635.