# 11.4: Scattering in one dimension - Square Well

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The square well potential has $$V (x < 0) = V (x > a) = 0; V (0 < x < a) = V_0$$. As with the step function, we can write the wavefunction as a plane wave in each of the three regions.

$\Phi (x < 0) = A \text{ exp}(ikx) + B \text{ exp}(−ikx) \nonumber$

$\Phi (0 < x < a) = F \text{ exp}(ik0x) + G \text{ exp}(−ik0x) \nonumber$

$\Phi (x > a) = C \text{ exp}(ikx) + D \text{ exp}(−ikx) \nonumber$

Once again there is no wave coming back from $$x = \infty (D = 0)$$.

There are now four boundary conditions from continuity of the wave function and its derivative at x=0 and x=a. The solving of four equations in four unknowns is straightforward but tedious. Eventually one can obtain ratios for reflected and transmitted flux:

$B/A = \frac{(k^2 − k^{'2} )(1 − e^{2ik'a} )}{(k + k' )^2 − (k − k' )^2 e^{2ik'a}} \nonumber$

$C/A = \frac{4kk' e^{i(k' − k)a}}{(k + k' )^2 − (k − k' )^2 e^{2ik'a}} \nonumber$

where $$k^2 = 2mE/\hbar^2$$ and $$k^{'2} = 2m(E − V_0)/\hbar^2$$. Since the wavenumber is the same on both sides of the barrier, the reflection and transmission coefficients are just:

$|B/A|^2 = \left[ 1 + \frac{4k^2 k^{'2}}{(k^2 − k^{'2} )^2 \sin^2 k' a} \right]^{−1} = \left[ 1 + \frac{4E(E − V_0)}{V^2_0 \sin^2 k' a} \right]^{−1} \nonumber$

$|C/A|^2 = \left[ 1 + \frac{(k^2 − k^{'2} )^2 \sin^2 k' a}{4k^2 k^{'2}} \right]^{−1} = \left[ 1 + \frac{V^2_0 \sin^2 k' a}{4E(E − V_0)} \right]^{-1} \nonumber$

We get complete transmission when $$k' a = n\pi$$, i.e. when an exact number of half waves fit in the well.

Assuming that $$E > V_0$$. Looking at the limits of this, we see that as $$E \rightarrow V_0$$ then $$\sin^2 (k' a) \rightarrow k' a$$ and the transmission coefficient

$|C/A|^2 \rightarrow \left[ 1 + \frac{mV_0a^2}{2\hbar^2} \right]^{-1} \nonumber$

As the incoming particle energy is increased, the transmission oscillates between $$\left[1 + \frac{V^2_0}{4E(E−V_0)} \right]^{−1}$$ and 1 at $$k' a = n\pi$$. The lower limit itself increases to 1 as E increases.

For the tunnelling case where $$E < V_0$$ we can use these solutions for B/A and C/A, except that $$k'$$ is now imaginary. This gives

$|C/A|^2 = \left[ 1 + \frac{4E(E − V_0)}{V^2_0 \sinh^2 |k' |a} \right]^{−1} \nonumber$

which decreases monotonically with decreasing E. Thus a small change in $$V_0$$ can give a large change in $$|C/A|^2$$. This is the principle on which the transistor and the tunnelling electron microscope are based.

Note that the transmitted wave $$\Phi (x > a) = C \text{ exp}(ikx)$$, differs from the incident wave only by a phase - it has the same wavevector. Thus the only effect of the potential on the transmitted particles is to change their phase, an idea we shall meet again.

This page titled 11.4: Scattering in one dimension - Square Well is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Graeme Ackland via source content that was edited to the style and standards of the LibreTexts platform.