## The Particle in a Box I: the Schrödinger Equation in One-dimension

In 1926, the Austrian physicist Erwin Schrödinger (1887-1961) made a fundamental mathematical discovery that had a profound impact on the study of the molecular world (in 1933, Schrödinger was awarded with the Nobel prize in Physics just 7 years later his breakthrough discovery). He discovered that a state of a quantum system composed by particles (such as electrons and nucleons) can be described by postulating the existence of a function of the particle coordinates and time, called state function or wave function ($\Psi$, psi function). This function are solution of a wave equation: the so-called the Schrödinger equation (SE). Although the SE equation can be solved analytically only for relatively simple cases, the development of computer and numerical methods has made possible the application of SE to study complex molecular.

Nowadays, computational quantum chemistry has surged to a level of a fundamental research discipline in chemistry and material science. In recognition of the contribution to the advancement of science, the pioneers of this discipline, Prof. Walter Kohn, and Prof. John A. Pople have been awarded the 2008 Nobel prize in chemistry.

In this primer, we are going to study analytical and numerical solutions of SE for simple quantum systems in stationary conditions, e.g. whose state does not change with the time. For this purpose, we are going to use the time-independent form of SE. This condition is an excellent approximation to describe the state of molecular systems.

The time-independent SE is given in his compact form by the equation.

${\bf \hat H }\Psi(x,y,z)=E\Psi(x,yz),$

where the $\bf \hat H$ (H hat) indicates the time-independent Hamiltonian operator used to calculate the total energy of the system. I just recall you that an operator is a mathematical transformation that applies to what follows the operator symbol, in this case, the wave function ($\Psi(x,y,z)$). We will see in a moment how this looks. The application of the operator produce as results the same wave function multiplied for a number $E.$ This number represented the total energy state of the system (or \em eigenstate\em). This equation is also called eigenvalue equation, and, in this case, $\Psi(x,y,z)$ is also said to be an eigenfunction of $\bf \hat H$ with the associated \em eigenvalue,\em $E.$

Shall we look more in detail to the Hamiltonian operator, and, for the sake of simplicity, shall we limit ourselves to consider only a mono-dimensional physical system, namely a system defined only along the reference axis x. Therefore, the wave function depends on only by  x or $\Psi(x).$ The Hamiltonian operator represents the total energy of the system, and, therefore, is composed by the summation of a kinetic and potential energy operator: $\bf \hat H = \hat K+ \hat V.$ The kinetic energy is expressed as

$E=p_x^2/2m$ where $p_x$  is the classic momentum that is associated with the partial differential operator  $p_x \rightarrow \left(\hbar/i \right)\left( \partial /\partial x\right).$ The potential energy is usually a function of the coordinates, and it is associated (in our case) with operator $\bf \hat x.$ Therefore we can write the resulting SE as:

$-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+V(x)\psi(x)=E\psi(x)$

This differential equation can be solved analytically for different types of potentials. A simple classical example is a particle in a one-dimension box within infinitely repulsive barriers.

In this case, the value of $V(x)$ is equal zero for and infinite in the walls. The wave function beyond the walls of the box is null. The wave function and energies of the particle inside the box are obtained by the solution of SE with $V(x)=0$ reported in the central part of the Figure 1.

## Derivation of the wavefunction

The one-dimensionSE is a linear second order differential equation with solutions

$\psi(x)=Ae^{ikx}+Be^{-ikx}$

using the Euler relation $e^{\pm ikx}=\cos(kx)\pm i\sin(kx)$

we can write

$\psi(x)=(A+B)\cos(kx)+(A-B)\sin(kx)$

or called $C=A+B$ and $D=A-B$ we can write

$\psi_k(x)=C\sin(kx)+D\cos(kx)$

by substituting in SE with $V(x)=0$ (e.g. particle inside the box)

$-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}=E\psi(x)$

we obtain

$-\frac{\hbar^2}{2m}\left( -C\sin(kx)-D\cos(kx)\right) = E_k\left(C\sin(kx)+D\cos(kx)\right)$

that gives

$E_k=\frac{k^2\hbar^2}{2m}$

the quantised total energy of the system.

Now we  estimate the values of  constant k, C, and  D of the solution.

By applying the boundary conditions

$\psi(0)=0; ~\psi(L)=0$

we obtain for x=0 that

$\psi(0) =C\sin(0)+D\cos(0)=D=0$

It is more convenient to represent the parameter k as a circular frequency with $\lambda$ the De Broglie wavelength

of the particle

$k=\frac{2\pi}{\lambda}$

For x=L the solution gives

$\psi(L)=C\sin \left( \frac{2\pi L}{\lambda} \right) =0$

This is accomplished for $\frac{2\pi L}{\lambda} = n\pi$ or

$\lambda =\frac{2L}{n}$

and

$\psi(x)=C\sin \left( \frac{n\pi x}{L} \right) \text{with} ~n=1,2, \dots$

We still need to calculate the value of the constant C.  As the square of the wave function $\psi(x)$ gives the probability density and its integral the probability to find an particle in a region of the box then if we integrate over all the x axis the probability should be equal to one:

$\int _{0}^{L}\psi^2(x) dx = C^2 \int_{0}^{L} \sin^2 \left( \frac{n \pi x}{L} \right) dx =1$ (1)

Using this integral we can find the value of C that normalize the function. The integral  $\int_{0}^{L} \sin^2 (Kx) dx = \int_{0}^{L} \sin (x) \sin(Kx)dx$ with $K=\frac{n\pi }{L}$ can be calculated by part

as $\int uw' dx = uw- \int u'w$ with in this case $u=sin(Kx); w'=K sin(Kx)$ therefore

$\int_{0}^L \sin^2 (x) dx = -\frac{\sin (Kx) \cos(Kx)}{K} \bigg|_{0}^L + K\int_0^L \cos^2(x) dx = -\frac{\sin (Kx) \cos(Kx)}{K} \bigg|_{0}^L + \int_0^L (1-\sin^2(Kx)) dx$

so we have  that

$\int_0^L \sin^2 (Kx) dx =-\frac{\sin (Kx) \cos(Kx)}{K} \bigg|_{0}^L + \int_0^L dx - \int_0^L sin^2(Kx) dx$

taking the last integral in the second member of the equation to the first member we have

$2 \int_0^L \sin^2 (Kx) dx = -\frac{\sin (Kx) \cos(Kx)}{K} \bigg|_{0}^L + x \bigg|_{0}^L$

but the first term in the second member is equal to zero  and the second equal to L, hence

$\int_0^L \sin^2 (Kx) dx = \frac{L}{2}$.

Therefore from Equ. (1), we have

$C^2 \frac{L}{2} =1$ and $C=\sqrt{\frac{2}{L}}$.

Therefore the wavefunctions of a particle in a box are given by

$\psi(x)=\sqrt{\frac{2}{L}} \sin \left( \frac{n\pi x}{L} \right) \text{with} ~n=1,2, \dots,~ \text {and} ~0\le x \le L$.

and the associated energies are

$E_n=\frac{n^2 h^2}{8mL^2}$.

The quantum number n is an integer associated with the different energy levels of the particle. Larger n values correspond to higher energy levels of the system. Also note in the graphs, the increase of the number of nodes in the wave function with the increase of the value of n.

In Figure 2, the wavefunctions for the first three energy level of the particle in a box are shown.

The separation between the adjacent energy level with quantum number $n$ and $n+1$ is

$\Delta E=E_{n+1}-E_n =(2n+1)\frac{h^2}{8mL^2}$

The probability density for a particle in a box is given by the square of the wavefunction

$\psi^2(x)=\frac{2}{L}\sin^2 \left(\frac{n\pi x}{L} \right)$

and the probability to find the particle in a region $[x_1,x_2]$ of the box can be calculated using the integral

$P=\int_{x_1}^{x_2} \psi^2(x) dx$.