# Particle in a Box Detailed Solution

By

Ryan Phillips

●

Published 2023-09-13

The particle in a box problem is a thought experiment made by Schrödinger to illustrate the basics of 'motion' in the quantum world. In this section, we will solve the time-independent SE for the particle in box problem¹.

The formal way to state the problem is to ask what happens to the wavefunction of a single particle along a 1D line when subjected to the following boundary conditions:

V(x)=\begin{cases}
0\text{ if }0<x<L,\\
\infty\text{ elsewhere}
\end{cases}

where V(x) is the potential at any point in space. For simplicity, we assumed V(x)=0 inside the box and that the particle's wavefunction does not change with time.

Next, we need the Schrödinger's equation, though it is a difficult equation to solve as it has derivatives in both time t and position x. If one is interested to find the "stationary solutions", such as standing waves, then it is beneficial to remove time from the SE.

Recall the Time-independent Schrödinger's equation states that

\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V(x)\right)\,\psi(x)=E\psi(x)

Inside the box 0<x<L where V(x)=0, we have

\begin{align}
-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi(x)}{\partial x^{2}}=E\psi(x)
\end{align}

and the wavefunction outside the box must obey V(x)=\infty,

\begin{align}
\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi(x)}{\partial x^{2}}+\infty \right)\psi(x) =E\psi(x) \nonumber
\end{align}

We can make an educated assumption that the wavefunction outside the box must be zero such that Schrodinger's equation still holds² (because 0=0). This also gives us the boundary conditions of the problem:

\begin{align}
\psi(0) & =0  \\
\psi(L) & =0 
\end{align}

Here, the wavefunction 'terminates' at the boundaries. But now, what about the wavefunction inside the box for 0 < x< L?

We can guess its solution; the second derivative of the wavefunction returns itself multiplied by a constant, which implies that the wavefunction must resemble something of form

\begin{align}
\psi(x)=ce^{\lambda x}
\end{align}

where \lambda is a complex value. If you've forgotten why, see 3.5.1 The Wavefunction and Schrödinger's equation.

Taking the first and second derivative,

\begin{align}
\psi'(x) &=\lambda ce^{\lambda x} \nonumber \\ 
\psi''(x)&=\lambda^2 ce^{\lambda x}  \nonumber 
\end{align}

Substituting into the SE, we obtain

\begin{align}
-\frac{\hbar^2}{2m} (\lambda^2 ce^{\lambda x}) &= E(ce^{\lambda x}) \nonumber\\ 
-\frac{\hbar^2}{2m} \lambda^2 &= E \nonumber\\
\lambda^2 &= -\frac{2mE}{\hbar^2} \nonumber\\
\lambda &= \pm\sqrt{-1}\sqrt{\frac{2mE}{\hbar^2}} \nonumber\\
\lambda &= \pm i\sqrt{\frac{2mE}{\hbar^2}}  \\ 
\end{align}

Putting the guess solution into the SE, we obtain the general solution to be

\begin{align}
\psi(x)=c_{1}e^{ikx}+c_{2}e^{-ikx} 
\end{align}

where c_1, c_2 are some constants and k = \sqrt{\frac{2mE}{\hbar^2}}.

Applying the boundary conditions Eq.(2) and (3), we will find that

\begin{align}
\psi(0)= c_{1}e^{0}+c_{2}e^{0} &= 0 \nonumber \\
c_{1} + c_{2} &= 0 \nonumber \\
c_{1} &= - c_{2} \nonumber \\
\end{align}

and thus

\begin{align}
\psi(L)= c_{1}e^{ikL}-c_{1}e^{-ikL}  = 0 \nonumber \\
c_{1}(e^{ikL}-e^{-ikL})  = 0  \\
\end{align}

Using Euler's identity³ and since \cos(x)=\cos(-x) while -\sin(x)=\sin(-x),

\begin{align}
e^{ix} = \cos(x) + i\sin(x) \nonumber \\
e^{-ix} = \cos(x) - i\sin(x) \nonumber \\
e^{ix}-e^{-ix}  = 2i \sin(x)  \nonumber \\
\end{align}

we can use these to simplify Eq.(7) such that

\begin{align}
c_{1}(e^{ikL}-e^{-ikL})  = 0 \nonumber \\
c_{1}[2i \sin(kL)] = 0
\end{align}

We don't have to worry about the complex constants 2ic_{1} and can simplify it later. What's more important is that for Eq.(8), the sine function only returns 0 when \underline{kL = n\pi}, where n takes positive integer values. See Figure 1 below.

At this point, we can already obtain the energy states equation. First, we rearrange the finding from Eq.(8):

\begin{align}
kL = n\pi  \nonumber \\
k = \frac{n\pi}{L}  \nonumber \\
\end{align}

and comparing with the value of k from Eq.(6),

\begin{align}
k &= \space \sqrt{\frac{2mE}{\hbar^2}} \nonumber \\
k^2 &= \frac{2mE}{\hbar^2} \nonumber \\
\frac{n^2 \pi^2}{L^2} &= \frac{2mE}{\hbar^2} \nonumber \\
E &= \frac{\hbar^{2}n^{2}\pi^{2}}{2mL^{2}}  \\
\end{align}

Unlike Bohr’s atomic model which postulated that discrete energy levels exist, Schrodinger’s derivation naturally shows that energy levels must take discrete values! Recall n takes only positive integer values, and is also known as the quantum number.

Consequently, from Eq.(8) we have

\psi(x)=A\sin kx

where A=2ic_1 is a complex constant.

Recall our normalization condition

\int_{-\infin}^{\infin} |\Psi|^2 \space dx= 1

hence

\begin{align}
\int_{-\infin}^{\infin} \psi(x)\psi^*(x) \space dx &= 1 \nonumber \\
\int_{-\infin}^{\infin} |A|^2 \sin^2(kx) \space dx &= 1 \nonumber \\
\int_{-\infin}^{\infin} \sin^2(kx) \space dx &= \frac{1}{A^2}  \\
\end{align}

Using the trigonometric identity,

\sin^2(x) = \frac{1-\cos(2x)}{2}

and adjusting the limits to the regions we are interested in, we can integrate Eq.(9) as follows:

\begin{align}
\int_{0}^{L} \frac{1-\cos(2kx)}{2} \space dx &= \frac{1}{A^2} \nonumber \\
\int_{0}^{L} 1-\cos(2kx) \space dx &= \frac{2}{A^2} \nonumber \\
\left[ x-\frac{\sin(2kx)}{2} \right]_0^L &= \frac{2}{A^2} \nonumber \\
\left[ L-\frac{\sin(2kL)}{2} -0 \right] &= \frac{2}{A^2}, \text{ where sin(2kL) = sin(2nπ) = 0} \nonumber \\
L &= \frac{2}{A^2} \nonumber \\
A &= \pm\sqrt{\frac{2}{L}} \nonumber \\

\end{align}

Finally the wavefunction of the particle in a box is

\begin{equation}
\psi(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi}{L}x\right)
\end{equation}

and it's allowed energy levels are given by Eq.(9).

Note that the wavefunction in Eq.(11) can take either positive or negative values of A. Wavefunctions have no physical meaning so it does not matter which we take; squared values are always positive.

We are only concerned with the square of the wavefunction as it gives us the Probability Density Function which tells us the probability of finding the particle within x_0 and x_1.

\begin{align}
P(x_0<x<x_1) &= \int_{x_0}^{x_1} \psi(x)\psi^*(x) \space dx \nonumber \\
&= \int_{x_0}^{x_1} |\psi(x)|^2 \space dx \\

\end{align}