Lecture 6: Schrödinger Equation

• • Contributed by Delmar Larsen
• Founder and Director at Libretexts

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:

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

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

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

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

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

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

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

$\hat{A}[c_1f_1(x)+c_2f_2(x)] = \ln [c_1f_1(x) + c_2f_2(x)]$
$c_1\hat{A}f_1(x) + c_2\hat{A}f_2(x) = c1 \ln f_1(x) + c_2 \ln f_2(x)$
$\neq \hat{A}[c_1f_1(x) + c_2f_2(x)]$