IV A first example: particle in the square well

The Schrödinger equation involves the potential energy $V(x)$, which depends on the physical circumstances and may be arbitrarily complicated. A simple situation is a particle that bounces between two hard walls at $x=-L/2$ and $x=L/2$. This problem is called the particle in the box, or the particle in the square well, and is one of the few cases where the stationary Schrödinger equation can be solved explicitly.

Classically, at a given energy $E$ the particle bounces between the walls with velocity $v=\sqrt{2E/m}$ when the particle moves to the right and $v=-\sqrt{2E/m}$ when it moves to the left. The corresponding potential energy is given by

	$V(x)=\mathrm{\infty }\mathit{\hspace{1em}}\text{elsewhere (in the walls)}.$		(36)

You may wish to draw the potential on a piece of paper.

Because the walls are infinitely high, we don’t expect to find the particle there at any time. This indeed follows from the stationary Schrödinger equation. In the walls, we have $E\psi (x)=({\widehat{p}}^2/2m+\mathrm{\infty })\psi (x)=\mathrm{\infty }\times \psi (x)$, which (for finite energy $E$) only can be fulfilled for

(37)

Between the walls, the stationary Schrödinger equation has the simple form

(38)

Even though in this notation it may look unfamiliar, this is just the wave equation for a vibrating string clamped between the walls (think of a guitar string, and see Young & Freedman chpt. 15.8.) The solutions are of the form

$$\psi (x)=a\cos kx+b\sin kx,\text{where }k=\sqrt{2mE}/\mathrm{\hslash }.$$

(39)

This is a periodic function of $x$, with period (wave length) $\lambda =2\pi /k$. However, not all values of $k$ are allowed, since the wave function must match continuously (smoothly) with the solution $\psi =0$ in the walls—the wavefunction cannot jump because this would correspond to an infinitely large kinetic energy (the kinetic energy is proportional to the second derivative of the wave function, see section II.2). This translates into the boundary conditions

$$\psi (-L/2)=0,\psi (L/2)=0.$$

(40)

They can only be fulfilled simultaneously if the distance $\lambda /2=\pi /k$ between the zeros of the oscillatory function Eq. (39) is a multiple of $L$, so that we can start with a zero at $x=-L/2$ and end up with another zero at $x=L/2$. The allowed values for $k$ hence are restricted to the discrete values

$$k_n=\frac{n\pi }{L},n\text{ an integer}.$$

(41)

We say that $k$ is quantised. Since $k=\sqrt{2mE}/\mathrm{\hslash }$, this translates directly into a quantisation of the energy,

$$E_n=\frac{k_n^2{\mathrm{\hslash }}^2}{2m}=\frac{{\pi }^2{\mathrm{\hslash }}^2}{2m{L}^2}{n}^2.$$

(42)

The quantised energy values are also called energy levels.

Note that these values depend on the assumed form of the potential energy $V(x)$; changing the potential results in a different Schrödinger equation, which has different solutions and leads to other energies. We will get acquainted with this by studying several more examples, including the hydrogen atom. In general, the level with the lowest energy characterises the ground state of the system. In difference to classical mechanics, the ground-state energy is not equal to the minimal potential energy $V_{\min }$, corresponding to the classical particle at rest. (For the square well, $V_{\min }=0$.) The difference $E_1-V_{\min }$ is called zero-point energy, and we will later see that it can be explained by Heisenberg’s uncertainty relation.

Based on the boundary conditions, you can confirm that the wave functions associated with the energy levels are

	${\psi }_n=a_n\cos k_{n}x\mathit{\hspace{1em}}\text{for }n\text{ odd},$		(43)
	${\psi }_n=b_n\sin k_{n}x\mathit{\hspace{1em}}\text{for }n\text{ even}.$		(44)

We restrict $n$ to be positive, since negative values do not deliver independent solutions.

You find diagrams of the energy levels and the associated wave functions of the particle in the square well in almost every book on quantum mechanics (e.g., Young and Freedman, Fig. 40.4; A Rae, Fig 2.1). You could sketch them by yourself!

Points to remember

• •

  For a particle in an infinite square well of width $L$, the stationary Schrödinger equation in the region between the walls is given by

  $$E\psi =-\frac{{\mathrm{\hslash }}^2}{2m}\frac{d^2\psi (x)}{d{x}^2}.$$

  (45)

  This is solved by $\psi =a\cos kx+b\sin kx$ provided $E={\mathrm{\hslash }}^2{k}^2/2m$.

• •

  At the infinitely high walls, the wave function has to vanish ($\psi (x=\mathrm{wall})=0$). This selects $k_n=n\pi /L$, where $n$ is an integer.

• •

  The associated energies are $E_n=\frac{{\pi }^2{\mathrm{\hslash }}^2}{2m{L}^2}{n}^2$.

Based on the specific solutions for the particle in the square well, we now can work to get a better feeling about the physical content of the wave function. What we already have agreed upon is to interpret

$$P(x)={|\psi (x)|}^2$$

(46)

as the probability density in a position measurement. We say density, because in order to get an actual probability to find the particle in some interval we still have to do an integral

(47)

If we do a position measurement, we would find the particle randomly located at various places, but after many experiments we would find the probability density as given in Eq. (46). We say the particle is delocalised in space. [Next to the wave functions for the first few levels of the particle in the square well, you may wish to sketch the corresponding probability densities $P(x)$.]

Since we will find the particle somewhere, we have to require

					(48)
	$\Rightarrow$		${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}𝑑x{|\psi (x)|}^2=1.$		(49)

This is called the normalisation condition for the wave function, and must be obeyed for any wave function $\psi (x)$ representing the state of a quantum-mechanical particle in a one-dimensional potential.

For the square well, the normalisation condition delivers suitable values of the constants $a_n$, $b_n$ in Eqs. (43,44). For $n$ odd, the normalisation integral is given by

$${\int }_{-L/2}^{L/2}𝑑x{|a_n|}^2{\cos }^2\frac{\pi nx}{L}={|a_n|}^2\frac{L}{2}.$$

(50)

The normalisation condition hence is fulfilled for $a_n=\sqrt{2/L}$ (this is the most convenient choice), and the normalised wave function in is

	${\psi }_n(x)$	$=$	$\sqrt{{\displaystyle \frac{2}{L}}}\cos {\displaystyle \frac{\pi nx}{L}}\mathit{\hspace{1em}}\text{for }n\text{ odd},$		(51)
	${\psi }_n(x)$	$=$	$\mathrm{ }\text{for }n\text{ even}.$		(52)

(Exercise: fill in the result for $n$ even.)
Points to remember

• •

  (recall) The probability density to find a particle at position $x$ is $P(x)={|\psi (x)|}^2$.

• •

  The probability to find the particle between two positions $x_1$ and $x_2$ is given by

  (53)

• •

  The probability density has to be normalised: ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}P(x)𝑑x=1$, hence ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|\psi (x)|}^2𝑑x=1$. This is also called the normalisation condition of the wave function.

What if instead of position we measure some other observable quantity of the particle in the square well, say its momentum or its energy?

Let’s start with momentum. Recall that classically the particle bounces between the walls with constant velocity. This translates to the values for momentum $p=\sqrt{2mE}$ when the particle moves right and $p=-\sqrt{2mE}$ when the particle moves left. Also recall that in the case of the freely moving particle with the wave function ${\mathrm{\Psi }}_k$ given in Eq. (7), we could read off the momentum from Eq. (13), $\widehat{p}{\mathrm{\Psi }}_k=p{\mathrm{\Psi }}_k$.

For the particle in the box, we cannot read off the momentum in this way since the wave function $\widehat{p}{\psi }_n$ is not proportional to ${\psi }_n$. E.g., for $n$ odd

$$-i\mathrm{\hslash }\frac{d}{dx}\cos k_{n}x=i\mathrm{\hslash }{k}_n\sin k_{n}x.$$

(54)

Indeed note that

$$\cos k_{n}x=\frac{1}{2}[\exp (i{k}_{n}x)+\exp (-i{k}_{n}x)]$$

(55)

is a superposition (sum) of wave functions that describe the particle moving right and left at the same time. So, the particle must be said to be delocalised in momentum. (Later we describe how to determine the probability that the particle has a given momentum $p$.)

On the other hand, the energy $E_n$ of the states ${\psi }_n$ fulfilling $\widehat{H}{\psi }_n=E_n{\psi }_n$ is well defined. In reality, subsequent energy measurements on particles prepared in a state ${\psi }_n$ all would yield the same value $E_n$, so that for such a state there is no uncertainty attached to energy.
Points to remember

• •

  Values of observables are not certain. A quantum particle can have finite probabilities to be at different places, and move into different directions, all at the same time. A classical particle is certainly not capable of showing such a weird behavior.