02. Wave Mechanics

2.1 The Schrödinger Equation

Schrödinger Equation

Consider single, non-relativistic particle of mass $m$ moving in a potential $V(x)$

Particle is described by a wave function $\Psi (\mathbf{x},t)$ governed by Schrödinger Equation

$$i\hslash \frac{\partial \Psi }{\partial t}=-\frac{{\hslash }^2}{2m}{\nabla }^2\Psi +V\Psi$$

where $\Psi (\mathbf{x},t)$ is de Broglie’s matter-wave

Note that the Schrödinger Equation is a linear partial differential equation for complex-value function $\Psi (\mathbf{x},t)$

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2.2 Stationary States

Stationary State Schrödinger Equation

Consider particle of mass $m$ and energy $E$ moving in potential $V=V(\mathbf{x})$

Stationary State Schrödinger Equation is

$$-\frac{{\hslash }^2}{2m}{\nabla }^2\psi +V\psi =E\psi$$

Note that it is also known as Time-Independent Schrödinger Equation

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Stationary State Wave Function of Energy $E$

Stationary State Function of Energy $E$ is

$$\Psi (\mathbf{x},t)=\psi (\mathbf{x})e^{-iEt/\hslash }$$

with angular frequency $\omega =\frac{E}{\hslash }$ so $E=\hslash \omega$

Proof

Consider separable solutions to Schrödinger Equation so

$$\Psi (\mathbf{x},t)=\psi (\mathbf{x})T(t)$$

Using Schrödinger Equation then

$$\begin{array}{rl}i\hslash \frac{\partial \Psi }{\partial t}& =-\frac{{\hslash }^2}{2m}{\nabla }^2\Psi +V\Psi \\ ⟹i\hslash \psi (\mathbf{x})\frac{\partial T}{\partial t}& =-\frac{{\hslash }^2}{2m}T(t){\nabla }^2\psi +V\psi T\\ ⟹\frac{i\hslash \frac{dT}{dt}}{T}& =\frac{-\frac{{\hslash }^2}{2m}{\nabla }^2\psi +V\psi }{\psi }\end{array}$$

As each side is independent of the other then it must be equal to a constant, suppose $E$

Hence

$$i\hslash \frac{dT}{dt}=ET⟹T(t)=e^{-iEt/\hslash }$$

Thus

$$\Psi (\mathbf{x},t)=\psi (\mathbf{x})e^{-iEt/\hslash }$$

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Full Schrödinger Equation

Find stationary states $\varphi (\mathbf{x})$ solving Stationary state schrödinger equation
As a particle can have different values of energy $E$ which is discrete so label as

$$E_n\text{ for }n\in N$$

Let

$$\Psi (\mathbf{x},t)=\sum _n{c}_n{\psi }_n(\mathbf{x})e^{-i{E}_{n}t/\hslash }$$

where $c_n\in \mathbb{C}$ are constants

By Linearity of Schrödinger Equation then the linear combination also satisfies it

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2.3 One-Dimensional Equations

One-Dimensional Schrödinger Equation

$$i\hslash \frac{\partial \Psi }{\partial t}=-\frac{{\hslash }^2}{2m}\frac{{\partial }^2\Psi }{\partial x^2}+V\Psi$$

with corresponding stationary state equation

$$-\frac{{\hslash }^2}{2m}\frac{d^2\psi }{d{x}^2}+V\psi =E\psi$$

and

$$\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$$

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2.4 Particle in a Box

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

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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

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Full Time-Dependent Wave Function for Particle in a Box

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

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2.5 Particle in a Three-Dimensional Box

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

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

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2.6 Degeneracy

$d$-fold Degenerate Energy Level

Energy Level $E$ is $d$-fold degenerate if

Space of Solutions to Stationary State Schrödinger Equation with energy $E$ has dimension $d$
where $d>1$

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Non-Degenerate Energy Level

Energy Level $E$ is non degenerate if

Space of Solutions to Stationary State Schrödinger Equation with energy $E$ has dimension $1$

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Finding Degeneracy of an Energy Level

Generally is the number of different ways to pick $n_1,n_2,\cdots ,n_k$ for a specific value of $E$
where $n_1,n_2,n_k\in {\mathbb{Z}}^{>0‘}$

Particle in a Three-Dimensional Box

As

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

So for $E=\frac{6{\pi }^2{\hslash }^2}{2m{a}^2}$ then

It is $3$-fold degenerate as it has solutions

$$(n_1,n_2,n_3)=(2,1,1)\text{ or }(1,2,1)\text{ or }(1,1,2)$$

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Degeneracy Energy Levels with Full Schrödinger Equation

Suppose energy level $E_n$ is $d_n$-fold degenerate (with $d_n\ge 1$) then

Full schrödinger equation can be written as

$$\Psi (\mathbf{x},t)=\sum _n\sum _{i=1}^{d_n}{c}_{n,i}{\psi }_{n,i}(\mathbf{x})e^{-i{E}_{n}t/\hslash }$$

with ${\varphi }_{n,i}(\mathbf{x})$ are linearly independent stationary states of energy $E_n,i=1,\cdots ,d_n$

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2.7 Particle on a Circle

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

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