Harmonic-Oscillator Number States

Number states are the normalized energy eigenstates of the quantum harmonic oscillator. They are eigenstates of the number operator $\hat N=\hat a^\dagger\hat a$, and the integer $n$ counts oscillator quanta.

For one oscillator,

$$\hat N\lvert n\rangle=n\lvert n\rangle, \qquad n=0,1,2,\ldots$$

The word “number” does not always mean literal particle number. For a mechanical oscillator it counts vibrational quanta. For a field mode it can become photon, phonon, or particle occupation number after additional physical interpretation.

Helpful background. Quantum Harmonic Oscillator to recognize oscillator energies, lengths, and eigenfunctions.

Energy Eigenvalues

The oscillator Hamiltonian is

$$\hat H =\hbar\omega\left(\hat N+\frac12\right).$$

Therefore

$$\hat H\lvert n\rangle =E_n\lvert n\rangle, \qquad E_n=\hbar\omega\left(n+\frac12\right).$$

The ground state is $\lvert0\rangle$, not $\lvert1\rangle$, and its energy is

$$E_0=\frac12\hbar\omega.$$

The levels are equally spaced:

$$E_{n+1}-E_n=\hbar\omega.$$

This equally spaced spectrum is why oscillator number states are the natural language for normal modes.

Orthonormal Basis

Number states are orthonormal:

$$\langle m\vert n\rangle=\delta_{mn}.$$

They form a complete basis for the oscillator Hilbert space:

$$\sum_{n=0}^{\infty} \lvert n\rangle\langle n\rvert =\hat I.$$

An arbitrary normalized oscillator state can be expanded as

$$\lvert\psi\rangle =\sum_{n=0}^{\infty}c_n\lvert n\rangle, \qquad \sum_{n=0}^{\infty}\lvert c_n\rvert^2=1.$$

If the oscillator energy is measured, the probability of obtaining the level $E_n$ is

$$P(n)=\lvert c_n\rvert^2.$$

When the following sum converges, the mean number is

$$\langle \hat N\rangle =\sum_{n=0}^{\infty}n\lvert c_n\rvert^2.$$

Ladder Actions

The annihilation and creation operators act as

$$\hat a\lvert n\rangle =\sqrt n\,\lvert n-1\rangle,$$

and

$$\hat a^\dagger\lvert n\rangle =\sqrt{n+1}\,\lvert n+1\rangle.$$

The ground state is defined by

$$\hat a\lvert0\rangle=0.$$

The normalized number states can be built from the ground state:

$$\lvert n\rangle =\frac{(\hat a^\dagger)^n}{\sqrt{n!}} \lvert0\rangle.$$

The operators themselves do not preserve norm: $\lVert a|n\rangle\rVert=\sqrt n$. The square-root factors tell how to normalize the resulting state and determine transition matrix elements.

Position And Momentum Matrix Elements

Using the oscillator length

$$\ell=\sqrt{\frac{\hbar}{m\omega}},$$

the position and momentum operators are

$$\hat x =\frac{\ell}{\sqrt2} \left(\hat a+\hat a^\dagger\right),$$

and

$$\hat p =\frac{\hbar}{i\ell\sqrt2} \left(\hat a-\hat a^\dagger\right).$$

Thus $\hat x$ and $\hat p$ connect only neighboring number states:

$$\langle m\vert \hat x\vert n\rangle =\frac{\ell}{\sqrt2} \left( \sqrt n\,\delta_{m,n-1} +\sqrt{n+1}\,\delta_{m,n+1} \right).$$

This nearest-neighbor structure is the algebraic source of many oscillator selection rules.

Wavefunction Representation

In position representation,

$$\psi_n(x)=\langle x\vert n\rangle.$$

With $\ell=\sqrt{\hbar/(m\omega)}$, the normalized wavefunctions are

$$\psi_n(x) =\frac{1}{\sqrt{2^n n!}} \left(\frac{1}{\pi\ell^2}\right)^{1/4} H_n\left(\frac{x}{\ell}\right) e^{-x^2/(2\ell^2)}.$$

Here $H_n$ is the $n$th Hermite polynomial. The parity is

$$\psi_n(-x)=(-1)^n\psi_n(x).$$

Even $n$ gives even wavefunctions; odd $n$ gives odd wavefunctions. The $n$th wavefunction has $n$ nodes.

Stationary But Not Classical

A number state has a time-dependent phase

$$\lvert n,t\rangle =e^{-iE_nt/\hbar}\lvert n\rangle.$$

Its probability density is time independent. Also,

$$\langle n\vert\hat x\vert n\rangle=0, \qquad \langle n\vert\hat p\vert n\rangle=0.$$

Thus a number state is not a particle moving back and forth along a classical oscillator trajectory. Classical-like oscillatory motion requires superpositions, with coherent states as the canonical example.

Later Uses

Number states reappear whenever harmonic modes are quantized:

• molecular vibrations and phonons;
• electromagnetic field modes and photons;
• trapped-ion motion;
• oscillator baths in open-system models;
• normal modes in many-body theory and QFT.

The many-mode occupation-number language is developed in Fock-Space Number States. The bridge from oscillator modes to field modes is summarized in Harmonic Oscillator to Fields.

Common Mistakes

• Starting the number label at $n=1$ instead of $n=0$.
• Forgetting the zero-point energy $\hbar\omega/2$.
• Treating $\hat a$ and $\hat a^\dagger$ as ordinary numbers that commute.
• Dropping the square-root factors in the ladder actions.
• Assuming a number state oscillates in position like a classical mass on a spring.
• Calling every oscillator quantum a particle without checking the physical interpretation of the mode.
• Confusing a single-oscillator number basis with many-mode Fock space.

Exercises

1. Show that $\lvert n\rangle=(\hat a^\dagger)^n\lvert0\rangle/\sqrt{n!}$ is normalized if $\lvert0\rangle$ is normalized.

Solution

Use the ladder action repeatedly:

$$\hat a^\dagger\lvert n-1\rangle =\sqrt n\,\lvert n\rangle.$$

Starting from $\lvert0\rangle$, repeated application gives

$$(\hat a^\dagger)^n\lvert0\rangle =\sqrt{n!}\,\lvert n\rangle.$$

Therefore dividing by $\sqrt{n!}$ produces the normalized state $\lvert n\rangle$.

2. Compute $\langle n\vert\hat x\vert n\rangle$ using ladder operators.

Solution

Since

$$\hat x =\frac{\ell}{\sqrt2} \left(\hat a+\hat a^\dagger\right),$$

we have

$$\langle n\vert\hat x\vert n\rangle =\frac{\ell}{\sqrt2} \left( \langle n\vert\hat a\vert n\rangle +\langle n\vert\hat a^\dagger\vert n\rangle \right).$$

But $\hat a\lvert n\rangle$ is proportional to $\lvert n-1\rangle$ and $\hat a^\dagger\lvert n\rangle$ is proportional to $\lvert n+1\rangle$, both orthogonal to $\lvert n\rangle$. Hence

$$\langle n\vert\hat x\vert n\rangle=0.$$

3. If $\lvert\psi\rangle=(\lvert0\rangle+\sqrt3\,\lvert1\rangle)/2$, what are the possible oscillator energies and their probabilities?

Solution

The coefficients are $c_0=1/2$ and $c_1=\sqrt3/2$. Therefore

$$P(0)=\frac14, \qquad P(1)=\frac34.$$

The possible energies are

$$E_0=\frac12\hbar\omega, \qquad E_1=\frac32\hbar\omega.$$

References

• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• M. O. Scully and M. S. Zubairy, Quantum Optics, Cambridge University Press, 1997.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.