From Phase Space to Canonical Quantization

Canonical quantization begins with a classical phase space, identifies its canonical brackets, and seeks operators whose commutators reproduce that elementary algebra. For finitely many mechanical degrees of freedom, the familiar bridge is

$$\{q^i,p_j\}_{\mathrm{PB}} = \delta^i_j \quad\longrightarrow\quad \left[ \hat q^i,\hat p_j \right] = i\hbar\delta^i_j.$$

For a field, the canonical label $i$ becomes a spatial point and possibly an internal or spinor index:

$$\bigl(q^i,p_i\bigr) \quad\longrightarrow\quad \bigl( \phi^a(\mathbf x), \pi_a(\mathbf x) \bigr).$$

The resulting equal-time algebra is the entry point to canonical QFT. It is not the whole theory. A regulator, Hamiltonian, state, representation, constraints, operator domains, and renormalized observables are still needed.

Phase-Space Formulation from QM to QFT also introduces Wigner functionals and phase-space integrals. Here the task is to derive field momenta and their brackets before choosing a quantum representation.

The Mechanical Recipe

For generalized coordinates $q^i$ with Lagrangian

$$L(q,\dot q,t),$$

the canonical momenta are

$$p_i := \frac{\partial L}{\partial\dot q^i}.$$

If the velocity–momentum relation is invertible, the Legendre transform gives

$$H(q,p,t) := \sum_i p_i\dot q^i-L.$$

The Poisson bracket

$$\{F,G\}_{\mathrm{PB}} := \sum_i \left( \frac{\partial F}{\partial q^i} \frac{\partial G}{\partial p_i} - \frac{\partial F}{\partial p_i} \frac{\partial G}{\partial q^i} \right)$$

encodes both the canonical structure and time evolution:

$$\dot F = \{F,H\}_{\mathrm{PB}} + \frac{\partial F}{\partial t}.$$

The detailed classical construction is developed in Hamiltonian Mechanics Review and Poisson Brackets. The important bridge is that a field theory repeats the construction with functions on space replacing finite coordinate lists.

Field Phase Space

Let the classical fields be $\phi^a(t,\mathbf x)$, with local Lagrangian density

$$\mathcal L \bigl( \phi^a, \partial_\mu\phi^a \bigr).$$

The action and Lagrangian are

$$\begin{aligned} S[\phi] &= \int dt\,L, \\ L &= \int d^d x\, \mathcal L. \end{aligned}$$

The canonical momentum field is defined pointwise:

$$\pi_a(t,\mathbf x) := \frac{\partial\mathcal L} {\partial(\partial_t\phi^a)}.$$

When the Legendre transform is nonsingular, the Hamiltonian density is

$$\mathcal H := \sum_a \pi_a\partial_t\phi^a - \mathcal L,$$

and

$$H[\phi,\pi] = \int d^d x\, \mathcal H.$$

The phase space is now a space of instantaneous field configurations:

$$\Gamma = \left\{ \bigl( \phi^a(\mathbf x), \pi_a(\mathbf x) \bigr) \right\}.$$

A point of $\Gamma$ is data on one spatial slice, not a complete spacetime history. Time evolution traces a trajectory through this infinite-dimensional phase space.

For functionals $F[\phi,\pi]$ and $G[\phi,\pi]$, the canonical field Poisson bracket is

$$\{F,G\}_{\mathrm{PB}} = \sum_a \int d^d x\, \left[ \frac{\delta F}{\delta\phi^a(\mathbf x)} \frac{\delta G}{\delta\pi_a(\mathbf x)} - \frac{\delta F}{\delta\pi_a(\mathbf x)} \frac{\delta G}{\delta\phi^a(\mathbf x)} \right].$$

The elementary brackets are

$$\begin{aligned} \{ \phi^a(\mathbf x), \pi_b(\mathbf y) \}_{\mathrm{PB}} &= \delta^a_b \delta^{(d)}(\mathbf x-\mathbf y), \\ \{ \phi^a(\mathbf x), \phi^b(\mathbf y) \}_{\mathrm{PB}} &= 0, \\ \{ \pi_a(\mathbf x), \pi_b(\mathbf y) \}_{\mathrm{PB}} &= 0. \end{aligned}$$

Hamilton’s field equations follow:

$$\begin{aligned} \partial_t\phi^a(\mathbf x) &= \frac{\delta H}{\delta\pi_a(\mathbf x)}, \\ \partial_t\pi_a(\mathbf x) &= - \frac{\delta H}{\delta\phi^a(\mathbf x)}. \end{aligned}$$

These formulas should be understood through a regulator, such as a spatial lattice or mode cutoff. On a finite spatial lattice, there are finitely many canonical pairs; the functional derivatives and Dirac delta emerge only in the continuum limit.

From Poisson Brackets to Equal-Time Operators

Canonical quantization promotes the elementary fields and momenta to operators and imposes, for bosonic variables,

$$\begin{aligned} \left[ \hat\phi^a(t,\mathbf x), \hat\pi_b(t,\mathbf y) \right] &= i\hbar \delta^a_b \delta^{(d)}(\mathbf x-\mathbf y), \\ \left[ \hat\phi^a(t,\mathbf x), \hat\phi^b(t,\mathbf y) \right] &= 0, \\ \left[ \hat\pi_a(t,\mathbf x), \hat\pi_b(t,\mathbf y) \right] &= 0. \end{aligned}$$

The word “equal-time” is essential. These relations specify the canonical algebra on one spatial slice. Unequal-time commutators follow from the Hamiltonian and equations of motion; they are not independent canonical postulates.

Because fields are operator-valued distributions, the smeared form is more fundamental. For test functions $f_a$ and $g^a$, define

$$\begin{aligned} \hat\phi(f) &= \sum_a \int d^d x\, f_a(\mathbf x) \hat\phi^a(\mathbf x), \\ \hat\pi(g) &= \sum_a \int d^d x\, g^a(\mathbf x) \hat\pi_a(\mathbf x). \end{aligned}$$

Then

$$\left[ \hat\phi(f), \hat\pi(g) \right] = i\hbar \sum_a \int d^d x\, f_a(\mathbf x)g^a(\mathbf x).$$

This version avoids pretending that $\hat\phi(\mathbf x)$ is an ordinary operator at a sharply specified point. Products such as $\hat\phi^2(\mathbf x)$ need a definition through regularization and renormalization.

The correspondence

$$\{F,G\}_{\mathrm{PB}} \quad\leadsto\quad \frac{1}{i\hbar} \left[ \hat F,\hat G \right]$$

is reliable as a guide for the elementary canonical variables. It cannot preserve the Poisson bracket of every classical observable exactly. Operator ordering already makes $q^2p^2$ ambiguous, and the Groenewold–Van Hove obstruction shows that no quantization map with all the desired bracket and product properties exists on the full polynomial algebra. Canonical quantization is a structured prescription, not a universal functor from every classical formula to one unique operator.

Worked Example: The Free Real Scalar Field

Use $c=1$, let $m$ denote inverse length (rest energy $\hbar m$), and impose periodic or sufficiently decaying spatial boundary conditions. Consider

$$\mathcal L = \frac12 (\partial_t\phi)^2 - \frac12 (\boldsymbol\nabla\phi)^2 - \frac12m^2\phi^2.$$

The conjugate momentum is

$$\pi = \frac{\partial\mathcal L} {\partial(\partial_t\phi)} = \partial_t\phi.$$

The Hamiltonian density is

$$\mathcal H = \frac12\pi^2 + \frac12(\boldsymbol\nabla\phi)^2 + \frac12m^2\phi^2,$$

so

$$H = \frac12 \int d^d x\, \left[ \pi^2 + (\boldsymbol\nabla\phi)^2 + m^2\phi^2 \right].$$

Hamilton’s equations give

$$\partial_t\phi = \pi,$$

and, after integrating the gradient variation by parts,

$$\partial_t\pi = \boldsymbol\nabla^2\phi - m^2\phi.$$

Combining them yields the Klein–Gordon equation

$$\left( \partial_t^2 - \boldsymbol\nabla^2 + m^2 \right) \phi = 0.$$

After quantization, impose

$$\left[ \hat\phi(t,\mathbf x), \hat\pi(t,\mathbf y) \right] = i\hbar \delta^{(d)}(\mathbf x-\mathbf y).$$

For operators without explicit Schrödinger-picture time dependence, the Heisenberg equation

$$\partial_t\hat O = \frac{1}{i\hbar} \left[ \hat O,\hat H \right]$$

then reproduces the operator Klein–Gordon equation. The algebra fixes kinematics, while the Hamiltonian fixes dynamics.

Functional Schrödinger Representation

The position representation of ordinary quantum mechanics has

$$\hat q\psi(q) = q\psi(q), \qquad \hat p\psi(q) = -i\hbar \frac{\partial\psi}{\partial q}.$$

Its field analogue acts on wavefunctionals $\Psi[\varphi]$:

$$\begin{aligned} \hat\phi(\mathbf x) \Psi[\varphi] &= \varphi(\mathbf x) \Psi[\varphi], \\ \hat\pi(\mathbf x) \Psi[\varphi] &= -i\hbar \frac{\delta}{\delta\varphi(\mathbf x)} \Psi[\varphi]. \end{aligned}$$

Formally, the scalar-field Schrödinger equation becomes

$$i\hbar\partial_t\Psi[\varphi,t] = \int d^d x\, \left[ - \frac{\hbar^2}{2} \frac{\delta^2}{\delta\varphi(\mathbf x)^2} + \frac12 (\boldsymbol\nabla\varphi)^2 + \frac12m^2\varphi^2 \right] \Psi[\varphi,t].$$

This is the formal flat-coordinate representation inherited from the lattice. There is no infinite-dimensional flat Lebesgue measure; a continuum Gaussian-measure representation can move terms between the momentum operator and the measure. The second functional derivative at one point is ultraviolet singular in the continuum. A lattice turns it into an ordinary sum of second derivatives and makes clear that the formal continuum Hamiltonian requires regularization. The wavefunctional perspective connects directly to vacuum preparation in From Euclidean Time to Euclidean QFT.

From Modes to Quanta

For a free field, spatial Fourier modes diagonalize the quadratic Hamiltonian. In natural units, a standard expansion is

$$\hat\phi(t,\mathbf x) = \int \frac{d^d k}{(2\pi)^d} \frac{1}{\sqrt{2\omega_{\mathbf k}}} \bigg[ \hat a_{\mathbf k} e^{-i\omega_{\mathbf k}t+i\mathbf k\cdot\mathbf x} + \hat a_{\mathbf k}^{\dagger} e^{i\omega_{\mathbf k}t-i\mathbf k\cdot\mathbf x} \bigg],$$

where

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

The equal-time field commutator is equivalent to

$$\left[ \hat a_{\mathbf k}, \hat a_{\mathbf q}^{\dagger} \right] = (2\pi)^d \delta^{(d)}(\mathbf k-\mathbf q),$$

with the other mode commutators vanishing. Restoring $\hbar$, the Hamiltonian in a finite box becomes a sum of oscillator Hamiltonians:

$$\hat H = \sum_{\mathbf k} \hbar\omega_{\mathbf k} \left( \hat a_{\mathbf k}^{\dagger} \hat a_{\mathbf k} + \frac12 \right).$$

The continuum zero-point sum diverges and requires interpretation or renormalization. Interactions couple the modes, so the oscillator decomposition does not make an interacting theory a set of independent particles.

The oscillator algebra and occupation-number interpretation have their canonical homes in Harmonic Oscillator to Fields and Bosonic Commutation Relations. Here they serve only to show that equal-time field brackets determine the mode normalization.

Fermionic Fields Require Graded Canonical Structure

The Dirac Lagrangian is first order in time derivatives. In natural units,

$$\mathcal L_D = \overline\psi \left( i\gamma^\mu\partial_\mu-m \right) \psi.$$

Its momentum conjugate to $\psi$ is proportional to $i\psi^\dagger$, while the momentum conjugate to $\psi^\dagger$ vanishes. The Legendre transform is therefore constrained rather than the nonsingular bosonic construction used above.

Quantization uses canonical anticommutation relations:

$$\begin{aligned} \left\{ \hat\psi_\alpha(t,\mathbf x), \hat\psi_\beta^\dagger(t,\mathbf y) \right\} &= \delta_{\alpha\beta} \delta^{(d)}(\mathbf x-\mathbf y), \\ \left\{ \hat\psi_\alpha(t,\mathbf x), \hat\psi_\beta(t,\mathbf y) \right\} &= 0. \end{aligned}$$

These relations are not obtained by blindly replacing every classical Poisson bracket with a commutator. A classical fermionic theory uses Grassmann-odd variables and a graded symplectic structure; its second-class constraints are handled with graded Dirac brackets before quantization. Relativistic spin–statistics then connects half-integer spin, locality, positivity, and the fermionic algebra.

The occupation-number consequences and mode-order signs are explained in Fermionic Anticommutation Relations. This page records only why the canonical starting point differs from the scalar case.

Constraints and Gauge Fields

The Legendre transform fails whenever the Hessian with respect to velocities is singular. Relations among coordinates and momenta then define a constraint surface

$$\chi_a(q,p) = 0$$

inside the original phase space.

For electromagnetism,

$$\mathcal L_{\mathrm{EM}} = - \frac14 F_{\mu\nu}F^{\mu\nu} - j_\mu A^\mu.$$

No time derivative of $A_0$ appears, so its conjugate momentum vanishes:

$$\pi^0 = 0.$$

This is a primary constraint. Requiring it to remain true under Hamiltonian evolution produces Gauss’s law:

$$\partial_i\pi^i - \rho = 0.$$

The field $A_0$ acts as a Lagrange multiplier rather than an independent propagating canonical coordinate. Gauge-related vector potentials label the same physical state, so quantizing all four components as unconstrained canonical pairs would overcount degrees of freedom.

Dirac’s constrained-Hamiltonian framework distinguishes:

• first-class constraints, whose brackets with all constraints vanish on the constraint surface; suitable combinations generate gauge transformations under the usual regularity assumptions;
• second-class constraints, whose mutual Poisson-bracket matrix is invertible.

For second-class constraints $\chi_a$, define

$$C_{ab} := \{\chi_a,\chi_b\}_{\mathrm{PB}}.$$

The Dirac bracket is

$$\{F,G\}_{D} = \{F,G\}_{\mathrm{PB}} - \{F,\chi_a\}_{\mathrm{PB}} \left(C^{-1}\right)^{ab} \{\chi_b,G\}_{\mathrm{PB}}.$$

It makes second-class constraints hold strongly and supplies the bracket to be quantized on the reduced phase space.

For first-class constraints, common strategies include:

• solve the constraints and quantize only gauge-invariant variables;
• quantize first and impose operator constraints such as

$$\widehat{\mathcal G}(\mathbf x) \lvert\Psi_{\mathrm{phys}}\rangle = 0;$$

• fix a gauge and include the associated determinant or ghost structure in a path-integral treatment;
• use BRST or related cohomological methods.

This is a preview. Constraint closure, anomalies, global gauge structure, boundary charges, and non-Abelian gauge fields require a full constrained-QFT treatment.

What Infinite Degrees of Freedom Change

For finitely many canonical pairs at fixed nonzero $\hbar$, the Stone–von Neumann theorem gives unitary equivalence of regular irreducible representations of the Weyl relations. That protection does not extend to systems with infinitely many degrees of freedom.

In QFT, the same abstract equal-time algebra can have unitarily inequivalent representations. Different vacua, thermal states, backgrounds, phases, or infinite-volume limits may therefore live in representations that cannot be connected by one unitary operator. A finite-volume spatial lattice temporarily restores finitely many degrees of freedom, but inequivalence can emerge as the regulator and volume limits are taken.

Other continuum cautions include:

• fields and momenta are operator-valued distributions;
• local products need composite-operator renormalization;
• the Hamiltonian and vacuum energy generally need renormalization;
• canonical transformations need not be unitarily implementable;
• the canonical algebra does not select a state or vacuum;
• interacting fields need not share the free-field Fock representation in a simple way.

Canonical commutators are therefore necessary kinematic data, not a complete nonperturbative definition of an interacting QFT.

Canonical Quantization Is Not Second Quantization

The terms describe related but distinct transitions:

Construction	Starting point	Main result
Canonical quantization	classical phase space $(q,p)$ or $(\phi,\pi)$	noncommuting operators and a quantum Hamiltonian
Second quantization	one-particle Hilbert space or mode basis	Fock space and creation/annihilation operators
Relativistic QFT	local fields plus relativistic, quantum, and renormalization structure	local operator algebra, states, correlations, and scattering data

A classical field can be canonically quantized, after which its normal modes acquire creation and annihilation operators. A nonrelativistic many-particle theory can instead be rewritten directly in second-quantized field notation. These paths often meet, but they are not definitions of one another.

Second Quantization: Bridge to QFT develops the Fock-space construction in detail.

Common Mistakes

• Treating a field configuration on one time slice as an entire spacetime history.
• Forgetting to derive the canonical momentum before imposing commutators.
• Writing equal-time delta functions without specifying spatial and Fourier conventions.
• Multiplying fields at the same point as if they were ordinary bounded operators.
• Assuming the Poisson-bracket-to-commutator rule uniquely quantizes every classical observable.
• Using commutators for fermionic canonical variables instead of the graded construction.
• Quantizing gauge potentials as unconstrained independent coordinates.
• Confusing canonical quantization with second quantization.
• Assuming the canonical algebra uniquely selects a QFT representation or vacuum.
• Removing the regulator before defining the Hamiltonian and local composite operators.

References

• P. A. M. Dirac, “Generalized Hamiltonian Dynamics,” Canadian Journal of Mathematics 2, 129–148, 1950, doi:10.4153/CJM-1950-012-1.
• P. A. M. Dirac, Lectures on Quantum Mechanics, Belfer Graduate School of Science, Yeshiva University, 1964.
• H. J. Groenewold, “On the Principles of Elementary Quantum Mechanics,” Physica 12, 405–460, 1946, doi:10.1016/S0031-8914(46)80059-4.
• M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, Princeton University Press, 1992.
• M. D. Schwartz, Quantum Field Theory and the Standard Model, Cambridge University Press, 2014.
• R. M. Wald, Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics, University of Chicago Press, 1994.
• S. Weinberg, The Quantum Theory of Fields, Volume I: Foundations, Cambridge University Press, 1995.

Exercises

1. Scalar Hamiltonian and field equation

Starting from

$$\mathcal L = \frac12\dot\phi^2 - \frac12(\boldsymbol\nabla\phi)^2 - \frac12m^2\phi^2,$$

derive $\pi$, $\mathcal H$, and both Hamilton field equations. Show that they reproduce the Klein–Gordon equation.

Solution

The momentum is

$$\pi = \frac{\partial\mathcal L}{\partial\dot\phi} = \dot\phi.$$

The Legendre transform gives

$$\begin{aligned} \mathcal H &= \pi\dot\phi-\mathcal L \\ &= \frac12\pi^2 + \frac12(\boldsymbol\nabla\phi)^2 + \frac12m^2\phi^2. \end{aligned}$$

The first Hamilton equation is

$$\dot\phi = \frac{\delta H}{\delta\pi} = \pi.$$

Varying the gradient term and integrating by parts gives

$$\frac{\delta H}{\delta\phi} = - \boldsymbol\nabla^2\phi + m^2\phi.$$

Therefore

$$\dot\pi = - \frac{\delta H}{\delta\phi} = \boldsymbol\nabla^2\phi - m^2\phi.$$

Using $\pi=\dot\phi$ yields

$$\left( \partial_t^2 - \boldsymbol\nabla^2 + m^2 \right) \phi = 0.$$

2. Lattice origin of the Dirac delta

On a spatial lattice of spacing $a$, let $\phi_{\mathbf n}$ be the field value and define the cell momentum by

$$p_{\mathbf n} = a^d\pi_{\mathbf n}.$$

Starting from $[\hat\phi_{\mathbf n},\hat p_{\mathbf m}]=i\hbar\delta_{\mathbf n\mathbf m}$, derive the lattice form of $[\hat\phi_{\mathbf n},\hat\pi_{\mathbf m}]$ and identify its continuum limit.

Solution

Since $\hat p_{\mathbf m}=a^d\hat\pi_{\mathbf m}$,

$$\left[ \hat\phi_{\mathbf n}, \hat\pi_{\mathbf m} \right] = \frac{i\hbar}{a^d} \delta_{\mathbf n\mathbf m}.$$

The lattice delta approximates the continuum distribution through

$$\frac{\delta_{\mathbf n\mathbf m}}{a^d} \longrightarrow \delta^{(d)}(\mathbf x-\mathbf y).$$

Thus

$$\left[ \hat\phi(\mathbf x), \hat\pi(\mathbf y) \right] = i\hbar \delta^{(d)}(\mathbf x-\mathbf y).$$

The factor $a^{-d}$ is essential; omitting it gives a continuum commutator with the wrong dimensions and normalization.

3. Smeared canonical algebra

For one scalar field, define

$$\hat\phi(f) = \int d^d x\, f(\mathbf x)\hat\phi(\mathbf x), \qquad \hat\pi(g) = \int d^d x\, g(\mathbf x)\hat\pi(\mathbf x).$$

Derive their commutator and explain why this form is mathematically preferable to the pointwise equation.

Solution

Insert the equal-time relation:

$$\begin{aligned} \left[ \hat\phi(f), \hat\pi(g) \right] &= \int d^d x\,d^d y\, f(\mathbf x)g(\mathbf y) \left[ \hat\phi(\mathbf x), \hat\pi(\mathbf y) \right] \\ &= i\hbar \int d^d x\,d^d y\, f(\mathbf x)g(\mathbf y) \delta^{(d)}(\mathbf x-\mathbf y) \\ &= i\hbar \int d^d x\, f(\mathbf x)g(\mathbf x). \end{aligned}$$

The smeared fields act on test functions and produce finite distributional pairings. This is the natural setting for operator-valued distributions; the unsmeared symbols are kernels rather than ordinary operators at individual points.

4. Recover the mode commutator

In natural units, insert the free-field mode expansions of $\hat\phi$ and $\hat\pi=\partial_t\hat\phi$ into the equal-time commutator. Show that

$$\left[ \hat a_{\mathbf k}, \hat a_{\mathbf q}^{\dagger} \right] = (2\pi)^d \delta^{(d)}(\mathbf k-\mathbf q)$$

reproduces $[\hat\phi(\mathbf x),\hat\pi(\mathbf y)]=i\delta^{(d)}(\mathbf x-\mathbf y)$.

Solution

At equal time, only the mixed $\hat a$–$\hat a^\dagger$ terms survive. Their two orderings give

$$\left[ \hat\phi(\mathbf x), \hat\pi(\mathbf y) \right] = \frac{i}{2} \int\frac{d^d k}{(2\pi)^d} \left[ e^{i\mathbf k\cdot(\mathbf x-\mathbf y)} + e^{-i\mathbf k\cdot(\mathbf x-\mathbf y)} \right].$$

Changing $\mathbf k\mapsto-\mathbf k$ in the second term makes the two integrals equal, so

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

The factor $1/\sqrt{2\omega_{\mathbf k}}$ in $\hat\phi$ and the factor $\omega_{\mathbf k}/\sqrt{2\omega_{\mathbf k}}$ from $\hat\pi$ are precisely what make the normalization work.

5. Electromagnetic primary and secondary constraints

Explain why the electromagnetic momentum $\pi^0$ vanishes. Why does preserving this condition under time evolution lead to Gauss’s law, and why does that obstruct treating $A_0$ as an ordinary dynamical coordinate?

Solution

The electromagnetic Lagrangian depends on spatial and temporal derivatives through $F_{\mu\nu}$. Because

$$F_{00}=0,$$

there is no $\partial_t A_0$ term. Hence

$$\pi^0 = \frac{\partial\mathcal L} {\partial(\partial_t A_0)} = 0.$$

In the Hamiltonian, $A_0$ multiplies the Gauss constraint. Requiring the primary constraint $\pi^0=0$ to be preserved gives

$$\dot\pi^0 = - \frac{\delta H}{\delta A_0} = 0,$$

which is equivalent to

$$\partial_i\pi^i-\rho = 0.$$

Thus $A_0$ enforces a constraint rather than carrying an independent canonical degree of freedom. Treating it as an unconstrained oscillator would include gauge-redundant and unphysical states.

6. A simple Dirac bracket

Take a four-dimensional phase space $(q_1,p_1,q_2,p_2)$ with second-class constraints

$$\chi_1=q_2, \qquad \chi_2=p_2.$$

Compute $C_{ab}=\{\chi_a,\chi_b\}_{\mathrm{PB}}$ and show that the Dirac bracket leaves $\{q_1,p_1\}_D=1$ while every Dirac bracket with $q_2$ or $p_2$ vanishes.

Solution

The constraint matrix and its inverse are

$$C = \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}, \qquad C^{-1} = \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}.$$

The variables $q_1$ and $p_1$ have zero Poisson brackets with both constraints, so the correction term in the Dirac bracket vanishes:

$$\{q_1,p_1\}_D = \{q_1,p_1\}_{\mathrm{PB}} = 1.$$

For any function $F$, the correction term cancels the ordinary bracket with either constrained variable:

$$\{F,q_2\}_D = 0, \qquad \{F,p_2\}_D = 0.$$

The Dirac bracket therefore removes the constrained canonical pair while preserving the physical pair $(q_1,p_1)$. This finite-dimensional example models the logic used before quantizing second-class field constraints.