2.3 The Infinite Square Well

We will now study the motion of a free particle that is confined to a finite region of space. This is known as the infinite square well because the situation is described by an infinitely deep potential well, or sometimes it is called the particle in a box problem.

2.3.1 Setting up the Schrodinger Equation

Our square well will go from $x=0$ to $x=L$ , then as stated above, the potential energy for this system takes the piecewise form

V(x)=\begin{cases}\infty&\text{for }x\leq 0\\ 0&\text{for }0<x<L\\ \infty&\text{for }x\geq L.\end{cases}

(2.53)

Inside the well, the TISE will therefore look just like it does for the free particle (since there are no forces acting inside the well), which is given by equation 2.1:

\frac{\operatorname{\mathrm{d}}\!^{2}u(x)}{\operatorname{\mathrm{d}}\!x^{2}}=-% \frac{2mE}{\hbar^{2}}u(x).

(2.54)

This equation may look the same as for the free particle, but the solutions take a slightly different form because the boundary conditions are different: equation 2.54 is only valid inside the well.

Outside the well, the potential is infinite, so the wavefunction cannot exist and must be zero.

2.3.2 Energy Eigenfunctions

Equation 2.54 has the general solution

u(x)=A\sin(kx)+B\cos(kx),

(2.55)

where the wavenumber is

k=\frac{\sqrt{2mE}}{\hbar}.

(2.56)

We could write this using complex exponentials like we did for the free particle, but using trigonometric functions simplifies things here as we will see shortly.

We now need to take into account the boundary conditions. Recall from section 1.3.1, that the wavefunction needs to be continuous over all space. For this to be the case, we must have that the wavefunction vanishes at the boundaries of the well.

u(0)=u(L)=0.

(2.57)

The first condition implies that $B$ must be zero, and the second implies

	$\displaystyle A\sin(kL)$	$\displaystyle=0$		(2.58)
	$\displaystyle\implies k_{n}$	$\displaystyle=\frac{n\pi}{L},$		(2.59)

where $n$ is a positive integer (negative integers give the same solutions multiplied by $-1$ since sine is an odd function, so we just ignore them, and $n=0$ makes the wavefunction zero everywhere, which is not interesting so we ignore that too). So the wavenumber can only take on a discrete set of values given by positive integer multiples of $\frac{\pi}{L}$ , which we have labelled $k_{n}$ with a subscript.

$A$ is the normalisation constant, which we can calculate by looking at the probability density:

\int_{-\infty}^{\infty}\lvert A\sin(k_{n}x)\rvert\operatorname{\mathrm{d}}\!x=% \lvert A\rvert^{2}\int_{0}^{L}\sin^{2}\left(\frac{n\pi}{L}x\right)% \operatorname{\mathrm{d}}\!x\overset{!}{=}1.

(2.60)

The integral is $\frac{L}{2}$ , which implies that $A=\sqrt{\frac{2}{L}}$ .

Thus, in contrast to the particle which is free over all space, the particle confined to a finite region has a discrete set of eigenfunctions, the spatial parts of which are given by

u_{n}(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi}{L}x\right),\quad n=1,2,3,\dots

(2.61)

The energy eigenvalues are given by equation 2.56:

E_{n}=\frac{\hbar^{2}k^{2}}{2m}=\frac{\pi^{2}\hbar^{2}}{2mL^{2}}n^{2}.

(2.62)

So the total energy is only allowed to be from a discrete set of values as well.

The full time-dependent energy eigenstates are

\psi_{n}(x,t)=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi}{L}x\right)e^{-\frac{\pi^% {2}\hbar}{2mL^{2}}n^{2}t},\quad n=1,2,3,\dots

(2.63)

In Dirac notation, these are written as $\lvert n\rangle$ . Although it can be ambiguous whether the wavefunction for a state $\lvert n\rangle$ (i.e. $\langle x\mathclose{}|\mathopen{}n\rangle$ ) is the full time-dependent eigenfunction $\psi_{n}$ or just the spatial part $u_{n}$ , it is usually either determined by context or it does not matter.

As $n$ increases, the spacing between the energy levels gets larger. Also note that the more narrow the well (smaller $L$ ), the higher the energy levels. Note that the lowest possible energy, what we call the ground state, is found when $n=1$ , so the particle can never have zero energy. These are manifestations of the uncertainty principle, specifically that localising a particle necessitates if having a higher momentum uncertainty as we saw in section 1.5. A particle which has any localisation at all can never have zero energy, since that would imply it has zero momentum for certain.

2.3.3 Probability Density

As we can see from equation 2.63, the wavefunction of the particle oscillates faster in both space and time the large $n$ is. This carries over to the spatial probability density, which is

\lvert\psi_{n}(x,t)\rvert^{2}=\frac{2}{L}\sin^{2}\left(\frac{n\pi}{L}x\right).

(2.64)

How does this compare to the probability density of a classical free particle in an infinite well? This would be like a particle sliding back and force on a frictionless surface between two walls, with elastic collisions at both ends. Since the particle’s speed is constant, it is equally likely to be found anywhere within the well, so we have

P_{\text{Cl}}=\frac{1}{L}.

(2.65)

The correspondence principle suggests that the quantum probability density would reduce to the classical probability density in the classical limit of large energy (large $n$ ). This does not appear to be the case here, but we need to look at the actual probability rather than just the values of probability density. It is the case that as $n\to\infty$ , $\lvert\psi_{n}\rvert^{2}\Delta x$ tends to $P_{\text{Cl}}\Delta x$ for any interval $\Delta x$ .

Why then does the quantum probability density oscillate with ever shorter wavelength for increasing $n$ ? The answer comes from looking at the expectation value of the particle’s kinetic energy. We saw that the kinetic energy operator was given by equation 1.12, so the expectation value for a general wavefunction $\psi(x,t)$ is

	$\displaystyle\langle\hat{T}\rangle$	$\displaystyle=\int_{-\infty}^{\infty}\psi^{\ast}\hat{T}\psi\operatorname{% \mathrm{d}}\!x$		(2.66)
		$\displaystyle=-\frac{\hbar^{2}}{2m}\int_{-\infty}^{\infty}\psi^{\ast}\frac{% \partial^{2}\psi}{\partial x^{2}}\operatorname{\mathrm{d}}\!x.$		(2.67)

Integrating by parts and using the same argument as in section 1.4.2 to justify that the surface terms vanish, we get

	$\displaystyle\langle\hat{T}\rangle$	$\displaystyle=-\frac{\hbar^{2}}{2m}\left[\left.\psi^{\ast}\frac{\partial\psi}{% \partial x}\right\|_{-\infty}^{\infty}-\int_{-\infty}^{\infty}\frac{\partial% \psi^{\ast}}{\partial x}\frac{\partial\psi}{\partial x}\operatorname{\mathrm{d% }}\!x\right]$		(2.68)
		$\displaystyle=\frac{\hbar^{2}}{2m}\int_{-\infty}^{\infty}\left\|\frac{\partial% \psi}{\partial x}\right\|^{2}\operatorname{\mathrm{d}}\!x.$		(2.69)

So the average kinetic energy is related to the “wiggliness” of the wavefunction. In particular, for the infinite square well we find that the expectation of kinetic energy for the energy eigenstates $\langle\hat{T}\rangle_{\psi_{n}}=E_{n}$ , so as $n$ increases the wavefunction oscillates with shorter wavelengths.