03 - Particles in Boxes or Circles

Particle in a Box

Consider a particle in a box on the $x$-axis
Particle moves freely inside some interval $[0,a]\subset \mathbb{R}$ where $a>0$ and cannot leave region

Modelled by potential function $V=V(x)$ defined by

$$V(x)=\{\begin{array}{ll}0& \text{if }0<x<a\\ +\infty & \text{otherwise}\end{array}$$

Hence solution is

$$\psi (x)={\psi }_n(x)=\{\begin{array}{ll}B\sin (\frac{n\pi x}{a})& 0<x<a\\ 0& \text{otherwise}\end{array}$$

with associated energy

$$E=E_n=\frac{n^2{\pi }^2{\hslash }^2}{2m{a}^2}$$

which is quantised as it is discrete

Proof

Inside box then One-dimensional Stationary State Schrödinger Equation rearranges to

$$\frac{d^2\psi }{d{x}^2}=-\frac{2mE}{{\hslash }^2}\psi$$

for $x\in (0,a)$

However for $x\notin (0,a)$ then $\psi =0$ (in order to satisfy Schrödinger Equation)

Hence general solution is

$$\psi (x)=\{\begin{array}{ll}A\cos \left(\frac{\sqrt{2mE}}{h}x\right)+B\sin \left(\frac{\sqrt{2mE}}{h}x\right)& E>>0\\ A+Bx& E=0\\ A\cosh \left(\frac{\sqrt{-2mE}}{h}x\right)+B\sinh \left(\frac{\sqrt{-2mE}}{h}x\right)& E<0\end{array}$$

Assuming $\varphi$ is continuous gives boundary condition $\varphi (0)=\varphi (a)=0$
Hence $\varphi (0)=0⟹A=0$ so

$$0=\varphi (a)=\{\begin{array}{ll}B\sin \left(\frac{\sqrt{2mE}}{h}a\right)& E>>0\\ Ba& E=0\\ B\sinh \left(\frac{\sqrt{-2mE}}{h}a\right)& E<0\end{array}$$

As $a>>0$, when $E\le 0$ then $B=0$ hence $\psi \equiv 0$ for $E\le 0$

For $E>>0$ then boundary condition $\varphi (a)=0$ implies

$$\frac{\sqrt{2mE}}{\hslash }=\frac{n\pi }{a}$$

for some integer $n\in \mathbb{Z}$
Note that ignoring cases $B=0$ and $\psi \equiv 0$ to avoid trivial solutions >

Hence solution is

$$\psi (x)={\psi }_n(x)=\{\begin{array}{ll}B\sin (\frac{n\pi x}{a})& 0<x<a\\ 0& \text{otherwise}\end{array}$$

with associated energy

$$E=E_n=\frac{n^2{\pi }^2{\hslash }^2}{2m{a}^2}$$

which is quantised as it is discrete

Ground State Energy

Consider the possible energies of quantum system which are discrete and bounded below

Ground State Energy (or Zero Point Energy) is the lowest possible energy
with corresponding ground state wave function

Higher energies in increasing order are $k$th excited energy with $k$th excited state wave function

Full Time-Dependent Wave Function for Particle in a Box

$${\Psi }_n(x,t)={\psi }_n(x)e^{-i{E}_{n}t/\hslash }=\{\begin{array}{ll}B\sin \left(\frac{n\pi x}{a}\right)e^{-i{n}^2{\pi }^2\hslash t/2m{a}^2}& 0<x<a\\ 0& \text{otherwise}\end{array}$$

Particle in a Three-Dimensional Box

Consider particle confined in box region given by

$$\{\mathbf{x}=(x,y,z)\in {\mathbb{R}}^3\mid 0\le x\le a,0\le y\le b,0\le z\le c\}\subset {\mathbb{R}}^3$$

where potential is zero inside box so

$$V(x,y,z)=\{\begin{array}{ll}0,& 0<x<a,0<y<b,0<z<c\\ +\infty ,& \text{otherwise}\end{array}$$

Stationary State Wave Function $\psi (\mathbf{x})=\psi (x,y,z)$ is zero on and outside the boundary of box

Inside box, Stationary State Schrödinger Equation reduces to

$$\frac{{\partial }^2\psi }{\partial x^2}+\frac{{\partial }^2\psi }{\partial y^2}+\frac{{\partial }^2\psi }{\partial z^2}=-\frac{2mE}{{\hslash }^2}\psi$$

Solving via separation of variables then Stationary State Wave Function is

$${\varphi }_{n_1,n_2,n_3}(x,y,z)=B\sin \left(\frac{n_1\pi x}{a}\right)\sin \left(\frac{n_2\pi y}{b}\right)\sin \left(\frac{n_3\pi z}{c}\right)$$

where quantum numbers $n_1,n_2,n_3\in {\mathbb{Z}}^{>0}$

With corresponding energies

$$E_{n_1,n_2,n_3}=\frac{{\pi }^2{\hslash }^2}{2m}\left(\frac{n_1^2}{a^2}+\frac{n_2^2}{b^2}+\frac{n_3^2}{c^2}\right)$$

Particle on a Circle

Consider free particle moving on a circle of radius $R$

Modelled by One-Dimensional Schrödinger Equation with potential zero

Spatial Coordinate $x$ is periodically identified with

$$x\sim x+2\pi R$$

and Wave Functions satifsying

$$\Psi (t,x)=\Psi (t,x+2\pi R)$$

Stationary State Schrödinger Equation is

$$-\frac{{\hslash }^2}{2m}{\partial }_{xx}\psi (x)=E\psi (X)$$

with periodicity

$$\psi (x)=\psi (x+2\pi R)$$

Hence (using same steps as for Particle in a box) then ground-state is

$${\psi }_{\text{g.s.}}=a,E_{\text{g.s.}}=0$$

So ground state is non-degenerate

Hence excited states

$$\psi (x)=A\cos \left(\frac{\sqrt{2mE}}{\hslash }x\right)+B\sin \left(\frac{\sqrt{2mE}}{\hslash }x\right),E=E_n=\frac{n^2{\hslash }^2}{2m{R}^2}(n\in {\mathbb{Z}}^{+})$$

So excited state is $2$-fold degenerate

Normalised Particle in Box

$${\Psi }_n(x,t)=\{\begin{array}{ll}\sqrt{\frac{2}{a}}\sin \left(\frac{n\pi x}{a}\right)e^{-i{n}^2{\pi }^2\hslash t/2m{a}^2}& 0<x<a\\ 0& \text{otherwise}\end{array}$$

Proof

$${\int }_{-\infty }^{\infty }|{\Psi }_n(x,t){|}^{2}dx=|B{|}^2{\int }_0^a{\sin }^2\left(\frac{n\pi x}{a}\right)dx=\frac{1}{2}a|B{|}^2$$

So $B=\sqrt{\frac{2}{a}}$ normalises

Probability Density and Distribution Function for Particle in a Box

Let $\Psi$ be wave function for Particle in a box

$${\rho }_n(x)=\frac{2}{a}{\sin }^2\left(\frac{n\pi x}{a}\right)=\frac{1}{a}\left(1-\cos \left(\frac{2n\pi x}{a}\right)\right)$$ $$F_n(x)\equiv {\int }_0^x{\rho }_n(y)dy=\frac{x}{a}-\frac{1}{2n\pi }\sin \left(\frac{2n\pi x}{a}\right)$$

Expectation of a Position for Particle in a Box

Let $\Psi$ be wave function satisfying Particle in a Box

$$\begin{array}{rl}{\mathbb{E}}_{{\Psi }_n}(x)& ={\int }_0^{a}x{\rho }_n(x)dx\\ & =[x{F}_n(x){]}_0^a-{\int }_0^a{F}_n(x)dx\\ & =a-{\left[\frac{x^2}{2a}+\frac{a}{4n^2{\pi }^2}\cos \left(\frac{2n\pi x}{a}\right)\right]}_0^a\\ & =\frac{1}{2}a\end{array}$$