Creation and Annihilation Operators

Creation and annihilation operators change occupation numbers in a chosen mode. If mode $i$ is occupied by $n_i$ particles, a creation operator raises that occupation and an annihilation operator lowers it.

For bosonic modes, the operators are usually written

$$a_i^\dagger, \qquad a_i.$$

For fermionic modes, a common notation is

$$c_i^\dagger, \qquad c_i.$$

A daggered operator adds one quantum when the state permits it; its adjoint removes one. These are linear operators used inside Hamiltonians and measurement models. Applying one to a state is not, by itself, a complete deterministic physical process.

Helpful background. Number States explains normalized occupation vectors and their bosonic factorials.

Raising and lowering mode occupations

Occupation-number notation describes number states such as

$$\lvert n_1,n_2,n_3,\ldots\rangle.$$

Operators that change these states are natural. Instead of rewriting symmetrized or antisymmetrized wavefunctions by hand every time a particle is added to a mode, one defines operators whose algebra handles the bookkeeping.

For example, a bosonic operator $a_2^\dagger$ raises the occupation of mode $2$:

$$\lvert n_1,n_2,n_3,\ldots\rangle_B \quad\longmapsto\quad \lvert n_1,n_2+1,n_3,\ldots\rangle_B,$$

up to a normalization factor. An annihilation operator lowers the occupation, and gives zero if the relevant mode is empty.

Bosonic Creation Operators

For bosons, mode occupations are nonnegative integers. The creation operator $a_i^\dagger$ acts as

$$a_i^\dagger \lvert\ldots,n_i,\ldots\rangle_B = \sqrt{n_i+1}\, \lvert\ldots,n_i+1,\ldots\rangle_B.$$

The square-root factor follows from the normalization of repeated creation and the bosonic commutator. The output is generally not normalized; its norm records the matrix element of the creation process.

In particular,

$$a_i^\dagger\lvert0\rangle = \lvert1_i\rangle_B.$$

Repeated creation gives

$$\lvert n_i\rangle_B = \frac{(a_i^\dagger)^{n_i}}{\sqrt{n_i!}} \lvert0\rangle$$

for a single mode.

Bosonic Annihilation Operators

The bosonic annihilation operator $a_i$ lowers the occupation:

$$a_i \lvert\ldots,n_i,\ldots\rangle_B = \sqrt{n_i}\, \lvert\ldots,n_i-1,\ldots\rangle_B.$$

If the mode is empty, the result is zero:

$$a_i\lvert\ldots,0_i,\ldots\rangle_B = 0.$$

For one mode,

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

This is the same algebraic pattern as the harmonic oscillator ladder operators.

Fermionic Creation Operators

For fermions, each mode has occupation $0$ or $1$. A fermionic creation operator $c_i^\dagger$ creates a fermion in mode $i$ only if that mode is empty.

Choose orthonormal complete modes in the order $1,2,\ldots$, with

$$|n_1,\ldots,n_M\rangle_F =(c_1^\dagger)^{n_1}\cdots(c_M^\dagger)^{n_M}|0\rangle.$$

Define

$$S_i = \sum_{k<i} n_k,$$

the number of occupied modes before mode $i$ in that ordering. Then

$$c_i^\dagger \lvert n_1,\ldots,0_i,\ldots\rangle_F = (-1)^{S_i} \lvert n_1,\ldots,1_i,\ldots\rangle_F.$$

If the mode is already occupied,

$$c_i^\dagger \lvert n_1,\ldots,1_i,\ldots\rangle_F = 0.$$

The sign $(-1)^{S_i}$ records the fermionic swaps needed to insert the new occupied mode into the chosen canonical ordering.

Fermionic Annihilation Operators

The fermionic annihilation operator removes a fermion from an occupied mode:

$$c_i \lvert n_1,\ldots,1_i,\ldots\rangle_F = (-1)^{S_i} \lvert n_1,\ldots,0_i,\ldots\rangle_F.$$

If the mode is empty,

$$c_i \lvert n_1,\ldots,0_i,\ldots\rangle_F = 0.$$

The same sign convention appears because removing the fermion also requires moving through the occupied modes that precede it in the chosen order.

Different books sometimes choose different ordering conventions. The physical content is invariant, but calculations and code must use one convention consistently.

Number Operators

The occupation of mode $i$ is measured by a number operator. For bosons,

$$N_i = a_i^\dagger a_i.$$

For fermions,

$$N_i = c_i^\dagger c_i.$$

On an occupation basis state,

$$N_i\lvert n_1,n_2,\ldots\rangle = n_i\lvert n_1,n_2,\ldots\rangle.$$

The total particle-number operator is the sum over modes:

$$N = \sum_i N_i.$$

For a normalized general state, the output norms are

$$\|a_i\Psi\|^2=\langle N_i\rangle,\quad \|a_i^\dagger\Psi\|^2=1+\langle N_i\rangle,$$ $$\|c_i\Psi\|^2=\langle N_i\rangle,\quad \|c_i^\dagger\Psi\|^2=1-\langle N_i\rangle.$$

When nonzero, divide by the corresponding norm to obtain a normalized conditional state. The bosonic statements require finite mean occupation; bosonic ladder operators are unbounded, whereas a normalized fermionic mode operator has norm one. On finite linear combinations of occupation states all these actions are well defined. Number Operators develops their counting identities.

When creation and annihilation operators are paired as $d_i^\dagger d_j$, they become the building blocks of one-body operators and move occupation from mode $j$ to mode $i$ without changing total particle number. When mode labels are replaced by position-space wavefunctions, the same operators enter mode expansions of field operators.

Relation to Harmonic Oscillator Ladder Operators

For a single bosonic mode, the formulas

$$a\lvert n\rangle = \sqrt n\,\lvert n-1\rangle, \qquad a^\dagger\lvert n\rangle = \sqrt{n+1}\,\lvert n+1\rangle$$

are identical in form to harmonic oscillator ladder-operator formulas.

The interpretation depends on the Hilbert space. In the ordinary one-particle oscillator, $\lvert n\rangle$ is the $n$th energy eigenstate of one particle in a potential. In Fock-space language, $\lvert n\rangle$ is a state with $n$ quanta occupying one mode. The same algebra is reused, but the physical meaning of the quanta depends on the system: photons, phonons, atoms in a trap mode, or field excitations.

Exercises

1. Compute $a^\dagger\lvert3\rangle$ for one bosonic mode.

Solution

Use

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

For $n=3$,

$$a^\dagger\lvert3\rangle = 2\lvert4\rangle.$$

2. With fermionic modes ordered as $1,2,3$, compute the sign in $c_3^\dagger\lvert1,0,0\rangle_F$.

Solution

For $i=3$, the occupied modes before mode $3$ are counted by

$$S_3=n_1+n_2=1+0=1.$$

Thus

$$c_3^\dagger\lvert1,0,0\rangle_F = -\lvert1,0,1\rangle_F.$$

3. Adding a fermion to a superposition. With modes ordered $1,2$, apply $c_2^\dagger$ to $|\Psi\rangle=(|1,0\rangle+|0,1\rangle)/\sqrt2$. Give its norm and the normalized result.

Solution

The occupied second-mode component is blocked, while the first component gives $c_2^\dagger|1,0\rangle=-|1,1\rangle$. Thus the result is $-|1,1\rangle/\sqrt2$, with norm squared $1/2=1-\langle N_2\rangle$. The normalized state is $-|1,1\rangle$; the overall minus sign is physically irrelevant here but relative signs matter in sums of processes.

References

• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.
• 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. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory, Addison-Wesley, 1995.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.