Mode Occupations

A discrete particle mode is a normalized one-particle state, such as an orbital or wavepacket. An orthonormal set of modes defines compatible occupation measurements. In oscillator and field descriptions, a mode similarly identifies an excitation pattern whose quanta can be counted; it need not be separately addressable in a particular experiment.

Occupation numbers are not particle labels. They are basis-dependent labels for a chosen set of one-particle modes. The same physical state may have definite occupation in one mode basis and be a superposition of occupations in another.

This page explains what modes are, how basis changes work in Fock space, and why “the particle is in this mode” is a statement about a chosen representation and measurement context.

Helpful background. Number States defines definite occupations; Creation and Annihilation Operators supplies the operator action used below.

Orthonormal modes and compatible occupations

Start with a one-particle Hilbert space $\mathcal h$. A discrete orthonormal mode basis is a set

$$\{\lvert\varphi_i\rangle\}$$

with

$$\langle\varphi_i\vert\varphi_j\rangle = \delta_{ij}.$$

In occupation notation, a basis state

$$\lvert n_1,n_2,\ldots\rangle$$

means mode $\varphi_i$ has occupation $n_i$.

The word “mode” is deliberately broad. Depending on the problem, modes may be:

• energy eigenstates of a trap;
• momentum modes in a box;
• localized lattice-site orbitals;
• spin-orbitals in atoms and molecules;
• polarization modes of light;
• wavepacket modes in scattering or quantum optics;
• normal modes of coupled oscillators;
• field modes in a nonrelativistic or relativistic field description.

Nonorthogonal wavepackets also define valid single-mode observables, but generally not independent entries in one occupation string. If $s=\langle f|g\rangle$, then $[a_f,a_g^\dagger]=s$ for bosons and $\{c_f,c_g^\dagger\}=s$ for fermions. Overlapping, nonidentical modes generally have noncommuting number operators; their occupations cannot be treated as independent bins whose sum is the total number.

Mode Creation Operators

Given a normalized one-particle mode $\lvert f\rangle$, the corresponding creation operator creates one particle or excitation in that mode:

$$a_f^\dagger\lvert0\rangle = \lvert1_f\rangle$$

for bosons, or

$$c_f^\dagger\lvert0\rangle = \lvert1_f\rangle$$

for fermions.

The mode number operator is

$$N_f=a_f^\dagger a_f$$

for a bosonic mode, and

$$N_f=c_f^\dagger c_f$$

for a fermionic mode. A number state with occupation $n_f$ satisfies

$$N_f\lvert\cdots,n_f,\cdots\rangle = n_f\lvert\cdots,n_f,\cdots\rangle.$$

For bosons, $n_f$ may be any nonnegative integer. For fermions, $n_f$ is $0$ or $1$.

Basis Changes

Suppose $\{\lvert\varphi_i\rangle\}$ and $\{\lvert\chi_\alpha\rangle\}$ are two orthonormal bases of the same one-particle Hilbert space, related by a unitary matrix $U$:

$$\lvert\chi_\alpha\rangle = \sum_i U_{i\alpha}\lvert\varphi_i\rangle.$$

The corresponding bosonic creation operators transform in the same way:

$$a_\alpha^\dagger = \sum_i U_{i\alpha}\,a_i^\dagger.$$

For fermions,

$$c_\alpha^\dagger = \sum_i U_{i\alpha}\,c_i^\dagger.$$

These transformations preserve the commutation or anticommutation relations because $U$ is unitary.

The vacuum is unchanged by ordinary one-particle basis changes:

$$a_i\lvert0\rangle=0 \quad \Longrightarrow \quad a_\alpha\lvert0\rangle=0.$$

Non-vacuum occupation labels do change. A one-particle state occupying mode $a$ can become a superposition of occupations in another mode basis.

One-Particle Example

Let

$$\lvert+\rangle = \frac{ \lvert a\rangle+\lvert b\rangle }{\sqrt2}, \qquad \lvert-\rangle = \frac{ \lvert a\rangle-\lvert b\rangle }{\sqrt2}.$$

Then

$$\lvert a\rangle = \frac{ \lvert+\rangle+\lvert-\rangle }{\sqrt2}.$$

In occupation notation,

$$\lvert1_a,0_b\rangle = \frac{ \lvert1_+,0_-\rangle + \lvert0_+,1_-\rangle }{\sqrt2}.$$

The state has definite occupation in the $a,b$ basis, but not in the $+,-$ basis. This is not a contradiction; occupation numbers are basis-dependent.

Two-Boson Basis Change

For bosons, repeated occupation makes basis changes especially visible. Let

$$a^\dagger = \frac{b^\dagger+c^\dagger}{\sqrt2}.$$

The two-boson state with both bosons in mode $a$ is

$$\lvert2_a\rangle = \frac{(a^\dagger)^2}{\sqrt2} \lvert0\rangle.$$

Substitute the basis change:

$$\lvert2_a\rangle = \frac{(b^\dagger+c^\dagger)^2}{2\sqrt2} \lvert0\rangle.$$

Using $b^\dagger c^\dagger=c^\dagger b^\dagger$, this becomes

$$\lvert2_a\rangle = \frac12\lvert2_b,0_c\rangle + \frac{1}{\sqrt2}\lvert1_b,1_c\rangle + \frac12\lvert0_b,2_c\rangle.$$

Thus a definite two-boson occupation in one mode can become a superposition of several occupation patterns in another basis.

Position Modes and Local Number

In continuum wave mechanics, position eigenstates are distributions rather than normalizable Hilbert-space vectors. A formal position-mode creation operator is often written

$$\psi^\dagger(\mathbf x),$$

where $\psi^\dagger(\mathbf x)\lvert0\rangle$ creates an idealized particle localized at $\mathbf x$. More physical localized modes are wavepackets:

$$a_f^\dagger = \int d^3x\, f(\mathbf x)\psi^\dagger(\mathbf x),$$

with

$$\int d^3x\,\lvert f(\mathbf x)\rvert^2=1.$$

For a scalar field (or with a sum over internal components), the number operator for a spatial region $R$ is

$$N_R = \int_R d^3x\, \psi^\dagger(\mathbf x)\psi(\mathbf x).$$

It counts particles in the region, not particles carrying private identity labels. This is one reason spatial regions often provide physically meaningful subsystems for identical particles.

The relation between these local fields and discrete mode operators is developed in Mode Expansions, while the meaning of the fields themselves is treated in Field Operators.

Momentum and Energy Modes

For a particle in a box with periodic boundary conditions, momentum modes are discrete plane waves. Hard-wall energy eigenfunctions are standing waves, so the boundary condition matters. A state such as

$$\lvert n_{\mathbf k_1},n_{\mathbf k_2},\ldots\rangle$$

specifies how many particles occupy each allowed momentum.

In free space, momentum labels become continuous and sums become integrals. One then needs delta-function normalization or wavepacket modes to write fully normalizable states.

Energy modes are eigenstates of a one-particle Hamiltonian. If

$$h\lvert\epsilon_i\rangle = \epsilon_i\lvert\epsilon_i\rangle,$$

then an occupation state in the energy basis records how many particles occupy each energy eigenmode. This is useful for traps, oscillator modes, and noninteracting gases. Interactions can make occupation of noninteracting energy modes nonconserved even when total particle number is conserved.

Spin-Orbitals

For electrons and many other fermions, a complete one-particle mode often includes both spatial and spin information. When spin and spatial dependence factorize, a spin-orbital has the form

$$\varphi_\alpha(\mathbf x,s) = \phi_n(\mathbf x)\chi_\sigma(s).$$

Spin–orbit coupling can require nonfactorized spinor wavefunctions instead. Fermionic occupation numbers still apply to complete normalized spin-orbitals:

$$n_\alpha\in\{0,1\}.$$

Two electrons may occupy the same spatial orbital only if their spin-orbitals differ. In occupation language, that means two different modes are occupied, not one mode with occupation $2$.

This is the Fock-space form of the Pauli exclusion principle.

Mode Entanglement Preview

Modes can define tensor factors or local observable algebras. For two bosonic modes $L$ and $R$,

$$\frac{ \lvert1_L,0_R\rangle + \lvert0_L,1_R\rangle }{\sqrt2}$$

is a one-particle state that is nonfactorizable across the $L\vert R$ mode split. It is mode-entangled in that representation, though its operational use depends on the available local operations and any particle-number superselection constraints.

By contrast,

$$\lvert1_L,1_R\rangle$$

has definite occupation in both modes and is not entangled across the $L\vert R$ occupation split. The first-quantized wavefunction is symmetrized or antisymmetrized if the particles are identical, but the mode occupation itself is a product.

The detailed cautions belong to Identical-Particle Entanglement Cautions.

Exercises

1. Region number. What does

$$N_R = \int_R d^3x\, \psi^\dagger(\mathbf x)\psi(\mathbf x)$$

measure?

Solution

It measures the number of particles in the spatial region $R$. It does not identify which private particle label is in the region. For identical particles, the region is physical; the particle labels are not observable identities.

2. Spin-orbital occupation. Can two electrons occupy the same spatial orbital without violating Pauli exclusion?

Solution

Yes, if they occupy different complete spin-orbitals. For example, the same spatial orbital paired with spin-up and spin-down states gives two distinct fermionic modes. Each complete mode has occupation $1$, not $2$.

3. Overlapping modes. In the one-particle space let $f=\lvert0\rangle$ and $g=(\lvert0\rangle+\lvert1\rangle)/\sqrt2$. Compute $[P_f,P_g]$. Why can their occupations not be independent number labels?

Solution

In the $0,1$ basis, $P_f=\operatorname{diag}(1,0)$ and $P_g=\tfrac12\begin{pmatrix}1&1\\1&1\end{pmatrix}$. Their commutator is $\tfrac12\begin{pmatrix}0&1\\-1&0\end{pmatrix}\ne0$. These are already incompatible occupation tests in the one-particle sector. Orthonormalizing the modes gives a different, compatible set of counting observables.

References

• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, McGraw-Hill, 1971.
• M. O. Scully and M. S. Zubairy, Quantum Optics, Cambridge University Press, 1997.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.