# What is a quantum harmonic oscillator

1. Jul 24, 2014

### Greg Bernhardt

Definition/Summary

This is the quantum-mechanical version of the classical harmonic oscillator. Like the classical one, the quantum harmonic oscillator appears in several places, and it also appears in the quantization of fields. This article will discuss the one-dimensional version, but it can readily be generalized to multiple dimensions and multiple modes.

One can find its wavefunctions with Hermite polynomials, a kind of orthogonal polynomial, but it has a very elegant development that uses ladder operators for moving up and down the eigenstates. Ladder operators are also useful in many-body problems and in quantum field theory.

Its ground state does not have zero energy, but half of a quantum of energy, and its position and momentum are spread out over a range of values that can easily be estimated from quantum-mechanical considerations.

Equations

Its Hamiltonian is
$H = \frac{p^2}{2m} + \frac12 m\omega^2 x^2$
where commutator$[x,p] = i\hbar$.

Its natural position and momentum scales are
$x_0 = \sqrt{\frac{\hbar}{m\omega}}$
$p_0 = \sqrt{\hbar m\omega}$
and its energy eigenvalues are
$E_n = \hbar\omega \left( n + \frac12 \right)$
where n is a nonnegative integer.

Using Hermite polynomials H(n,x), its wavefunction solution is
$\psi_n(x) = \frac{1}{\sqrt{2^n n! \sqrt{\pi} x_0}} e^{-x^2/(2x_0^2)}H_n(x/x_0)$

in position space, with a momentum-space solution being found by replacing x and x0 with p and p0.

Its ground state has energy
$E_0 = \frac12 \hbar\omega$

and wavefunction
$\psi_0(x) = \frac{1}{\sqrt{\sqrt{\pi} x_0}} e^{-x^2/(2x_0^2)}$
The spread of x and p in the ground state are thus approximately x0 and p0.

Extended explanation

The ladder-operator formulation of the quantum-harmonic-oscillator problem is very elegant; its raising and lowering operators for the value of n can be interpreted as creation and annihilation operators for quanta of n. Expressing operators x and p in terms of a and its Hermitian conjugate a+,
$x = \frac{x_0}{\sqrt{2}}(a + a^\dagger)$
$p = - \frac{p_0}{\sqrt{2}}(a - a^\dagger)$
we get the commutator $[a,a^\dagger] = 1$

Construct a number operator, $N = a^\dagger a$, and the Hamiltonian becomes
$H = \hbar\omega \left( N + \frac12 \right)$

N satisfies commutation relations
$[N,a^\dagger] = a^\dagger ,\ [N,a] = - a$

Define a ground state, |0>, by making it satisfy
$a |0> = 0$

Construct a state |n> as
$|n> ~ (a^\dagger)^n |0>$

It can readily be shown from the commutation relations that
$N|n> = n|n>$

Imposing the normalization condition <n1|n2> = 1 if n1 = n2 and 0 otherwise, we find the normalized energy eigenstates:
$|n> = \frac{1}{\sqrt{n!}} (a^\dagger)^n |0>$

By finding a and a+ in terms of x and p, using differential operators of these variables as appropriate, and using Hermite-polynomial identities, one can recover the wavefunction forms for x and p.

One can also find eigenstates of lowering operator a. However, that operator has a continuous spectrum of possible eigenvalues:
$a|(\lambda)> = \lambda|(\lambda)>$

and its eigenstates are
$|(\lambda)> = e^{-|\lambda|^2/2} \sum_n \frac{\lambda^n}{\sqrt{n!}} |n>$

Since a does not commute with the Hamiltonian, this eigenstate is not an energy eigenstate.

* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!