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

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

The daggered operator creates one particle or excitation in the mode. The undaggered operator annihilates one. The word “annihilate” means “remove from the occupation state,” not “destroy a tiny classical object.”

Motivation from Occupation Numbers

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 is not decoration. It is required so that normalized occupation states remain compatible with the bosonic inner product and the commutation relation stated on the next page.

In particular,

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

Repeated creation gives

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

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 a fixed ordering of modes $1,2,\ldots$. 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:

$$N = \sum_i N_i.$$

Detailed number-operator identities belong to Number Operators; the essential point here is that creation and annihilation operators raise and lower the eigenvalues of $N_i$.

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.

Bosons Versus Fermions

Bosonic and fermionic operators look similar because both alter mode occupations, but their algebra is different.

Bosonic creation operators can be applied repeatedly to the same mode:

$$(a_i^\dagger)^n\lvert0\rangle \ne 0 \qquad \text{for }n=1,2,\ldots .$$

Fermionic creation operators cannot:

$$(c_i^\dagger)^2\lvert0\rangle = 0.$$

This is Pauli exclusion in operator form. The bosonic case is made precise by bosonic commutation relations; the fermionic case is made precise by fermionic anticommutation relations.

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.

Common Mistakes

• Treating creation operators as basis-independent before specifying modes.
• Forgetting bosonic square-root normalization factors.
• Applying a fermionic creation operator twice to the same mode and expecting a nonzero state.
• Dropping fermionic sign factors from the chosen mode ordering.
• Confusing oscillator ladder operators for one particle with creation operators for particles in a Fock space.
• Thinking “annihilation” means a nonunitary physical destruction process rather than an operator action on a basis state.

Cross-Links

• Occupation-Number Basis
• Vacuum State
• Mode Occupations
• Number States
• Bosonic Fock Space
• Fermionic Fock Space
• Number Operators
• Mode Expansions
• Field Operators
• One-Body Operators
• Normal Ordering
• Fock Space Examples
• Fock Space Exercises
• Bosonic Commutation Relations
• Fermionic Anticommutation Relations
• Pauli Exclusion Principle
• Ladder-Operator Solution
• Reference Bridge: Second Quantization
• Harmonic Oscillator to Fields

References

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

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. Compute $a\lvert0\rangle$ for one bosonic mode.

Solution

The annihilation operator lowers occupation with a factor $\sqrt n$. For $n=0$,

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

There is no state with negative occupation.

3. 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

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

4. Why does $(c_i^\dagger)^2\lvert0\rangle=0$ express Pauli exclusion?

Solution

The first $c_i^\dagger$ creates a fermion in mode $i$. The second tries to create another fermion in the same complete one-particle mode. Fermionic occupation of a mode can only be $0$ or $1$, so the result is zero.