# Lecture 6: Schrödinger Equation

- Page ID
- 38871

Recap of 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.

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

Recall the wave equation

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

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

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

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 \(\omega = 2\pi \nu\).

Using the de Broglie formula \(\lambda = h/p\), we solve for \(p\). The momentum \(p\) is given by \(p = mv\). Recalling that \(KE = 0.5\,mv^2\), \(KE\) can be written in terms of \(p\) by \(KE = p^2/2m\). \(E_{total}\) can now 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

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

The kinetic operator in one dimension is

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

In three dimensions, the operator is

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

The Laplacian operator

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

\[ \nabla^2 = \left ( \frac {\partial ^2}{\partial x^2} + \dfrac {\partial ^2}{\partial y^2} + \dfrac {\partial ^2}{\partial z^2} \right ) \label {3-20}\]

with the del operator,

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

also is used in Quantum Mechanics. 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 waves in three dimensions this is can be expanded to

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

The use of the time-dependent Schrödinger Equations will always be valid, albeit necessary 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 it is **trivial**

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

where \(\psi(\vec{r})\) is a wavefunction dependent (or time-independent) wavefuction that only depends on space 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, eg, \(\hat{H}\). So 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 energy, which is a number, (a quantity or *observable*), times the wavefunction.

## Fundamental Properties of 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.

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

## Linearity of Operators

Almost all operators encountered in quantum mechanics are *linear operators*.

An operator \(\hat{A}\) is *linear *if

\[ \hat{A}[c_1f(x)+c_2f_2(x)]= c_1\hat{A}f_1(x)+c_2\hat{A}f_2(x) \]

and the operator is *nonlinear *if

\[\hat{A}[c_1f_1(x)+c_2f_2(x)] \neq c_1\hat{A}f_1(x)+c_2\hat{A}f_2(x)\]

Example \(\PageIndex{1}\): Linear or NOT?

Determine from the following operators which are linear and nonlinear:

- \(\hat{A}f(x)= f(x)^2\) [square f(x)]
- \(\hat{A}f(x)= f^*(x)\) [form the complex conjugate of f(x)]
- \(\hat{A}f(x)= 0\) [multiply f(x) by zero]
- \(\hat{A}f(x)= [f(x)]^{-1}\) [take the reciprocal of f(x)]
- \(\hat{A}f(x)= f(0)\) [evaluate f(x) at x=0]
- \(\hat{A}f(x)= \ln f(x)\) [take the log of f(x)]

**Strategy:**-
It is important to note that an operator \(\hat{A}\) is

*linear*if\[ \underbrace{\hat{A}[c_1f(x)+c_2f_2(x)]}_{\text{left side}}= \underbrace{c_1\hat{A}f_1(x)+c_2\hat{A}f_2(x) }_{\text{right side}}\]

and the operator is

*nonlinear*if\[ \underbrace{ \hat{A}[c_1f_1(x)+c_2f_2(x)]}_{\text{left side}} \neq \underbrace{ c_1\hat{A}f_1(x)+c_2\hat{A}f_2(x) }_{\text{right side}}\]

**Answer a:**-
Evaluate the left side

\[\begin{align} \hat{A}[c_1f(x)+c_2f_2(x)] &= [c_1f_1(x)+c_2f_2(x)]^2 \\ &= c_1^2 f_1(x)^2+2c_1f_1(x) c_2f_2(x)+c_2^2f_2(x)^2 \end{align}\]

Evaluate the right side

\[c_1 \hat{A} f_1(x)+c_2\hat{A}f_2(x)=c_1[f_1(x)]^2+c_2[f_2(x)]^2 \neq \hat{A}[c_1f_1(x)+c_2f_2(x)] \]

*This operator is nonlinear* **Answer b:**-
Evaluate the left side

\[\hat{A}[c_1f_1(x)+c_2f_2(x)] = c_1^*f_1^*(x) + c_2^*f_2(x)\]

Evaluate the right side

\[c_1\hat{A}f_1(x) + c_2\hat{A}f_2(x) = c_1f_1^*(x) + c_2f_2^*(x)\]

\[ = \hat{A}[c_1f_1(x) + c_2f_2(x)]\]

*This operator is linear* **Answer c:**-
Evaluate the left side

\[ \hat{A}[c_1f_1(x)+c_2f_2(x)] = 0\]

Evaluate the right side

\[c_1\hat{A}f_1(x) + c_2\hat{A}f_2(x) = c_1f_1(x) + c_2f_2(x) = 0\]

\[ = \hat{A}[c_1f_1(x) + c_2f_2(x)]\]

*This operator is linear* **Answer d:**-
Evaluate the left side

\[\hat{A}[c_1f_1(x)+c_2f_2(x)] = \dfrac{1}{c_1f_1(x) + c_2f_2(x)}\]

Evaluate the right side

\[c_1\hat{A}f_1(x) + c_2\hat{A}f_2(x) = \dfrac{c_1}{f_1(x)} + \dfrac{c_2}{f_2(x)} \]

\[ \neq \hat{A}[c_1f_1(x) + c_2f_2(x)]\]

*This operator is nonlinear* **Answer e:**-
Evaluate the left side

\[\hat{A}[c_1f_1(x)+c_2f_2(x)] = c_1f_1(0) + c_2f_2(0)\]

Evaluate the right side

\[ = c_1\hat{A}f_1(x) + c_2\hat{A}f_2(x)\]

*This operator is linear* **Answer f:**-
Evaluate the left side

\[\hat{A}[c_1f_1(x)+c_2f_2(x)] = \ln [c_1f_1(x) + c_2f_2(x)]\]

Evaluate the right side

\[c_1\hat{A}f_1(x) + c_2\hat{A}f_2(x) = c1 \ln f_1(x) + c_2 \ln f_2(x)\]

\[ \neq \hat{A}[c_1f_1(x) + c_2f_2(x)]\]

*This operator is nonlinear*