Comprehensive Study Notes

Particle in a Rectangular Well

Consider a particle in a one-dimensional box with walls of finite height (Fig. 2.5a). The potential-energy function is $V=V{0}$ for $x<0, V=0$ for $0 \leq x \leq l$, and $V=V{0}$ for $x>l$. There are two cases to examine, depending on whether the particle's energy $E$ is less than or greater than $V_{0}$.

FIGURE 2.5 (a) Potential energy for a particle in a one-dimensional rectangular well. (b) The groundstate wave function for this potential. (c) The first excited-state wave function. (a) (b) (c)

We first consider $E<V{0}$. The Schrödinger equation (1.19) in regions I and III is $d^{2} \psi / d x^{2}+\left(2 m / \hbar^{2}\right)\left(E-V{0}\right) \psi=0$. This is a linear homogeneous differential equation with constant coefficients, and the auxiliary equation (2.7) is $s^{2}+\left(2 m / \hbar^{2}\right)\left(E-V{0}\right)=0$ with roots $s= \pm\left(2 m / \hbar^{2}\right)^{1 / 2}\left(V{0}-E\right)^{1 / 2}$. Therefore,

\( \begin{aligned} \psi{\mathrm{I}} & =C \exp \left[\left(2 m / \hbar^{2}\right)^{1 / 2}\left(V{0}-E\right)^{1 / 2} x\right]+D \exp \left[-\left(2 m / \hbar^{2}\right)^{1 / 2}\left(V{0}-E\right)^{1 / 2} x\right] \ \psi{\text {III }} & =F \exp \left[\left(2 m / \hbar^{2}\right)^{1 / 2}\left(V{0}-E\right)^{1 / 2} x\right]+G \exp \left[-\left(2 m / \hbar^{2}\right)^{1 / 2}\left(V{0}-E\right)^{1 / 2} x\right] \end{aligned} \)

where $C, D, F$, and $G$ are constants. As in Section 2.3, we must prevent $\psi{\mathrm{I}}$ from becoming infinite as $x \rightarrow-\infty$. Since we are assuming $E<V{0}$, the quantity $\left(V{0}-E\right)^{1 / 2}$ is a real, positive number, and to keep $\psi{\mathrm{I}}$ finite as $x \rightarrow-\infty$, we must have $D=0$. Similarly, to keep $\psi_{\text {III }}$ finite as $x \rightarrow+\infty$, we must have $F=0$. Therefore,

\( \psi{\mathrm{I}}=C \exp \left[\left(2 m / \hbar^{2}\right)^{1 / 2}\left(V{0}-E\right)^{1 / 2} x\right], \quad \psi{\mathrm{III}}=G \exp \left[-\left(2 m / \hbar^{2}\right)^{1 / 2}\left(V{0}-E\right)^{1 / 2} x\right] \)

In region II, $V=0$, the Schrödinger equation is (2.10) and its solution is (2.15):

\( \begin{equation} \psi_{\mathrm{II}}=A \cos \left[\left(2 m / \hbar^{2}\right)^{1 / 2} E^{1 / 2} x\right]+B \sin \left[\left(2 m / \hbar^{2}\right)^{1 / 2} E^{1 / 2} x\right] \tag{2.32} \end{equation} \)

To complete the problem, we must apply the boundary conditions. As with the particle in a box with infinite walls, we require the wave function to be continuous at $x=0$ and at $x=l$; so $\psi{\mathrm{I}}(0)=\psi{\mathrm{II}}(0)$ and $\psi{\mathrm{II}}(l)=\psi{\mathrm{III}}(l)$. The wave function has four arbitrary constants, so more than these two boundary conditions are needed. As well as requiring $\psi$ to be continuous, we shall require that its derivative $d \psi / d x$ be continuous everywhere. To justify this requirement, we note that if $d \psi / d x$ changed discontinuously at a point, then its derivative (its instantaneous rate of change) $d^{2} \psi / d x^{2}$ would become infinite at that point. However, for the particle in a rectangular well, the Schrödinger equation $d^{2} \psi / d x^{2}=\left(2 m / \hbar^{2}\right)(V-E) \psi$ does not contain anything infinite on the right side, so $d^{2} \psi / d x^{2}$ cannot become infinite. [For a more rigorous argument, see D. Branson, Am. J. Phys., 47, 1000 (1979).] Therefore, $d \psi{\mathrm{I}} / d x=d \psi{\mathrm{II}} / d x$ at $x=0$ and $d \psi{\mathrm{II}} / d x=d \psi{\mathrm{III}} / d x$ at $x=l$.

From $\psi{\mathrm{I}}(0)=\psi{\mathrm{II}}(0)$, we get $C=A$. From $\psi{\mathrm{I}}^{\prime}(0)=\psi{\mathrm{II}}^{\prime}(0)$, we get (Prob. 2.21a) $B=\left(V{0}-E\right)^{1 / 2} A / E^{1 / 2}$. From $\psi{\mathrm{II}}(l)=\psi_{\mathrm{III}}(l)$, we get a complicated equation that allows $G$ to be found in terms of $A$. The constant $A$ is found by normalization.

Taking $\psi{\mathrm{II}}^{\prime}(l)=\psi{\mathrm{III}}^{\prime}(l)$, dividing it by $\psi{\mathrm{II}}(l)=\psi{\mathrm{III}}(l)$, and expressing $B$ in terms of $A$, we get the following equation for the energy levels (Prob. 2.21b):

\( \begin{equation} \left(2 E-V{0}\right) \sin \left[(2 m E)^{1 / 2} l / \hbar\right]=2\left(V{0} E-E^{2}\right)^{1 / 2} \cos \left[(2 m E)^{1 / 2} l / \hbar\right] \tag{2.33} \end{equation} \)

\( \begin{equation} \varepsilon \equiv E / V{0} \quad \text { and } \quad b \equiv\left(2 m V{0}\right)^{1 / 2} l / \hbar \tag{2.34} \end{equation} \)

we divide (2.33) by $V_{0}$ to get

\( \begin{equation} (2 \varepsilon-1) \sin \left(b \varepsilon^{1 / 2}\right)-2\left(\varepsilon-\varepsilon^{2}\right)^{1 / 2} \cos \left(b \varepsilon^{1 / 2}\right)=0 \tag{2.35} \end{equation} \)

Only the particular values of $E$ that satisfy (2.33) give a wave function that is continuous and has a continuous derivative, so the energy levels are quantized for $E<V{0}$. To find the allowed energy levels, we can plot the left side of (2.35) versus $\varepsilon$ for $0<\varepsilon<1$ and find the points where the curve crosses the horizontal axis (see also Prob. 4.31c). A detailed study (Merzbacher, Section 6.8) shows that the number of allowed energy levels with $E<V{0}$ is $N$, where $N$ satisfies

\( \begin{equation} N-1<b / \pi \leq N, \quad \text { where } b \equiv\left(2 m V_{0}\right)^{1 / 2} l / \hbar \tag{2.36} \end{equation} \)

For example, if $V_{0}=h^{2} / m l^{2}$, then $b / \pi=2\left(2^{1 / 2}\right)=2.83$, and $N=3$.

Figure 2.5 shows $\psi$ for the lowest two energy levels. The wave function is oscillatory inside the box and dies off exponentially outside the box. It turns out that the number of nodes increases by one for each higher level.

So far we have considered only states with $EV{0}$, the quantity $\left(V{0}-E\right)^{1 / 2}$ is imaginary, and instead of dying off to zero as $x$ goes to $\pm \infty, \psi{\text {I }}$ and $\psi{\text {III }}$ oscillate (similar to the free-particle $\psi$ ). We no longer have any reason to set $D$ in $\psi{\mathrm{I}}$ and $F$ in $\psi{\text {III }}$ equal to zero, and with these additional constants available to satisfy the boundary conditions on $\psi$ and $\psi^{\prime}$, one finds that $E$ need not be restricted to obtain properly behaved wave functions. Therefore, all energies above $V_{0}$ are allowed.

A state in which $\psi \rightarrow 0$ as $x \rightarrow \infty$ and as $x \rightarrow-\infty$ is called a bound state. For a bound state, significant probability for finding the particle exists in only a finite region of space. For an unbound state, $\psi$ does not go to zero as $x \rightarrow \pm \infty$ and is not normalizable. For the particle in a rectangular well, states with $EV_{0}$ are unbound. For the particle in a box with infinitely high walls, all states are bound. For the free particle, all states are unbound.

For an online simulation of the particle in a well, go to www.falstad.com/qm1d and choose Finite Well in the Setup box. You can vary the well width and depth and see the effect on the energy levels and wave functions.