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1.29: The Finite Square Well

  • Page ID
    50135
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    When the confining potential is finite, we can no longer assume that the wavefunction is zero at the boundaries of the well. For a finite confining potential, the wavefunction penetrates into the barrier: the lower the confining potential, the greater the penetration.

    From Equation (1.27.1) the general solution for the wavefunction within the well was

    \[ \psi(x)=A\ cos(kx)+B\ sin(kx) , \nonumber \]

    but from the solutions for the infinite well, we can see that, within the well, the wavefunction looks like

    \[ \psi(x)=A\ cos(kx) \nonumber \]

    or

    \[ \psi(x)=A\ sin(kx) \nonumber \]

    The simplification is possible because the well potential is symmetric around x = 0. Thus, the probability density \(|ψ(x)|^{2}\) should also be symmetric,\(^{†}\) and indeed both ψ(x) = sin(kx) and ψ(x) = cos(kx) have symmetric probability distributions even though ψ(x) = sin(kx) is an antisymmetric wavefunction.

    We‟ll consider the symmetric (cos(kx)) and antisymmetric (sin(kx)) wavefunction solutions separately.

    Symmetric wavefunction

    We can assume a solution of the form:

    \[ \psi(x)=\left\{\begin{array}{ll}
    e^{\alpha x} & \text { for } x \leq-L / 2 \\
    A \cos (k x) & \text { for }-L / 2 \leq x \leq L / 2 \\
    e^{-\alpha x} & \text { for } x \geq L / 2
    \end{array}\right. \nonumber \]

    where

    \[ k = \sqrt{\frac{2mE}{\hbar^{2}}} \nonumber \]

    and

    \[ \alpha = \sqrt{\frac{2m(V_{0}-E)}{\hbar^{2}}} \nonumber \]

    Note the solution as written is not normalized. We can normalize it later.

    The next step is to evaluate the constant A by matching the piecewise solutions at the edge of the well. We need only consider one edge, because we have already fixed the symmetry of the solution.

    At the right edge, equating the amplitude of the wavefunction gives

    \[ \psi(L/2)=A\ cos(kL/2) = \text{exp}[-\alpha L/2] \nonumber \]

    Equating the slope of the wavefunction gives

    \[ \psi’(L/2)=-kA\ sin(kL/2) = -\alpha\ \text{exp}[-\alpha L/2] \nonumber \]

    Dividing Equation (1.29.8) by Equation (1.29.7) to eliminate A gives

    \[ tan(kL/2)=\alpha /k \nonumber \]

    But \(\alpha\) and k are both functions of energy

    \[ tan(\frac{\pi}{2}\sqrt{\frac{E}{E_{L}}}) = \sqrt{\frac{V_{0}-E}{E}} \nonumber \]

    where we have defined the infinite square well ground state energy

    \[ E_{L}=\frac{\hbar^{2}\pi^{2}}{2mL^{2}} \nonumber \]

    As in the infinite square well case, we find that only certain, discrete, values of energy give a solution. Once again, the energies of electron states in the well are quantized. To obtain the energies we need to solve Equation (1.29.10). Unfortunately, this is a transcendental equation, and must be solved numerically or graphically. We plot the solutions in Figure 1.29.1.

    Antisymmetric wavefunction

    Antisymmetric solutions are found in a similar manner to the symmetric solutions. We first assume an antisymmetric solution of the form:

    \[ \psi(x)=\left\{\begin{array}{ll}
    e^{\alpha x} & \text { for } x \leq-L / 2 \\
    A \sin (k x) & \text { for }-L / 2 \leq x \leq L / 2 \\
    -e^{-\alpha x} & \text { for } x \geq L / 2
    \end{array}\right. \nonumber \]

    Then, we evaluate the constant A by matching the piecewise solutions at the edge of the well. Again, we need only consider one edge, because we have already fixed the symmetry of the solution.

    At the right edge, equating the amplitude of the wavefunction gives

    \[ \psi(L/2)=A\ sin(kL/2)=-\text{exp}[-\alpha L/2] \nonumber \]

    Equating the slope of the wavefunction gives

    \[ \psi’(L/2)=kA\ cos(kL/2)=\alpha\ \text{exp}[-\alpha L/2] \nonumber \]

    Dividing Equation (1.29.8) by Equation (1.29.7) to eliminate A gives

    \[ cot(kL/2)=-\alpha /k \nonumber \]

    Expanding \(\alpha\) and k in terms of energy

    \[ -cot(\frac{\pi}{2}\sqrt{\frac{E}{E_{L}}}) = \sqrt{\frac{V_{0}-E}{E}} . \nonumber \]

    In Figure 1.29.1, we solve for the energy. Note that there is always at least one bound solution no matter how shallow the well. In Figure 1.29.2 we plot the solutions for a confining potential \(V_{0}=5.E_{L}\).

    Screenshot 2021-04-15 at 11.35.36.png
    Figure \(\PageIndex{1}\): A graphical solution for the energy in the finite quantum well. The green and blue curves are the LHS of Equations (1.132) and (1.138), respectively. The red curves are the RHS for different values of the confining potential \(V_{0}\). Solutions correspond to the intersections between the red lines and the green or blue curves.
    Screenshot 2021-04-15 at 11.36.40.png
    Figure \(\PageIndex{2}\): The three bound states for electrons in a well with confining potential \(V_{0} = 5.E_{L}\). Note that the higher the energy, the lower the effective confining potential, and the greater the penetration into the barriers.

    This page titled 1.29: The Finite Square Well is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Marc Baldo (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.