Normal Ordering

Normal ordering is a convention for writing products of creation and annihilation operators with all creation operators placed to the left of all annihilation operators. It is denoted by colons:

$$:O:.$$

The rule is simple to state but easy to misuse:

• for bosons, move creation operators left using commutation relations, with no sign from swaps;
• for fermions, move creation operators left using anticommutation relations, with one minus sign for each odd swap;
• the operation is defined relative to a chosen vacuum and a chosen creation-annihilation split.

Normal ordering is not the same as ordinary algebraic simplification. It is a prescription: reorder the operators, keep the operator part, and separate any c-number terms generated by commutators or anticommutators.

Creation Operators to the Left

For one bosonic mode,

$$:a^\dagger a: = a^\dagger a, \qquad :a a^\dagger: = a^\dagger a.$$

The first product is already normal ordered. The second is not, because the annihilation operator appears to the left of the creation operator.

The operator identity is

$$a a^\dagger = a^\dagger a+1.$$

Thus

$$a a^\dagger = :a a^\dagger: +1.$$

The difference between the original product and the normal-ordered product is the c-number term produced by the commutator.

For several bosonic modes,

$$:a_i a_j^\dagger: = a_j^\dagger a_i,$$

and

$$a_i a_j^\dagger = :a_i a_j^\dagger: +\delta_{ij}.$$

Bosonic Normal Ordering

For bosons, creation operators commute with each other, and annihilation operators commute with each other:

$$[a_i^\dagger,a_j^\dagger]=0, \qquad [a_i,a_j]=0.$$

Normal ordering therefore only has to move annihilation operators past creation operators. For example,

$$:a_i a_j^\dagger a_k: = a_j^\dagger a_i a_k.$$

The original product is related to the normal-ordered product by

$$a_i a_j^\dagger a_k = a_j^\dagger a_i a_k +\delta_{ij}a_k.$$

The term $\delta_{ij}a_k$ is not part of the normal-ordered product; it is produced by the commutator needed to reorder the operators.

For bosonic field operators,

$$:\psi(\mathbf x)\psi^\dagger(\mathbf y): = \psi^\dagger(\mathbf y)\psi(\mathbf x),$$

and

$$\psi(\mathbf x)\psi^\dagger(\mathbf y) = :\psi(\mathbf x)\psi^\dagger(\mathbf y): +\delta^{(3)}(\mathbf x-\mathbf y).$$

The delta function is the continuum version of the Kronecker delta.

Fermionic Signs

For fermions, each swap of two fermionic operators introduces a minus sign. For one pair,

$$:c_i c_j^\dagger: = -c_j^\dagger c_i.$$

Using

$$\{c_i,c_j^\dagger\} = \delta_{ij},$$

one obtains

$$c_i c_j^\dagger = \delta_{ij} -c_j^\dagger c_i = :c_i c_j^\dagger: +\delta_{ij}.$$

The contraction term looks similar to the bosonic case, but the normal-ordered operator carries a sign.

For a longer fermionic product, normal ordering means applying the parity of the permutation required to move all creation operators left. For example,

$$:c_i c_j^\dagger c_k: = -c_j^\dagger c_i c_k.$$

If two identical fermionic creation operators or two identical fermionic annihilation operators appear, the product may vanish by anticommutation:

$$(c_i^\dagger)^2=0, \qquad c_i^2=0.$$

The signs are not optional bookkeeping. They are the operator form of antisymmetry.

Vacuum Expectation Values

Let $\lvert0\rangle$ be the vacuum annihilated by every annihilation operator:

$$d_i\lvert0\rangle=0.$$

Any nonconstant normal-ordered monomial has zero vacuum expectation value:

$$\langle0\vert :O:\vert0\rangle=0,$$

provided $:O:$ contains at least one creation or annihilation operator. The reason is that after normal ordering, every annihilation operator is on the right and kills the vacuum ket. If the product contains only creation operators, the vacuum bra kills it after taking the adjoint interpretation of the Fock inner product.

For example,

$$\langle0\vert a a^\dagger\vert0\rangle = 1,$$

but

$$\langle0\vert :a a^\dagger:\vert0\rangle = \langle0\vert a^\dagger a\vert0\rangle = 0.$$

This is why normal ordering is often described as subtracting vacuum contractions.

Number Operators and Vacuum Energy

The number operator is already normal ordered:

$$N = a^\dagger a = :a^\dagger a:.$$

The opposite ordering is not:

$$a a^\dagger = N+1, \qquad :a a^\dagger: = N.$$

For a harmonic oscillator,

$$H = \hbar\omega \left( a^\dagger a+\frac12 \right).$$

Normal ordering relative to the oscillator vacuum gives

$$:H: = \hbar\omega a^\dagger a.$$

This removes the zero-point energy in that chosen convention. In ordinary nonrelativistic problems, shifting all energies by a constant is often harmless. In field theory, infinite sums of zero-point energies require more careful interpretation and cannot always be dismissed by a slogan.

Hamiltonians and Interaction Terms

Many-particle Hamiltonians are often written already in normal-ordered form:

$$H = \sum_{ij} h_{ij}d_i^\dagger d_j + \frac12 \sum_{ijkl} V_{ij;kl} d_i^\dagger d_j^\dagger d_l d_k.$$

The one-body term $d_i^\dagger d_j$ is normal ordered. The two-body term $d_i^\dagger d_j^\dagger d_l d_k$ is also normal ordered.

In field notation, the density

$$n(\mathbf x) = \psi^\dagger(\mathbf x)\psi(\mathbf x)$$

is normal ordered relative to the empty-particle vacuum. But products such as $\psi(\mathbf x)\psi^\dagger(\mathbf y)$ are not; reordering them produces delta functions.

Normal ordering is therefore a way to separate operator products into a normal-ordered part plus contraction terms. The full systematic expansion of products into normal-ordered products and contractions is the content of Wick-type formulas.

QFT Preview and Caveats

Normal ordering is indispensable in many-body theory and QFT, but its meaning depends on context.

In the nonrelativistic Fock vacuum, all particles are absent. In a filled Fermi sea, a more useful reference state may treat states below the Fermi surface as filled and define particle-hole excitations relative to that sea. In relativistic QFT, the split into creation and annihilation operators is tied to a choice of modes, vacuum, and sometimes spacetime background.

The induced zero-, one-, and residual many-body terms for filled, coherent, and Bogoliubov references are developed in Normal Ordering in Many-Body QM.

Three caveats are especially important:

• Normal ordering is relative to a reference vacuum.
• Normal ordering removes vacuum expectation values by convention, not by deriving new physics.
• Renormalization is not merely normal ordering; it handles scale dependence and parameter definitions in interacting theories.

For the bridge from quantum mechanics to QFT, the practical lesson is modest: learn to move operators into normal order, track fermionic signs, and recognize the c-number terms produced by commutators or anticommutators.

Common Mistakes

• Forgetting the minus sign when normal ordering fermionic products.
• Treating normal ordering as if it were basis-independent.
• Dropping c-number terms without saying that normal ordering has been applied.
• Thinking normal ordering proves that vacuum energy is never observable.
• Applying normal ordering relative to the empty vacuum when the relevant reference state is a filled Fermi sea.
• Confusing normal ordering with Hermitian conjugation or time ordering.

Cross-Links

• Creation and Annihilation Operators
• Bosonic Commutation Relations
• Fermionic Anticommutation Relations
• Number Operators
• Field Operators
• Many-Particle Hamiltonians
• Wick’s Theorem Preview
• Second Quantization: Bridge to QFT
• Formula Sheet
• Reference Bridge: Second Quantization
• Harmonic Oscillator to Fields

References

• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, McGraw-Hill, 1971.
• J. W. Negele and H. Orland, Quantum Many-Particle Systems, Addison-Wesley, 1988.
• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.
• G. D. Mahan, Many-Particle Physics, 3rd ed., Kluwer Academic/Plenum, 2000.
• M. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory, Addison-Wesley, 1995.
• M. Srednicki, Quantum Field Theory, Cambridge University Press, 2007.

Exercises

1. Bosonic normal ordering. Normal order $a_i a_j^\dagger a_k$ and identify the extra term in the original product.

Solution

Move $a_j^\dagger$ left:

$$:a_i a_j^\dagger a_k: = a_j^\dagger a_i a_k.$$

Using $a_i a_j^\dagger=a_j^\dagger a_i+\delta_{ij}$,

$$a_i a_j^\dagger a_k = a_j^\dagger a_i a_k +\delta_{ij}a_k.$$

Thus the extra term is $\delta_{ij}a_k$.

2. Fermionic sign. Normal order $c_i c_j^\dagger$.

Solution

One swap is required to move $c_j^\dagger$ left, so

$$:c_i c_j^\dagger: = -c_j^\dagger c_i.$$

The original product is

$$c_i c_j^\dagger = \delta_{ij} -c_j^\dagger c_i = :c_i c_j^\dagger: +\delta_{ij}.$$

3. Vacuum expectation. Compute $\langle0\vert a a^\dagger\vert0\rangle$ and $\langle0\vert :a a^\dagger:\vert0\rangle$.

Solution

Since $a a^\dagger=a^\dagger a+1$,

$$\langle0\vert a a^\dagger\vert0\rangle = 1.$$

But

$$:a a^\dagger: = a^\dagger a,$$

so

$$\langle0\vert :a a^\dagger:\vert0\rangle = 0.$$

4. Oscillator Hamiltonian. Normal order $H=\hbar\omega(a^\dagger a+\frac12)$.

Solution

The operator $a^\dagger a$ is already normal ordered, and the constant is removed if one defines normal ordering by subtracting the vacuum expectation value:

$$:H: = \hbar\omega a^\dagger a.$$

This is a convention that shifts the zero of energy.

5. Field delta term. For bosonic fields, show that

$$\psi(\mathbf x)\psi^\dagger(\mathbf y) = :\psi(\mathbf x)\psi^\dagger(\mathbf y): +\delta^{(3)}(\mathbf x-\mathbf y).$$

Solution

The field commutator is

$$[\psi(\mathbf x),\psi^\dagger(\mathbf y)] = \delta^{(3)}(\mathbf x-\mathbf y).$$

Therefore

$$\psi(\mathbf x)\psi^\dagger(\mathbf y) = \psi^\dagger(\mathbf y)\psi(\mathbf x) +\delta^{(3)}(\mathbf x-\mathbf y).$$

The normal-ordered product is

$$:\psi(\mathbf x)\psi^\dagger(\mathbf y): = \psi^\dagger(\mathbf y)\psi(\mathbf x),$$

which gives the stated identity.