# Finding finite square well width and depth from transmission resonance

For an electron incident a one-dimensional finite square well the transmission probability is $\approx$1 for electron energies $E_1=0.6 \textbf{ eV}$, $E_2=1.9 \textbf{ eV}$ and $E_3=3.4 \textbf{ eV}$. How can the potential well width and depth be calculated?

I know that an expression for the transmission coefficient $T$ can be found from deriving the wave-equation for the finite potential well, where the probability of transmission is the probability amplitude ratio of the incident wave to the transmitted wave, i.e. $T=\frac{|F|^2}{|A|^2}$ where $A$ is the amplitude of the incident wave and $F$ is the amplitude of the transmitted wave. By Properly normalizing these amplitudes from the boundary conditions (continuity in regard to $\psi(x)$ and $\frac{d}{dt}\psi(x)$ at the boundaries) the ratio can be described as $$T^{-1} = 1 + \frac{V_0 ^2}{4E(E+V_0)} \sin^2(\frac{2a}{\hbar}\sqrt{2m(E+V_0)})$$ where $a$ is the width of the well, $m$ the mass of the electron and $V_0$ the absolute value of the well depth.

The transmission probability is equal to one if the sine is zero, which is for $$\frac{2a}{\hbar}\sqrt{2m(E+V_0)}=n\pi$$ Where $n$ is any integer.

The solution seems simple enough, just plug in the values for the energies and solve for the width $a$ and the depth $V_0$. However this leads me to an unsolvable system of equations since I do not know which $n$ corresponds to which resonant energy value. How can one approach this problem more successfully?

• have you tried with the $n$ s corresponding to the $E_n$ s? Apr 6, 2015 at 16:42
• @danimal That was my initial approach. This however led to a negative $V_0$ value. The resonance at $E_1$ does not have to be given by $n=1$. Apr 6, 2015 at 16:52
• ok i see... can you make an approximation using the infinite square well $E\propto (n/a)^2$? Apr 6, 2015 at 16:54

The problem can be solved by introducing the relations $$n_2-n_1=1,\:\:\:n_3-n_2=1$$ Since the resonance occurs for the energies $E_1$,$E_2$, and $E_3$ in a sequence the argument for the sine $n \pi$ also occur in a sequence in this case.
This leads to one of possible solvable equations for the well potential $$2 \sqrt{E_2+V_0} = \sqrt{E_1+V_0}+\sqrt{E_3+V_0}$$