Relativistic Spectrum and Oscillator Modes

Relativistic processes immediately expose a limitation of fixed-particle-number quantum mechanics. In neutron beta decay,

$$n\longrightarrow p+e^-+\bar\nu_e,$$

one incoming neutron is replaced by a proton, an electron, and an antineutrino. Energy, momentum, electric charge, and the other exact charges remain conserved, but neither the particle species nor the total particle count is fixed. A relativistic quantum theory therefore needs a Hilbert space and operators that can connect sectors with different particle content.

To build that language, begin with the simplest free case. A single massive neutral spin-zero bosonic field is a collection of positive-frequency harmonic oscillators, one for each allowed momentum. A single oscillator has a tower of states $|0\rangle,|1\rangle,|2\rangle,\ldots$; in field theory the integer $n_{\mathbf p}$ becomes the number of particles carrying momentum $\mathbf p$.

This is not just a suggestive analogy. The relativistic energy of one spinless particle is

$$\omega_{\mathbf p}=\sqrt{\mathbf p^2+m^2},$$

and the energy of noninteracting particles is additive. The harmonic oscillator is precisely the quantum system whose spectrum is generated by adding identical bosonic quanta. The first construction in this course marries these facts: attach a harmonic oscillator to every momentum mode and let its quanta be particles.

The result is the seed of Fock space, perturbation theory, propagators, and scattering amplitudes. Interactions will later mix the modes, create particles, annihilate particles, and shift the location of poles, but the free oscillator basis is the table on which all those calculations are set.

A useful way to read this page is to separate two ideas that often get blended together. First, a harmonic oscillator has a rigorous algebra of creation, annihilation, and number operators. Second, a free relativistic scalar field assigns one such oscillator to each allowed momentum. The particles are the oscillator quanta; they are not extra objects added on top of the field. Fermionic fields require anticommuting creation and annihilation operators and will be treated separately.

Relativistic particles and additive spectra

Periodic box and discrete momenta

The course conventions from the overview are in force. In this page we temporarily put space in a large periodic box of volume $V=L^3$ so that momenta are discrete,

$$\mathbf p=\frac{2\pi}{L}\mathbf n, \qquad \mathbf n\in\mathbb Z^3.$$

Then $\delta_{\mathbf p\mathbf q}$ is an ordinary Kronecker delta. The box still has infinitely many modes. Whenever a vacuum constant is retained below, first restrict the sum to a finite, inversion-symmetric set $K_\Lambda$, for example $|\mathbf p|\leq\Lambda$; this ultraviolet cutoff is additional to the box. The regulated scalar occupation basis makes that distinction explicit. At fixed cutoff, with the same momentum restriction on both sides, the infinite-volume limit replaces

$$\sum_{\mathbf p}\longrightarrow V\int\frac{d^3p}{(2\pi)^3}.$$

For a single oscillator, $q$ denotes the coordinate, $\pi$ its canonical momentum, and $\omega>0$ its frequency. The symbol $m$ is reserved for the relativistic particle mass. We take $m>0$ for the noncompact real scalar here, so even its constant spatial mode has positive frequency. At $m=0$, that periodic constant mode instead has a free-particle Hamiltonian and no normalizable oscillator vacuum; it requires the separate zero-mode treatment.

Lorentz invariance fixes the invariant mass of a one-particle state by

$$p_\mu p^\mu=m^2.$$

On the positive-energy branch,

$$p^0=\omega_{\mathbf p}=\sqrt{\mathbf p^2+m^2}.$$

Measuring the one-particle and additive energies relative to the selected free vacuum, a one-particle momentum eigenstate satisfies

$$H|\mathbf p\rangle=\omega_{\mathbf p}|\mathbf p\rangle, \qquad \mathbf P|\mathbf p\rangle=\mathbf p|\mathbf p\rangle.$$

For noninteracting particles, the Hamiltonian is additive. A state with momenta $\mathbf p_1,\ldots,\mathbf p_N$ has

$$H|\mathbf p_1,\ldots,\mathbf p_N\rangle = \left(\omega_{\mathbf p_1}+\cdots+\omega_{\mathbf p_N}\right) |\mathbf p_1,\ldots,\mathbf p_N\rangle.$$

This formula has the same structure as a set of independent counters. For each momentum $\mathbf p$, we only need to know how many times $\omega_{\mathbf p}$ appears in the total energy. That observation is the doorway from wave mechanics to fields.

For the free scalar field considered here, each allowed momentum $\mathbf p$ labels an oscillator with frequency $\omega_{\mathbf p}=\sqrt{\mathbf p^2+m^2}$. Occupying that oscillator creates particles on the positive-energy mass shell.

For the bosonic scalar field considered here, “free field” therefore means a system whose normal modes are independent oscillators and whose mode frequencies are determined by relativistic kinematics.

The harmonic oscillator as a quantum counter

Start with one classical oscillator of positive frequency $\omega>0$,

$$L=\frac12\dot q^2-\frac12\omega^2q^2.$$

Its canonical momentum is

$$\pi=\frac{\partial L}{\partial\dot q}=\dot q,$$

and the Hamiltonian is

$$H=\pi\dot q-L =\frac12\pi^2+\frac12\omega^2q^2.$$

Canonical quantization promotes $q$ and $\pi$ to operators obeying

$$[q,\pi]=i.$$

Define ladder operators by

$$a=\sqrt{\frac{\omega}{2}}q+\frac{i}{\sqrt{2\omega}}\pi, \qquad a^\dagger=\sqrt{\frac{\omega}{2}}q-\frac{i}{\sqrt{2\omega}}\pi.$$

Equivalently,

$$q=\frac{a+a^\dagger}{\sqrt{2\omega}}, \qquad \pi=-i\sqrt{\frac{\omega}{2}}(a-a^\dagger).$$

Using $[q,\pi]=i$, the ladder operators obey

$$[a,a^\dagger]=1.$$

Substituting the inverse relations into the Hamiltonian gives

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

The constant $\omega/2$ is the zero-point energy of this specified, unshifted quadratic Hamiltonian. It appears because $a$ and $a^\dagger$ do not commute; choosing another overall energy origin does not remove that algebraic term from the original Hamiltonian. For a field, summing these constants over an unbounded set of modes produces the ultraviolet vacuum-energy divergence.

Define the number operator

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

Its commutators with the ladder operators are

$$[N,a^\dagger]=a^\dagger, \qquad [N,a]=-a.$$

So if

$$N|n\rangle=n|n\rangle,$$

then

$$N(a^\dagger|n\rangle)=(n+1)a^\dagger|n\rangle, \qquad N(a|n\rangle)=(n-1)a|n\rangle.$$

The eigenvalues of $N$ are nonnegative, since

$$\langle n|N|n\rangle = \langle n|a^\dagger a|n\rangle = \|a|n\rangle\|^2\ge 0.$$

Repeated lowering must therefore stop. The state where it stops is the oscillator vacuum,

$$a|0\rangle=0.$$

The normalized excited states are

$$|n\rangle=\frac{(a^\dagger)^n}{\sqrt{n!}}|0\rangle, \qquad n=0,1,2,\ldots.$$

The operators act as

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

Thus

$$H|n\rangle=\omega\left(n+\frac12\right)|n\rangle.$$

The harmonic oscillator spectrum. The operator $a^\dagger$ raises the occupation number by one and increases the energy by $\omega$; the operator $a$ lowers the occupation number by one and decreases the energy by $\omega$.

In the basis $|0\rangle,|1\rangle,|2\rangle,\ldots$, the matrix elements are

$$(a^\dagger)_{mn}=\langle m|a^\dagger|n\rangle =\sqrt{n+1}\,\delta_{m,n+1},$$

and

$$a_{mn}=\langle m|a|n\rangle =\sqrt n\,\delta_{m,n-1}.$$

So, schematically,

$$a^\dagger= \begin{pmatrix} 0&0&0&0&\cdots\\ 1&0&0&0&\cdots\\ 0&\sqrt2&0&0&\cdots\\ 0&0&\sqrt3&0&\cdots\\ \vdots&\vdots&\vdots&\vdots&\ddots \end{pmatrix}, \qquad a= \begin{pmatrix} 0&1&0&0&\cdots\\ 0&0&\sqrt2&0&\cdots\\ 0&0&0&\sqrt3&\cdots\\ 0&0&0&0&\cdots\\ \vdots&\vdots&\vdots&\vdots&\ddots \end{pmatrix}.$$

The number operator is diagonal:

$$N= \begin{pmatrix} 0&0&0&\cdots\\ 0&1&0&\cdots\\ 0&0&2&\cdots\\ \vdots&\vdots&\vdots&\ddots \end{pmatrix}.$$

This matrix picture is humble but important. A free bosonic scalar field repeats this same algebra once for every momentum mode.

From one oscillator to momentum modes

Attach an oscillator to each allowed momentum $\mathbf p$:

$$a_{\mathbf p}, \qquad a_{\mathbf p}^\dagger, \qquad N_{\mathbf p}=a_{\mathbf p}^\dagger a_{\mathbf p}.$$

The independent oscillator algebra is

$$[a_{\mathbf p},a_{\mathbf q}^\dagger]=\delta_{\mathbf p\mathbf q}, \qquad [a_{\mathbf p},a_{\mathbf q}]=0, \qquad [a_{\mathbf p}^\dagger,a_{\mathbf q}^\dagger]=0.$$

The frequency of the oscillator labeled by $\mathbf p$ is the relativistic one-particle energy,

$$\omega_{\mathbf p}=\sqrt{\mathbf p^2+m^2}.$$

Before subtracting its vacuum constant, the regulated quadratic free Hamiltonian is

$$H_{\mathrm{free}} = \sum_{\mathbf p}\omega_{\mathbf p}\left(a_{\mathbf p}^\dagger a_{\mathbf p}+\frac12\right).$$

The vacuum energy is

$$E_{\mathrm{vac}}=\frac12\sum_{\mathbf p}\omega_{\mathbf p}.$$

Without the ultraviolet cutoff, this sum already diverges at fixed $L$: along $\mathbf n=(j,0,0)$, the positive frequencies grow with $|j|$. A finite box is not an ultraviolet regulator. At fixed finite cutoff the vacuum energy is finite; as $V\to\infty$ its total is extensive while its density has a finite thermodynamic limit. Removing the cutoff is a separate operation.

For the selected free vacuum on this fixed background, subtract the finite regulated constant to define the excitation Hamiltonian. The notation below means normal ordering the underlying unsimplified quadratic oscillator presentation, not applying colons to the already reduced expression $\omega_{\mathbf p}(N_{\mathbf p}+1/2)$ as though scalar constants disappeared:

$$:H_{\mathrm{free}}: = \sum_{\mathbf p}\omega_{\mathbf p}a_{\mathbf p}^\dagger a_{\mathbf p}.$$

In subsequent field-mode formulas, $H_{\mathrm{free}}$ without colons denotes this excitation Hamiltonian. The finite subtraction changes neither the energy gaps nor the creator commutators in the same background. In the continuum, the excitation Hamiltonian is defined directly on its operator domain, not by subtracting infinity from an operator; see the regulated Hamiltonian and vacuum constant. Choosing an energy origin in this fixed-background calculation does not settle gravitational or boundary-dependent vacuum-energy questions.

Before taking $V\to\infty$, it is worth keeping the finite-volume dictionary visible:

$$\sum_{\mathbf p}\longrightarrow V\int\frac{d^3p}{(2\pi)^3}, \qquad \delta_{\mathbf p\mathbf q}\longrightarrow \frac{(2\pi)^3}{V}\delta^{(3)}(\mathbf p-\mathbf q).$$

Different continuum normalizations move factors of $V$ and $(2\pi)^3$ between $a_{\mathbf p}$, $a(\mathbf p)$, and $|\mathbf p\rangle$. The physics is unchanged, but the bookkeeping is not. This is why early calculations are safest in a box, and why later relativistic pages will state their state normalization explicitly.

The normalized occupation-number state is

$$|\{n_{\mathbf p}\}\rangle = \prod_{\mathbf p}\frac{(a_{\mathbf p}^\dagger)^{n_{\mathbf p}}}{\sqrt{n_{\mathbf p}!}}|0\rangle, \qquad n_{\mathbf p}=0,1,2,\ldots,$$

where only finitely many occupation numbers are nonzero for an ordinary finite-particle state. It obeys

$$N_{\mathbf p}|\{n_{\mathbf q}\}\rangle =n_{\mathbf p}|\{n_{\mathbf q}\}\rangle.$$

For the normal-ordered free Hamiltonian,

$$H_{\mathrm{free}}|\{n_{\mathbf p}\}\rangle = \left(\sum_{\mathbf p}n_{\mathbf p}\omega_{\mathbf p}\right) |\{n_{\mathbf p}\}\rangle,$$

while the momentum operator is

$$\mathbf P=\sum_{\mathbf p}\mathbf p\,a_{\mathbf p}^\dagger a_{\mathbf p},$$

so

$$\mathbf P|\{n_{\mathbf p}\}\rangle = \left(\sum_{\mathbf p}n_{\mathbf p}\mathbf p\right) |\{n_{\mathbf p}\}\rangle.$$

The integer $n_{\mathbf p}$ is now a particle number: it counts how many quanta occupy the mode with momentum $\mathbf p$.

Creation operators and conserved charges

The interpretation of $a_{\mathbf p}^\dagger$ as a particle-creation operator follows from its commutators with conserved generators.

For the normal-ordered free Hamiltonian,

$$H=\sum_{\mathbf q}\omega_{\mathbf q}a_{\mathbf q}^\dagger a_{\mathbf q},$$

we compute

$$[H,a_{\mathbf p}^\dagger] = \sum_{\mathbf q}\omega_{\mathbf q} [a_{\mathbf q}^\dagger a_{\mathbf q},a_{\mathbf p}^\dagger].$$

Using $[AB,C]=A[B,C]+[A,C]B$ and $[a_{\mathbf q}^\dagger,a_{\mathbf p}^\dagger]=0$,

$$[a_{\mathbf q}^\dagger a_{\mathbf q},a_{\mathbf p}^\dagger] = a_{\mathbf q}^\dagger[a_{\mathbf q},a_{\mathbf p}^\dagger] = a_{\mathbf q}^\dagger\delta_{\mathbf q\mathbf p}.$$

Therefore

$$[H,a_{\mathbf p}^\dagger] =\omega_{\mathbf p}a_{\mathbf p}^\dagger.$$

Similarly,

$$[\mathbf P,a_{\mathbf p}^\dagger]=\mathbf p\,a_{\mathbf p}^\dagger.$$

If $|\Psi\rangle$ has energy $E$ and momentum $\mathbf P_\Psi$, then $a_{\mathbf p}^\dagger|\Psi\rangle$ has energy $E+\omega_{\mathbf p}$ and momentum $\mathbf P_\Psi+\mathbf p$, provided the resulting state is nonzero. This is the precise sense in which $a_{\mathbf p}^\dagger$ creates one particle of momentum $\mathbf p$: it shifts the eigenvalues of the conserved generators by the one-particle energy and momentum. This definition is more robust than picturing a particle as a tiny object being inserted into space.

The vacuum of the free field is the state annihilated by every lowering operator:

$$a_{\mathbf p}|0\rangle=0 \qquad \text{for every }\mathbf p.$$

A one-particle state is

$$|\mathbf p\rangle=a_{\mathbf p}^\dagger|0\rangle,$$

and a two-particle state is

$$|\mathbf p,\mathbf q\rangle=a_{\mathbf p}^\dagger a_{\mathbf q}^\dagger|0\rangle.$$

If both particles occupy the same mode, the normalized state is

$$|2_{\mathbf p}\rangle=\frac{(a_{\mathbf p}^\dagger)^2}{\sqrt{2!}}|0\rangle.$$

The factor $1/\sqrt{2!}$ is not cosmetic. It is the first glimpse of the combinatorics that later becomes Wick’s theorem and the symmetry factors of Feynman diagrams.

A two-mode check

Suppose only two distinct momentum modes are occupied, $\mathbf p$ and $\mathbf q$, with $\mathbf p\ne\mathbf q$. Consider

$$|\Psi\rangle = \frac{(a_{\mathbf p}^\dagger)^2}{\sqrt{2!}}a_{\mathbf q}^\dagger|0\rangle.$$

This state has occupation numbers

$$n_{\mathbf p}=2, \qquad n_{\mathbf q}=1,$$

with all other $n_{\mathbf k}=0$. Its normal-ordered energy is

$$E_\Psi=2\omega_{\mathbf p}+\omega_{\mathbf q},$$

and its momentum is

$$\mathbf P_\Psi=2\mathbf p+\mathbf q.$$

This small computation is the prototype for free-particle kinematics in QFT. The state is not described by giving permanent names to three particles; it is described by saying which modes are occupied and how many times.

Summary

The massive free relativistic bosonic scalar considered here is a system of independent positive-frequency oscillators. The oscillator labeled by $\mathbf p$ has frequency

$$\omega_{\mathbf p}=\sqrt{\mathbf p^2+m^2},$$

and the integer eigenvalue of $N_{\mathbf p}=a_{\mathbf p}^\dagger a_{\mathbf p}$ is interpreted as the number of particles in that momentum mode.

The free Hamiltonian and momentum are

$$H=\sum_{\mathbf p}\omega_{\mathbf p}a_{\mathbf p}^\dagger a_{\mathbf p}, \qquad \mathbf P=\sum_{\mathbf p}\mathbf p\,a_{\mathbf p}^\dagger a_{\mathbf p},$$

with energies measured relative to the selected free vacuum after the regulated subtraction. The commutators

$$[H,a_{\mathbf p}^\dagger]=\omega_{\mathbf p}a_{\mathbf p}^\dagger, \qquad [\mathbf P,a_{\mathbf p}^\dagger]=\mathbf p\,a_{\mathbf p}^\dagger$$

show that $a_{\mathbf p}^\dagger$ creates one quantum with energy $\omega_{\mathbf p}$ and momentum $\mathbf p$.

The conceptual shift is already visible: a field quantum is not a little classical ball with a permanent label. It is an excitation of a mode, and the occupation number of that mode is the observable information. The next page turns this into the formal language of identical particles and Fock space.

Common pitfalls

Confusing particle mass with oscillator frequency

The particle mass $m$ is fixed, but the oscillator frequency depends on momentum:

$$\omega_{\mathbf p}=\sqrt{\mathbf p^2+m^2}.$$

Only the zero-momentum mode has frequency $m$.

Forgetting the vacuum energy

For the unshifted quadratic Hamiltonian of one oscillator,

$$H=\omega\left(N+\frac12\right),$$

not $\omega N$. In flat-spacetime scattering theory one often normal-orders and subtracts $E_{\mathrm{vac}}$, but this is a convention for energy differences, not a proof that vacuum energy is meaningless.

Treating continuum delta functions like Kronecker deltas

In a finite box,

$$[a_{\mathbf p},a_{\mathbf q}^\dagger]=\delta_{\mathbf p\mathbf q}.$$

In infinite volume, after a conventional rescaling of operators,

$$[a(\mathbf p),a^\dagger(\mathbf q)] =(2\pi)^3\delta^{(3)}(\mathbf p-\mathbf q).$$

The continuum notation is compact but hides normalization factors. Finite-volume notation is safer at the beginning.

Thinking of particles as permanently labeled objects

The occupation number $n_{\mathbf p}$ counts quanta of a mode. It does not assign permanent identities to individual particles. This distinction becomes essential for identical bosons and fermions.

Exercises

Exercise 1: deriving the oscillator algebra

With

$$a=\sqrt{\frac{\omega}{2}}q+\frac{i}{\sqrt{2\omega}}\pi, \qquad a^\dagger=\sqrt{\frac{\omega}{2}}q-\frac{i}{\sqrt{2\omega}}\pi,$$

and $[q,\pi]=i$, show that $[a,a^\dagger]=1$. Then show that

$$H=\frac12\pi^2+\frac12\omega^2q^2 = \omega\left(a^\dagger a+\frac12\right).$$

Solution

First compute

$$[a,a^\dagger] = \left[\sqrt{\frac{\omega}{2}}q+\frac{i}{\sqrt{2\omega}}\pi, \sqrt{\frac{\omega}{2}}q-\frac{i}{\sqrt{2\omega}}\pi\right].$$

The $[q,q]$ and $[\pi,\pi]$ terms vanish. The two cross terms give

$$[a,a^\dagger] = -\frac{i}{2}[q,\pi]+\frac{i}{2}[\pi,q].$$

Since $[q,\pi]=i$ and $[\pi,q]=-i$,

$$[a,a^\dagger] = -\frac{i}{2}(i)+\frac{i}{2}(-i)=1.$$

Next invert the definitions:

$$q=\frac{a+a^\dagger}{\sqrt{2\omega}}, \qquad \pi=-i\sqrt{\frac{\omega}{2}}(a-a^\dagger).$$

Substitution gives

$$\frac12\pi^2+\frac12\omega^2q^2 = \frac{\omega}{4}\left(-(a-a^\dagger)^2+(a+a^\dagger)^2\right).$$

Expanding the squares carefully,

$$-(a-a^\dagger)^2+(a+a^\dagger)^2 =2aa^\dagger+2a^\dagger a.$$

Thus

$$H=\frac{\omega}{2}(aa^\dagger+a^\dagger a).$$

Using $aa^\dagger=a^\dagger a+1$,

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

Exercise 2: normalization of number states

Let $a|0\rangle=0$ and $\langle0|0\rangle=1$. Prove that

$$|n\rangle=\frac{(a^\dagger)^n}{\sqrt{n!}}|0\rangle$$

is normalized.

Solution

We need

$$\langle n|n\rangle = \frac{1}{n!}\langle0|a^n(a^\dagger)^n|0\rangle.$$

Using $[a,a^\dagger]=1$,

$$a(a^\dagger)^n=(a^\dagger)^n a+n(a^\dagger)^{n-1}.$$

Acting on $|0\rangle$, the first term vanishes, so

$$a(a^\dagger)^n|0\rangle=n(a^\dagger)^{n-1}|0\rangle.$$

Repeating this $n$ times gives

$$a^n(a^\dagger)^n|0\rangle=n!|0\rangle.$$

Therefore

$$\langle n|n\rangle = \frac{1}{n!}\langle0|n!|0\rangle=1.$$

Exercise 3: energy added by a creation operator

Let

$$H=\sum_{\mathbf q}\omega_{\mathbf q}a_{\mathbf q}^\dagger a_{\mathbf q}, \qquad [a_{\mathbf q},a_{\mathbf p}^\dagger]=\delta_{\mathbf q\mathbf p}.$$

Show that $[H,a_{\mathbf p}^\dagger]=\omega_{\mathbf p}a_{\mathbf p}^\dagger$. If $H|\Psi\rangle=E|\Psi\rangle$, what is the energy of $a_{\mathbf p}^\dagger|\Psi\rangle$?

Solution

Use $[AB,C]=A[B,C]+[A,C]B$:

$$[H,a_{\mathbf p}^\dagger] = \sum_{\mathbf q}\omega_{\mathbf q} [a_{\mathbf q}^\dagger a_{\mathbf q},a_{\mathbf p}^\dagger].$$

Since $[a_{\mathbf q}^\dagger,a_{\mathbf p}^\dagger]=0$,

$$[a_{\mathbf q}^\dagger a_{\mathbf q},a_{\mathbf p}^\dagger] = a_{\mathbf q}^\dagger[a_{\mathbf q},a_{\mathbf p}^\dagger] = a_{\mathbf q}^\dagger\delta_{\mathbf q\mathbf p}.$$

Thus

$$[H,a_{\mathbf p}^\dagger] =\omega_{\mathbf p}a_{\mathbf p}^\dagger.$$

Now

$$H(a_{\mathbf p}^\dagger|\Psi\rangle) =([H,a_{\mathbf p}^\dagger]+a_{\mathbf p}^\dagger H)|\Psi\rangle =(\omega_{\mathbf p}+E)a_{\mathbf p}^\dagger|\Psi\rangle.$$

So, provided $a_{\mathbf p}^\dagger|\Psi\rangle\ne0$, it has energy $E+\omega_{\mathbf p}$.

Exercise 4: finite-volume and continuum commutators

In a box, suppose

$$[a_{\mathbf p},a_{\mathbf q}^\dagger]=\delta_{\mathbf p\mathbf q}.$$

Two continuum normalizations are common. Let

$$a_1(\mathbf p)=\sqrt{\frac{V}{(2\pi)^3}}\,a_{\mathbf p}, \qquad a_2(\mathbf p)=\sqrt V\,a_{\mathbf p}.$$

Using

$$\delta_{\mathbf p\mathbf q}\longrightarrow\frac{(2\pi)^3}{V}\delta^{(3)}(\mathbf p-\mathbf q),$$

compute the continuum commutators for $a_1$ and $a_2$. Which convention gives

$$[a(\mathbf p),a^\dagger(\mathbf q)] =(2\pi)^3\delta^{(3)}(\mathbf p-\mathbf q)?$$

Solution

For $a_1$,

$$[a_1(\mathbf p),a_1^\dagger(\mathbf q)] = \frac{V}{(2\pi)^3}[a_{\mathbf p},a_{\mathbf q}^\dagger] = \frac{V}{(2\pi)^3}\delta_{\mathbf p\mathbf q}.$$

Taking the continuum limit gives

$$[a_1(\mathbf p),a_1^\dagger(\mathbf q)] \longrightarrow \delta^{(3)}(\mathbf p-\mathbf q).$$

For $a_2$,

$$[a_2(\mathbf p),a_2^\dagger(\mathbf q)] =V\delta_{\mathbf p\mathbf q} \longrightarrow (2\pi)^3\delta^{(3)}(\mathbf p-\mathbf q).$$

Thus the common convention

$$[a(\mathbf p),a^\dagger(\mathbf q)] =(2\pi)^3\delta^{(3)}(\mathbf p-\mathbf q)$$

corresponds to $a(\mathbf p)=\sqrt V\,a_{\mathbf p}$. The alternative normalization is also valid, but then factors of $(2\pi)^3$ move into the integration measure or state normalization.

References and further reading

• Sidney Coleman, Lectures of Sidney Coleman on Quantum Field Theory, Chapters 1–3.
• Mark Srednicki, Quantum Field Theory, Chapters 2–3.
• Steven Weinberg, The Quantum Theory of Fields, Volume I, Chapters 2 and 5.
• Michael E. Peskin and Daniel V. Schroeder, An Introduction to Quantum Field Theory, Chapter 2.
• Matthew D. Schwartz, Quantum Field Theory and the Standard Model, Chapters 2–3.