02 - Schrödinger Equations

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 $\Psi (\mathbf{x},t)$

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

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

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

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

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

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

Normalisation for Stationary State Wave Functions of energy $E$

Let $\Psi$ be a Stationary state wave function of energy $E$

Then

$$\rho (\mathbf{x},t)=|\Psi (\mathbf{x},t){|}^2=|\psi (\mathbf{x})e^{-iEt/\hslash }{|}^2=|\psi (\mathbf{x}){|}^2$$

Hence $\Psi (\mathbf{x},t)$ is normalised for all time if $\varphi (\mathbf{x})$ is normalised for all time so requires

$${\iiint }_{{\mathbb{R}}^3}|\psi (\mathbf{x}){|}^2{d}^{3}x=1$$

Continuity Equation for Schrödinger Equation

Schrödinger Equation implies continuity equation

$$\frac{\partial \rho }{\partial t}+\nabla \cdot \mathbf{j}=0$$

where $\rho (\mathbf{x},t)=|\Psi (\mathbf{x},t){|}^2$ and vector field

$$\mathbf{j}(\mathbf{x},t)\equiv \frac{i\hslash }{2m}(\Psi (\mathbf{x},t)\cdot \overline{(\nabla \Psi )(\mathbf{x},t)}-\overline{\Psi (\mathbf{x},t)}\cdot (\nabla \Psi )(\mathbf{x},t))$$

is known as the probability current $\mathbf{j}$

Proof

$$\begin{array}{rl}\frac{\partial }{\partial t}|\Psi (\mathbf{x},t){|}^2& =\left(\frac{\partial }{\partial t}\overline{\Psi (\mathbf{x},t)}\right)\Psi (\mathbf{x},t)+\overline{\Psi (\mathbf{x},t)}\frac{\partial }{\partial t}\Psi (\mathbf{x},t)\\ & =\overline{\left[-\frac{i}{\hslash }\left(-\frac{{\hslash }^2}{2m}{\nabla }^2\Psi +V\Psi \right)\right]}\Psi +\overline{\Psi }\left[-\frac{i}{\hslash }\left(-\frac{{\hslash }^2}{2m}{\nabla }^2\Psi +V\Psi \right)\right]\\ & =\frac{i}{\hslash }\left(-\frac{{\hslash }^2}{2m}{\nabla }^2\overline{\Psi }+V\overline{\Psi }\right)\Psi -\overline{\Psi }\frac{i}{\hslash }\left(-\frac{{\hslash }^2}{2m}{\nabla }^2\Psi +V\Psi \right)\\ & =\frac{i\hslash }{2m}(\overline{\Psi }{\nabla }^2\Psi -\Psi {\nabla }^2\overline{\Psi })\\ & =\frac{i\hslash }{2m}\nabla \cdot (\overline{\Psi }\nabla \Psi -\Psi \nabla \overline{\Psi })=-\nabla \cdot \mathbf{j}\end{array}$$

Conservation of Probability

Suppose for all time $t$,

$\mathbf{j}(\mathbf{x},t)$ satisfies boundary condition that it tends to zero faster than $\frac{1}{|\mathbf{x}{|}^2}$ as $|\mathbf{x}|=r\to \infty$

Hence

$${\iiint }_{{\mathbb{R}}^3}|\Psi (\mathbf{x},t){|}^2{d}^{3}x\text{ is independent of }t$$

In particular, if $\Psi$ is normalised at some time $t=t_0$ then it is normalised for all $t$

Proof

Let $S$ be a closed surface that bounds region $D\subset {\mathbb{R}}^3$ then

$$\begin{array}{rl}\frac{d}{dt}{\iiint }_D|\Psi (\mathbf{x},t){|}^2{d}^{3}x& ={\iiint }_D\frac{\partial \rho }{\partial t}{d}^{3}x\\ & ={\iiint }_D(-\nabla \cdot \mathbf{j})d^{3}x\\ & =-{\iint }_S\mathbf{j}\cdot d\mathbf{S}\end{array}$$

Suppose $S$ is a sphere of radius $r>0$, centred on origin, so $D$ is a ball
Then outward pointing unit normal vector to $S$ is

$$\mathbf{n}=\frac{\mathbf{x}}{r}$$

Hence

$$\mathbf{j}\cdot d\mathbf{S}=\mathbf{j}\cdot \mathbf{n}{r}^{2}d{S}_{\text{unit}}$$

where
$d{S}_{\text{unit}}=\sin \theta d\delta d\varphi$ is area element on unit radius sphere in spherical polar coordinates

As $\mathbf{j}\cdot \mathbf{n}=o\left(\frac{1}{r^2}\right)$ uniformly in angular coordinates then

Hence as $r\to \infty$ then surface integral tends to zero hence

$$\frac{d}{dt}{\iiint }_{{\mathbb{R}}^3}|\Psi (\mathbf{x},t){|}^2{d}^{3}x=0$$

Thus

$${\iiint }_{{\mathbb{R}}^3}|\Psi (\mathbf{x},t){|}^2{d}^{3}x\text{ is independent of }t$$

Conditions for Solutions to Schrödinger Equation

1. Wave Function $\Psi (\mathbf{x},t)$ should be continuous, single-valued function
   Ensures probability density function $\rho =|\Psi {|}^2$ is single-valued with no discontinuities

2. $\Psi$ should be normalisable
   Ensures integral of $|\Psi {|}^2$ over all space should be finite and non-zero
   Note that this condition may be relaxed for free particles

3. $\nabla \Psi$ should be continuous everywhere, except at an infinite discontinuity in potential $V$

Initial Value Problem for Schrodinger's Equation for Particle in Box

Consider $\Psi (x,0)=f(x)$ where $f:[0,a]\to \mathbb{C}$ with $f(0)=f(a)=0$

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

where

$$c_m={\int }_0^{a}f(x){\psi }_m(x)dx$$

Proof $f(0)=f(a)=0$ then $f$ can be expanded as a Fourier Sine Series

As

$$f(x)=\sum _{n=1}^{\infty }{c}_n\sqrt{\frac{2}{a}}\sin \left(\frac{n\pi x}{a}\right)=\sum _{n=1}^{\infty }{c}_n{\psi }_n(x)$$

for appropriate $c_n\in \mathbb{C}$

As ${\psi }_n(x)=\sqrt{\frac{2}{a}}\sin \left(\frac{n\pi x}{a}\right)$ are orthonormal

$${\int }_0^a{\psi }_m(x){\psi }_n(x)dx={\delta }_{mn}$$

Hence

$${\int }_0^{a}f(x){\psi }_m(x)dx=\sum _{n=1}^{\infty }{c}_n{\int }_0^a{\psi }_n(x){\psi }_m(x)dx=c_m$$

Parity of a Wave Function

In one dimension, stationary state wave functions satisfying

$$\psi (-x)=\pm \psi (x)$$

describes an

$$\left\{\begin{array}{c}\text{even}\\ \text{odd}\end{array}\right\}\text{ parity state}$$

Classical Forbidden Region

In classical system then

$$E=\frac{{\rho }^2}{2m}+V(\mathbf{x})\ge V(\mathbf{x})$$

Then

$\{x\in {\mathbb{R}}^3:E<V(\mathbf{x})\}$ is known as the classically forbidden region