Canonical Quantization of a Field

Derivation

Canonical Quantization of a Field

D-377 Home PU-402 Threads fields · matter Depends on Hamiltonian Formulation for Fields, canonical-commutation-relations

Statement

Starting from a classical field theory in Hamiltonian form with field \(\phi(\mathbf{x})\) and conjugate momentum density \(\pi(\mathbf{x}) = \partial\mathcal{L}/\partial\dot\phi\), canonical quantization promotes the fields to operator-valued distributions on a Hilbert space and replaces the classical equal-time Poisson brackets by commutators via \(\{\,\cdot\,,\cdot\,\}_{\mathrm{PB}} \to \tfrac{1}{i\hbar}[\,\cdot\,,\cdot\,]\). This yields the equal-time canonical commutation relations \[ [\hat\phi(\mathbf{x}),\hat\pi(\mathbf{y})] = i\hbar\,\delta^{3}(\mathbf{x}-\mathbf{y}), \qquad [\hat\phi(\mathbf{x}),\hat\phi(\mathbf{y})] = [\hat\pi(\mathbf{x}),\hat\pi(\mathbf{y})] = 0, \] which turn a classical field into a quantum operator field (in natural units \(\hbar=1\) the right-hand side is \(i\,\delta^{3}(\mathbf{x}-\mathbf{y})\)).

Why it matters

This single step is the bridge from classical field theory to quantum field theory. Every particle in the Standard Model is a quantum of a field whose existence follows from imposing exactly these commutators: the Fourier modes of \(\hat\phi\) become creation and annihilation operators, the field energy becomes a sum of harmonic-oscillator Hamiltonians, and the discrete "particle" spectrum emerges from the continuous classical field.

It also fixes the physical dimension of the theory. The Dirac delta on the right-hand side is the field-theoretic generalization of the mechanics relation \([\hat q,\hat p]=i\hbar\), and its \(\delta^{3}\) structure encodes that independent points of space carry independent, mutually commuting degrees of freedom — the property that makes locality and microcausality possible.

Assumptions

The classical theory admits a well-defined Hamiltonian formulation.If the Legendre transform \(\pi=\partial\mathcal{L}/\partial\dot\phi\) is non-invertible (constrained/singular Lagrangian, e.g. gauge fields), the naive brackets are wrong and one needs Dirac brackets before quantizing.

Equal-time quantization on a fixed spatial slice is meaningful.Requires a global time coordinate and a Cauchy foliation; in a general curved spacetime with no timelike Killing vector the split \((\mathbf{x},t)\) and hence the equal-time bracket loses its preferred status.

The map \(\{\,,\}\to\tfrac{1}{i\hbar}[\,,]\) is consistent (no operator-ordering pathology at this level).Groenewold–van Hove forbids a bracket-preserving map for all observables; if we demanded it beyond the basic canonical pair the prescription would be inconsistent, so it is imposed only on \(\phi,\pi\) and their linear functionals.

The field is a genuine operator-valued distribution, not a function.If one insists \(\hat\phi(\mathbf{x})\) is an ordinary operator at a sharp point, products like \(\hat\phi(\mathbf{x})^2\) diverge; smearing \(\hat\phi(f)=\int f\,\hat\phi\) is required for the relations to be well defined.

Derivation

\[ H = \int d^{3}x\,\big[\pi(\mathbf{x})\dot\phi(\mathbf{x}) - \mathcal{L}\big] = \int d^{3}x\,\mathcal{H}\big(\phi,\pi,\nabla\phi\big) \]

Import the Hamiltonian-field-formulation result: the Legendre transform gives \(H\) as a spatial integral of the Hamiltonian density with independent canonical pair \((\phi,\pi)\). A

\[ \{\phi(\mathbf{x}),\pi(\mathbf{y})\}_{\mathrm{PB}} = \delta^{3}(\mathbf{x}-\mathbf{y}), \quad \{\phi,\phi\}=\{\pi,\pi\}=0 \]

State the classical fundamental equal-time Poisson brackets. The functional bracket \(\{A,B\}=\int d^{3}z\big(\frac{\delta A}{\delta\phi(\mathbf z)}\frac{\delta B}{\delta\pi(\mathbf z)}-\frac{\delta A}{\delta\pi(\mathbf z)}\frac{\delta B}{\delta\phi(\mathbf z)}\big)\) applied to the pair yields \(\delta^{3}\). A

\[ \phi(\mathbf{x}) \longmapsto \hat\phi(\mathbf{x}), \qquad \pi(\mathbf{x}) \longmapsto \hat\pi(\mathbf{x}) \]

Promotion: replace the classical fields by self-adjoint operator-valued distributions acting on a common dense domain of a separable Hilbert space \(\mathcal{H}\). This is a postulate of quantization, justified a posteriori by experiment. B

\[ \{A,B\}_{\mathrm{PB}} \;\longmapsto\; \frac{1}{i\hbar}\,[\hat A,\hat B] \]

Dirac's correspondence: the Lie-algebra structure of Poisson brackets (antisymmetry, Jacobi identity, Leibniz rule) is preserved to leading order by the commutator, since \(\tfrac{1}{i\hbar}[\,,]\) is itself an antisymmetric derivation. B

\[ \frac{1}{i\hbar}[\hat\phi(\mathbf{x}),\hat\pi(\mathbf{y})] = \delta^{3}(\mathbf{x}-\mathbf{y}) \]

Apply the map of step 4 to the fundamental bracket of step 2: the c-number \(\delta^{3}\) on the right multiplies the identity operator, so the equation is an operator identity with distributional right-hand side. B

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

Multiply both sides by \(i\hbar\). Symbols only; \(i\hbar\) is a constant so it commutes through. A

\[ [\hat\phi(\mathbf{x}),\hat\phi(\mathbf{y})] = 0, \qquad [\hat\pi(\mathbf{x}),\hat\pi(\mathbf{y})] = 0 \]

Apply step 4 to the vanishing classical brackets \(\{\phi,\phi\}=\{\pi,\pi\}=0\). Fields of the same type at equal time are simultaneously diagonalizable — the "coordinate representation" of the field. A

\[ \hat\phi(f)=\int d^{3}x\, f(\mathbf x)\hat\phi(\mathbf x),\ \ [\hat\phi(f),\hat\pi(g)] = i\hbar\!\int d^{3}x\, f(\mathbf x)g(\mathbf x)\,\hat{\mathbb 1} \]

Smear against test functions \(f,g\in\mathcal{S}(\mathbb R^3)\) to obtain bounded, well-defined operator relations; the \(\delta^{3}\) collapses one integral, confirming the point relation is the distributional kernel of a sensible operator statement. C

Result

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

Reading. At one instant of time, the field amplitude at a point and its conjugate momentum at the same point are incompatible observables obeying an uncertainty relation, while the amplitude at one point commutes perfectly with the amplitude and the momentum at every other point. Space is treated as a continuum of independent quantum mechanical degrees of freedom, one canonical pair \((\hat\phi,\hat\pi)\) per point, glued together only by the Hamiltonian's gradient terms.

Units check. In SI, \([\phi]\) for a scalar is \(\mathrm{J^{1/2}\,m^{-1/2}}\) and the momentum density \([\pi]=[\mathcal{L}]\cdot[\dot\phi]^{-1}\cdot\text{(time)}\) works out so that \([\phi\,\pi]=\mathrm{J\,s\,m^{-3}}=[\hbar]\cdot\mathrm{m^{-3}}\). Since \([\delta^{3}(\mathbf x-\mathbf y)]=\mathrm{m^{-3}}\), the right side \(i\hbar\,\delta^{3}\) carries \(\mathrm{J\,s\,m^{-3}}\), matching the left side dimension by dimension. In natural units \(\hbar=1\) and the relation reads \([\hat\phi,\hat\pi]=i\delta^{3}\).

Limiting cases

Classical limit \(\hbar\to0\): the right-hand side vanishes, \([\hat\phi,\hat\pi]\to0\), and the operators become commuting c-number fields — classical field theory is recovered, with commutators degenerating back into Poisson brackets.
Single-point / finite-mode truncation: replace \(\int d^3x\to\sum_n\) on a lattice of spacing \(a\); then \(\delta^{3}(\mathbf x-\mathbf y)\to a^{-3}\delta_{mn}\) and each site is an ordinary quantum particle with \([\hat q_n,\hat p_m]=i\hbar\,\delta_{nm}\), recovering point-particle quantum mechanics.
Free field, mode expansion: Fourier transforming turns the relation into \([\hat a_{\mathbf k},\hat a^\dagger_{\mathbf k'}]=(2\pi)^3\delta^{3}(\mathbf k-\mathbf k')\), the ladder algebra whose Fock space contains the particle states.
Non-relativistic reduction: for a Schrödinger field \(\psi\), the same prescription gives \([\hat\psi(\mathbf x),\hat\psi^\dagger(\mathbf y)]=\delta^{3}(\mathbf x-\mathbf y)\) — the "second quantization" of many-body quantum mechanics.

Breaks when

Constrained / gauge systems. For the electromagnetic field the Lagrangian is singular (\(\pi^0\equiv0\)), the Legendre transform is non-invertible, and imposing \([\hat A_\mu,\hat\pi^\nu]=i\hbar\,\delta^\nu_\mu\,\delta^3\) is inconsistent with Gauss's law. One must use Dirac brackets (fix a gauge) or the BRST/Gupta–Bleuler machinery instead.
Fermionic fields. Spin-statistics forbids commutators for half-integer-spin fields: quantizing the Dirac field with \([\,\cdot\,,\cdot\,]\) gives a Hamiltonian unbounded below and violates microcausality. The correct rule is the anticommutator \(\{\hat\psi_a(\mathbf x),\hat\psi_b^\dagger(\mathbf y)\}=\delta_{ab}\,\delta^{3}(\mathbf x-\mathbf y)\).
Interacting fields at coincident points (Haag's theorem). The naive equal-time relations, together with a Fock vacuum, cannot survive interactions unitarily equivalent to the free theory; products \(\hat\phi(\mathbf x)^n\) need regularization/renormalization or the relation is formally divergent.
Genuinely curved / time-dependent spacetime. Without a preferred time slicing there is no canonical "equal-time" surface, the vacuum becomes observer-dependent (Unruh, Hawking), and the equal-time commutator loses its status as a Lorentz-invariant statement.

Failure modes

Dropping the delta function. Writing \([\hat\phi(\mathbf x),\hat\pi(\mathbf y)]=i\hbar\) (a pure number) instead of \(i\hbar\,\delta^{3}(\mathbf x-\mathbf y)\). The delta is mandatory: it makes the two sides dimensionally consistent and encodes that different points are independent.
Using \(\pi=\dot\phi\). The conjugate momentum is \(\pi=\partial\mathcal{L}/\partial\dot\phi\), which equals \(\dot\phi\) only for the canonically normalized free scalar; for a field with a nontrivial kinetic prefactor \(Z\), \(\pi=Z\dot\phi\) and the commutator picks up factors of \(Z\).
Mixing unequal times. The relation holds at equal time \(x^0=y^0\). Students wrongly apply it to \([\hat\phi(x),\hat\pi(y)]\) at different times, where the answer is instead governed by the (nonlocal) field equations and the Pauli–Jordan function.
Quantizing spinors with commutators. Applying the \(\{,\}_{\mathrm{PB}}\to[,]\) rule blindly to the Dirac field; the spin-statistics theorem demands anticommutators for spin-\(\tfrac12\).
Treating \(\hat\phi(\mathbf x)\) as an ordinary operator. Assuming \(\hat\phi(\mathbf x)\) has finite matrix elements at a sharp point; it is a distribution, and only smeared \(\hat\phi(f)\) are honest operators.
Sign/factor of \(i\). Writing \(-i\hbar\) or dropping the \(i\); the \(i\) is fixed by requiring \(\hat\phi,\hat\pi\) self-adjoint (Hermitian) with a real Poisson bracket, exactly as in \([\hat q,\hat p]=+i\hbar\).

Discussion

The equal-time commutator is the defining axiom of canonical quantum field theory: once it is imposed, the entire particle content follows. Expanding the free scalar in modes, \(\hat\phi(\mathbf x)=\int\frac{d^3k}{(2\pi)^3}\frac{1}{\sqrt{2\omega_k}}\big(\hat a_{\mathbf k}e^{i\mathbf k\cdot\mathbf x}+\hat a^\dagger_{\mathbf k}e^{-i\mathbf k\cdot\mathbf x}\big)\), the single relation \([\hat\phi,\hat\pi]=i\hbar\delta^3\) is algebraically equivalent to \([\hat a_{\mathbf k},\hat a^\dagger_{\mathbf k'}]=(2\pi)^3\delta^3(\mathbf k-\mathbf k')\). The field Hamiltonian then becomes an infinite collection of decoupled harmonic oscillators, one per momentum mode, and \(\hat a^\dagger_{\mathbf k}\) creates a particle of momentum \(\mathbf k\). The "quantum" of the field is thus a direct consequence of promoting one Poisson bracket to one commutator.

The \(\delta^{3}(\mathbf x-\mathbf y)\) is not a cosmetic detail; it is the statement of locality at the kinematic level. Because \(\hat\phi\) at one point commutes with everything at spatially distinct points on the same slice, no instantaneous measurement at \(\mathbf x\) can disturb an observable at \(\mathbf y\ne\mathbf x\). Promoted to a Lorentz-covariant statement, this becomes microcausality: \([\hat\phi(x),\hat\phi(y)]=0\) for spacelike-separated \(x,y\), the precise sense in which relativistic quantum fields respect causality. The choice between commutators and anticommutators is then forced by this same requirement combined with a Hamiltonian bounded below — the content of the spin–statistics theorem.

Canonical quantization is one of several routes to the same theory. The path integral reaches the same physics through the generating functional, and the two are related because the time-ordered products of the operator formalism are exactly the correlation functions computed by the path integral. Canonical quantization's virtue is that it makes the operator/Hilbert-space structure and the particle interpretation manifest; its cost is manifest Lorentz covariance, which the equal-time slicing breaks (though final S-matrix elements remain covariant).

At a deeper level the prescription is not a theorem but a well-motivated postulate, and its limits are instructive. Groenewold–van Hove shows no map can turn every classical observable's Poisson bracket into the corresponding commutator without contradiction, so the correspondence is exact only on the linear (Heisenberg) subalgebra generated by \(\phi,\pi\) and \(\mathbb 1\); everything higher — products, powers, composite operators — requires an ordering choice and, in the interacting theory, renormalization. Haag's theorem sharpens the warning: in the continuum the interaction-picture representation built on these free-field commutators is unitarily inequivalent to the free one, so rigorous constructive QFT treats the canonical relations as a formal starting point to be tamed by regularization. That the resulting perturbation series nonetheless predicts the electron \(g\!-\!2\) to twelve figures is the strongest evidence that promoting brackets to commutators captures something true about nature.

Common misconceptions. The relation does not say the field and its momentum can never be known together anywhere — it says they are incompatible only at the same point; at different points they commute. And \([\hat\phi,\hat\pi]=i\hbar\delta^3\) is an equal-time statement — it is not the propagator, and it says nothing directly about correlations between different instants.

Worked examples

Free real scalar: verify the mode algebra reproduces the canonical relation.

Take \(\hat\phi(\mathbf x)=\int\!\frac{d^3k}{(2\pi)^3}\frac{1}{\sqrt{2\omega_k}}(\hat a_{\mathbf k}e^{i\mathbf k\cdot\mathbf x}+\hat a^\dagger_{\mathbf k}e^{-i\mathbf k\cdot\mathbf x})\) and \(\hat\pi=\dot{\hat\phi}=\int\!\frac{d^3k}{(2\pi)^3}(-i)\sqrt{\tfrac{\omega_k}{2}}(\hat a_{\mathbf k}e^{i\mathbf k\cdot\mathbf x}-\hat a^\dagger_{\mathbf k}e^{-i\mathbf k\cdot\mathbf x})\), with \([\hat a_{\mathbf k},\hat a^\dagger_{\mathbf k'}]=(2\pi)^3\delta^3(\mathbf k-\mathbf k')\). B

\[ [\hat\phi(\mathbf x),\hat\pi(\mathbf y)] = \int\!\frac{d^3k\,d^3k'}{(2\pi)^6}\frac{-i}{2}\sqrt{\frac{\omega_{k'}}{\omega_k}}\,[\hat a_{\mathbf k}e^{i\mathbf k\cdot\mathbf x}+\hat a^\dagger_{\mathbf k}e^{-i\mathbf k\cdot\mathbf x},\ \hat a_{\mathbf k'}e^{i\mathbf k'\cdot\mathbf y}-\hat a^\dagger_{\mathbf k'}e^{-i\mathbf k'\cdot\mathbf y}] \]

Only cross terms \([\hat a,\hat a^\dagger]\) and \([\hat a^\dagger,\hat a]\) survive; substitute the ladder algebra, which sets \(\mathbf k'=\mathbf k\) and \(\omega_{k'}=\omega_k\). B

\[ = \int\!\frac{d^3k}{(2\pi)^3}\frac{-i}{2}\Big(-e^{i\mathbf k\cdot(\mathbf x-\mathbf y)}-e^{-i\mathbf k\cdot(\mathbf x-\mathbf y)}\Big)(-1)= \int\!\frac{d^3k}{(2\pi)^3}\,i\cdot\tfrac12\cdot 2\cos[\mathbf k\cdot(\mathbf x-\mathbf y)] \]

Combine exponentials; the \(\omega_k\) factors cancel exactly (this is why \(\pi\) carries \(\sqrt{\omega_k}\)). Numbers now: \(\tfrac12\cdot2=1\). A

\[ [\hat\phi(\mathbf x),\hat\pi(\mathbf y)] = i\int\!\frac{d^3k}{(2\pi)^3}e^{i\mathbf k\cdot(\mathbf x-\mathbf y)} = i\,\delta^{3}(\mathbf x-\mathbf y) \]

Reading. The oscillator algebra of the modes is exactly equivalent to the canonical field commutator (with \(\hbar=1\)); the \(\omega_k\) weighting in \(\hat\phi,\hat\pi\) is engineered precisely so the frequency factors cancel and leave a clean \(\delta^{3}\). Units check. \(\int d^3k/(2\pi)^3\) of a dimensionless exponential gives \(\mathrm{m^{-3}}\), matching \(\delta^3\).

Lattice regularization: a scalar field on a cubic lattice, spacing \(a=0.10\ \mathrm{nm}\).

Discretize \(\mathbf x\to a\mathbf n\), define site variables \(\hat q_{\mathbf n}=a^{3/2}\hat\phi(a\mathbf n)\), \(\hat p_{\mathbf n}=a^{3/2}\hat\pi(a\mathbf n)\). Then \(\delta^3(\mathbf x-\mathbf y)\to a^{-3}\delta_{\mathbf n\mathbf m}\). B

\[ [\hat q_{\mathbf n},\hat p_{\mathbf m}] = a^{3}\,[\hat\phi(a\mathbf n),\hat\pi(a\mathbf m)] = a^{3}\cdot i\hbar\,a^{-3}\delta_{\mathbf n\mathbf m} = i\hbar\,\delta_{\mathbf n\mathbf m} \]

Substitute the canonical relation and the lattice delta; the \(a^{3}\) prefactors cancel, leaving an ordinary QM commutator per site. Symbols first. A

\[ a^{-3} = (0.10\times10^{-9}\ \mathrm{m})^{-3} = (1.0\times10^{-10})^{-3}\ \mathrm{m^{-3}} = 1.0\times10^{30}\ \mathrm{m^{-3}} \]

Numbers: evaluate the effective on-site delta strength that a continuum measurement would see. \(\hbar=1.055\times10^{-34}\ \mathrm{J\,s}\). A

\[ [\hat q_{\mathbf n},\hat p_{\mathbf m}] = i\hbar\,\delta_{\mathbf n\mathbf m},\qquad [\hat\phi,\hat\pi]\big|_{\mathbf n=\mathbf m}\sim i\hbar\,a^{-3}=1.06\times10^{-4}\,i\ \mathrm{J\,s\,m^{-3}} \]

Reading. Each lattice site is an independent quantum oscillator obeying textbook \([\hat q,\hat p]=i\hbar\); the continuum field commutator is the \(a\to0\) limit in which the on-site strength \(i\hbar a^{-3}\) diverges into the Dirac delta. This makes concrete why \(\hat\phi(\mathbf x)\) is a distribution. Units check. \(\hbar\,a^{-3}=\mathrm{J\,s}\cdot\mathrm{m^{-3}}\), the required dimension of \([\hat\phi,\hat\pi]\); \([\hat q,\hat p]=\mathrm{J\,s}\) matches \(\hbar\).

Problems

(Grade 1) State the equal-time canonical commutation relations for a real scalar field and give the dimension of each side in SI.
Solution
\([\hat\phi(\mathbf x),\hat\pi(\mathbf y)]=i\hbar\,\delta^3(\mathbf x-\mathbf y)\), \([\hat\phi,\hat\phi]=[\hat\pi,\hat\pi]=0\). Dimensions: \([\hbar]=\mathrm{J\,s}\), \([\delta^3]=\mathrm{m^{-3}}\), so both sides are \(\mathrm{J\,s\,m^{-3}}\). Consistency requires \([\hat\phi\,\hat\pi]=\mathrm{J\,s\,m^{-3}}\), i.e. the product of the field and its conjugate momentum density carries an action per unit volume.
(Grade 2) A scalar has Lagrangian density \(\mathcal{L}=\tfrac{Z}{2}\dot\phi^2-\tfrac12(\nabla\phi)^2-\tfrac12 m^2\phi^2\) with \(Z=4\). Find the conjugate momentum and the commutator \([\hat\phi(\mathbf x),\hat{\dot\phi}(\mathbf y)]\).
Solution
\(\pi=\partial\mathcal L/\partial\dot\phi=Z\dot\phi=4\dot\phi\). The canonical relation is with \(\pi\), not \(\dot\phi\): \([\hat\phi,\hat\pi]=i\hbar\delta^3\). Hence \([\hat\phi(\mathbf x),\hat{\dot\phi}(\mathbf y)]=\tfrac1Z[\hat\phi,\hat\pi]=\tfrac{i\hbar}{4}\delta^3(\mathbf x-\mathbf y)\). The nonstandard normalization \(Z\) suppresses the velocity commutator by \(1/Z\).
(Grade 3) Using \([\hat a_{\mathbf k},\hat a^\dagger_{\mathbf k'}]=(2\pi)^3\delta^3(\mathbf k-\mathbf k')\), compute \([\hat\phi(\mathbf x),\hat\phi(\mathbf y)]\) at equal time and confirm it vanishes.
Solution
With \(\hat\phi=\int\frac{d^3k}{(2\pi)^3}\frac{1}{\sqrt{2\omega_k}}(\hat a_{\mathbf k}e^{i\mathbf k\cdot\mathbf x}+\hat a^\dagger_{\mathbf k}e^{-i\mathbf k\cdot\mathbf x})\), the commutator gets contributions \([\hat a_{\mathbf k},\hat a^\dagger_{\mathbf k'}]\) (\(\to+e^{i\mathbf k\cdot(\mathbf x-\mathbf y)}\)) and \([\hat a^\dagger_{\mathbf k},\hat a_{\mathbf k'}]\) (\(\to-e^{-i\mathbf k\cdot(\mathbf x-\mathbf y)}\)), each weighted by \(\frac{1}{2\omega_k}\). They give \(\int\frac{d^3k}{(2\pi)^3}\frac{1}{2\omega_k}(e^{i\mathbf k\cdot(\mathbf x-\mathbf y)}-e^{-i\mathbf k\cdot(\mathbf x-\mathbf y)})\); sending \(\mathbf k\to-\mathbf k\) in the second term shows the two cancel, so \([\hat\phi(\mathbf x),\hat\phi(\mathbf y)]=0\). Physically, field amplitudes at one instant are simultaneously measurable.
(Grade 4) Show that \(\hat\pi(\mathbf x)\) generates spatial-independent shifts of \(\hat\phi\): evaluate \(e^{\frac{i}{\hbar}\int d^3y\,\epsilon(\mathbf y)\hat\pi(\mathbf y)}\,\hat\phi(\mathbf x)\,e^{-\frac{i}{\hbar}\int d^3y\,\epsilon(\mathbf y)\hat\pi(\mathbf y)}\) to first order in \(\epsilon\).
Solution
Let \(\hat G=\frac{1}{\hbar}\int d^3y\,\epsilon(\mathbf y)\hat\pi(\mathbf y)\). By the BCH/Hadamard lemma the transformed operator is \(\hat\phi+i[\hat G,\hat\phi]+O(\epsilon^2)\). Now \(i[\hat G,\hat\phi(\mathbf x)]=\frac{i}{\hbar}\int d^3y\,\epsilon(\mathbf y)[\hat\pi(\mathbf y),\hat\phi(\mathbf x)]=\frac{i}{\hbar}\int d^3y\,\epsilon(\mathbf y)(-i\hbar)\delta^3(\mathbf y-\mathbf x)=\epsilon(\mathbf x)\). Thus \(\hat\phi(\mathbf x)\to\hat\phi(\mathbf x)+\epsilon(\mathbf x)\): the momentum density is the generator of translations of the field configuration, the field-theory analogue of \(\hat p\) translating \(\hat q\).
(Grade 5) Explain why applying \(\{,\}\to\frac{1}{i\hbar}[,]\) to the Dirac field is inconsistent, and state the correct relation. Illustrate with the sign of the mode-number contribution to the energy.
Solution
If one imposed \([\hat b_{\mathbf p},\hat b^\dagger_{\mathbf p'}]=(2\pi)^3\delta^3(\mathbf p-\mathbf p')\) for the Dirac field, the Hamiltonian \(H=\int\frac{d^3p}{(2\pi)^3}\sum_s E_p(\hat b^\dagger_s\hat b_s-\hat d_s\hat d^\dagger_s)\) would, after commuting \(\hat d\hat d^\dagger=\hat d^\dagger\hat d+\text{const}\), contain \(-\hat d^\dagger_s\hat d_s\): antiparticle occupation would lower the energy without bound, so no stable ground state exists. The spin–statistics theorem requires anticommutators: \(\{\hat\psi_a(\mathbf x),\hat\psi_b^\dagger(\mathbf y)\}=\delta_{ab}\delta^3(\mathbf x-\mathbf y)\), giving \(\hat d\hat d^\dagger=-\hat d^\dagger\hat d+\text{const}\), so both terms enter \(H\) with \(+E_p\) and the energy is bounded below. This also yields the Pauli exclusion principle, since \(\{\hat d^\dagger,\hat d^\dagger\}=0\Rightarrow(\hat d^\dagger)^2=0\).