Normal Ordering and Vacuum Terms

Normal ordering relative to a selected free vacuum is a linear prescription for polynomial expressions in creation and annihilation operators: place every creator to the left of every annihilator without adding the commutator terms that an operator identity would require. For the regulated free scalar, this turns the symmetrized quadratic Hamiltonian into the nonnegative excitation Hamiltonian and removes its finite zero-point constant. It does not change energy gaps or free Heisenberg equations, and it does not eliminate vacuum fluctuations.

The discussion uses the standard massive real scalar in $d$-dimensional Minkowski spacetime, first in a periodic spatial box of volume $V$ with a finite, inversion-symmetric momentum set $K_\Lambda$. This makes every reordering and vacuum term finite before the continuum operator is defined directly by second quantization. The prescription remains tied to the chosen free Fock representation. Renormalized coincident products, interacting operator mixing, locally covariant Wick powers, and gravitational vacuum energy require additional constructions and lie outside this page’s scope.

Required background. Fock Space, Vacuum, and Particle Number supplies the selected vacuum, occupation-number basis, number operators, finite-mode regulator, and continuum second-quantization domains used below.

Normal ordering in the selected free Fock representation

For $\mathbf k,\mathbf l\in K_\Lambda$, let

$$[b_{\mathbf k},b^\dagger_{\mathbf l}] = \delta_{\mathbf k\mathbf l}, \qquad b_{\mathbf k}|0\rangle=0 .$$

All polynomial calculations take place on the invariant algebraic core

$$\mathcal D_{\Lambda,\mathrm{alg}} = \operatorname{span} \left\{ |\{n_{\mathbf k}\}\rangle: n_{\mathbf k}\in\mathbb N_0 \right\}.$$

The set $K_\Lambda$ is finite, but each oscillator still has arbitrarily high occupation, so $b_{\mathbf k}$ and $b^\dagger_{\mathbf k}$ are unbounded. The core keeps the formulas below meaningful without pretending that a mode cutoff makes these operators bounded.

The method has five parts.

• Input. Specify a Fock vacuum, the associated annihilator–creator split, a polynomial presentation, and enough smearing or regulation to make its products meaningful.
• Procedure. Expand each word in creators and annihilators, then move all creators left of all annihilators without inserting commutator terms. Extend this rule linearly and leave scalar multiples of the identity unchanged.
• Output. The notation ${:}\mathcal O{:}_0$ denotes the expression normal ordered relative to $|0\rangle$. The subscript records the reference state.
• Validation. Every positive-degree normal-ordered monomial has zero vacuum expectation, while constants survive. For a Hamiltonian, also check its action on occupation states and its commutators with creators.
• Stop rule. If removing a regulator requires local counterterms, scale or scheme choices, operator mixing, or a state-independent locally covariant definition, normal ordering alone has reached its limit.

For example,

$${:}b_{\mathbf k}b^\dagger_{\mathbf l}{:}_0 = b^\dagger_{\mathbf l}b_{\mathbf k},$$

whereas the canonical commutator gives the operator identity

$$b_{\mathbf k}b^\dagger_{\mathbf l} = {:}b_{\mathbf k}b^\dagger_{\mathbf l}{:}_0 + \delta_{\mathbf k\mathbf l}I .$$

These are different instructions: normal ordering omits the contraction, while algebraic reordering retains it. In particular,

$$\begin{aligned} {:}b_{\mathbf k}b^\dagger_{\mathbf l}{:}_0 &= b^\dagger_{\mathbf l}b_{\mathbf k}, \\ {:}\bigl( b^\dagger_{\mathbf l}b_{\mathbf k} + \delta_{\mathbf k\mathbf l}I \bigr){:}_0 &= b^\dagger_{\mathbf l}b_{\mathbf k} + \delta_{\mathbf k\mathbf l}I . \end{aligned}$$

Thus the colon prescription acts on a specified polynomial presentation; it is not an algebra homomorphism on operator identities. Applying colons to both sides of a commutation relation can therefore destroy a true equality. This presentation dependence, and the fact that scalar terms remain scalar terms, are emphasized in Coleman 2019, § 4.5, pp. 73–75.

For any nonconstant normal-ordered word $M$,

$$\langle0|{:}M{:}_0|0\rangle=0 .$$

If the word contains an annihilator, the rightmost annihilation part kills the ket. If it contains only creators, the bra vacuum kills the leftmost creation part. The identity is the exception: $\langle0|{:}cI{:}_0|0\rangle=c$. Normal ordering removes contractions from the chosen presentation; it is not the instruction “subtract the vacuum expectation from every operator.”

A scalar product separates into a normal product and a contraction

For each retained momentum, define

$$\begin{aligned} \omega_{\mathbf k} &= \sqrt{\mathbf k^2+m^2}, \\ u_{\mathbf k}(x) &= \frac{ e^{-i\omega_{\mathbf k}x^0+i\mathbf k\cdot\mathbf x} }{ \sqrt{2\omega_{\mathbf k}V} }. \end{aligned}$$

The regulated field then splits into an annihilation and a creation part:

$$\phi^{(+)}_\Lambda(x) = \sum_{\mathbf k\in K_\Lambda} b_{\mathbf k}u_{\mathbf k}(x).$$ $$\phi^{(-)}_\Lambda(x) = \bigl(\phi^{(+)}_\Lambda(x)\bigr)^\dagger.$$ $$\phi_\Lambda(x) = \phi^{(+)}_\Lambda(x) + \phi^{(-)}_\Lambda(x).$$

Only one term in $\phi_\Lambda(x)\phi_\Lambda(y)$ has an annihilator to the left of a creator. Moving that creator left but omitting the commutator gives

$$\begin{aligned} {:}\phi_\Lambda(x)\phi_\Lambda(y){:}_0 ={}& \phi^{(-)}_\Lambda(x)\phi^{(-)}_\Lambda(y) \\ &+ \phi^{(-)}_\Lambda(x)\phi^{(+)}_\Lambda(y) \\ &+ \phi^{(-)}_\Lambda(y)\phi^{(+)}_\Lambda(x) \\ &+ \phi^{(+)}_\Lambda(x)\phi^{(+)}_\Lambda(y). \end{aligned}$$

The omitted commutator is the regulated positive-frequency two-point function,

$$\begin{aligned} W^+_\Lambda(x,y) &= \langle0| \phi_\Lambda(x)\phi_\Lambda(y) |0\rangle \\ &= \sum_{\mathbf k\in K_\Lambda} \frac{ e^{-i\omega_{\mathbf k}(x^0-y^0)} e^{i\mathbf k\cdot(\mathbf x-\mathbf y)} }{ 2\omega_{\mathbf k}V }. \end{aligned}$$

Consequently the exact regulated identity is

$$\begin{aligned} \phi_\Lambda(x)\phi_\Lambda(y) &= {:}\phi_\Lambda(x)\phi_\Lambda(y){:}_0 \\ &\quad+ W^+_\Lambda(x,y)I . \end{aligned}$$

This is the two-factor instance of the contraction rule, not a proof of the general combinatorics. With the site’s correlator convention, time ordering instead gives

$$\begin{aligned} \mathrm T\!\left\{ \phi_\Lambda(x)\phi_\Lambda(y) \right\} &= {:}\phi_\Lambda(x)\phi_\Lambda(y){:}_0 \\ &\quad+ D_{F,\Lambda}(x-y)I . \end{aligned}$$

with no extra factor of $i$ multiplying $D_{F,\Lambda}$. Higher products and their pairings are developed on Wick’s Theorem and Free Gaussian Factorization. The two-field formula and its contraction are derived in Schwartz 2014, § 7.A, pp. 100–102.

The subscript on the colons matters. Suppose another pure quasifree Fock vacuum $|\Omega\rangle$ supplies a different annihilator–creator split and regulated two-point function $W^+_{\Omega,\Lambda}$. Since both decompositions reproduce the same field product,

$$\begin{gathered} {:}\phi_\Lambda(x)\phi_\Lambda(y){:}_\Omega \\ = {:}\phi_\Lambda(x)\phi_\Lambda(y){:}_0 \\ {}\quad+ \bigl[ W^+_{0,\Lambda}(x,y) - W^+_{\Omega,\Lambda}(x,y) \bigr]I . \end{gathered}$$

A unitary change of one-particle packet basis that preserves the same positive-frequency subspace changes neither vacuum nor normal ordering. A Bogoliubov transformation that mixes the old creators and annihilators generally changes both. Normal ordering is therefore representation-relative, not a state-independent property of the abstract field algebra.

The regulated scalar Hamiltonian loses one vacuum constant

For the finite mode set, retain the symmetrized quadratic presentation of the free Hamiltonian,

$$H^{\mathrm{sym}}_\Lambda = \frac12 \sum_{\mathbf k\in K_\Lambda} \omega_{\mathbf k} \left( b^\dagger_{\mathbf k}b_{\mathbf k} + b_{\mathbf k}b^\dagger_{\mathbf k} \right).$$

The commutator separates this operator into an excitation part and a finite vacuum constant:

$$\begin{aligned} H^{\mathrm{sym}}_\Lambda &= H_{\mathrm{exc},\Lambda} + E_{0,\Lambda}I, \\ H_{\mathrm{exc},\Lambda} &= \sum_{\mathbf k\in K_\Lambda} \omega_{\mathbf k}N_{\mathbf k}, \\ E_{0,\Lambda} &= \frac12 \sum_{\mathbf k\in K_\Lambda} \omega_{\mathbf k}, \qquad N_{\mathbf k} = b^\dagger_{\mathbf k}b_{\mathbf k}. \end{aligned}$$

Normal ordering the displayed symmetrized quadratic word gives

$${:}H^{\mathrm{sym}}_\Lambda{:}_0 = H_{\mathrm{exc},\Lambda} = H^{\mathrm{sym}}_\Lambda-E_{0,\Lambda}I .$$

The first expression specifies what the colons act on. Because constants survive normal ordering, putting colons around the already simplified identity $H^{\mathrm{sym}}_\Lambda=H_{\mathrm{exc},\Lambda}+E_{0,\Lambda}I$ would not remove $E_{0,\Lambda}I$. The familiar notation ${:}H{:}$ is safe only when its unsimplified field or oscillator presentation has been declared.

Three independent checks fix the interpretation. First,

$$H_{\mathrm{exc},\Lambda}|0\rangle=0, \qquad H_{\mathrm{exc},\Lambda}\geq0 .$$

Second, an occupation vector obeys

$$H_{\mathrm{exc},\Lambda} |\{n_{\mathbf k}\}\rangle = \left( \sum_{\mathbf k\in K_\Lambda} \omega_{\mathbf k}n_{\mathbf k} \right) |\{n_{\mathbf k}\}\rangle .$$

Third, the removed scalar commutes with every operator, so

$$\begin{aligned} [H^{\mathrm{sym}}_\Lambda,b^\dagger_{\mathbf k}] &= [H_{\mathrm{exc},\Lambda},b^\dagger_{\mathbf k}] \\ &= \omega_{\mathbf k}b^\dagger_{\mathbf k}. \end{aligned}$$

At finite cutoff, their unitary groups differ only by a scalar phase,

$$e^{-itH^{\mathrm{sym}}_\Lambda} = e^{-itE_{0,\Lambda}} e^{-itH_{\mathrm{exc},\Lambda}}.$$

Thus fixed-background free dynamics, excitation energies, and transition frequencies agree. The zero-point term and the warning that it cannot simply be ignored once gravity is dynamical are discussed in Coleman 2019, § 4.5, pp. 73–75.

The continuum excitation Hamiltonian is not obtained by writing the undefined difference $H-\infty$. Let $E$ be multiplication by $E_{\mathbf p}=\sqrt{\mathbf p^2+m^2}$ on the one-particle space. Define

$$H_{\mathrm{exc}} = \mathrm d\Gamma(E)$$

directly. On the $n$-particle sector,

$$(H_{\mathrm{exc}}\Psi)_n = \left( \sum_{j=1}^{n}E_{\mathbf p_j} \right) \psi_n,$$

Write the sector energy as

$$E^{(n)} = \sum_{j=1}^{n}E_{\mathbf p_j} , \qquad E^{(0)}=0 .$$

Its operator domain consists exactly of vectors $\Psi=(\psi_0,\psi_1,\ldots)$ satisfying

$$\psi_n\in\mathcal D(E^{(n)}) \qquad (n\geq0)$$

and

$$\sum_{n=0}^{\infty} \|E^{(n)}\psi_n\|^2 < \infty .$$

The empty sum vanishes on the vacuum sector. This direct definition preserves positivity and self-adjointness without treating a divergent zero-point sum as an operator. For a complex scalar, the same free-vacuum prescription removes the Hamiltonian’s zero-point term, while the particle–antiparticle signs in the conserved charge come from the charge itself, as explained in Complex Scalars, Antiparticles, and Conserved Charge.

What normal ordering does not remove

At finite regulator, taking $y=x$ in the two-field identity gives

$$\begin{aligned} \phi_\Lambda(x)^2 &= {:}\phi_\Lambda(x)^2{:}_0 + C_\Lambda I, \\ C_\Lambda &= W^+_\Lambda(x,x) = \frac{1}{2V} \sum_{\mathbf k\in K_\Lambda} \frac{1}{\omega_{\mathbf k}}. \end{aligned}$$

This is a legitimate finite-mode identity. As the ultraviolet cutoff is removed, $C_\Lambda$ diverges in the usual spacetime dimensions, and the unsmeared field is an operator-valued distribution rather than an operator at a point. The regulated calculation motivates a free vacuum-relative Wick square; it does not by itself define the continuum coincident product. That construction continues on Free Wick Products and Point Splitting.

Normal ordering is also more structured than subtracting one expectation value. For a single regulated oscillator, set $X=b+b^\dagger$. Repeated use of $[b,b^\dagger]=1$ gives

$$\begin{aligned} X^2 &= {:}X^2{:}_0+I, \\ X^4 &= {:}X^4{:}_0 + 6{:}X^2{:}_0 + 3I, \end{aligned}$$

and hence

$${:}X^4{:}_0 = X^4-6X^2+3I .$$

Merely replacing $X^4$ by $X^4-\langle0|X^4|0\rangle I=X^4-3I$ would make its vacuum expectation zero, but it would not produce the normal-ordered quartic. The lower-degree terms encode all self-contractions of the selected presentation.

Nor does normal ordering make the vacuum fluctuation-free. Even at finite regulator,

$$\langle0| {:}\phi_\Lambda(x)^2{:}_0 {:}\phi_\Lambda(y)^2{:}_0 |0\rangle = 2\bigl[W^+_\Lambda(x,y)\bigr]^2,$$

which is generally nonzero. Normal ordering removes contractions within each normal-ordered factor; contractions between separate factors remain.

In an interacting theory, insertions of local products can require counterterms that mix several operators with the same quantum numbers. Their finite definitions can depend on a renormalization scale and scheme. Moving free creators left does not determine those counterterms; continue to Renormalized Composite-Operator Insertions and Operator Mixing and Renormalization Matrices.

On curved spacetime, subtraction relative to one preferred global vacuum is generally unavailable and would not meet local covariance. The singular structure shared by Hadamard states is universal, and differences between their two-point functions are smooth. Locally covariant point splitting therefore subtracts a locally constructed Hadamard parametrix rather than the full two-point function of a preferred state; finite curvature ambiguities remain constrained by covariance and scaling. This stronger problem is treated in Wick Polynomials and Point Splitting and Wick Polynomials and Microlocal Conditions. The distinction between reference-state normal ordering and locally covariant Wick powers is developed in Hollands and Wald 2015, arXiv v2, § 3.1, pp. 38–46, Open PDF.

Finally, the finite-cutoff phase argument concerns nongravitational dynamics on a fixed background. It does not settle how vacuum contributions enter a renormalized stress tensor or the cosmological constant. That question is deferred to Vacuum Energy and the Cosmological Constant.

Common pitfalls

Treating colons as an equality-preserving algebraic operation. The prescription acts on a specified word and omits commutator terms. Normal ordering both sides of $bb^\dagger=b^\dagger b+1$ therefore does not preserve the identity.

Saying that normal ordering sends every vacuum expectation to zero. It sends positive-degree normal-ordered monomials to zero in the reference vacuum. Constants remain, and products of separate normal-ordered factors can have nonzero vacuum expectation.

Writing the continuum Hamiltonian as $H-\infty$. A divergent scalar is not an operator that can be subtracted formally. Define the excitation Hamiltonian directly as $\mathrm d\Gamma(E)$ on its energy domain.

Calling free normal ordering composite-operator renormalization. The free prescription uses one chosen annihilator–creator split. Interacting local insertions, operator mixing, scale dependence, and locally covariant curved spacetime products require additional renormalization conditions.

Check your understanding

1. Retrieval. State the input and reordering rule for ${:}\mathcal O{:}_0$. Why must the reference vacuum or annihilator split be named?
2. Distinction. Compare $b_{\mathbf k}b^\dagger_{\mathbf l}$, ${:}b_{\mathbf k}b^\dagger_{\mathbf l}{:}_0$, and ${:}(b^\dagger_{\mathbf l}b_{\mathbf k} +\delta_{\mathbf k\mathbf l}I){:}_0$.
3. Derivation. Starting from $H^{\mathrm{sym}}_\Lambda$, derive $E_{0,\Lambda}$ and the excitation energy of $|\{n_{\mathbf k}\}\rangle$. Check the commutator with one creator.
4. Failure diagnosis. What is wrong with defining the continuum result as $H_{\mathrm{exc}}=H-\infty$? Give the correct definition and its domain condition.
5. Transfer. Decide whether normal ordering changes under (a) a unitary rotation among positive-frequency wave packets and (b) a Bogoliubov transformation that mixes creators with annihilators.
6. Handoff. For each problem, choose the needed continuation: a general free-field pairing formula, a continuum free Wick square, interacting operator mixing, or a locally covariant curved-spacetime Wick square.

Answers and repair routes

1. The input is a selected free Fock representation, its vacuum and annihilator–creator split, a polynomial presentation, and sufficient regulation or smearing. Move creators left without adding commutators and keep constants. The reference must be named because a change that mixes creators and annihilators changes the prescription. Repair the representation step at Fock Space, Vacuum, and Particle Number.
2. The operator identity is $b_{\mathbf k}b^\dagger_{\mathbf l} =b^\dagger_{\mathbf l}b_{\mathbf k} +\delta_{\mathbf k\mathbf l}I$. Normal ordering the left-hand word yields only $b^\dagger_{\mathbf l}b_{\mathbf k}$, whereas normal ordering the already reordered right-hand presentation keeps the scalar delta. Re-read Normal ordering in the selected free Fock representation.
3. Replacing $b_{\mathbf k}b^\dagger_{\mathbf k}$ by $b^\dagger_{\mathbf k}b_{\mathbf k}+I$ gives $E_{0,\Lambda}=\frac12\sum_{\mathbf k}\omega_{\mathbf k}$ and $H_{\mathrm{exc},\Lambda}=\sum_{\mathbf k}\omega_{\mathbf k}N_{\mathbf k}$. Its occupation eigenvalue is $\sum_{\mathbf k}\omega_{\mathbf k}n_{\mathbf k}$, and $[H_{\mathrm{exc},\Lambda},b^\dagger_{\mathbf l}] =\omega_{\mathbf l}b^\dagger_{\mathbf l}$. Repair the occupation algebra at Fock Space, Vacuum, and Particle Number.
4. Infinity is not a scalar defining an operator difference. The correct continuum Hamiltonian is $\mathrm d\Gamma(E)$, defined sector by sector on vectors for which the squared norms of the sector energies are summable. Repair the domain argument at Fock Space, Vacuum, and Particle Number.
5. A unitary packet rotation within the same positive-frequency one-particle space preserves the vacuum and normal ordering. A creator–annihilator mixing generally selects a different vacuum and a different normal-ordering map. Repair the distinction at Fock Space, Vacuum, and Particle Number.
6. Use Wick’s Theorem and Free Gaussian Factorization for the general pairing formula; Free Wick Products and Point Splitting for a continuum free Wick square; Operator Mixing and Renormalization Matrices for interacting mixing; and Wick Polynomials and Point Splitting for the curved-spacetime construction.

References

• Coleman, Sidney. Lectures of Sidney Coleman on Quantum Field Theory. Edited by Bryan Gin-ge Chen, David Derbes, David Griffiths, Brian Hill, Richard Sohn, and Yuan-Sen Ting. World Scientific, 2019. doi:10.1142/9371.
• Hollands, Stefan, and Robert M. Wald. “Quantum Fields in Curved Spacetime.” Physics Reports 574 (2015): 1–35. doi:10.1016/j.physrep.2015.02.001. Open PDF, arXiv:1401.2026v2.
• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge University Press, 2014. doi:10.1017/9781139540940.