2.2 Particle in a One-Dimensional Box
This section solves the time-independent Schrödinger equation for a particle in a onedimensional box. By this we mean a particle subjected to a potential-energy function that is infinite everywhere along the $x$ axis except for a line segment of length $l$, where the potential energy is zero. Such a system may seem physically unreal, but this model can be applied with some success to certain conjugated molecules; see Prob. 2.17. We put the origin at the left end of the line segment (Fig. 2.1).
FIGURE 2.1 Potential energy function $V(x)$ for the particle in a one-dimensional box. 
We have three regions to consider. In regions I and III, the potential energy $V$ equals infinity and the time-independent Schrödinger equation (1.19) is
\( -\frac{\hbar^{2}}{2 m} \frac{d^{2} \psi}{d x^{2}}=(E-\infty) \psi \)
Neglecting $E$ in comparison with $\infty$, we have
\( \frac{d^{2} \psi}{d x^{2}}=\infty \psi, \quad \psi=\frac{1}{\infty} \frac{d^{2} \psi}{d x^{2}} \)
and we conclude that $\psi$ is zero outside the box:
\( \begin{equation} \psi{\mathrm{I}}=0, \quad \psi{\mathrm{III}}=0 \tag{2.9} \end{equation} \)
For region II, $x$ between zero and $l$, the potential energy $V$ is zero, and the Schrödinger equation (1.19) becomes
\( \begin{equation} \frac{d^{2} \psi{\mathrm{II}}}{d x^{2}}+\frac{2 m}{\hbar^{2}} E \psi{\mathrm{II}}=0 \tag{2.10} \end{equation} \)
where $m$ is the mass of the particle and $E$ is its energy. We recognize (2.10) as a linear homogeneous second-order differential equation with constant coefficients. The auxiliary equation (2.7) gives
\( \begin{gather} s^{2}+2 m E \hbar^{-2}=0 \ s= \pm(-2 m E)^{1 / 2} \hbar^{-1} \tag{2.11}\ s= \pm i(2 m E)^{1 / 2} / \hbar \tag{2.12} \end{gather} \)
where $i=\sqrt{-1}$. Using (2.8), we have
\( \begin{equation} \psi{\mathrm{II}}=c{1} e^{i(2 m E)^{1 / 2} x / \hbar}+c_{2} e^{-i(2 m E)^{1 / 2} x / \hbar} \tag{2.13} \end{equation} \)
Temporarily, let
\( \begin{aligned} \theta & \equiv(2 m E)^{1 / 2} x / \hbar \ \psi{\text {III }} & =c{1} e^{i \theta}+c_{2} e^{-i \theta} \end{aligned} \)
We have $e^{i \theta}=\cos \theta+i \sin \theta\left[\right.$ Eq. (1.28)] and $e^{-i \theta}=\cos (-\theta)+i \sin (-\theta)=\cos \theta-$ $i \sin \theta$, since
\( \begin{equation} \cos (-\theta)=\cos \theta \quad \text { and } \quad \sin (-\theta)=-\sin \theta \tag{2.14} \end{equation} \)
Therefore,
\( \begin{aligned} \psi{\mathrm{II}} & =c{1} \cos \theta+i c{1} \sin \theta+c{2} \cos \theta-i c{2} \sin \theta \ & =\left(c{1}+c{2}\right) \cos \theta+\left(i c{1}-i c_{2}\right) \sin \theta \ & =A \cos \theta+B \sin \theta \end{aligned} \)
where $A$ and $B$ are new arbitrary constants. Hence,
\( \begin{equation} \psi_{\mathrm{II}}=A \cos \left[\hbar^{-1}(2 m E)^{1 / 2} x\right]+B \sin \left[\hbar^{-1}(2 m E)^{1 / 2} x\right] \tag{2.15} \end{equation} \)
Now we find $A$ and $B$ by applying boundary conditions. It seems reasonable to postulate that the wave function will be continuous; that is, it will make no sudden jumps in 
FIGURE 2.2 Lowest four energy levels for the particle in a one-dimensional box. value (see Fig. 3.4). If $\psi$ is to be continuous at the point $x=0$, then $\psi{\mathrm{I}}$ and $\psi{\mathrm{II}}$ must approach the same value at $x=0$ :
\( \begin{aligned} \lim {x \rightarrow 0} \psi{\mathrm{I}} & =\lim {x \rightarrow 0} \psi{\mathrm{II}} \ 0 & =\lim _{x \rightarrow 0}\left{A \cos \left[\hbar^{-1}(2 m E)^{1 / 2} x\right]+B \sin \left[\hbar^{-1}(2 m E)^{1 / 2} x\right]\right} \ 0 & =A \end{aligned} \)
since
\( \begin{equation} \sin 0=0 \quad \text { and } \quad \cos 0=1 \tag{2.16} \end{equation} \)
With $A=0$, Eq. (2.15) becomes
\( \begin{equation} \psi_{\mathrm{II}}=B \sin \left[(2 \pi / h)(2 m E)^{1 / 2} x\right] \tag{2.17} \end{equation} \)
Applying the continuity condition at $x=l$, we get
\( \begin{equation} B \sin \left[(2 \pi / h)(2 m E)^{1 / 2} l\right]=0 \tag{2.18} \end{equation} \)
$B$ cannot be zero because this would make the wave function zero everywhere-we would have an empty box. Therefore,
\( \sin \left[(2 \pi / h)(2 m E)^{1 / 2} l\right]=0 \)
The zeros of the sine function occur at $0, \pm \pi, \pm 2 \pi, \pm 3 \pi, \ldots= \pm n \pi$. Hence,
\( \begin{equation} (2 \pi / h)(2 m E)^{1 / 2} l= \pm n \pi \tag{2.19} \end{equation} \)
The value $n=0$ is a special case. From (2.19), $n=0$ corresponds to $E=0$. For $E=0$, the roots (2.12) of the auxiliary equation are equal and (2.13) is not the complete solution of the Schrödinger equation. To find the complete solution, we return to (2.10), which for $E=0$ reads $d^{2} \psi{\mathrm{II}} / d x^{2}=0$. Integration gives $d \psi{\mathrm{II}} / d x=c$ and $\psi{\mathrm{II}}=c x+d$, where $c$ and $d$ are constants. The boundary condition that $\psi{\text {II }}=0$ at $x=0$ gives $d=0$, and the condition that $\psi{\mathrm{II}}=0$ at $x=l$ then gives $c=0$. Thus, $\psi{\mathrm{II}}=0$ for $E=0$, and therefore $E=0$ is not an allowed energy value. Hence, $n=0$ is not allowed.
Solving (2.19) for $E$, we have
\( \begin{equation} E=\frac{n^{2} h^{2}}{8 m l^{2}} \quad n=1,2,3, \ldots \tag{2.20} \end{equation} \)
Only the energy values (2.20) allow $\psi$ to satisfy the boundary condition of continuity at $x=l$. Application of a boundary condition has forced us to the conclusion that the values of the energy are quantized (Fig. 2.2). This is in striking contrast to the classical result that the particle in the box can have any nonnegative energy. Note that there is a minimum value, greater than zero, for the energy of the particle. The state of lowest energy is called the ground state. States with energies higher than the ground-state energy are excited states. (In classical mechanics, the lowest possible energy of a particle in a box is zero. The classical particle sits motionless inside the box with zero kinetic energy and zero potential energy.)
EXAMPLE
A particle of mass $2.00 \times 10^{-26} \mathrm{~g}$ is in a one-dimensional box of length 4.00 nm . Find the frequency and wavelength of the photon emitted when this particle goes from the $n=3$ to the $n=2$ level.
By conservation of energy, the energy $h \nu$ of the emitted photon equals the energy difference between the two stationary states [Eq. (1.4); see also Section 9.9]:
\( \begin{gathered} h \nu=E{\text {upper }}-E{\text {lower }}=\frac{n{u}^{2} h^{2}}{8 m l^{2}}-\frac{n{l}^{2} h^{2}}{8 m l^{2}} \ \nu=\frac{\left(n{u}^{2}-n{l}^{2}\right) h}{8 m l^{2}}=\frac{\left(3^{2}-2^{2}\right)\left(6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)}{8\left(2.00 \times 10^{-29} \mathrm{~kg}\right)\left(4.00 \times 10^{-9} \mathrm{~m}\right)^{2}}=1.29 \times 10^{12} \mathrm{~s}^{-1} \end{gathered} \)
where $u$ and $l$ stand for upper and lower. Use of $\lambda \nu=c$ gives $\lambda=2.32 \times 10^{-4} \mathrm{~m}$. (A common student error is to set $h \nu$ equal to the energy of one of the states instead of the energy difference between states.)
EXERCISE For an electron in a certain one-dimensional box, the longest-wavelength transition occurs at 400 nm . Find the length of the box. (Answer: 0.603 nm .)
Substitution of (2.19) into (2.17) gives for the wave function
\( \begin{equation} \psi_{\mathrm{II}}=B \sin \left(\frac{n \pi x}{l}\right), \quad n=1,2,3, \ldots \tag{2.21} \end{equation} \)
The use of the negative sign in front of $n \pi$ does not give us another independent solution. Since $\sin (-\theta)=-\sin \theta$, we would simply get a constant, -1 , times the solution with the plus sign.
The constant $B$ in Eq. (2.21) is still arbitrary. To fix its value, we use the normalization requirement, Eqs. (1.24) and (1.22):
\( \begin{gather} \int{-\infty}^{\infty}|\Psi|^{2} d x=\int{-\infty}^{\infty}|\psi|^{2} d x=1 \ \int{-\infty}^{0}\left|\psi{\mathrm{I}}\right|^{2} d x+\int{0}^{l}\left|\psi{\mathrm{II}}\right|^{2} d x+\int{l}^{\infty}\left|\psi{\mathrm{III}}\right|^{2} d x=1 \ |B|^{2} \int_{0}^{l} \sin ^{2}\left(\frac{n \pi x}{l}\right) d x=1=|B|^{2} \frac{l}{2} \tag{2.22} \end{gather} \)
where the integral was evaluated by using Eq. (A.2) in the Appendix. We have
\( |B|=(2 / l)^{1 / 2} \)
Note that only the absolute value of $B$ has been found. $B$ could be $-(2 / l)^{1 / 2}$ as well as $(2 / l)^{1 / 2}$. Moreover, $B$ need not be a real number. We could use any complex number with absolute value $(2 / l)^{1 / 2}$. All we can say is that $B=(2 / l)^{1 / 2} e^{i \alpha}$, where $\alpha$ is the phase of $B$ and could be any value in the range 0 to $2 \pi$ (Section 1.7). Choosing the phase to be zero, we write as the stationary-state wave functions for the particle in a box
\( \begin{equation} \psi_{\mathrm{II}}=\left(\frac{2}{l}\right)^{1 / 2} \sin \left(\frac{n \pi x}{l}\right), \quad n=1,2,3, \ldots \tag{2.23} \end{equation} \)
Graphs of the wave functions and the probability densities are shown in Figs. 2.3 and 2.4.
The number $n$ in the energies (2.20) and the wave functions (2.23) is called a quantum number. Each different value of the quantum number $n$ gives a different wave function and a different state.
FIGURE 2.3 Graphs of $\psi$ for the three lowest-energy particle-in-a-box states. 
FIGURE 2.4 Graphs of $|\psi|^{2}$ for the lowest particle-in-a-box states.
The wave function is zero at certain points; these points are called nodes. For each increase of one in the value of the quantum number $n, \psi$ has one more node. The existence of nodes in $\psi$ and $|\psi|^{2}$ may seem surprising. Thus, for $n=2$, Fig. 2.4 says that there is zero probability of finding the particle in the center of the box at $x=l / 2$. How can the particle get from one side of the box to the other without at any time being found in the center? This apparent paradox arises from trying to understand the motion of microscopic particles using our everyday experience of the motions of macroscopic particles. However, as noted in Chapter 1, electrons and other microscopic "particles" cannot be fully and correctly described in terms of concepts of classical physics drawn from the macroscopic world.
Figure 2.4 shows that the probability of finding the particle at various places in the box is quite different from the classical result. Classically, a particle of fixed energy in a box bounces back and forth elastically between the two walls, moving at constant speed. Thus it is equally likely to be found at any point in the box. Quantum mechanically, we find a maximum in probability at the center of the box for the lowest energy level. As we go to higher energy levels with more nodes, the maxima and minima of probability come closer together, and the variations in probability along the length of the box ultimately become undetectable. For very high quantum numbers, we approach the classical result of uniform probability density.
This result, that in the limit of large quantum numbers quantum mechanics goes over into classical mechanics, is known as the Bohr correspondence principle. Since Newtonian mechanics holds for macroscopic bodies (moving at speeds much less than the speed of light), we expect nonrelativistic quantum mechanics to give the same answer as classical mechanics for macroscopic bodies. Because of the extremely small size of Planck's constant, quantization of energy is unobservable for macroscopic bodies. Since the mass of the particle and the length of the box squared appear in the denominator of Eq. (2.20), a macroscopic object in a macroscopic box having a macroscopic energy of motion would have a huge value for $n$, and hence, according to the correspondence principle, would show classical behavior.
We have a whole set of wave functions, each corresponding to a different energy and characterized by the quantum number $n$, which is a positive integer. Let the subscript $i$ denote a particular wave function with the value $n_{i}$ for its quantum number:
\( \begin{gathered} \psi{i}=\left(\frac{2}{l}\right)^{1 / 2} \sin \left(\frac{n{i} \pi x}{l}\right), \quad 0<x<l \ \psi_{i}=0 \quad \text { elsewhere } \end{gathered} \)
Since the wave function has been normalized, we have
\( \begin{equation} \int{-\infty}^{\infty} \psi{i}^{} \psi_{j} d x=1 \quad \text { if } i=j \tag{2.24} \end{equation} \)
We now ask for the value of this integral when we use wave functions corresponding to different energy levels:
\( \int{-\infty}^{\infty} \psi{i}^{*} \psi{j} d x=\int{0}^{l}\left(\frac{2}{l}\right)^{1 / 2} \sin \left(\frac{n{i} \pi x}{l}\right)\left(\frac{2}{l}\right)^{1 / 2} \sin \left(\frac{n{j} \pi x}{l}\right) d x, \quad n{i} \neq n{j} \)
Use of Eq. (A.5) in the Appendix gives
\( \begin{equation} \int{-\infty}^{\infty} \psi{i}^{} \psi{j} d x=\frac{2}{l}\left[\frac{\sin \left[\left(n{i}-n{j}\right) \pi\right]}{2\left(n{i}-n{j}\right) \pi / l}-\frac{\sin \left[\left(n{i}+n{j}\right) \pi\right]}{2\left(n{i}+n_{j}\right) \pi / l}\right]=0 \tag{2.25} \end{equation} \)
since $\sin m \pi=0$ for $m$ an integer. We thus have
\( \begin{equation} \int{-\infty}^{\infty} \psi{i}^{} \psi_{j} d x=0, \quad i \neq j \tag{2.26} \end{equation} \)
When (2.26) holds, the functions $\psi{i}$ and $\psi{j}$ are said to be orthogonal to each other for $i \neq j$. We can combine (2.24) and (2.26) by writing
\( \begin{equation} \int{-\infty}^{\infty} \psi{i}^{} \psi{j} d x=\delta{i j} \tag{2.27} \end{equation} \)
The symbol $\delta_{i j}$ is called the Kronecker delta (after a mathematician). It equals 1 when the two indexes $i$ and $j$ are equal, and it equals 0 when $i$ and $j$ are unequal:
\( \delta_{i j} \equiv \begin{cases}0 & \text { for } i \neq j \tag{2.28}\ 1 & \text { for } i=j\end{cases} \)
The property (2.27) of the wave functions is called orthonormality. We proved orthonormality only for the particle-in-a-box wave functions. We shall prove it more generally in Section 7.2.
You might be puzzled by Eq. (2.26) and wonder why we would want to multiply the wave function of one state by the wave function of a different state. We will later see (Section 7.3, for example) that it is often helpful to use equations that contain a sum involving all the wave functions of a system, and such equations can lead to integrals like that in (2.26).
A more rigorous way to look at the particle in a box with infinite walls is to first treat the particle in a box with a finite jump in potential energy at the walls and then take the limit as the jump in $V$ becomes infinite. The results, when the limit is taken, will be the same as (2.20) and (2.23) (see Prob. 2.22).
We have considered only the stationary states of the particle in a one-dimensional box. For an example of a nonstationary state of this system, see the example near the end of Section 7.8.
Some online computer simulations of the particle in a box can be found at www .chem.uci.edu/undergraduate/applets/dwell/dwell.htm (shows the effects on the wave functions and energy levels when a barrier of variable height and width is introduced into the middle of the box); web.williams.edu/wp-etc/chemistry/dbingemann/Chem153/particle .html (shows quantization by plotting the solution to the Schrödinger equation as the energy is varied and as the box length is varied); and falstad.com/qm1d/ (shows both timeindependent and time-dependent states; see Prob. 7.47).