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4. Bound states of the square well

One of the simplest potentials to study the properties of is the so-called square well potential,

V = $\displaystyle \left\{\vphantom{\begin{array}{ll}0& \vert x\vert>a \\  -V_0 & \vert x\vert<a\end{array}}\right.$$\displaystyle \begin{array}{ll}0& \vert x\vert>a \\  -V_0 & \vert x\vert<a\end{array}$ $\displaystyle \left.\vphantom{\begin{array}{ll}0& \vert x\vert>a \\  -V_0 & \vert x\vert<a\end{array}}\right.$. (4.1)

  
Figure 4.1: The square well potential
\includegraphics[width=10cm]{Figures/square_well.eps}

We define three areas, from left to right I, II and III. In areas I and III we have the Schrödinger equation

- $\displaystyle {\frac{\hbar^2}{2m}}$$\displaystyle {\frac{d^2}{d x^2}}$$\displaystyle \psi$(x) = E$\displaystyle \psi$(x) (4.2)

whereas in area II we have the equation

- $\displaystyle {\frac{\hbar^2}{2m}}$$\displaystyle {\frac{d^2}{d x^2}}$$\displaystyle \psi$(x) = (E + V0)$\displaystyle \psi$(x) (4.3)

Solution to a few ODE's. In this class we shall quite often encounter the ordinary differential equations

$\displaystyle {\frac{d^{2}}{dx^{2}}}$f (x) = - $\displaystyle \alpha^{2}_{}$f (x) (4.4)

which has as solution

f (x) = A1cos($\displaystyle \alpha$x) + B1sin($\displaystyle \alpha$x) = C1ei$\scriptstyle \alpha$x + D1e-i$\scriptstyle \alpha$x, (4.5)

and

$\displaystyle {\frac{d^{2}}{dx^{2}}}$g(x) = + $\displaystyle \alpha^{2}_{}$g(x) (4.6)

which has as solution

g(x) = A2cosh($\displaystyle \alpha$x) + B2sinh($\displaystyle \alpha$x) = C2e$\scriptstyle \alpha$x + D2e- $\scriptstyle \alpha$x. (4.7)

Let us first look at E > 0. In that case the equation in regions I and III can be written as

$\displaystyle {\frac{d^2}{d x^2}}$$\displaystyle \psi$(x) = - $\displaystyle {\frac{2m}{\hbar^2}}$E$\displaystyle \psi$(x) = - k2$\displaystyle \psi$(x), (4.8)

where

k = $\displaystyle \sqrt{\frac{2m}{\hbar^2}E}$. (4.9)

The solution to this equation is a sum of sines and cosines of kx, which cannot be normalised: Write $ \psi_{{III}}^{}$(x) = Acos(kx) + Bsin(kx) (A, B, complex) and calculate the part of the norm originating in region III,
$\displaystyle \int_{a}^{\infty}$|$\displaystyle \psi$(x)|2dx = $\displaystyle \int_{a}^{\infty}$| A|2cos2kx + | B|2sin2kx + 2$\displaystyle \Re$(AB*)sin(kx)cos(kx)dx  
  = $\displaystyle \lim_{N\rightarrow\infty}^{}$N$\displaystyle \int_{a}^{{2\pi}/k}$| A|2cos2(kx) + | B|2sin2(kx)  
  = $\displaystyle \lim_{N\rightarrow\infty}^{}$N(| A|2/2 + | B|2/2) = $\displaystyle \infty$. (4.10)

We also find that the energy cannot be less than - V0, since we vannot construct a solution for that value of the energy. We thus restrict ourselves to - V0 < E < 0. We write

E = - $\displaystyle {\frac{\hbar^2 k^2}{2m}}$,      E + V0 = $\displaystyle {\frac{\hbar^2 \kappa^2}{2m}}$. (4.11)

The solutions in the areas I and III are of the form (i = 1, 3)

$\displaystyle \psi$(x) = Aiekx + Bie-kx. (4.12)

In region II we have the oscillatory solution

$\displaystyle \psi$(x) = A2cos($\displaystyle \kappa$x) + B2sin($\displaystyle \kappa$x). (4.13)

Now we have to impose the conditions on the wave functions we have discussed before, continuity of $ \psi$ and its derivatives. Actually we also have to impose normalisability, which means that B1 = A3 = 0 (exponentially growing functions can not be normalised). As we shall see we only have solutions at certain energies. Continuity implies that
A1e-ka + B1eka = A2cos($\displaystyle \kappa$a) - B2sin($\displaystyle \kappa$a)  
A3eka + B3e-ka = A2cos($\displaystyle \kappa$a) + B2sin($\displaystyle \kappa$a)  
kA1eka - kB1eka = $\displaystyle \kappa$A2sin($\displaystyle \kappa$a) + $\displaystyle \kappa$B2cos($\displaystyle \kappa$a)  
kA3eka - kB3e-ka = - $\displaystyle \kappa$A2sin($\displaystyle \kappa$a) + $\displaystyle \kappa$B2cos($\displaystyle \kappa$a) (4.14)

After imposing the normalisability condition we realise we can take the ratio of the first and third and second and fourth equation:
 
k = $\displaystyle {\frac{\kappa [A_2\sin(\kappa a) + B_2\cos(\kappa a)]}{A_2 \cos(\kappa a) - B_2 \sin(\kappa a)}}$  
k = $\displaystyle {\frac{\kappa [A_2\sin(\kappa a) - B_2\cos(\kappa a)]}{A_2 \cos(\kappa a) + B_2 \sin(\kappa a)}}$ (4.15)

We can combine these two equations to a single one by equating the right-hand sides. After deleting the common factor $ \kappa$, and multiplying with the denominators we find
[A2cos($\displaystyle \kappa$a)+B2sin($\displaystyle \kappa$a)][A2sin($\displaystyle \kappa$a)-B2cos($\displaystyle \kappa$a)] & = [A2sin($\displaystyle \kappa$a) + B2cos($\displaystyle \kappa$a)][A2cos($\displaystyle \kappa$a) - B2sin($\displaystyle \kappa$a)],

which simplifies to

A2B2 = 0 (4.16)

We thus have two families of solutions, those characterised by B2 = 0 and those that have A2 = 0.



 
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© 1998 Niels Walet, UMIST
Email Niels.Walet@umist.ac.uk