Creation, Annihilation, and Second Quantization

Second quantization expresses many-particle quantum mechanics using operators that add or remove a quantum from a mode. It is especially useful when particle numbers vary or when explicit sums over permutations become unwieldy. The name is historical: it does not mean quantizing an already quantum particle a second time.

Helpful background. Fock Space and Occupation Number supplies the state space. Identical particles explains the exchange symmetry encoded by the operator algebra.

Enter this chapter

Step	Pages	What to learn
Act on occupation states	Creation and annihilation, number operators	Normalization factors, fermionic signs, and changes of particle number
Use the statistics algebra	Bosonic commutators, fermionic anticommutators	How exchange symmetry controls reordering
Change representation	Mode expansions, field operators	How mode and position descriptions represent the same operator
Construct observables and dynamics	One-body operators, two-body operators, many-particle Hamiltonians	Matrix elements, operator order, and pair-counting factors
Simplify products	Normal ordering, Wick’s theorem	Reference states, contractions, and the Gaussian-state assumptions
Continue to fields	Bridge to QFT	What relativistic field theory adds to nonrelativistic many-body language

The square roots and signs have physical consequences

For one bosonic mode,

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

These coefficients are matrix elements between normalized number states; the operators themselves do not preserve a state’s norm. For example, $\|a^\dagger|n\rangle\|^2=n+1$. Their difference gives $[a,a^\dagger]=I$ on the finite-particle domain. A finite occupation cutoff modifies this identity at its upper boundary.

For fermions, a mode is either empty or occupied, and $(c_r^\dagger)^2=0$. Fixing a mode order gives a sign $(-1)^{\sum_{s<r}n_s}$ when an operator acts on mode $r$. This sign records anticommutation past the occupied modes before it. Losing it changes interference and Hamiltonian matrix elements, not merely notation.

Build a Hamiltonian from its matrix elements

Let $h_{rs}=\langle r|h|s\rangle$ in an orthonormal one-particle basis. Its many-particle form is

$$\widehat h=\sum_{rs}h_{rs}d_r^\dagger d_s,$$

where $d=a$ for bosons and $d=c$ for fermions. The term $d_r^\dagger d_s$ removes a particle from mode $s$ and places it in mode $r$. It preserves total particle number.

For a symmetric pair interaction, define unsymmetrized product-basis matrix elements $V_{rs;tu}=\langle r,s|V|t,u\rangle$. Then

$$\widehat V=\frac12\sum_{rstu}V_{rs;tu} d_r^\dagger d_s^\dagger d_u d_t.$$

The reversed annihilation order matches the chosen input indices. The factor 1/2 accounts for unordered physical pairs. Antisymmetrized matrix-element conventions have different prefactors, so they must be translated consistently. The linked one- and two-body pages derive these expressions and their field forms.

A one-particle basis change transforms both the coefficient matrices and the mode operators. An ordinary unitary change preserves the bilinear or quartic operator structure. It does not turn a one-body operator into a genuine two-body interaction. Also distinguish particle count from spatial range: the Hubbard interaction is two-body but acts on one lattice site.

Normal ordering and Wick’s theorem answer different questions

Normal ordering prescribes an arrangement of operators relative to a stated reference. In the empty bosonic vacuum, $aa^\dagger=a^\dagger a+1$. The extra term comes from the algebra. Normal ordering is not, in general, the operation of subtracting one vacuum expectation value from an arbitrary product.

Wick’s theorem relates higher moments of operators linear in creation and annihilation operators to two-point contractions for an appropriate Gaussian reference state. Generic interacting states do not have that property. Even a noninteracting Hamiltonian can be used with a non-Gaussian state, such as a fixed-number bosonic state. Specify the state as well as the Hamiltonian before applying Gaussian factorization.

Exercises

A two-mode hopping term. For bosons, apply $a_1^\dagger a_2$ to $|n_1,n_2\rangle$. When does the result vanish?

Solution

Annihilation first gives $\sqrt{n_2}|n_1,n_2-1\rangle$; creation then gives $\sqrt{(n_1+1)n_2}|n_1+1,n_2-1\rangle$. The result vanishes when mode 2 is empty. The total occupation remains $n_1+n_2$.

A fermionic sign. With $|1,1\rangle=c_1^\dagger c_2^\dagger|0\rangle$, evaluate $c_2|1,1\rangle$.

Solution

Anticommuting $c_2$ past $c_1^\dagger$ gives $-c_1^\dagger c_2c_2^\dagger|0\rangle=-c_1^\dagger|0\rangle=-|1,0\rangle$. The occupied mode before mode 2 supplies the minus sign.

References

• Bruus, H., and Flensberg, K., Many-Body Quantum Theory in Condensed Matter Physics, Oxford University Press (2004).
• Fetter, A. L., and Walecka, J. D., Quantum Theory of Many-Particle Systems, Dover (2003).
• Negele, J. W., and Orland, H., Quantum Many-Particle Systems, Westview Press (1998).