advanced33 min studyLesson 654 in Physics

Charged scalar fields and antiparticles

Quantize a complex Klein–Gordon field with two oscillator families and distinguish positive energy, signed charge, and particle-number conventions.

Before you begin

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The question for today

What you’ll understand

Identify the independent canonical pairs of a complex scalar field.
Derive particle and antiparticle charge signs from a declared symmetry convention.
Separate conserved charge from total free-particle number.

A complex scalar field contains two real degrees of freedom and produces two distinguishable excitation families. Both families have positive energy and the same mass, but they carry opposite charges under a global phase symmetry. The negative-frequency coefficient becomes an antiparticle creation operator. That fact combines the field equation, canonical normalization, and charge generator; it is not a prescription to keep a negative physical energy. We first discuss the global free theory, before introducing any dynamical electromagnetic gauge field.

Complex conjugation is not a second copy of the same ladder

Write a complex scalar as two real fields divided by square root of two. The standard complex-field Lagrangian then becomes the sum of two standard real-field Lagrangians. There is no extra one-half multiplying the complex expression. Treat the field and its conjugate as independent variables in the variational calculation. Their canonical momenta are the time derivatives of the opposite conjugate fields, so the canonical pairing differs from a naive field-with-its-own-derivative assignment.

The mode expansion has one annihilator multiplying the positive-frequency phase and a different family's creator multiplying the negative-frequency phase. Hermitian conjugation interchanges which families are created and destroyed. Commutators within each family have the usual positive bosonic normalization, and cross-family commutators vanish. A real scalar is the neutral self-conjugate special case, but imposing Hermiticity on a genuinely charged field removes its independent charged degrees of freedom.

\mathcal L=\partial_\mu\Phi^{\dagger}\partial^\mu\Phi-m^2\Phi^{\dagger}\Phi,\quad \pi_\Phi=\dot\Phi^{\dagger},\quad \Phi(x)=\int\frac{d^3p}{(2\pi)^3\sqrt{2E_{\mathbf p}}}\left(a_{\mathbf p}e^{-ip\cdot x}+b_{\mathbf p}^{\dagger}e^{ip\cdot x}\right)

We retain ℏ = c = 1, η = diag(1,−1,−1,−1), p·x = Ept−p·x, and Ep = √(p²+m²) > 0. Φ has energy dimension one. Both [aₚ,a†q] and [bₚ,b†q] equal (2π)³δ³(p−q); cross brackets vanish.↔ Scroll wide equations horizontally.

The charge generator determines the sign

Choose a global transformation generated by the exponential of positive imaginary alpha times charge, with the field transforming by an exponential of negative imaginary q alpha. The Noether current then has a definite sign. The resulting normal-ordered charge counts a quanta with plus q and b quanta with minus q. Stating the phase convention prevents a harmless relabeling of particles from becoming an inconsistent algebraic sign change halfway through a calculation.

The field operator itself lowers the total charge by q: it either destroys a positive-charge particle or creates a negative-charge antiparticle. Its adjoint raises the charge. This distinguishes the transformation of a field from the charge of the particle created by its adjoint. A matrix element of a charged field connects different charge sectors, while a neutral bilinear can act within one sector. The term antiparticle refers to the opposite-charge excitation of the same relativistic field, not a negative-norm state.

j^\mu=iq\left(\Phi^{\dagger}\partial^\mu\Phi-(\partial^\mu\Phi^{\dagger})\Phi\right),\quad :Q:=q\int\frac{d^3p}{(2\pi)^3}(a_{\mathbf p}^{\dagger}a_{\mathbf p}-b_{\mathbf p}^{\dagger}b_{\mathbf p}),\quad [Q,\Phi]=-q\Phi

The convention is U(α) = exp(iαQ), UΦU⁻¹ = exp(−iqα)Φ. q and Q are dimensionless in these natural units, and αq is dimensionless. The current has energy dimension three. Charges are defined with the vacuum reference removed.↔ Scroll wide equations horizontally.

Positive energy and conserved number answer different questions

The free normal-ordered Hamiltonian adds the positive energies of particles and antiparticles. Charge subtracts their counts instead. Creating a neutral pair therefore costs positive energy while leaving charge unchanged. The free theory conserves each family number separately because its modes do not interact. This stronger conservation is not implied by global phase symmetry alone: an interacting phase-invariant process may change total number while preserving the difference.

Particle and antiparticle labels are clean for free modes and suitable asymptotic states. A time-dependent background can mix frequency notions, and an interacting finite-time field need not have an exact free occupation interpretation. Introducing electromagnetism further imposes Gauss constraints and makes charged observables and their localization more subtle. The global free construction is therefore a precise starting model, not a demonstration that every charged field operator is a local gauge-invariant measurable quantity.

:H:=\int\frac{d^3p}{(2\pi)^3}E_{\mathbf p}(a_{\mathbf p}^{\dagger}a_{\mathbf p}+b_{\mathbf p}^{\dagger}b_{\mathbf p}),\qquad Q=q(N_a-N_b),\quad N_{\mathrm{tot}}=N_a+N_b

H has energy units; Nₐ and Nᵦ count free quanta and are dimensionless. The displayed formulas are free Fock-space operators, understood through packets or regulated box modes. Only charge conservation is guaranteed by the stated global U(1) symmetry in a general invariant interaction.↔ Scroll wide equations horizontally.

Work it through

Add energy without changing charge

A free charged scalar state has two a quanta and three b quanta, each with mode energy 5 in natural units. Find its normal-ordered energy and charge. Then add one particle–antiparticle pair at that energy.

Both species contribute positively, giving $H=(2+3)(5)=25$ .
The charge is $Q=q(2-3)=-q$ . The excess antiparticle determines the sign.
Adding a pair changes the counts to three and four. Energy increases by 10 to 35 while Q remains −q, because the added charges cancel.

The result

The original state has energy 25 and charge −q. After adding a neutral pair it has energy 35 with the same charge.

The living notebook

Explore the model

Interactive lab

Interactive lab · probability

A signed count differs from an unsigned count

Reset and set BOTH Success probability p to 0.4 and Trials n to 10; predict mean successes 4. Then Reset and set BOTH p to 0.6 and n to 10; predict mean 6. If successes and failures receive charges +q and −q, their inferred mean net charges are −2q and +2q at fixed total count. This exact binomial accounting does not model a scalar field, charge conservation dynamics, or pair creation.

Exact probabilityHighlighted outcomeMean np

Mean · np: 5
Variance · np(1−p): 2.5
Total probability: 1

Success probability p0.5Trials n10

t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA total count adds both species, whereas a signed charge weights them oppositely.

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Read the checkpoint explanation

The a term destroys charge +q and the b† term creates charge −q. Both lower the total charge by q, giving [Q,Φ] = −qΦ.

Your turn

Try it yourself

A free scalar state has total particle-plus-antiparticle number six and charge +2q. Find both species counts. If all particles are in rest modes of mass 3, find its normal-ordered energy.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Add and subtract the equations for total count and charge difference.

Compare with the solution

The equations are $N_a+N_b=6$ and $N_a-N_b=2$ , so Nₐ = 4 and Nᵦ = 2. Each rest excitation has positive energy m = 3, giving H = 18. Charge alone would not determine this energy because neutral pairs can be added at fixed charge. The stated total count and rest-mode condition supply the extra information.

Keep the idea

Take it with you

A complex scalar requires independent particle and antiparticle ladder families.
The chosen symmetry phase fixes the relation between field transformation and particle charge.
Both families carry positive energy, while their signed number difference gives the conserved charge.

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