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.

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

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

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 square-root factors are not decoration. They preserve normalization 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.

Where This Is Used

• Quantum Harmonic Oscillator gives the full canonical model.
• Ladder-Operator Solution: First Encounter derives the ladder actions and spectrum.
• Differential-Equation Solution derives the Hermite-function wavefunctions.
• Hermite Functions gives a focused reference for the coordinate-space eigenfunctions.
• Zero-Point Energy explains why the $n=0$ state still has energy $\hbar\omega/2$.
• Coherent States uses number states to build classical-like oscillator states.
• Displaced Oscillator shows why a shifted ground state is a coherent superposition in the unshifted number basis.
• Squeezed States: First Encounter shows how nonclassical oscillator states mix even number states.
• Stationary States explains why energy eigenstate probabilities are time independent.

References

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

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

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