Particle in a Box

Setup

Suppose we have a particle moving freely in a box, whose Wave function and Energy we want to calculate.

1. We can model this as a Quantum state $\psi (x),x\in [0,L{]}^n$.
2. The Wave function is $0$ on the boundary $\partial [0,L{]}^n$ of the domain.
3. Since we are working with a free particle, the Hamiltonian $H$ is given by $H=\frac{P^2}{2m}=-\frac{{\hslash }^2}{2m}{\nabla }^2$, where
   1. $m$ is the Particle’s mass.
   2. $P$ is the Momentum Operator.

Solution

Using the Time-independent Schrodinger equation, we get

$$\begin{array}{rll}E|\psi ⟩& =H|\psi ⟩& \text{By definition of our Hamiltonian H}\\ & =\frac{P^2}{2m}|\psi ⟩& \text{By definition of the Momentum Operator}\\ & =-\frac{{\hslash }^2}{2m}{\nabla }^2|\psi ⟩& \end{array}$$

Using the ansatz $\psi (x)={\prod }_i{\psi }_i(x_i),x_i\in \mathbb{R}$, we can split the equation above into $n$ independent 2nd order differential equations $E_i{\psi }_i=-\frac{{\hslash }^2}{2m}\frac{{\partial }^2}{\partial x_i^2}{\psi }_i$, which have the generic solution ${\psi }_i(x_i)={\alpha }_i\sin (\frac{\sqrt{2m{E}_i}}{\hslash }{x}_i)+{\beta }_i\cos (\frac{\sqrt{2m{E}_i}}{\hslash }{x}_i)$. Plugging in the boundary condition ${\psi }_i(0)=0$ gives us the constraint ${\beta }_i=0$, simplifying our solution to ${\psi }_i(x_i)={\alpha }_i\sin (\frac{\sqrt{2m{E}_i}}{\hslash }{x}_i)$. Plugging in the boundary condition $0={\psi }_i(L)$ we get

$$\begin{array}{rcll}0& =& {\psi }_i(L)& \text{Plugging in our ansatz}\\ & =& {\alpha }_i\sin (\frac{\sqrt{2m{E}_i}}{\hslash }L)& \\ 0& \stackrel{{\alpha }_i\ne 0}{=}& \sin (\frac{\sqrt{2m{E}_i}}{\hslash }L)& \\ \frac{\sqrt{2m{E}_i}}{\hslash }L& \stackrel{n_i\in \mathbb{Z}}{=}& 2\pi n_i& \\ \sqrt{2m{E}_i}& =& \frac{2\pi n_i\hslash }{L}& \end{array}$$

Plugging the result above we get ${\psi }_i(x_i)={\alpha }_i\sin (\frac{\sqrt{2m{E}_i}}{\hslash }{x}_i)={\alpha }_i\sin (\frac{2\pi n_i\hslash }{\hslash L}{x}_i)={\alpha }_i\sin (\frac{2\pi n_i}{L}{x}_i)$. Since $\psi$ needs to be Normalizable, we have

$$\begin{array}{rll}1& =\Vert \psi {\Vert }^2& x\in {\mathbb{R}}^n\\ & ={\int }_{[0,L{]}^n}\psi (x)dx& \\ & ={\int }_{[0,L{]}^n}{\prod }_i{\psi }_i(x_i)dx& \\ & ={\int }_{[0,L{]}^n}{\prod }_i|{\alpha }_i{|}^2{\sin }^2(\frac{2\pi n_i}{L}{x}_i)dx& \\ & ={\prod }_i|{\alpha }_i{|}^2{\int }_{[0,L{]}^n}{\prod }_i{\sin }^2(\frac{2\pi n_i}{L}{x}_i)dx& \alpha :={\prod }_i{\alpha }_i\\ & =|\alpha {|}^2{\int }_{[0,L{]}^n}{\prod }_i{\sin }^2(\frac{2\pi n_i}{L}{x}_i)dx& {\int }_{{\mathbb{R}}^n}{\prod }_i^n{f}_i(x_i)dx={\prod }_{{\mathbb{R}}^n}{\int }_{\mathbb{R}}{f}_i(x_i)d{x}_i\\ & =|\alpha {|}^2{\prod }_i{\int }_0^L{\sin }^2(\frac{2\pi n_i}{L}{x}_i)d{x}_i& \int {\sin }^2(\alpha x)dx=\frac{x}{2}-\frac{\sin (2ax)}{4a}\\ & =|\alpha {|}^2{\prod }_i{\left[\frac{x}{2}-\frac{1}{4\frac{2\pi n_i}{L}x}\sin (\frac{4\pi n_i}{L}x)\right]}_0^L& \\ & =|\alpha {|}^2{\prod }_i[\frac{L}{2}-0]& \\ & =|\alpha {|}^2{\left(\frac{L}{2}\right)}^n& \end{array}$$

Since the Global Phase of $\psi$ does not impact the solution, we can assume that $\alpha \in {\mathbb{R}}^{+}$, giving us that $1=|\alpha {|}^2{\left(\frac{L}{2}\right)}^n\Rightarrow \alpha ={\left(\frac{L}{2}\right)}^{-n}$. Hence the solution is given by $\psi (x)={\left(\frac{L}{2}\right)}^{-\frac{n}{2}}{\prod }_i\sin (\frac{2\pi n_i}{L}{x}_i),n_i\in {\mathbb{Z}}_0^{+}$. Since $\psi$ is normalized, the Energy associated with this solution is given by $⟨\psi |H|\psi ⟩$.

$$\begin{array}{rl}⟨\psi |H|\psi ⟩& ={\int }_{[0,L{]}^n}\overline{\psi (x)}H\psi (x)dx\\ & ={\int }_{[0,L{]}^n}\overline{{\left(\frac{L}{2}\right)}^{-\frac{n}{2}}{\prod }_i\sin (\frac{2\pi n_i}{L}{x}_i)}(-\frac{{\hslash }^2}{2m}{\nabla }^2){\left(\frac{L}{2}\right)}^{-\frac{n}{2}}{\prod }_i\sin (\frac{2\pi n_i}{L}{x}_i)dx\\ & =-\frac{{\hslash }^2}{2m}{\left(\frac{L}{2}\right)}^{-n}{\int }_{[0,L{]}^n}\overline{{\prod }_i\sin (\frac{2\pi n_i}{L}{x}_i)}({\nabla }^2){\prod }_i\sin (\frac{2\pi n_i}{L}{x}_i)dx\\ & =-\frac{{\hslash }^2}{2m}{\left(\frac{L}{2}\right)}^{-n}{\int }_{[0,L{]}^n}\overline{{\prod }_i\sin (\frac{2\pi n_i}{L}{x}_i)}({\sum }_j\frac{{\partial }^2}{\partial x_j^2}){\prod }_i\sin (\frac{2\pi n_i}{L}{x}_i)dx\\ & =-\frac{{\hslash }^2}{2m}{\left(\frac{L}{2}\right)}^{-n}{\int }_{[0,L{]}^n}\overline{{\prod }_i\sin (\frac{2\pi n_i}{L}{x}_i)}({\sum }_j-{\left(\frac{2\pi n_j}{L}\right)}^2){\prod }_i\sin (\frac{2\pi n_i}{L}{x}_i)dx\\ & =-\frac{{\hslash }^2}{2m}{\left(\frac{L}{2}\right)}^{-n}{\sum }_j\left[-{\left(\frac{2\pi n_j}{L}\right)}^2{\int }_{[0,L{]}^n}\overline{{\prod }_i\sin (\frac{2\pi n_i}{L}{x}_i)}{\prod }_i\sin (\frac{2\pi n_i}{L}{x}_i)dx\right]\\ & =\frac{2{\pi }^2{\hslash }^2}{m{L}^2}{\left(\frac{L}{2}\right)}^{-n}{\sum }_j\left[n_j^2{\int }_{[0,L{]}^n}\overline{{\prod }_i\sin (\frac{2\pi n_i}{L}{x}_i)}{\prod }_i\sin (\frac{2\pi n_i}{L}{x}_i)dx\right]\\ & =\frac{2{\pi }^2{\hslash }^2}{m{L}^2}{\left(\frac{L}{2}\right)}^{-n}{\sum }_j\left[n_j^2{\int }_{[0,L{]}^n}{\prod }_i{\sin }^2(\frac{2\pi n_i}{L}{x}_i)dx\right]\\ & =\frac{2{\pi }^2{\hslash }^2}{m{L}^2}{\left(\frac{L}{2}\right)}^{-n}{\sum }_j\left[n_j^2{\prod }_i{\int }_0^L{\sin }^2(\frac{2\pi n_i}{L}{x}_i)d{x}_i\right]\\ & =\frac{2{\pi }^2{\hslash }^2}{m{L}^2}{\left(\frac{L}{2}\right)}^{-n}{\sum }_j\left[n_j^2{\prod }_i\frac{L}{2}\right]\\ & =\frac{2{\pi }^2{\hslash }^2}{m{L}^2}{\left(\frac{L}{2}\right)}^{-n}{\sum }_j\left[n_j^2{\left(\frac{L}{2}\right)}^n\right]\\ & =\frac{2{\pi }^2{\hslash }^2}{m{L}^2}{\left(\frac{L}{2}\right)}^{-n}{\left(\frac{L}{2}\right)}^n{\sum }_j\left[n_j^2\right]\\ & =\frac{2{\pi }^2{\hslash }^2}{m{L}^2}{\sum }_j{n}_j^2\end{array}$$

Where we needed to lookup the Table of integrals to find ${\int }_0^L{\sin }^2(\frac{2\pi n_i}{L}{x}_i)d{x}_i$.

• Correct the above