Yes — but with an important distinction:$$\boxed{\text{The field equations are often linear for free fields, but interacting QFT is not linear.}}$$The field equations are often linear for free fields, but interacting QFT is not linear.​

Let’s do a few concrete Weyl-vector examples.

1. Expand $p_\mu\sigma^\mu$pμ​σμ

Use$$\sigma^\mu=(I,\sigma_x,\sigma_y,\sigma_z)$$σμ=(I,σx​,σy​,σz​)

and, with metric $(+,-,-,-)$(+,−,−,−),$$p_\mu=(E,-p_x,-p_y,-p_z)$$pμ​=(E,−px​,−py​,−pz​)

So$$p_\mu\sigma^\mu = E I – p_x\sigma_x – p_y\sigma_y – p_z\sigma_z$$pμ​σμ=EI−px​σx​−py​σy​−pz​σz​

Example: particle moving in the $z$z-direction:$$p^\mu=(E,0,0,p)$$pμ=(E,0,0,p)

Then$$p_\mu\sigma^\mu = E I-p\sigma_z$$pμ​σμ=EI−pσz​ $$= \begin{pmatrix} E-p & 0\\ 0 & E+p \end{pmatrix}$$=(E−p0​0E+p​)

Similarly,$$p_\mu\bar{\sigma}^\mu = E I+p\sigma_z = \begin{pmatrix} E+p & 0\\ 0 & E-p \end{pmatrix}$$pμ​σˉμ=EI+pσz​=(E+p0​0E−p​)

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2. Massless right-handed Weyl equation

For a massless right-handed spinor,$$p_\mu\sigma^\mu \psi_R=0$$pμ​σμψR​=0

Using the $z$z-direction result:$$\begin{pmatrix} E-p & 0\\ 0 & E+p \end{pmatrix} \begin{pmatrix} a\\ b \end{pmatrix} =0$$(E−p0​0E+p​)(ab​)=0

For a massless particle,$$E=p$$E=p

so$$\begin{pmatrix} 0 & 0\\ 0 & 2E \end{pmatrix} \begin{pmatrix} a\\ b \end{pmatrix} =0$$(00​02E​)(ab​)=0

This gives$$2E b=0$$2Eb=0

so$$b=0$$b=0

Therefore$$\boxed{ \psi_R = \begin{pmatrix} 1\\ 0 \end{pmatrix} }$$ψR​=(10​)​

up to normalization.

This is spin-up along the direction of motion.

So a right-handed massless Weyl spinor has positive helicity:$$\boxed{ h=+\frac12}$$h=+21​​

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3. Massless left-handed Weyl equation

For left-handed spinors,$$p_\mu\bar{\sigma}^\mu \psi_L=0$$pμ​σˉμψL​=0

Using$$p_\mu\bar{\sigma}^\mu = \begin{pmatrix} E+p & 0\\ 0 & E-p \end{pmatrix}$$pμ​σˉμ=(E+p0​0E−p​)

and $E=p$E=p,$$\begin{pmatrix} 2E & 0\\ 0 & 0 \end{pmatrix} \begin{pmatrix} a\\ b \end{pmatrix} =0$$(2E0​00​)(ab​)=0

This gives$$a=0$$a=0

so$$\boxed{ \psi_L = \begin{pmatrix} 0\\ 1 \end{pmatrix} }$$ψL​=(01​)​

This is spin-down along the direction of motion.

So a left-handed massless Weyl spinor has negative helicity:$$\boxed{ h=-\frac12}$$h=−21​​

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4. Massive Dirac case: left and right are coupled

The chiral Dirac equations are$$p_\mu\sigma^\mu\psi_R=m\psi_L$$pμ​σμψR​=mψL​ $$p_\mu\bar{\sigma}^\mu\psi_L=m\psi_R$$pμ​σˉμψL​=mψR​

Again take motion along $z$z:$$p_\mu\sigma^\mu = \begin{pmatrix} E-p&0\\ 0&E+p \end{pmatrix}$$pμ​σμ=(E−p0​0E+p​) $$p_\mu\bar{\sigma}^\mu = \begin{pmatrix} E+p&0\\ 0&E-p \end{pmatrix}$$pμ​σˉμ=(E+p0​0E−p​)

Suppose spin-up, so use$$\psi_R= \begin{pmatrix} A\\ 0 \end{pmatrix}, \qquad \psi_L= \begin{pmatrix} B\\ 0 \end{pmatrix}$$ψR​=(A0​),ψL​=(B0​)

Then the first equation gives$$(E-p)A=mB$$(E−p)A=mB

The second gives$$(E+p)B=mA$$(E+p)B=mA

From the second,$$B=\frac{m}{E+p}A$$B=E+pm​A

Using$$E^2-p^2=m^2$$E2−p2=m2

we also have$$\frac{m}{E+p} = \frac{E-p}{m}$$E+pm​=mE−p​

So$$\boxed{ \psi_L= \frac{m}{E+p}\psi_R }$$ψL​=E+pm​ψR​​

For very high energy,$$E\gg m$$E≫m

so$$\frac{m}{E+p}\approx \frac{m}{2E}\ll 1$$E+pm​≈2Em​≪1

Therefore a high-energy spin-up massive fermion is mostly right-handed, with a small left-handed component.

That is why chirality and helicity become nearly the same at high energy.

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Does This Mean QFT Is Linear Like QM?

Partly yes, but mostly no.

Free QFT is linear

The free Klein–Gordon equation is linear:$$(\Box+m^2)\phi=0$$(□+m2)ϕ=0

The free Dirac equation is linear:$$(i\gamma^\mu\partial_\mu-m)\psi=0$$(iγμ∂μ​−m)ψ=0

The free Weyl equation is linear:$$i\sigma^\mu\partial_\mu\psi_R=0$$iσμ∂μ​ψR​=0

So for free particles, QFT resembles ordinary quantum mechanics: superpositions work cleanly.

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But interacting QFT is generally nonlinear

Once interactions are included, the equations are no longer simple linear wave equations.

Example scalar interaction:$$\mathcal L = \frac12(\partial\phi)^2 – \frac12m^2\phi^2 – \frac{\lambda}{4!}\phi^4$$L=21​(∂ϕ)2−21​m2ϕ2−4!λ​ϕ4

The equation of motion becomes$$(\Box+m^2)\phi + \frac{\lambda}{3!}\phi^3 =0$$(□+m2)ϕ+3!λ​ϕ3=0

That $\phi^3$ϕ3 term makes it nonlinear.

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QED is also interacting

For the electron field interacting with electromagnetism:$$(i\gamma^\mu D_\mu-m)\psi=0$$(iγμDμ​−m)ψ=0

where$$D_\mu=\partial_\mu+ieA_\mu$$Dμ​=∂μ​+ieAμ​

So the electron field couples to the photon field.

The Maxwell equation also gets a source term:$$\partial_\mu F^{\mu\nu}=e\bar{\psi}\gamma^\nu\psi$$∂μ​Fμν=eψˉ​γνψ

That source contains products of fields.

So the full coupled theory is nonlinear.

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The clean answer

$$\boxed{ \text{Quantum states still evolve linearly in Hilbert space.} }$$Quantum states still evolve linearly in Hilbert space.​

But:$$\boxed{ \text{the field dynamics of interacting QFT is not linear in the fields.} }$$the field dynamics of interacting QFT is not linear in the fields.​

So QFT keeps the linear superposition principle of quantum mechanics, but interactions make the field equations and scattering structure much richer.