This is actually in my quantum mechanics textbook (pure math question though), and I just cannot see why this equality is true. Any help would be greatly appreciated!

Let $F$ and $A$ be nonzero complex numbers, and let $a, k, \text{ and}\ l$ be positive real numbers. Assume that

\begin{equation} F = \frac{e^{-2ika} A}{\cos(2la) - i \frac{k^2 + l^2}{2kl} \sin(2la)}. \end{equation}

Define $T := |F|^2 / |A|^2.$ Then

\begin{equation} T^{-1} = 1 + \left(\frac{k^2 - l^2}{2kl}\right)^2 \sin^2(2la). \end{equation}

Thanks so much!

  • 1
    $\begingroup$ For those who are curious, this is the equation for the transmission coefficient for the finite potential well of length $2a$ centered at the origin with potential $-V_0$ inside the well and $0$ outside. The particle is being thrown in from $-\infty$ with energy $E > 0$. $k = \frac{\sqrt{2mE}}{hbar}$ and $l = \frac{\sqrt{2m(E-V_0)}}{hbar}.$ $T$ gives the probability that the particle passes beyond (transmits through) the potential well. Observe that whenever the argument of sin^2 is 0, the particle will always transmit! The energies E allowing this are exactly those of the infinite well! $\endgroup$ – Doug Feb 8 '13 at 7:50
  • $\begingroup$ I worked this exact problem out in my Modern Physics class, Spring 1990. $\endgroup$ – Ron Gordon Feb 8 '13 at 8:17


$$\left ( \frac{k^2+\ell^2}{2 k \ell} \right )^2 = 1 + \left ( \frac{k^2-\ell^2}{2 k \ell} \right )^2$$


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