errata in WKB connection formulas

File in which the problem appears

Lecture notes Applications of Quantum Mechanics
WKB approximation
Connection formulas

1.Missing parantheses

Problematic sentence

"Around the turning point, the Schrodinger equation then becomes:"
-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \psi(x) + E + V'(0) x\, \psi(x) = E \psi(x)

Correct version

"Around the turning point, the Schrodinger equation then becomes:"
-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \psi(x) + (E + V'(0) x)\, \psi(x) = E \psi(x)

2.Incorrect exponents in Ai and Bi and Airy function definition with factor 1/2 is not consistent in document

Problematic sentence

"For the first Airy function Ai(x) this reads "
\frac{1}{2\sqrt{\pi}(-z)^{1/4}}\sin\left( \frac{2}{3}(-z)^{2/3} + \frac{\pi}{4} \right)
\frac{1}{\sqrt{\pi}z^{1/4}}e^{-\frac{2}{3}z^{2/3}}

"The asymptotic behavior of the second Airy function Bi(x) is given by "
\frac{1}{\sqrt{\pi}(-z)^{1/4}}\cos\left( \frac{2}{3}(-z)^{2/3} + \frac{\pi}{4} \right),
\frac{1}{\sqrt{\pi}z^{1/4}}e^{\frac{2}{3}z^{2/3}},

Correct version

"For the first Airy function Ai(x) this reads "
\frac{1}{\sqrt{\pi}(-z)^{1/4}}\sin\left( \frac{2}{3}(-z)^{3/2} + \frac{\pi}{4} \right)
\frac{1}{2\sqrt{\pi}z^{1/4}}e^{-\frac{2}{3}z^{3/2}}

"The asymptotic behavior of the second Airy function Bi(x) is given by "
\frac{1}{\sqrt{\pi}(-z)^{1/4}}\cos\left( \frac{2}{3}(-z)^{3/2} + \frac{\pi}{4} \right),
\frac{1}{\sqrt{\pi}z^{1/4}}e^{\frac{2}{3}z^{3/2}},

3.Not sure about this, but z instead of x in figures for clarity

Problematic sentence

Figures of Ai(x) and Bi(x)

Correct version

Figures of Ai(z) and Bi(z)

4.alpha to the 2/3 instead of 3/2

Problematic sentence

"Using the linear approximation for the potential in the WKB wave function we find"
p(x) \approx \sqrt{2m(E-(E+V^{'}(0)x ))} = \hbar \alpha^{2/3} \sqrt{-x}.

Correct version

"Using the linear approximation for the potential in the WKB wave function we find"
p(x) \approx \sqrt{2m(E-(E+V^{'}(0)x ))} = \hbar \alpha^{3/2} \sqrt{-x}.

5.x instead of alpha under squareroot

Problematic sentence

"Inserting this into the WKB wave function we arrive at. "
= \frac{C}{\sqrt{(\alpha x)^{1/4}\sqrt{\hbar x}}}e^{-\frac{2}{3}(\alpha x)^{3/2}} + \frac{D}{\sqrt{(\alpha x)^{1/4}\sqrt{\hbar x}}}e^{+\frac{2}{3}(\alpha x)^{3/2}}

Correct version

"Inserting this into the WKB wave function we arrive at "
= \frac{C}{\sqrt{(\alpha x)^{1/4}\sqrt{\hbar \alpha}}}e^{-\frac{2}{3}(\alpha x)^{3/2}} + \frac{D}{\sqrt{(\alpha x)^{1/4}\sqrt{\hbar \alpha}}}e^{+\frac{2}{3}(\alpha x)^{3/2}}

6.Missing 1 over h-bar in sin and cos

Problematic sentence

"We can redefine the coefficients to instead write ψI(x) in terms of sin and cos: "
\psi_{I}(x) = \frac{A}{\sqrt{p(x)}}\sin\left( \int_x^0 p(x') dx'+ \theta \right) + \frac{B}{\sqrt{p(x)}}\cos\left(\int_x^0 p(x') dx'+ \theta \right)

Correct version

"We can redefine the coefficients to instead write ψI(x) in terms of sin and cos: "
\psi_{I}(x) = \frac{A}{\sqrt{p(x)}}\sin\left(\frac{1}{\hbar} \int_x^0 p(x') dx'+ \theta \right) + \frac{B}{\sqrt{p(x)}}\cos\left(\frac{1}{\hbar}\int_x^0 p(x') dx'+ \theta \right)

7.Extra factor of half in cos term not mentioned before which is already used in Ai and not mentioned in following equations

Problematic sentence

"We can redefine the coefficients to instead write ψI(x) in terms of sin and cos: "
\psi_{II}(x) = \frac{a}{\sqrt{\pi}(-z)^{1/4}}\sin\left( \frac{2}{3}(-z)^{2/3} + \frac{\pi}{4} \right) + \frac{b}{2\sqrt{\pi}(-z)^{1/4}}\cos\left( \frac{2}{3}(-z)^{2/3} + \frac{\pi}{4} \right).

Correct version

"We can redefine the coefficients to instead write ψI(x) in terms of sin and cos: "
\psi_{II}(x) = \frac{a}{\sqrt{\pi}(-z)^{1/4}}\sin\left( \frac{2}{3}(-z)^{2/3} + \frac{\pi}{4} \right) + \frac{b}{\sqrt{\pi}(-z)^{1/4}}\cos\left( \frac{2}{3}(-z)^{2/3} + \frac{\pi}{4} \right).

8.Integration bounds of connection formules for potential with negative slope

Problematic sentence

"For a potential with a negative slope, "
Side E<V(x)
(decaying) \frac{C}{\sqrt{\|p(x)\|}} e^{-\frac{1}{\hbar} \int_{x_t}^x \|p(x')\|dx'}
(blowing up) \frac{D}{\sqrt{\|p(x)\|}} e^{\frac{1}{\hbar} \int_{x_t}^x \|p(x')\|dx'}

Side E>V(x)
(decaying) \frac{2C}{\sqrt{p(x)}} \sin\left( \frac{1}{\hbar} \int_x^{x_t} p(x') dx' + \frac{\pi}{4} \right)
(blowing up) \frac{D}{\sqrt{p(x)}} \cos\left( \frac{1}{\hbar} \int_x^{x_t} p(x') dx' + \frac{\pi}{4} \right)

Correct version

"For a potential with a negative slope, "
Side E<V(x)
(decaying) \frac{C}{\sqrt{\|p(x)\|}} e^{-\frac{1}{\hbar} \int_{x}^{x_t} \|p(x')\|dx'}
(blowing up) \frac{D}{\sqrt{\|p(x)\|}} e^{\frac{1}{\hbar} \int_{x}^{x_t} \|p(x')\|dx'}

Side E>V(x)
(decaying) \frac{2C}{\sqrt{p(x)}} \sin\left( \frac{1}{\hbar} \int_{x_t}^{x} p(x') dx' + \frac{\pi}{4} \right)
(blowing up) \frac{D}{\sqrt{p(x)}} \cos\left( \frac{1}{\hbar} \int_{x_t}^{x} p(x') dx' + \frac{\pi}{4} \right)