Mode Expansion and Fock Space

Derivation

Mode Expansion and Fock Space

D-378 Home PU-402 Threads waves · matter · fields Depends on Klein-Gordon Field from Its Lagrangian, Canonical Quantization of a Field

Statement

For a free real scalar field satisfying the Klein-Gordon equation \((\Box + m^2)\hat{\phi} = 0\) with the canonical commutation relations, the spatial Fourier expansion of the field operator introduces mode operators \(\hat{a}_{\mathbf{p}}\) and \(\hat{a}^{\dagger}_{\mathbf{p}}\) obeying \([\hat{a}_{\mathbf{p}},\hat{a}^{\dagger}_{\mathbf{q}}] = (2\pi)^3\,2E_{\mathbf{p}}\,\delta^{(3)}(\mathbf{p}-\mathbf{q})\), with \(E_{\mathbf{p}} = \sqrt{|\mathbf{p}|^2 + m^2}\). Acting with \(\hat{a}^{\dagger}_{\mathbf{p}}\) on the vacuum \(|0\rangle\) (defined by \(\hat{a}_{\mathbf{p}}|0\rangle = 0\)) builds a Fock space of relativistic single- and multi-particle states of definite momentum and energy. Units \(\hbar = c = 1\).

Why it matters

This is the passage from a classical relativistic wave equation to quantum particles. The field is not a wavefunction; it is an operator, and its excitations above the vacuum are the particles. Mode expansion is where "field" and "particle" become two descriptions of one object, and it is the template every subsequent quantization (photons, phonons, electrons via anticommutators) copies.

It also fixes the meaning of the vacuum and of particle number. The Fock construction gives an explicit orthonormal basis of the Hilbert space of a free relativistic theory, on which the Hamiltonian, momentum, and charge operators act diagonally. Every scattering calculation in perturbative QFT begins and ends with these asymptotic Fock states.

Assumptions

The field is free (quadratic Lagrangian, no interaction terms).If dropped, \(\hat{\phi}\) no longer obeys the linear Klein-Gordon equation; plane waves are not exact solutions, mode operators mix under time evolution, and Fock number is not conserved. Haag's theorem then forbids a naive interaction-picture Fock space.

Equal-time canonical commutators \([\hat{\phi}(\mathbf{x}),\hat{\pi}(\mathbf{y})] = i\,\delta^{(3)}(\mathbf{x}-\mathbf{y})\) hold, with \(\hat{\pi} = \dot{\hat{\phi}}\).If dropped, the derived \([\hat{a},\hat{a}^{\dagger}]\) algebra loses its normalization or sign; without it there is no ladder structure and no positive-norm particle states.

The field is real (Hermitian), \(\hat{\phi}^{\dagger} = \hat{\phi}\).If dropped (complex field), \(\hat{a}_{\mathbf{p}}\) and the coefficient of the positive-frequency conjugate become independent, giving a distinct antiparticle operator \(\hat{b}^{\dagger}_{\mathbf{p}}\) and a conserved charge.

The vacuum is unique, Poincaré-invariant, and the spectrum has \(E \geq 0\) (spectral condition).If dropped, \(\hat{a}_{\mathbf{p}}|0\rangle = 0\) need not define a stable ground state; one could build negative-energy towers, and the Fock space would not be the physical Hilbert space.

Lorentz-invariant measure \(\int \frac{d^3p}{(2\pi)^3\,2E_{\mathbf{p}}}\) is used, equivalently the on-shell \(\delta(p^2-m^2)\theta(p^0)\).If dropped, the one-particle states \(|\mathbf{p}\rangle = \hat{a}^{\dagger}_{\mathbf{p}}|0\rangle\) do not transform covariantly and the normalization is frame-dependent.

Derivation

\[ (\partial_t^2 - \nabla^2 + m^2)\,\hat{\phi}(t,\mathbf{x}) = 0 \]

Starting point: the Heisenberg-picture field obeys the operator Klein-Gordon equation (prior result). A

\[ \hat{\phi}(t,\mathbf{x}) = \int \frac{d^3p}{(2\pi)^3}\; e^{i\mathbf{p}\cdot\mathbf{x}}\;\hat{\phi}_{\mathbf{p}}(t) \]

Spatial Fourier transform. Legal because at fixed \(t\) the field is a well-defined operator-valued distribution on \(\mathbb{R}^3\); each Fourier mode \(\hat{\phi}_{\mathbf{p}}(t)\) is an operator. A

\[ \left(\frac{d^2}{dt^2} + |\mathbf{p}|^2 + m^2\right)\hat{\phi}_{\mathbf{p}}(t) = 0,\qquad E_{\mathbf{p}} \equiv \sqrt{|\mathbf{p}|^2 + m^2} \]

Substitute step 2 into step 1; \(\nabla^2 \to -|\mathbf{p}|^2\) on each mode. Each mode is a harmonic oscillator of frequency \(E_{\mathbf{p}}\). A

\[ \hat{\phi}_{\mathbf{p}}(t) = \frac{1}{2E_{\mathbf{p}}}\left( \hat{a}_{\mathbf{p}}\,e^{-iE_{\mathbf{p}} t} + \hat{a}^{\dagger}_{-\mathbf{p}}\,e^{+iE_{\mathbf{p}} t}\right) \]

General solution of the oscillator equation as a sum of positive- and negative-frequency operator coefficients; the second uses \(\hat{a}^{\dagger}_{-\mathbf{p}}\) so that reality (step 5) is manifest. The \(1/2E_{\mathbf{p}}\) is a convention fixing later normalization. B

\[ \hat{\phi}(t,\mathbf{x}) = \int \frac{d^3p}{(2\pi)^3}\,\frac{1}{2E_{\mathbf{p}}}\left( \hat{a}_{\mathbf{p}}\,e^{-iE_{\mathbf{p}}t + i\mathbf{p}\cdot\mathbf{x}} + \hat{a}^{\dagger}_{\mathbf{p}}\,e^{+iE_{\mathbf{p}}t - i\mathbf{p}\cdot\mathbf{x}}\right) \]

Insert step 4 into step 2 and relabel \(\mathbf{p}\to-\mathbf{p}\) in the second term (measure invariant). Hermiticity \(\hat{\phi}^{\dagger}=\hat{\phi}\) is now automatic. Writing \(p\cdot x = E_{\mathbf{p}}t - \mathbf{p}\cdot\mathbf{x}\), the two exponentials are \(e^{-ip\cdot x}\) and \(e^{+ip\cdot x}\), manifestly Lorentz-covariant. B

\[ \hat{\pi}(t,\mathbf{x}) = \dot{\hat{\phi}} = \int \frac{d^3p}{(2\pi)^3}\,\frac{-i}{2}\left( \hat{a}_{\mathbf{p}}\,e^{-ip\cdot x} - \hat{a}^{\dagger}_{\mathbf{p}}\,e^{+ip\cdot x}\right) \]

Conjugate momentum for the free scalar is \(\hat{\pi}=\dot{\hat{\phi}}\); differentiate step 5 in \(t\), and the \(E_{\mathbf{p}}\) from \(\partial_t\) cancels one factor in \(1/2E_{\mathbf{p}}\). B

\[ \hat{a}_{\mathbf{p}} = \int d^3x\; e^{-i\mathbf{p}\cdot\mathbf{x}}\left( E_{\mathbf{p}}\,\hat{\phi}(t,\mathbf{x}) + i\,\hat{\pi}(t,\mathbf{x})\right)e^{+iE_{\mathbf{p}}t} \]

Invert steps 5-6 for the mode operators. Legal: the combination \(E_{\mathbf{p}}\hat{\phi}+i\hat{\pi}\) projects onto the positive-frequency piece; the \(\mathbf{x}\)-integral of \(e^{i(\mathbf{p}-\mathbf{q})\cdot\mathbf{x}}\) gives \((2\pi)^3\delta^{(3)}(\mathbf{p}-\mathbf{q})\), killing the \(\hat{a}^{\dagger}\) term. The result is \(t\)-independent. C

\[ [\hat{\phi}(t,\mathbf{x}),\hat{\pi}(t,\mathbf{y})] = i\,\delta^{(3)}(\mathbf{x}-\mathbf{y}),\quad [\hat{\phi},\hat{\phi}]=[\hat{\pi},\hat{\pi}]=0 \]

Impose the equal-time canonical commutators (prior result), the quantization postulate. A

\[ [\hat{a}_{\mathbf{p}},\hat{a}^{\dagger}_{\mathbf{q}}] = \int d^3x\,d^3y\; e^{-i\mathbf{p}\cdot\mathbf{x}}\,e^{+i\mathbf{q}\cdot\mathbf{y}}\,e^{i(E_{\mathbf{p}}-E_{\mathbf{q}})t}\Big( E_{\mathbf{q}}\,[\hat{\phi}_x,\,i\hat{\pi}_y] + E_{\mathbf{p}}\,[i\hat{\pi}_x,\,\hat{\phi}_y] + E_{\mathbf{p}}E_{\mathbf{q}}[\hat{\phi}_x,\hat{\phi}_y] + [\hat{\pi}_x,\hat{\pi}_y]\Big) \]

Substitute step 7 (and its dagger) into the bracket and expand bilinearly. Only cross terms \([\hat{\phi},\hat{\pi}]\) survive by step 8; \([\hat{\phi},\hat{\phi}]\) and \([\hat{\pi},\hat{\pi}]\) vanish. C

\[ [\hat{a}_{\mathbf{p}},\hat{a}^{\dagger}_{\mathbf{q}}] = (E_{\mathbf{p}}+E_{\mathbf{q}})\int d^3x\; e^{-i(\mathbf{p}-\mathbf{q})\cdot\mathbf{x}}\cdot\tfrac{1}{2}\cdot\frac{2}{E_{\mathbf{p}}+E_{\mathbf{q}}}\cdots \]

Carry out the \(\hat{\pi}\)-\(\hat{\phi}\) contractions using step 8: each contributes \(i\cdot(-i)=1\) times \(\delta^{(3)}\). The \(\mathbf{y}\)-integral collapses one delta; on the support \(\mathbf{p}=\mathbf{q}\) so \(E_{\mathbf{p}}+E_{\mathbf{q}}=2E_{\mathbf{p}}\). C

\[ [\hat{a}_{\mathbf{p}},\hat{a}^{\dagger}_{\mathbf{q}}] = (2\pi)^3\,2E_{\mathbf{p}}\,\delta^{(3)}(\mathbf{p}-\mathbf{q}),\qquad [\hat{a}_{\mathbf{p}},\hat{a}_{\mathbf{q}}] = [\hat{a}^{\dagger}_{\mathbf{p}},\hat{a}^{\dagger}_{\mathbf{q}}] = 0 \]

Collect the remaining \(\mathbf{x}\)-integral \(\int d^3x\,e^{-i(\mathbf{p}-\mathbf{q})\cdot\mathbf{x}} = (2\pi)^3\delta^{(3)}(\mathbf{p}-\mathbf{q})\). The relativistic \(2E_{\mathbf{p}}\) normalization is the promised algebra. C

\[ \hat{H} = \int \frac{d^3p}{(2\pi)^3\,2E_{\mathbf{p}}}\; E_{\mathbf{p}}\,\hat{a}^{\dagger}_{\mathbf{p}}\hat{a}_{\mathbf{p}} \;+\; (\text{c-number}) \]

Insert step 5 into \(\hat{H}=\int d^3x\,\tfrac{1}{2}(\hat{\pi}^2+(\nabla\hat{\phi})^2+m^2\hat{\phi}^2)\), normal-order using step 11, and drop the infinite zero-point constant. \(\hat{H}\) is diagonal in the modes. C

\[ \hat{a}_{\mathbf{p}}|0\rangle = 0\quad\forall\,\mathbf{p};\qquad |\mathbf{p}_1,\dots,\mathbf{p}_n\rangle \equiv \hat{a}^{\dagger}_{\mathbf{p}_1}\cdots\hat{a}^{\dagger}_{\mathbf{p}_n}|0\rangle \]

Define the vacuum as the state annihilated by all lowering operators (bounded-below spectrum, step 12), then generate the Fock basis. \([\hat{H},\hat{a}^{\dagger}_{\mathbf{p}}]=E_{\mathbf{p}}\hat{a}^{\dagger}_{\mathbf{p}}\) makes each \(\hat{a}^{\dagger}_{\mathbf{p}}\) raise energy by \(E_{\mathbf{p}}\): these are particles. B

Result

\[ \hat{\phi}(x) = \int \frac{d^3p}{(2\pi)^3\,2E_{\mathbf{p}}}\left(\hat{a}_{\mathbf{p}}e^{-ip\cdot x} + \hat{a}^{\dagger}_{\mathbf{p}}e^{+ip\cdot x}\right),\qquad [\hat{a}_{\mathbf{p}},\hat{a}^{\dagger}_{\mathbf{q}}] = (2\pi)^3\,2E_{\mathbf{p}}\,\delta^{(3)}(\mathbf{p}-\mathbf{q}) \]

Reading. The quantized field is a superposition, over all momenta, of operators that destroy (\(\hat{a}\)) or create (\(\hat{a}^{\dagger}\)) a relativistic quantum of energy \(E_{\mathbf{p}}=\sqrt{|\mathbf{p}|^2+m^2}\). Their oscillator-like algebra means the Hilbert space is a Fock space: a vacuum, one-particle states \(\hat{a}^{\dagger}_{\mathbf{p}}|0\rangle\), and symmetric \(n\)-particle towers. The commutator's \(2E_{\mathbf{p}}\) is exactly the factor that makes single-particle states Lorentz-covariantly normalized, \(\langle\mathbf{p}|\mathbf{q}\rangle=(2\pi)^3 2E_{\mathbf{p}}\delta^{(3)}(\mathbf{p}-\mathbf{q})\).

Units check. With \(\hbar=c=1\), \([\hat{\phi}]=\) mass\(^1\) (from \([\hat{\phi}^2 m^2]=[\text{energy density}]=\)mass\(^4\)). In \([\hat{a},\hat{a}^{\dagger}]=(2\pi)^3 2E\,\delta^{(3)}\): \([\delta^{(3)}(\mathbf{p})]=\)mass\(^{-3}\), \([E]=\)mass\(^{1}\), so the right side is mass\(^{-2}\), giving \([\hat{a}]=\)mass\(^{-1}\). Then in the field expansion \([d^3p/(2E)]=\)mass\(^{2}\), \([e^{ipx}]=1\), \([\hat{a}]=\)mass\(^{-1}\), product mass\(^{1}=[\hat{\phi}]\). Consistent.

Limiting cases

Non-relativistic / heavy field \(|\mathbf{p}|\ll m\): \(E_{\mathbf{p}}\approx m + |\mathbf{p}|^2/2m\); the rest energy \(m\) factors out and the ladder algebra reduces to the Schrödinger-field (second-quantized) case with kinetic term \(|\mathbf{p}|^2/2m\).
Massless limit \(m\to 0\): \(E_{\mathbf{p}}=|\mathbf{p}|\); modes are those of a massless scalar (template for photon polarizations), gapless spectrum, long-range correlations.
Single mode / box normalization \(V<\infty\): the integral becomes a sum, \(\delta^{(3)}\to V\delta_{\mathbf{p}\mathbf{q}}/(2\pi)^3\)-like, and each mode is a textbook harmonic oscillator with discrete \(\hat{n}_{\mathbf{p}}=\hat{a}^{\dagger}_{\mathbf{p}}\hat{a}_{\mathbf{p}}\).
Classical limit (large occupation, coherent state): \(\langle\hat{\phi}\rangle\) recovers a classical Klein-Gordon wave; \(\hbar\) restored, the commutator \(\propto\hbar\to 0\).

Breaks when

Interactions are switched on. With a \(\lambda\hat{\phi}^4\) term the plane waves are no longer solutions; \(\hat{a}_{\mathbf{p}}(t)\) evolve non-trivially, particle number is not conserved, and Haag's theorem says no unitary map connects the free and interacting Fock representations. The mode operators survive only as asymptotic (in/out) fields via the LSZ construction.
Curved or time-dependent spacetime. There is no invariant split into positive/negative frequency, so "particle" is observer-dependent. Different Bogoliubov-inequivalent vacua give Unruh/Hawking radiation; \(\hat{a}_{\mathbf{p}}|0\rangle=0\) picks out only one of infinitely many inequivalent Fock spaces.
Finite temperature / thermal state. The physical state is not the Fock vacuum but a thermal density matrix; the relevant representation (thermo-field dynamics) is unitarily inequivalent to the zero-temperature Fock space.
Spontaneous symmetry breaking / non-Fock condensates. For a field with \(\langle\hat{\phi}\rangle\neq 0\) the naive vacuum is unstable; one must expand about the shifted field, and the infinite-volume broken vacuum lies outside the original Fock space.

Failure modes

Dropping the \(2E_{\mathbf{p}}\). Using \([\hat{a},\hat{a}^{\dagger}]=(2\pi)^3\delta^{(3)}\) (Schrödinger normalization) and then claiming Lorentz-covariant states; the two conventions differ by \(\sqrt{2E_{\mathbf{p}}}\) rescalings and mixing them corrupts every matrix element.
Treating \(\hat{\phi}\) as a wavefunction. Interpreting \(\hat{\phi}(x)\) as a probability amplitude and \(|\hat{\phi}|^2\) as a density; \(\hat{\phi}\) is an operator, and single-particle position "wavefunctions" are not sharply localizable (Newton-Wigner subtleties).
Forgetting normal ordering. Keeping the zero-point term and reporting a divergent vacuum energy as physical, instead of recognizing it as an additive c-number fixed by convention (gravitationally subtle, but flat-space observables use differences).
Sign/frequency swap. Attaching \(\hat{a}\) to \(e^{+iE t}\) instead of \(e^{-iEt}\); this makes \(\hat{a}^{\dagger}\) lower the energy, inverts the spectrum, and destroys stability of the vacuum.
Confusing \(\hat{a}^{\dagger}_{-\mathbf{p}}\) relabeling. Failing to flip \(\mathbf{p}\to-\mathbf{p}\) in the negative-frequency term, producing a non-Hermitian "field" for a supposedly real scalar.
Bose vs Fermi. Using commutators for a would-be spin-\(\tfrac12\) field; the spin-statistics theorem forces anticommutators there, and the Fock space is antisymmetric.

Discussion

The mode expansion is the precise sense in which quantum field theory unifies waves and particles. The field \(\hat{\phi}(x)\) is a continuum of coupled harmonic oscillators, one per momentum mode; the "waves" are the classical solutions \(e^{\pm ip\cdot x}\), while the "particles" are the discrete quanta counted by \(\hat{n}_{\mathbf{p}}=\hat{a}^{\dagger}_{\mathbf{p}}\hat{a}_{\mathbf{p}}\). Nothing in the formalism is added by hand to make particles: they fall out of the operator algebra the moment canonical commutators are imposed. This is why the same machinery describes photons in a cavity, phonons in a crystal, and Higgs bosons at the LHC.

The Fock space it builds is a very specific Hilbert space: the direct sum \(\mathcal{H} = \bigoplus_{n=0}^{\infty}\mathrm{Sym}^n(\mathcal{H}_1)\) of totally symmetric \(n\)-particle sectors, symmetric because the creation operators commute. The vacuum is not "nothing"; it is the lowest-energy state, Lorentz-invariant, yet seething with the correlations that give rise to the Casimir effect and to virtual particles in perturbation theory. The one-particle sector \(\mathcal{H}_1\) carries the irreducible Poincaré representation of mass \(m\) and spin \(0\), which is exactly Wigner's classification realized concretely.

The connections run outward in every direction. The commutator's \(2E_{\mathbf{p}}\) is dictated by demanding a Lorentz-invariant measure on the mass shell, tying this construction to relativistic kinematics. The same operators define the Feynman propagator \(\langle 0|T\hat{\phi}(x)\hat{\phi}(y)|0\rangle\), the building block of every Feynman diagram. And promoting the coefficients from commuting to anticommuting objects yields the fermionic Fock space, so this single derivation is the common ancestor of both statistics.

At the deepest level the Fock representation is a choice, not an inevitability. The Stone-von Neumann theorem, which guarantees a unique representation of the canonical commutation relations in quantum mechanics with finitely many degrees of freedom, fails for fields (infinitely many modes). There are uncountably many unitarily inequivalent representations of the field CCR; the Fock/Minkowski vacuum is privileged only because Poincaré symmetry and the spectral condition single it out. Change the symmetry structure (an accelerating frame, a black hole, a broken phase, finite temperature) and a different, inequivalent representation is physically correct — the Rindler wedge sees a thermal bath where the inertial observer sees the vacuum. The particle concept is therefore representation-dependent, and "how many particles are there" has no observer-independent answer.

Common misconceptions. (i) The vacuum is not empty space of classical intuition; it is a definite quantum state with nonzero field fluctuations. (ii) \(\hat{a}^{\dagger}_{\mathbf{p}}|0\rangle\) is not a normalizable state — it is a distribution (delta-normalized), so physical states are wave-packet superpositions \(\int d^3p\,f(\mathbf{p})\hat{a}^{\dagger}_{\mathbf{p}}|0\rangle\). (iii) "Second quantization" is a misnomer: nothing is quantized twice; the field, not a pre-existing wavefunction, is what is quantized once.

Worked examples

Energy and momentum of a two-particle state. Take a real scalar of mass \(m=0.14\ \text{GeV}\) (a toy "pion") with two quanta of momenta \(\mathbf{p}_1 = (0.30,0,0)\ \text{GeV}\) and \(\mathbf{p}_2=(-0.30,0,0)\ \text{GeV}\). Find the total energy of \(|\mathbf{p}_1,\mathbf{p}_2\rangle=\hat{a}^{\dagger}_{\mathbf{p}_1}\hat{a}^{\dagger}_{\mathbf{p}_2}|0\rangle\) and its total momentum.

Set up: eigenvalue of \(\hat{H}\) is the sum of \(E_{\mathbf{p}}\) over quanta because \([\hat{H},\hat{a}^{\dagger}_{\mathbf{p}}]=E_{\mathbf{p}}\hat{a}^{\dagger}_{\mathbf{p}}\). A

\[ E_{\mathbf{p}} = \sqrt{|\mathbf{p}|^2+m^2},\quad |\mathbf{p}_1|=|\mathbf{p}_2|=0.30\ \text{GeV} \]

Symbols first: both quanta have equal \(|\mathbf{p}|\), so equal energies. A

\[ E_{\mathbf{p}} = \sqrt{(0.30)^2 + (0.14)^2} = \sqrt{0.09+0.0196}=\sqrt{0.1096}=0.3311\ \text{GeV} \]

Numbers with units. A

\[ E_{\text{tot}} = 2E_{\mathbf{p}} = 0.6621\ \text{GeV},\qquad \mathbf{P}_{\text{tot}} = \mathbf{p}_1+\mathbf{p}_2 = 0 \]

Additivity of energy; momentum operator eigenvalue is the vector sum, which cancels. A

\[ E_{\text{tot}} = 0.662\ \text{GeV},\qquad \mathbf{P}_{\text{tot}} = \mathbf{0},\qquad \sqrt{s} = 0.662\ \text{GeV} \]

Reading. The state is a zero-total-momentum two-particle state; its invariant mass \(\sqrt{s}=E_{\text{tot}}=0.662\) GeV exceeds \(2m=0.28\) GeV by the kinetic energy of the pair. This is the center-of-momentum frame configuration used to define scattering thresholds.

Vacuum fluctuation normalization. Verify the one-particle inner product for a scalar with the derived algebra, and evaluate it for \(m=1\ \text{GeV}\), \(\mathbf{p}=\mathbf{q}=(0,0,2)\ \text{GeV}\) in the smeared (finite-volume) sense with box volume \(V=(10\ \text{GeV}^{-1})^3\).

Set up: \(\langle\mathbf{p}|\mathbf{q}\rangle = \langle 0|\hat{a}_{\mathbf{p}}\hat{a}^{\dagger}_{\mathbf{q}}|0\rangle\); move \(\hat{a}\) right using the commutator, \(\hat{a}_{\mathbf{p}}|0\rangle=0\). B

\[ \langle\mathbf{p}|\mathbf{q}\rangle = \langle 0|[\hat{a}_{\mathbf{p}},\hat{a}^{\dagger}_{\mathbf{q}}]|0\rangle = (2\pi)^3\,2E_{\mathbf{p}}\,\delta^{(3)}(\mathbf{p}-\mathbf{q}) \]

Only the c-number commutator survives between vacua; \(\langle 0|0\rangle=1\). Symbols first. B

\[ (2\pi)^3\delta^{(3)}(\mathbf{p}-\mathbf{q}) \longrightarrow V\ \text{at}\ \mathbf{p}=\mathbf{q}, \qquad E_{\mathbf{p}}=\sqrt{2^2+1^2}=\sqrt{5}=2.236\ \text{GeV} \]

Finite-volume regularization: \((2\pi)^3\delta^{(3)}(0)\to V\), the box volume, and compute the on-shell energy. B

\[ \langle\mathbf{p}|\mathbf{p}\rangle = 2E_{\mathbf{p}}\,V = 2(2.236)(1000\ \text{GeV}^{-3}) = 4472\ \text{GeV}^{-2} \]

Insert \(V=10^3\ \text{GeV}^{-3}\) and \(2E_{\mathbf{p}}\). Units: mass \(\times\) mass\(^{-3}\) = mass\(^{-2}\). B

\[ \langle\mathbf{p}|\mathbf{p}\rangle = 2E_{\mathbf{p}}V \approx 4.47\times 10^{3}\ \text{GeV}^{-2} \]

Reading. The single-particle norm is \(2E_{\mathbf{p}}\) times the quantization volume, confirming the relativistic normalization: a faster (higher-\(E\)) particle carries a larger norm, precisely compensating the Lorentz contraction of the volume so that \(\langle\mathbf{p}|\mathbf{p}\rangle/V \propto E_{\mathbf{p}}\) transforms as a time component of a four-vector density. The infinite continuum limit \(V\to\infty\) recovers the delta-function normalization.

Problems

Show that \([\hat{H},\hat{a}^{\dagger}_{\mathbf{p}}] = E_{\mathbf{p}}\,\hat{a}^{\dagger}_{\mathbf{p}}\) using \(\hat{H}=\int\frac{d^3p}{(2\pi)^3 2E_{\mathbf{p}}}E_{\mathbf{p}}\hat{a}^{\dagger}_{\mathbf{p}}\hat{a}_{\mathbf{p}}\) and the commutator algebra. Interpret the result physically.
Solution
Write \([\hat{H},\hat{a}^{\dagger}_{\mathbf{q}}] = \int\frac{d^3p}{(2\pi)^3 2E_{\mathbf{p}}}E_{\mathbf{p}}\,[\hat{a}^{\dagger}_{\mathbf{p}}\hat{a}_{\mathbf{p}},\hat{a}^{\dagger}_{\mathbf{q}}]\). Using \([\hat{A}\hat{B},\hat{C}]=\hat{A}[\hat{B},\hat{C}]+[\hat{A},\hat{C}]\hat{B}\), and \([\hat{a}_{\mathbf{p}},\hat{a}^{\dagger}_{\mathbf{q}}]=(2\pi)^3 2E_{\mathbf{p}}\delta^{(3)}(\mathbf{p}-\mathbf{q})\), only \(\hat{a}^{\dagger}_{\mathbf{p}}[\hat{a}_{\mathbf{p}},\hat{a}^{\dagger}_{\mathbf{q}}]\) survives (\([\hat{a}^{\dagger}_{\mathbf{p}},\hat{a}^{\dagger}_{\mathbf{q}}]=0\)). Thus \([\hat{H},\hat{a}^{\dagger}_{\mathbf{q}}]=\int\frac{d^3p}{(2\pi)^3 2E_{\mathbf{p}}}E_{\mathbf{p}}\hat{a}^{\dagger}_{\mathbf{p}}(2\pi)^3 2E_{\mathbf{p}}\delta^{(3)}(\mathbf{p}-\mathbf{q}) = E_{\mathbf{q}}\hat{a}^{\dagger}_{\mathbf{q}}\). Physically, \(\hat{a}^{\dagger}_{\mathbf{q}}\) is a raising operator that adds exactly \(E_{\mathbf{q}}=\sqrt{|\mathbf{q}|^2+m^2}\) to the energy: it creates one particle of momentum \(\mathbf{q}\).
A real scalar has \(m=0.5\ \text{GeV}\). Compute the energy \(E_{\mathbf{p}}\) and group velocity \(\mathbf{v}=\partial E/\partial\mathbf{p}\) of a quantum with \(\mathbf{p}=(0.5,0,0)\ \text{GeV}\). Comment on the ratio \(|\mathbf{v}|\) to the speed of light.
Solution
\(E_{\mathbf{p}}=\sqrt{|\mathbf{p}|^2+m^2}=\sqrt{0.25+0.25}=\sqrt{0.5}=0.7071\) GeV. Group velocity: \(\mathbf{v}=\partial E/\partial\mathbf{p}=\mathbf{p}/E_{\mathbf{p}}\), so \(|\mathbf{v}|=|\mathbf{p}|/E=0.5/0.7071=0.7071\). In units \(c=1\), the quantum moves at \(0.707c\). Since \(|\mathbf{p}|=m\) here, the particle is mildly relativistic (\(\gamma = E/m = 1.414\)). The group velocity equals \(pc^2/E\) with units restored and is always \(<c\) for \(m>0\).
Verify the units check independently: given \([\hat{\phi}]=\text{mass}\) in \(\hbar=c=1\), and the field expansion of the Result box, deduce \([\hat{a}_{\mathbf{p}}]\) and confirm it matches the value implied by the commutator normalization.
Solution
From \(\hat{\phi}=\int\frac{d^3p}{(2\pi)^3 2E_{\mathbf{p}}}(\hat{a}_{\mathbf{p}}e^{-ipx}+\text{h.c.})\): \([d^3p]=\text{mass}^3\), \([2E_{\mathbf{p}}]=\text{mass}\), so \([d^3p/2E]=\text{mass}^2\); \([e^{-ipx}]=\text{mass}^0\). For the whole integral to have \([\hat{\phi}]=\text{mass}^1\), need \(\text{mass}^2\cdot[\hat{a}]=\text{mass}^1\), giving \([\hat{a}_{\mathbf{p}}]=\text{mass}^{-1}\). Cross-check via the commutator: \([\hat{a},\hat{a}^{\dagger}]=(2\pi)^3 2E\delta^{(3)}\) has dimension \([\hat{a}]^2=\text{mass}\cdot\text{mass}^{-3}=\text{mass}^{-2}\), so \([\hat{a}]=\text{mass}^{-1}\). The two agree.
Coherent state and the classical limit. Define \(|\alpha\rangle = \exp\!\big(\int\frac{d^3p}{(2\pi)^3 2E_{\mathbf{p}}}(\alpha_{\mathbf{p}}\hat{a}^{\dagger}_{\mathbf{p}} - \alpha^*_{\mathbf{p}}\hat{a}_{\mathbf{p}})\big)|0\rangle\). Show that \(\hat{a}_{\mathbf{q}}|\alpha\rangle = \alpha_{\mathbf{q}}|\alpha\rangle\) and hence that \(\langle\alpha|\hat{\phi}(x)|\alpha\rangle\) is a classical solution of the Klein-Gordon equation.
Solution
The displacement operator \(\hat{D}=\exp(\int_p(\alpha_p\hat{a}^{\dagger}_p-\alpha^*_p\hat{a}_p))\) satisfies \(\hat{D}^{\dagger}\hat{a}_{\mathbf{q}}\hat{D}=\hat{a}_{\mathbf{q}}+\alpha_{\mathbf{q}}\) (from \([\hat{a}_{\mathbf{q}},\int_p\alpha_p\hat{a}^{\dagger}_p]=\alpha_{\mathbf{q}}\) using the algebra, plus BCH). Then \(\hat{a}_{\mathbf{q}}|\alpha\rangle=\hat{a}_{\mathbf{q}}\hat{D}|0\rangle=\hat{D}(\hat{D}^{\dagger}\hat{a}_{\mathbf{q}}\hat{D})|0\rangle=\hat{D}(\hat{a}_{\mathbf{q}}+\alpha_{\mathbf{q}})|0\rangle=\alpha_{\mathbf{q}}\hat{D}|0\rangle=\alpha_{\mathbf{q}}|\alpha\rangle\), since \(\hat{a}_{\mathbf{q}}|0\rangle=0\). Therefore \(\langle\alpha|\hat{\phi}(x)|\alpha\rangle=\int\frac{d^3p}{(2\pi)^3 2E_{\mathbf{p}}}(\alpha_{\mathbf{p}}e^{-ipx}+\alpha^*_{\mathbf{p}}e^{+ipx})\), a real superposition of on-shell plane waves — i.e. a classical solution of \((\Box+m^2)\phi=0\). The coherent state is the quantum state whose field expectation reproduces classical field theory; large \(|\alpha|\) is the classical (many-quanta) limit.
Micro-causality. Using the Result-box expansion, show that the commutator \([\hat{\phi}(x),\hat{\phi}(y)]\) is a c-number and argue it vanishes for spacelike separation \((x-y)^2<0\).
Solution
Expand both fields and use the algebra; only \([\hat{a}_{\mathbf{p}},\hat{a}^{\dagger}_{\mathbf{q}}]\) and \([\hat{a}^{\dagger}_{\mathbf{p}},\hat{a}_{\mathbf{q}}]\) contribute: \([\hat{\phi}(x),\hat{\phi}(y)]=\int\frac{d^3p}{(2\pi)^3 2E_{\mathbf{p}}}\big(e^{-ip(x-y)}-e^{+ip(x-y)}\big)\equiv i\Delta(x-y)\), a c-number (no operators remain). Call it \(\Delta(x-y)\). Each term is Lorentz-invariant (invariant measure, on-shell \(p\)). For spacelike \(x-y\) one can choose a frame where \(x-y\to-(x-y)\) (a rotation, connected to the identity, reverses the spatial separation while leaving \(x^0-y^0=0\)); under it the two exponential terms swap, so \(\Delta(x-y)=-\Delta(x-y)=0\). Hence spacelike-separated field operators commute: measurements at spacelike separation cannot interfere, enforcing relativistic causality. For timelike separation no such rotation exists and the commutator is nonzero.