# 6: Schrödinger Equation (Lecture)

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Lecture 5

Schrödinger Equation is a wave equation that is used to describe quantum mechanical system and is akin to Newtonian mechanics in classical mechanics. The Schrödinger Equation is an eigenvalue/eigenvector problem. To use it we have to recognize that observables are associated with linear operators that "operate" on the wavefunction.

## Oscillating Membrane (2D Waves)

Solving for the function \(u(x,y,t)\) in a vibrating, rectangular membrane is done in a similar fashion by separation of variables, and setting boundary conditions. The solved function is very similar, where

\[u(x,y,t) = A_{nm} \cos \left(\omega_{nm} t + \phi_{nm}\right) \sin \left(\dfrac {n_x \pi x}{\ell_a}\right) \sin \left(\dfrac {n_y\pi y}{\ell_b}\right)\]

where

- \(\ell_a\) is the length of the rectangular membrane and \(\ell_b\) is the width.
- \(n_x\) and \(n_y\) are two quantum numbers (one in each dimension)

Now there are now **two "quantum numbers"** for a 2D standing wave instead of one for the 1D version.

The (1,1) Mode | The (1,2) Mode | The (2,1) Mode | The (2,2) Mode |
---|---|---|---|

Key principle: Degeneracy

These solutions introduces the concept of *degeneracy*. Two or more different states of a quantum mechanical system are said to be *degenerate *if they give the same value of energy upon measurement. The number of different states corresponding to a particular energy level is known as the degree of degeneracy of the level

*The (2,1) and (1,2) normal modes and combinations (all solutions) have the same energy (if the box width and length are identical). from Daniel A. Russell.*

If the boundary condition is not a square or rectangle, but a circle, similar solutions can be obtained!

*Images developed by Dan Russell, Graduate Program in Acoustics, The Pennsylvania State University*

## The Schrödinger Equation: A Better Quantum Approach to Quantum

Recall the wave equation

\[\dfrac {\partial^2 u}{\partial x^2} = \dfrac {1}{v^2} \cdot \dfrac {\partial^2 u}{\partial t^2}.\]

Through separation of variables, \(u(x,t) = \psi (x) \cos(\omega t)\) where \(\psi (x)\) is the spatial amplitude. Replacing \(u(x,t)\) with \(\psi (x) \cos(\omega t)\), the classical wave equation becomes

\[\dfrac {d^2 \psi (x)}{d x^2} + \dfrac {\omega^2}{v^2} \psi (x) = 0\]

With manipulation, the equation can be rewritten to

\[\dfrac {d^2 \psi (x)}{d x^2} + \dfrac {4\pi^2}{\lambda^2} \psi (x) = 0\]

from the relation \(\omega = 2\pi \nu\).

Now introduce a little quantum mechanics via the de Broglie formula

\[\lambda = h/p\]

and solve for momentum \(p\), which is also given by \(p = mv\). Recalling that

\[KE = \dfrac{1}{2} mv^2\]

and \(KE\) can be written in terms of \(p\) by

\[KE = p^2/2m\]

Now \(E_{total}\) can be solved to be \(p^2/2m + V(x)\). Solving for \(p\) results in

\[p = \sqrt{2m(E_{total}-V(x))}\]

Now that \(p\) is known, and \(h = 6.626 \times 10^{-34}\), \(\lambda\) can be solved to be equal to \(\dfrac {h}{\sqrt{2m(E-V(x))}}\). Through squaring that equation and taking the reciprocal of it, the result is

\[\dfrac {1}{\lambda^2} = \dfrac {2m(E-V(x))}{h^2}\]

This can be put into

\[\dfrac {d^2 \psi (x)}{d x^2} + \dfrac {4\pi^2}{\lambda^2} \psi (x) = 0\]

giving

\[\dfrac {d^2 \psi (x)}{d x^2} + \dfrac {2m}{\hbar^2}[E-V(x)] \psi (x) = 0\]

This can be rearranged to make the **Schrödinger equation**

\[\underbrace{\dfrac {-\hbar^2}{2m} \dfrac {d^2 \psi (x)}{d x^2}}_{\text{kinetic energy}} + \underbrace{V(x)\psi (x)}_{\text{potential energy}} = \underbrace{E\psi (x)}_{\text{total energy}}\]

The **kinetic energy operator** in one dimension is

\[KE_{1D}=\dfrac {-\hbar^2}{2m} \dfrac{d^2}{dx^2} \nonumber\]

and the **potential energy operator **is

\[V(x) \nonumber\]

The Laplacian operator

In three dimensions, the kinetic energy operator is

\[KE_{3D}=\dfrac {-\hbar^2}{2m} \nabla^2\]

The three second derivatives in parentheses together are called the **Laplacian operator**, or del-squared,

\[\nabla^2 f \triangleq \nabla\cdot\left(\nabla f\right) \label{m0099_eLaplaceDef}\]

with the del operator,

\[\nabla = \left ( \vec {x} \frac {\partial}{\partial x} + \vec {y} \frac {\partial}{\partial y} + \vec {z} \frac {\partial }{\partial z} \right ) \label{3-21}\]

The symbols with arrows over them are unit vectors.

## Two Flavors of Schrödinger Equations

We have the time-dependent Schrödinger Equation, which is the complete and full description of the system (**both spatial and temporal**)

\[ \textcolor{red} { \underbrace{ i\hbar\dfrac{\partial\Psi(x,t)}{\partial t}=\left[-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}+V(x)\right]\Psi(x,t) }_{\text{time-dependent Schrödinger equation in 1D}}}\label{3.1.16}\]

For quantum waves in three dimensions, the time-dependent Schrödinger Equation is expressed as:

\[\textcolor{red}{ \underbrace{ i\hbar\dfrac{\partial}{\partial t}\Psi(\vec{r},t)=\left[-\dfrac{\hbar^2}{2m}\nabla^2+V(\vec{r})\right]\Psi(\vec{r},t)}_{\text{time-dependent Schrödinger equation in 3D}}}\label{3.1.17}\]

The use of the time-dependent Schrödinger Equations will always be valid, albeit unnecessary in many situations.

For *conservative ***(stationary)*** *systems, the energy is a constant, and the time-dependence can be separated from the space-only factor (via the *Separation of Variables* technique)

\[ \textcolor{red}{ \underbrace{\left[-\dfrac{\hbar^2}{2m}\nabla^2+V(\vec{r})\right]\psi(\vec{r})=E\psi(\vec{r})} _{\text{time-independent Schrödinger equation}}} \label{3.1.19}\]

The use of the time-independent Schrödinger Equations will valid when the potential \(V(x\) is NOT a function of time).

Using the time-independent Schrödinger Equation does NOT mean that there is no time-dependence to the resulting wavefunction. It just means that the time-dependence is trivial in this case:

\[\Psi(\vec{r},t)=\psi(\vec{r})e^{-iEt / \hbar}\label{3.1.18}\]

where \(\psi(\vec{r})\) is a time-independent wavefunction that only depends on spatial coordinates.

## Operators

An operator is a generalization of the concept of a function applied to a function. Whereas a function is a rule for turning one number into another, an operator is a rule for turning one function into another.

For the time-independent Schrödinger Equation, the operator of relevance is the Hamiltonian operator (often just called the Hamiltonian) and is the most ubiquitous operator in quantum mechanics.

\[ \hat{H} = -\dfrac{\hbar^2}{2m}\nabla^2+V(\vec{r})\]

We often (but not always) indicate that an object is an operator by placing a `hat' over it, e.g., \(\hat{H}\). So the time-independent Schrödinger Equation (Equation \ref{3.1.19}) can then be simplified to

\[ \hat{H} \psi(\vec{r}) = E\psi(\vec{r}) \label{simple}\]

Equation \ref{simple} says that the Hamiltonian operator operates on the wavefunction to produce the total energy, which is a number (i.e., a quantity or *observable*) times the wavefunction.

## Fundamental Properties of (Linear) Operators

Most properties of operators are straightforward, but they are summarized below for completeness.

The sum and difference of two operators \(\hat{A} \) and \(\hat{B} \) are given by

\[ (\hat{A} \pm \hat{B}) f = \displaystyle \hat{A} f \pm \hat{B} f \]

The product of two operators is defined by

\[ \hat{A} \hat{B} f \equiv \hat{A} [ \hat{B} f ] \]

Two operators are equal if

\[\hat{A} f = \hat{B} f \]

for **all **functions \(f\).

The identity operator \(\hat{1}\) does nothing (or multiplies by 1)

\[ {\hat 1} f = f \]

The *associative law* holds for operators

\[ \hat{A}(\hat{B}\hat{C}) = (\hat{A}\hat{B})\hat{C} \]

The *commutative law* does **not **generally hold for operators. In general,

\[ \hat{A} \hat{B} \neq \hat{B}\hat{A}\]

It is convenient to define the **commutator **of \(\hat{A} \) and \(\hat{B} \)

\[ [\hat{A}, \hat{B}] \equiv \hat{A} \hat{B} - \hat{B} \hat{A} \]

Note that the order matters, so that

\[[\hat{A} ,\hat{B} ] = - [\hat{B} ,\hat{A} ] .\]

If \(\hat{A} \) and \(\hat{B} \) commute, then

\[[\hat{A} ,\hat{B} ] = 0.\]

The \(n\)-th power of an operator \(\hat{A}^n \) is defined as \(n\) successive applications of the operator, e.g.

\[ \hat{A}^2 f = \hat{A} \hat{A} f \]