Particle in Infinite Square Potential Well

Consider a particle trapped in a one-dimensional square potential well, of infinite depth, which is such that

$\begin{displaymath}U(x) = \left\{ \begin{array}{lll} 0&\mbox{\hspace{0.5cm}}&0\l... ... \leq a\\ [0.5ex] \infty &&\mbox{otherwise} \end{array}\right..\end{displaymath}$

(4.72)

The particle is excluded from the region or , so $\psi=0$ in this region (i.e., there is zero probability of finding the particle outside the well). Within the well, a particle of definite energy has a stationary wavefunction, $\psi(x)$ , that satisfies

$\displaystyle -\frac{\hbar^{2}}{2\,m}\,\frac{d^{2}\psi}{d x^{2}} = E\,\psi.$

(4.73)

[See Equation (4.71).] The boundary conditions are

$\displaystyle \psi(0) = \psi(a) = 0.$

(4.74)

This follows because $\psi=0$ in the region or , and $\psi(x)$ must be continuous [because a discontinuous wavefunction would generate a singular term (i.e., the term involving $d^{2}\psi/dx^{2}$ ) in the time-independent Schrödinger equation, (4.71), that could not be balanced, even by an infinite potential].

Let us search for solutions to Equation (4.73) of the form

$\displaystyle \psi(x) = \psi_0\,\sin(k\,x),$

(4.75)

where $\psi_0$ is a constant. It follows that

$\displaystyle \frac{\hbar^{2}\,k^{2}}{2\,m} = E.$

(4.76)

The solution (4.75) automatically satisfies the boundary condition $\psi(0)=0$ . The second boundary condition, $\psi(a)=0$ , leads to a quantization of the wavenumber; that is,

$\displaystyle k= n\,\frac{\pi}{a},$

(4.77)

where $n=1,\,2,\,3,$ et cetera. (A “quantized” quantity is one that can only take certain discrete values.) According to Equation (4.76), the energy is also quantized. In fact, , where

$\displaystyle E_n = n^{2}\,\frac{\hbar^{2}\,\pi^{2}}{2\,m\,a^{2}}.$

(4.78)

Thus, the allowed wavefunctions for a particle trapped in a one-dimensional square potential well of infinite depth are

$\displaystyle \psi_n(x,t) = A_n\,\sin\left(n\,\pi\,\frac{x}{a}\right)\,\exp\left(-{\rm i}\,n^{2}\,\frac{E_1}{\hbar}\,t\right),$

(4.79)

where is a positive integer, and a constant. We cannot have , because, in this case, we obtain a null wavefunction—that is, $\psi=0$ , everywhere—which corresponds to a non-existent state. Furthermore, if takes a negative integer value then it generates exactly the same wavefunction as the corresponding positive integer value (assuming $A_{-n}=-A_n$ ).

The constant , appearing in the previous wavefunction, can be determined from the constraint that the wavefunction be properly normalized. For the case under consideration, the normalization condition (4.27) reduces to

$\displaystyle \int_0^a\,\vert\psi(x)\vert^{\,2}\,dx = 1.$

(4.80)

It follows from Equation (4.79) that $\vert A_n\vert^{2}=2/a$ . Hence, the properly normalized version of the wavefunction (4.79) is

$\displaystyle \psi_n(x,t) = \left(\frac{2}{a}\right)^{1/2}\,\sin\left(n\,\pi\,\frac{x}{a}\right)\,\exp\left(-{\rm i}\,n^{2}\,\frac{E_1}{\hbar}\,t\right).$

(4.81)

Figure 4.4 shows the first four properly normalized stationary wavefunctions for a particle trapped in a one-dimensional square potential well of infinite depth; that is, $\psi_n(x)= (2/a)^{1/2}\,\sin(n\,\pi\,x/a)$ , for to .

**Figure 4.4:** First four stationary wavefunctions for a particle trapped in a one-dimensional square potential well of infinite depth.
$\includegraphics[width=0.9\textwidth]{Chapter05/fig11_05.eps}$

At first sight, it seems rather strange that the lowest possible energy for a particle trapped in a one-dimensional potential well is not zero, as would be the case in classical mechanics, but rather $E_1= \hbar^{2}\,\pi^{2}/(2\,m\,a^{2})$ . In fact, as explained in the following, this residual energy is a direct consequence of Heisenberg's uncertainty principle. A particle trapped in a one-dimensional well of width is likely to be found anywhere inside the well. Thus, the uncertainty in the particle's position is ${\mit\Delta} x\sim a$ . It follows from the uncertainty principle, (4.65), that

$\displaystyle {\mit\Delta} p \gtrsim \frac{\hbar}{2\,{\mit\Delta} x}\sim \frac{\hbar}{a}.$

(4.82)

In other words, the particle cannot have zero momentum. In fact, the particle's momentum must be at least $p\sim \hbar/a$ . However, for a free particle, $E=p^{2}/2\,m$ . Hence, the residual energy associated with the particle's residual momentum is

$\displaystyle E \sim \frac{p^{2}}{m}\sim \frac{\hbar^{2}}{m\,a^{2}}\sim E_1.$

(4.83)

This type of residual energy, which often occurs in quantum mechanical systems, and has no equivalent in classical mechanics, is called zero-point energy.

The most general wavefunction for a particle trapped in a one-dimensional square potential well, of infinite depth, is a superposition of all of the possible stationary states. That is,

$\displaystyle \psi(x,t) = \sum_{n=1,\infty} a_n\,\psi_n(x,t),$

(4.84)

where the are complex numbers, and the $\psi_n(x,t)$ are specified in Equation (4.81). Consider

$\displaystyle \int_0^a\vert\psi(x,t)\vert^2\,dx$	$\displaystyle = \sum_{n,m=1,\infty}a_n\,a_m^\ast \int_0^a\psi_n(x,t)\,\psi_m^\ast(x,t)\,dx$
	$\displaystyle =\sum_{n,m=1,\infty} a_n\,a_m^\ast\, \frac{2}{a}\int_0^a \sin\left(n\,\pi\,\frac{x}{a}\right)\,\sin\left(m\,\pi\,\frac{x}{a}\right)dx$
	$\displaystyle =\sum_{n,m=1,\infty} a_n\,a_m^\ast\,\frac{2}{\pi}\int_0^\pi \sin(n\,\theta)\,\sin(n\,\theta)\,d\theta.$	(4.85)

However,

$\displaystyle \frac{2}{\pi}\int_0^\pi \sin(n\,\theta)\,\sin(n\,\theta)\,d\theta=\delta_{nm},$

(4.86)

where $\delta_{nm}$ , which is known as a Kronecker delta, takes the value 1 if , and 0 otherwise. Hence, we deduce that

$\displaystyle \int_0^a\vert\psi(x,t)\vert^2\,dx =\sum_{n=1,\infty} \vert a_n\vert^2.$

(4.87)

Thus, the wavefunction (4.84) is properly normalized provided

$\displaystyle \sum_{n=1,\infty} \vert a_n\vert^2 = 1.$

(4.88)

Suppose that we make a measurement of the energy of a particle whose wavefunction is specified by Equation (4.84). Given that the wavefunction is a superposition of stationary states associated with the quantized energies [see Equation (4.78)], it seems reasonable to assume that the measurement will result in one of these energies. In fact, according to quantum mechanics, the probability that a measurement of the particle's energy will give the result is $\vert a_n\vert^2$ . Thus, we can see that the normalization condition (4.88) ensures that the sum of all of these probabilities is unity. This must be the case, because a measurement of the particle's energy is certain to give one of the allowed energies. Suppose that we make a measurement of the particle's energy, and obtain the result . A second measurement, made immediately after the first, must yield the same result. In other words, immediately after the first measurement, the particle's wavefunction must be such that a measurement of its energy is certain to give the result , and has no chance of giving the result , where $m\neq n$ . This implies that $\vert a_n\vert^2 =1$ and $\vert a_m\vert^2=0$ , where $m\neq n$ . We conclude that, after the first measurement, the particle's wavefunction is $\psi_n(x,t)$ . This is another example of the collapse of a wavefunction consequent on a measurement. (See Section 4.2.8.)