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# 6.6: Solving the 1D Semi-Infinite Square Well

Imagine a particle trapped in a one-dimensional well of length L. Inside the well there is no potential energy. However, the “right-hand wall” of the well (and the region beyond this wall) has a finite potential energy. This means that it is possible for the particle to escape the well if it had enough energy.

Again, since this potential is a piece-wise function, Schrödinger’s equation must be solved in the three regions separately.
In the region x < 0, we have already seen that since the potential is infinite there is no chance of finding the particle in this region. Thus, $$\Psi$$ = 0 in this region.
In the region 0 < x < L, the equation and solution should look familiar:

\begin{align} - \dfrac{\hbar}{2m}\dfrac{d^2}{dx^2}\Psi(x) + U(x)\Psi(x) &= E\Psi(x) \\ \nonumber - \dfrac{\hbar}{2m}\dfrac{d^2}{dx^2}\Psi(x) + (0)\Psi(x) &= E\Psi(x) \\ \nonumber - \dfrac{\hbar}{2m}\dfrac{d}{dx^2}\Psi(x) &= E\Psi(x) \\ \nonumber \dfrac{d^2}{dx^2}\Psi(x) &= -\dfrac{2mE}{\hbar^2}\Psi(x) \end{align}

The general solution to this equation is

$\Psi(x) = A\sin(kx) + B\cos(kx) \text{with} k = \sqrt{\dfrac{2mE}{\hbar^2}}$

In order for this solution to be continuous with the solution for x < 0, the coefficient B must equal zero. Thus,

$\Psi(x) = A\sin(kx)$

In the region x > L, the equation is:

\begin{align} & - \dfrac{\hbar}{2m}\dfrac{d^2}{dx^2}\Psi(x) + U(x)\Psi(x) = E\Psi(x) \\ \nonumber & - \dfrac{\hbar}{2m}\dfrac{d^2}{dx^2}\Psi(x) = (E - U)\Psi(x) \\ \nonumber & \dfrac{d^2}{dx^2}\Psi(x) = \dfrac{2m(U-E)}{\hbar^2}\Psi(x) \end{align}

The general solution is:

\begin{align} & \Psi(x) = Ce^{aX} + De^{-aX} \text{ with } \alpha = \sqrt{\dfrac{2m(U - E)}{\hbar^2}}\end{align}

Since this region contains the point x = +∞, C must equal zero or the wavefunction will diverge. Therefore,

$\Psi(x) = De^{-aX}$

The wave function, and the derivative of the wave function, must be continuous across the boundary at x = L. Forcing continuity leads to:

\begin{align} & \Psi(x = L^-) = \Psi(x = L^+) \\ \nonumber & A\sin(kL) = De^{-aL} \end{align}
and forcing the continuity of the derivative leads to:

\begin{align} & \Psi(x = L^-) = \Psi(x = L^+) \\ \nonumber & kA\cos(kL) = -\alpha De^{-aL} \end{align}

Substituting the first equation into the second equation yields:

\begin}align} & kA\cos(kL) = - \alpha( A \sin( kL)) \\ \nonumber & \tan( kL) = - \dfrac{k}{\alpha} \\ \nonumber & \tan \bigg( \sqrt{\dfrac{2mE}{\hbar^2}L \bigg) = - \dfrac{ This last result is a transcendental equation for the allowed energy levels. If the potential energy and width of the well are known, the allowed energy levels can be determined by using a solver or graphing the function. # The 1D Semi-Infinite Well Determine the allowed energy levels for a proton trapped in a semi-infinite square well of width 5.0 fm and depth 60 MeV. Applying the previous result: \[ \tan \bigg \sqrt{\frac{2mc^2L^2E}{(\hbar c)^2}} \bigg) = - \sqrt{\frac{E}{U-E}}
$\tan \bigg \sqrt{\frac{2(938 \text{ MeV} (5.0 \text{ fm})^2 E}{(194.7 \text{ MeV fm})^2}} \bigg) = - \sqrt{\frac{E}{60-E}}$

$\tan \bigg \sqrt{1.204 E} \bigg) = - \sqrt{\frac{E}{60-E}}$
with E in MeV.
The solutions to this equation, which represent the allowed energy levels for the proton, are 6.53, 25.75, and 55.08 MeV.