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WKB approximation

The wave function is expressed as the exponential of a function: Ψ (x) = e Φ (x), {\displaystyle \Psi (x)=e^{\Phi (x)},} where Φ ″ (x) + Φ ′ (x) 2 = 2 m ℏ 2 (V (x) − E). {\displaystyle \Phi ''(x)+\Phi '(x)^{2}={\frac {2m}{\hbar ^{2}}}\left(V(x)-E\right).} Φ ′ (x) {\displaystyle \Phi '(x)} is then separated into real and imaginary parts: Φ ′ (x) = A (x) + i B (x), {\displaystyle \Phi '(x)=A(x)+iB(x),} where A (x) and B (x) are real-valued functions.

Substituting the second equation into the first and using the fact that the real part needs to be 0 results in:

A ′ (x) + A (x) 2 − B (x) 2 = 2 m ℏ 2 (V (x) − E). {\displaystyle A'(x)+A(x)^{2}-B(x)^{2}={\frac {2m}{\hbar ^{2}}}\left(V(x)-E\right).}

To solve this equation using the semiclassical approximation, each function must be expanded as a power series in ℏ {\displaystyle \hbar }. From the equations, the power series must start with at least an order of ℏ − 1 {\displaystyle \hbar ^{-1}} to satisfy the real part of the equation; for a good classical limit starting with the highest power of the Planck constant possible is preferable, which leads to A (x) = 1 ℏ ∑ k = 0 ∞ ℏ k A k (x) {\displaystyle A(x)={\frac {1}{\hbar }}\sum _{k=0}^{\infty }\hbar ^{k}A_{k}(x)} and B (x) = 1 ℏ ∑ k = 0 ∞ ℏ k B k (x), {\displaystyle B(x)={\frac {1}{\hbar }}\sum _{k=0}^{\infty }\hbar ^{k}B_{k}(x),} with the following constraints on the lowest order terms, A 0 (x) 2 − B 0 (x) 2 = 2 m (V (x) − E) {\displaystyle A_{0}(x)^{2}-B_{0}(x)^{2}=2m\left(V(x)-E\right)} and A 0 (x) B 0 (x) = 0. {\displaystyle A_{0}(x)B_{0}(x)=0.}

At this point two extreme cases can be considered.

Case 1

If the amplitude varies slowly as compared to the phase A 0 (x) = 0 {\displaystyle A_{0}(x)=0} and B 0 (x) = ± 2 m (E − V (x)) {\displaystyle B_{0}(x)=\pm {\sqrt {2m\left(E-V(x)\right)}}} which corresponds to classical motion. Resolving the next order of expansion yields Ψ (x) ≈ C e i ∫ d x 2 m ℏ 2 (E − V (x)) + θ 2 m ℏ 2 (E − V (x)) 4 {\displaystyle \Psi (x)\approx C{\frac {e^{i\int dx{\sqrt {{\frac {2m}{\hbar ^{2}}}\left(E-V(x)\right)}}+\theta }}{\sqrt{{\frac {2m}{\hbar ^{2}}}\left(E-V(x)\right)}}}}

Case 2

If the phase varies slowly as compared to the amplitude, B 0 (x) = 0 {\displaystyle B_{0}(x)=0} and A 0 (x) = ± 2 m (V (x) − E) {\displaystyle A_{0}(x)=\pm {\sqrt {2m\left(V(x)-E\right)}}} which corresponds to tunneling. Resolving the next order of the expansion yields Ψ (x) ≈ C + e + ∫ d x 2 m ℏ 2 (V (x) − E) + C − e − ∫ d x 2 m ℏ 2 (V (x) − E) 2 m ℏ 2 (V (x) − E) 4 {\displaystyle \Psi (x)\approx {\frac {C_{+}e^{+\int dx{\sqrt {{\frac {2m}{\hbar ^{2}}}\left(V(x)-E\right)}}}+C_{-}e^{-\int dx{\sqrt {{\frac {2m}{\hbar ^{2}}}\left(V(x)-E\right)}}}}{\sqrt{{\frac {2m}{\hbar ^{2}}}\left(V(x)-E\right)}}}}

In both cases it is apparent from the denominator that both these approximate solutions are bad near the classical turning points E = V (x) {\displaystyle E=V(x)}. Away from the potential hill, the particle acts similar to a free and oscillating wave; beneath the potential hill, the particle undergoes exponential changes in amplitude. By considering the behaviour at these limits and classical turning points a global solution can be made.

To start, a classical turning point, x 1 {\displaystyle x_{1}} is chosen and 2 m ℏ 2 (V (x) − E) {\displaystyle {\frac {2m}{\hbar ^{2}}}\left(V(x)-E\right)} is expanded in a power series about x 1 {\displaystyle x_{1}}: 2 m ℏ 2 (V (x) − E) = v 1 (x − x 1) + v 2 (x − x 1) 2 + ⋯ {\displaystyle {\frac {2m}{\hbar ^{2}}}\left(V(x)-E\right)=v_{1}(x-x_{1})+v_{2}(x-x_{1})^{2}+\cdots }

Keeping only the first order term ensures linearity: 2 m ℏ 2 (V (x) − E) = v 1 (x − x 1). {\displaystyle {\frac {2m}{\hbar ^{2}}}\left(V(x)-E\right)=v_{1}(x-x_{1}).}

Using this approximation, the equation near x 1 {\displaystyle x_{1}} becomes a differential equation: d 2 d x 2 Ψ (x) = v 1 (x − x 1) Ψ (x). {\displaystyle {\frac {d^{2}}{dx^{2}}}\Psi (x)=v_{1}(x-x_{1})\Psi (x).}

This can be solved using Airy functions as solutions. Ψ (x) = C A A i (v 1 3 (x − x 1)) + C B B i (v 1 3 (x − x 1)) {\displaystyle \Psi (x)=C_{A}Ai\left({\sqrt{v_{1}}}(x-x_{1})\right)+C_{B}Bi\left({\sqrt{v_{1}}}(x-x_{1})\right)}

Taking these solutions for all classical turning points, a global solution can be formed that links the limiting solutions. Given the two coefficients on one side of a classical turning point, the two coefficients on the other side of a classical turning point can be determined by using this local solution to connect them.

Hence, the Airy function solutions will asymptote into sine, cosine and exponential functions in the proper limits. The relationships between C, θ {\displaystyle C,\theta } and C +, C − {\displaystyle C_{+},C_{-}} are C + = 1 2 C cos (θ − π 4) {\displaystyle C_{+}={\frac {1}{2}}C\cos {\left(\theta -{\frac {\pi }{4}}\right)}} and

With the coefficients found, the global solution can be found. Therefore, the transmission coefficient for a particle tunneling through a single potential barrier is T (E) = e − 2 ∫ x 1 x 2 d x 2 m ℏ 2, {\displaystyle T(E)=e^{-2\int _{x_{1}}^{x_{2}}dx{\sqrt {{\frac {2m}{\hbar ^{2}}}\left}}},} where x 1, x 2 {\displaystyle x_{1},x_{2}} are the two classical turning points for the potential barrier.

For a rectangular barrier, this expression simplifies to: T (E) = e − 2 2 m ℏ 2 (V 0 − E) (x 2 − x 1). {\displaystyle T(E)=e^{-2{\sqrt {{\frac {2m}{\hbar ^{2}}}(V_{0}-E)}}(x_{2}-x_{1})}.}