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Physics LibreTexts

5.6: Solving the 1D Semi-Infinite Square Well

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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, Ψ = 0 in this region.
In the region 0 < x < L, the equation and solution should look familiar:

2md2dx2Ψ(x)+U(x)Ψ(x)=EΨ(x)2md2dx2Ψ(x)+(0)Ψ(x)=EΨ(x)2mddx2Ψ(x)=EΨ(x)d2dx2Ψ(x)=2mE2Ψ(x)

The general solution to this equation is

Ψ(x)=Asin(kx)+Bcos(kx)withk=2mE2

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

Ψ(x)=Asin(kx)

In the region x > L, the equation is:

2md2dx2Ψ(x)+U(x)Ψ(x)=EΨ(x)2md2dx2Ψ(x)=(EU)Ψ(x)d2dx2Ψ(x)=2m(UE)2Ψ(x)

The general solution is:

Ψ(x)=CeaX+DeaX with α=2m(UE)2

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

Ψ(x)=DeaX

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

Ψ(x=L)=Ψ(x=L+)Asin(kL)=DeaL
and forcing the continuity of the derivative leads to:

Ψ(x=L)=Ψ(x=L+)kAcos(kL)=αDeaL

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:

\boldsymbol{\tan \bigg \sqrt{\frac{2mc^2L^2E}{(\hbar c)^2}} \bigg) = - \sqrt{\frac{E}{U-E}}}
\boldsymbol{\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}}}

\boldsymbol{\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.


This page titled 5.6: Solving the 1D Semi-Infinite Square Well is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Paul D'Alessandris.

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