13.1 The WKB Method The WKB (Wentzel—Kramers—Brillouin) method provides approximate solutions to the one-dimensional Schrodinger equation when the potential varies slowly compared to the de Broglie wavelength.
The ansatz ψ ( x ) = A ( x ) e i S ( x ) / ℏ \psi(x) = A(x)e^{iS(x)/\hbar} ψ ( x ) = A ( x ) e i S ( x ) /ℏ substituted into − ℏ 2 2 m ψ " ′ + V ψ = E ψ -\frac{\hbar^2}{2m}\psi"' + V\psi = E\psi − 2 m ℏ 2 ψ " ′ + V ψ = E ψ gives, to leading order in ℏ \hbar ℏ :
S ( x ) = ± ∫ x p ( x ′ ) d x ′ , p ( x ) = 2 m [ E − V ( x ) ] S(x) = \pm\int^x p(x')\,dx', \quad p(x) = \sqrt{2m[E - V(x)]} S ( x ) = ± ∫ x p ( x ′ ) d x ′ , p ( x ) = 2 m [ E − V ( x )]
The WKB wavefunctions:
ψ ( x ) ≈ C p ( x ) exp ( ± i ℏ ∫ x p ( x ′ ) d x ′ ) (classically allowed, E > V) \psi(x) \approx \frac{C}{\sqrt{p(x)}}\exp\!\left(\pm\frac{i}{\hbar}\int^x p(x')\,dx'\right) \quad \text{(classically allowed, E > V\text{)}} ψ ( x ) ≈ p ( x ) C exp ( ± ℏ i ∫ x p ( x ′ ) d x ′ ) (classically allowed, E > V )
ψ ( x ) ≈ C ∣ p ( x ) ∣ exp ( ± 1 ℏ ∫ x ∣ p ( x ′ ) ∣ d x ′ ) (classically forbidden, E < V) \psi(x) \approx \frac{C}{\sqrt{|p(x)|}}\exp\!\left(\pm\frac{1}{\hbar}\int^x |p(x')|\,dx'\right) \quad \text{(classically forbidden, E < V\text{)}} ψ ( x ) ≈ ∣ p ( x ) ∣ C exp ( ± ℏ 1 ∫ x ∣ p ( x ′ ) ∣ d x ′ ) (classically forbidden, E < V )
At a classical turning point (E = V ( x 0 ) E = V(x_0) E = V ( x 0 ) ), the WKB approximation breaks down. The Airy function connects the oscillating and decaying solutions:
2 p ( x ) cos ( 1 ℏ ∫ x x 0 p ( x ′ ) d x ′ − π 4 ) ⟷ 1 ∣ p ( x ) ∣ exp ( − 1 ℏ ∫ x 0 x ∣ p ( x ′ ) ∣ d x ′ ) \frac{2}{\sqrt{p(x)}}\cos\!\left(\frac{1}{\hbar}\int_x^{x_0} p(x')\,dx' - \frac{\pi}{4}\right) \longleftrightarrow \frac{1}{\sqrt{|p(x)|}}\exp\!\left(-\frac{1}{\hbar}\int_{x_0}^x |p(x')|\,dx'\right) p ( x ) 2 cos ( ℏ 1 ∫ x x 0 p ( x ′ ) d x ′ − 4 π ) ⟷ ∣ p ( x ) ∣ 1 exp ( − ℏ 1 ∫ x 0 x ∣ p ( x ′ ) ∣ d x ′ )
13.3 Bohr—Sommerfeld Quantisation The WKB quantisation condition for a bound state in a potential well with turning points a a a and b b b :
∫ a b p ( x ) d x = ( n + 1 2 ) π ℏ , n = 0 , 1 , 2 , … \int_a^b p(x)\,dx = \left(n + \frac{1}{2}\right)\pi\hbar, \quad n = 0, 1, 2, \ldots ∫ a b p ( x ) d x = ( n + 2 1 ) π ℏ , n = 0 , 1 , 2 , …
The factor of 1 / 2 1/2 1/2 (Maslov index) accounts for the phase loss at each turning point.
Application: Harmonic oscillator. V ( x ) = 1 2 m ω 2 x 2 V(x) = \frac{1}{2}m\omega^2 x^2 V ( x ) = 2 1 m ω 2 x 2 . Turning points at x = ± 2 E / ( m ω 2 ) x = \pm\sqrt{2E/(m\omega^2)} x = ± 2 E / ( m ω 2 ) .
∫ − A A 2 m E − m 2 ω 2 x 2 d x = π E ω = ( n + 1 2 ) π ℏ \int_{-A}^{A}\sqrt{2mE - m^2\omega^2 x^2}\,dx = \frac{\pi E}{\omega} = \left(n + \frac{1}{2}\right)\pi\hbar ∫ − A A 2 m E − m 2 ω 2 x 2 d x = ω π E = ( n + 2 1 ) π ℏ
E n = ( n + 1 2 ) ℏ ω E_n = \left(n + \frac{1}{2}\right)\hbar\omega E n = ( n + 2 1 ) ℏ ω
The WKB gives the exact result for the harmonic oscillator --- a fortunate coincidence due to the quadratic potential.
Application: Power-law potential. For V ( x ) = V 0 ∣ x / a ∣ α V(x) = V_0|x/a|^\alpha V ( x ) = V 0 ∣ x / a ∣ α :
E n ∝ ( n + 1 2 ) 2 α / ( α + 2 ) E_n \propto \left(n + \frac{1}{2}\right)^{2\alpha/(\alpha+2)} E n ∝ ( n + 2 1 ) 2 α / ( α + 2 )
Worked Example 13.1: WKB Tunnelling Through a Barrier For a potential barrier V ( x ) = V 0 ( 1 − x 2 / a 2 ) V(x) = V_0(1 - x^2/a^2) V ( x ) = V 0 ( 1 − x 2 / a 2 ) for ∣ x ∣ < a |x| < a ∣ x ∣ < a With E < V 0 E < V_0 E < V 0 The WKB transmission probability is:
T ≈ exp ( − 2 ℏ ∫ − a 0 a 0 2 m ( V 0 ( 1 − x 2 / a 2 ) − E ) d x ) T \approx \exp\!\left(-\frac{2}{\hbar}\int_{-a_0}^{a_0}\sqrt{2m(V_0(1 - x^2/a^2) - E)}\,dx\right) T ≈ exp ( − ℏ 2 ∫ − a 0 a 0 2 m ( V 0 ( 1 − x 2 / a 2 ) − E ) d x )
Where a 0 = a 1 − E / V 0 a_0 = a\sqrt{1 - E/V_0} a 0 = a 1 − E / V 0 is the classical turning point.
T ≈ exp ( − 2 ℏ 2 m V 0 ∫ − a 0 a 0 1 − E / V 0 − x 2 / a 2 d x ) T \approx \exp\!\left(-\frac{2}{\hbar}\sqrt{2mV_0}\int_{-a_0}^{a_0}\sqrt{1 - E/V_0 - x^2/a^2}\,dx\right) T ≈ exp ( − ℏ 2 2 m V 0 ∫ − a 0 a 0 1 − E / V 0 − x 2 / a 2 d x )
= exp ( − 2 ℏ 2 m V 0 ⋅ π a 2 2 a ( 1 − E / V 0 ) ) = \exp\!\left(-\frac{2}{\hbar}\sqrt{2mV_0}\cdot\frac{\pi a^2}{2a}(1 - E/V_0)\right) = exp ( − ℏ 2 2 m V 0 ⋅ 2 a π a 2 ( 1 − E / V 0 ) )
= exp ( − π a ℏ 2 m V 0 ( 1 − E V 0 ) ) = \exp\!\left(-\frac{\pi a}{\hbar}\sqrt{2mV_0}\left(1 - \frac{E}{V_0}\right)\right) = exp ( − ℏ π a 2 m V 0 ( 1 − V 0 E ) )
For alpha decay (V 0 ≈ 25 V_0 \approx 25 V 0 ≈ 25 MeV, a ≈ 30 a \approx 30 a ≈ 30 fm, E = 5 E = 5 E = 5 MeV, m = 4 × 931.5 m = 4 \times 931.5 m = 4 × 931.5 MeV/c 2 c^2 c 2 ):
\frac{\pi a}{\hbar c}\sqrt{2mc^2 V_0}\left(1 - \frac{E}{V_0}\right) = \frac{\pi \times 30\,\text{fm}{197\,\text{MeV}\cdot\text{fm}\sqrt{2 \times 3726 \times 25}\times 0.8}}
= 0.479 × 432.6 × 0.8 = 165.7 = 0.479 \times 432.6 \times 0.8 = 165.7 = 0.479 × 432.6 × 0.8 = 165.7
T ≈ e − 165.7 ≈ 5 × 10 − 73 T \approx e^{-165.7} \approx 5 \times 10^{-73} T ≈ e − 165.7 ≈ 5 × 1 0 − 73
This extremely small probability explains the enormously long half-lives of alpha-emitting nuclei (∼ 10 9 \sim 10^9 ∼ 1 0 9 years for 238 ^{238} 238 U). The Geiger—Nuttall law relates log T 1 / 2 \log T_{1/2} log T 1/2 to E − 1 / 2 E^{-1/2} E − 1/2 Consistent with the WKB exponential dependence.