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Section 4.3 Semi-infinite Square Well Potential and Quantum Tunneling

If we want to determine the wavefunction solutions for a particle confined in a region where the potential energy is some known function \(U(x)\text{,}\) then we perform a procedure similar to what we did for the example of the infinite square well potential. However, this time the potential is not zero (or a constant) but changes with position. Solving Schrödinger's equation for the case of a general potential energy function \(U(x)\) can be very complicated and certainly this is beyond the scope of this course. In fact, there are very few potential energy functions for which the Schrödinger equation can be solved exactly.

However, we can give a slightly more complicated example of solving Schrödinger's equation where the potential energy has different constant values in two different regions of space, as shown in Figure 4.7. In this case, the potential differs from the infinite square well potential in that the boundary at \(x = L\) is now set to the finite value \(U(x) = U_0\) for \(x > L\) instead of being infinite. If we consider a solution for a total energy \(E\) which is less than the potential energy step \(U_0\text{,}\) then we see that Schrödinger's equation is different in the two regions \((0 \lt x \lt L)\) and \((x > L)\text{.}\) Rearranging Schrödinger's equation, we obtain

Figure 4.7. The semi-infinite square well potential energy function.
\begin{equation} \frac{d^2\psi(x)}{dx^2} = \frac{2m}{\hbar^2}\left(U(x)-E\right)\psi(x)\text{.}\tag{4.14} \end{equation}

We see that in the region \(0 \lt x \lt L\text{,}\) Schrödinger's equation reduces again to (4.1) so we should expect wavefunction solutions in that region to be similar to sinusoidal functions as given in (4.3). However, in the region \((x > L)\) the total energy is less than the potential energy (\(E \lt U_0\)) and Schrödinger's equation becomes

\begin{equation} \frac{d^2\psi(x)}{dx^2} = \frac{2m}{\hbar^2}\left(U_0-E\right)\psi(x) \hspace{0.5in}\mbox{for }\text{.}\label{eq_SchrodingerE_U}\tag{4.15} \end{equation}

As we did before, we must determine a mathematical function \(\psi(x)\) that is consistent with (4.15), remembering that the quantity \(\left[U_0 - E\right]\) is a positive quantity. This time the second derivative of \(\psi(x)\) is equal to a positive number times the same function. In this case, sinusoidal functions will not work, but again, if we think back to our calculus course, we recall that another set of functions we know is the right choice here — the exponential functions \(C e^{+\kappa x}\) and \(D e^{-\kappa x}\text{,}\) where \(C, D\text{,}\) and \(\kappa\) are constants, with \(\kappa > 0\text{.}\)

Given the potential shown in Figure 4.7, show that the function \(C e^{\kappa x}\) is a solution to Schrödinger's equation (4.15) and determine what the constant \(\kappa\) must be for this to be a solution.

Solution.

To see if \(\psi(x) = C e^{+\kappa x}\) is a solution, we first evaluate the left hand side of (4.15) by calculating the second derivative of \(\psi(x)\text{:}\)

\begin{align*} \frac{d \psi(x)}{dx} =\mathstrut \amp \kappa C e^{\kappa x}\\ \frac{d^2 \psi(x)}{dx^2} =\mathstrut \amp \kappa^2 C e^{\kappa x}\text{.} \end{align*}

Inserting this into Schrödinger's equation (4.15) we find

\begin{equation} \kappa^2 C e^{\kappa x} = \frac{2m}{\hbar^2}\left( U_0 - E \right) C e^{\kappa x}\text{.}\tag{4.16} \end{equation}

In order for \(\psi(x) = C e^{+\kappa x}\) to be a solution, the left side of this equation must equal the right side, and therefore

\begin{equation} \kappa^2 = \frac{2m}{\hbar^2}\left( U_0 - E \right)\text{.}\tag{4.17} \end{equation}

The complete wavefunction for the semi-infinite square well is made by combining the three solutions to Schrödinger's equation in the three regions, (\(x\lt 0\)), (\(0 \lt x \lt L\)), and (\(x > L\)):

\begin{equation} \psi(x) = \left\{\begin{array}{ll} 0 \amp \mbox{for \(x\lt 0\)} \\ A \sin(kx) + B \cos(kx) \amp \mbox{for \(0\lt x\lt L\)} \\ C e^{+\kappa x} + D e^{- \kappa x} \amp \mbox{for \(x > L\)} \end{array} \right.\text{.}\tag{4.18} \end{equation}

As we did in the previous section, we can further simplify this solution by matching this solution to what the wavefunction should be at \(x = 0\) and as \(x \rightarrow \infty\text{.}\) At \(x = 0\text{,}\) the wavefunction should be zero so that it matches \(\psi(x)=0\) for \((x \lt 0)\text{.}\) This requires that the constant \(B = 0\) so the function \(\cos{(kx)}\) does not appear. As \(x \rightarrow \infty\text{,}\) the function \(e^{+ \kappa x}\) would become infinite and is therefore a not well-behaved solution. So we choose \(C = 0\) to eliminate that function from our solution and the final solution becomes

\begin{equation} \psi(x) = \left\{\begin{array}{ll} 0 \amp \mbox{for } \\ A \sin(kx) \amp \mbox{for r} \\ D e^{- \kappa x} \amp \mbox{for } \end{array} \right.\text{.}\label{eq_two_regions_wavefunction}\tag{4.19} \end{equation}

The wavefunctions in the two regions given in (4.19) must match at position \(x = L\) such the the wavefunction and its first derivative make a smooth transition from the region \(x\lt L\) to the region \(x>L\text{.}\) The mathematics is beyond the scope of this course, but there are only certain values of \(k\text{,}\) and hence \(E\text{,}\) for which the wavefunctions make a smooth transition. Graphs of the wavefunctions for the two lowest energy states \(\psi_1(x)\) and \(\psi_2(x)\) are shown in Figure 4.10.

Figure 4.10. The lowest two energy states of a particle in a semi-infinite potential well.

In some of the problems at the end of the chapter you will use an Excel spreadsheet to determine solutions to the semi-infinite square well using a numerical method.

An examination of Figure 4.10 leaves us the following important conclusions:

  • Wavefunctions in the region where \(E > U\) are sinusoidal functions similar to those in the infinite square well potential. The wavelength depends on the difference \(\left|E - U\right|\text{.}\)

  • Wavefunctions in the region where \(E \lt U\) are exponential functions.

  • In the case of the semi-infinite well, the length of the well \(L\) is not exactly an integer or half-integer multiple of the wavelength as in the infinite square well (see Figure 4.3).

Note also that in the region where \(E \lt U\) the wavefunction \(\psi(x)\) is not zero. This might not seem at first glance to be all that surprising until you consider that a particle's kinetic energy \(K = E - U\text{,}\) and the region where \(E \lt U\) is a classically forbidden region (i.e., a particle can't ever be found in that region, classically) since classical mechanics forbids a particle ever from having a negative kinetic energy. But even though classical mechanics says that a particle can never enter a region where \(E \lt U\text{,}\) we can see from this example that quantum theory predicts a non-zero probability for a particle to be found in this region.

Figure 4.11. For this potential, a particle with energy \(E\) in the region \(0 \lt x \lt L_1\) would remain trapped forever, according to classical physics. But quantum theory predicts that the particle can “tunnel” through the thin barrier and escape.

But what if there is a potential well such as the one shown in Figure 4.11? If a particle confined in the region \(0 \lt x \lt L_1\) has an energy \(E \lt U_0\text{,}\) then according to classical theory, that particle will remain trapped between \(x = 0\) and \(x = L\) forever, since it can never pass through the classically-forbidden region \(L_1 \lt x \lt L_2\text{.}\) But we just saw in our previous example that Schrödinger's Equation predicts a non-zero probability of a particle being found in a classically-forbidden region, so presumably, there is a non-zero probability that the particle in Figure 4.11 can be found in the region \(L_1 \lt x \lt L_2\text{.}\) But if it can make it into that region, then there is a non-zero probability that the particle can be found outside the confined region, i.e. for \(x > L_2\text{!}\) In other words, a classically-trapped particle can “tunnel” out of the trap in quantum mechanics.

We won't calculate the probabilities for tunneling. But examination of Figure 4.10 indicates that the probabilities tail off the wider the classically-forbidden region is. So, the probabilities can become quite small unless the barrier is very narrow and unless the energy \(E\) is close to the barrier height \(U_0\text{.}\) Despite these limitations, quantum mechanical tunneling can play a very important role in various processes, and has already found its way into some useful applications:

  • Scanning tunneling microscopes (STMs) have been developed that use tunneling to produce images of surfaces that can resolve individual atoms.

  • Josephson junctions and tunneling diodes have become tools of modern electronics.

  • Tunneling is being proposed as a new mechanism for developing transistors, which could improve the performance of integrated circuits in the future.