Quantum Mechanics/Operators and Commutators

Why Operators?

When we work with the Schrödinger Equation, or in any other formulation of Quantum Mechanics, exact values of properties can not be used. Instead we use operators.

The simplest way of expressing Schrödinger's Equation is

$\widehat{ϵ}\mathrm{\Psi }=\frac{{\hat{p}}^2}{2m}\mathrm{\Psi }+V(x)\mathrm{\Psi }$

Showing that energy = kinetic + potential.

Now this could look deceptively simple if we didn't use operators for energy and momentum. We could simply divide by the wavefunction Ψ. But like most things, it's never simple. The energy operator acts on the wavefunction, as does the momentum operator. So we need to find the wavefunction in order to make any sense of this equation.

A Look at a Few Common Operators

Although we could theoretically come up with an infinite number of operators, in practice there are a few which are much more important than any others.

Momentum

The Momentum operator is

$\hat{p}=-i\hslash \frac{\partial }{\partial x}$

So the relative part of the Schrödinger Equation will become

$-\frac{{\hslash }^2}{2m}\frac{{\partial }^2}{{\partial x}^2}\mathrm{\Psi }$

Energy

Energy Operator

$\widehat{ϵ}=i\hslash \frac{\partial }{\partial t}$

So Schrödinger's now looks like

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

As you can see this is now a differential equation, which may or may not be easily solved depending on the potential $V(x)$.

Note that this is the one dimensional form of Schrödinger's Equation, it does become more complex for higher dimensions.

Hamiltonian

We often call the Right Hand Side of this equation the Hamiltonian Operator.

$\hat{H}=-\frac{{\hslash }^2}{2m}\frac{{\partial }^2}{{\partial x}^2}+V(x)$

and it represents the total energy of the particle of mass m in the Potential Field V.

Expectation Values

In Quantum Mechanics, everything is probabilistic, the probability of finding a particle is the square of the amplitude of the wavefunction for instance. So we often want to know the expected value of position, or momentum, or anything else. And there is quite a nice method of doing this.

$⟨\hat{K}⟩=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\mathrm{\Psi }}^{*}\hat{K}\mathrm{\Psi }dx$

So for instance if you knew the wavefunction and wanted to find the expectation value of momentum you'd use the equation

$⟨\hat{p}⟩=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\mathrm{\Psi }}^{*}-i\hslash \frac{\partial }{\partial x}\mathrm{\Psi }dx$

Dirac Notation (Bra-Ket Notation)

The way we've been writing these above is fine until you have to do it thirty times, so when Quantum was being developed, Paul Dirac came up with a new way of writing this:

$|n⟩\equiv {\mathrm{\Psi }}_n...⟨n|\equiv {\mathrm{\Psi }}_n^{*}$

So the expression for the expectation value of momentum can now be written as

$⟨\hat{p}⟩=⟨n|\hat{p}|n⟩$

Which neatens it up substantially

For more on Dirac Notation see Wikipedia:Bra-ket_notation

Commutators

"And you know why four plus minus one plus ten is fourteen minus one? 'Cause addition is commutative, right." --Tom Lehrer

Commutators are very important in Quantum Mechanics, as well as being how Heisenberg Discovered the Uncertainty Principle they are often used in particle physics. It is known that you can not know the value of two physical values at the same time if they do not commute.

Mathematical Definition of Commutator

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

This is equal to 0 if they commute and something else if they don't.

As you can probably see all natural numbers will commute for instance

$[6,7]=6*7-7*6=42-42=0$

But if we look at Momentum and Position things start to get interesting

$\begin{array}{ccc}[\hat{p},\hat{x}]\mathrm{\Psi }& =& (\hat{p}\hat{x}-\hat{x}\hat{p})\mathrm{\Psi }\\ & =& -i\hslash \frac{\partial }{\partial x}(x\mathrm{\Psi })+ix\hslash \frac{\partial }{\partial x}(\mathrm{\Psi })\\ & =& -i\hslash \mathrm{\Psi }-ix\hslash \frac{\partial \mathrm{\Psi }}{\partial x}+ix\hslash \frac{\partial \mathrm{\Psi }}{\partial x}\\ & =& (-i\hslash )\mathrm{\Psi }\end{array}$ $[\hat{p},\hat{x}]=-i\hslash$

This tells us a few things, one of the most important is that you can't know position and momentum precisely at the same time.

--Frontier 12:33, 28 Apr 2005 (UTC)