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​)

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​​

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​​

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.

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.

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.

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.

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.

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For a free Schrödinger particle, start with the time-dependent Schrödinger equation:$$i\hbar\frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2}$$iℏ∂t∂ψ​=−2mℏ2​∂x2∂2ψ​

Assume a plane-wave solution:$$\psi(x,t) = A e^{i(kx-\omega t)}$$ψ(x,t)=Aei(kx−ωt)

Step 1: Compute the derivatives

Time derivative:$$\frac{\partial \psi}{\partial t} = -i\omega \psi$$∂t∂ψ​=−iωψ

Therefore$$i\hbar\frac{\partial \psi}{\partial t} = \hbar\omega \psi$$iℏ∂t∂ψ​=ℏωψ

Spatial second derivative:$$\frac{\partial^2 \psi}{\partial x^2} = -k^2 \psi$$∂x2∂2ψ​=−k2ψ

Therefore$$-\frac{\hbar^2}{2m} \frac{\partial^2\psi}{\partial x^2} = \frac{\hbar^2k^2}{2m}\psi$$−2mℏ2​∂x2∂2ψ​=2mℏ2k2​ψ

Substituting into Schrödinger’s equation gives$$\hbar\omega\psi = \frac{\hbar^2k^2}{2m}\psi$$ℏωψ=2mℏ2k2​ψ

Cancelling $\psi$ψ,$$\boxed{ \omega=\frac{\hbar k^2}{2m} }$$ω=2mℏk2​​

This is the Schrödinger dispersion relation.

Visualizing the dispersion relation

The relation is quadratic in $k$k:$$\omega \propto k^2$$ω∝k2

$\omega=\frac{\hbar k^2}{2m}$ω=2mℏk2​

Unlike light waves, where$$\omega = ck$$ω=ck

the Schrödinger particle’s frequency grows as the square of the wavenumber.

Connection to momentum and energy

Using de Broglie’s relations:$$p=\hbar k$$p=ℏk $$E=\hbar\omega$$E=ℏω

Substituting into the dispersion relation:$$E = \hbar\left(\frac{\hbar k^2}{2m}\right) = \frac{\hbar^2k^2}{2m}$$E=ℏ(2mℏk2​)=2mℏ2k2​

Since $p=\hbar k$p=ℏk,$$\boxed{ E=\frac{p^2}{2m} }$$E=2mp2​​

which is exactly the classical nonrelativistic kinetic energy.

Thus the Schrödinger dispersion relation is simply the wave version of Newtonian mechanics.

Phase velocity

The phase velocity is$$v_p=\frac{\omega}{k}$$vp​=kω​

Substituting the dispersion relation:$$v_p = \frac{\hbar k}{2m} = \frac{p}{2m}$$vp​=2mℏk​=2mp​

Since$$v=\frac{p}{m}$$v=mp​

we obtain$$\boxed{ v_p=\frac{v}{2} }$$vp​=2v​​

The phase of the wave moves at half the particle velocity.

Group velocity

A particle is represented by a wave packet, not a single plane wave.

The packet moves at the group velocity:$$v_g = \frac{d\omega}{dk}$$vg​=dkdω​

Differentiating,$$v_g = \frac{\hbar k}{m}$$vg​=mℏk​

Using $p=\hbar k$p=ℏk,$$v_g = \frac{p}{m}$$vg​=mp​

Therefore$$\boxed{ v_g=v }$$vg​=v​

The group velocity equals the particle’s classical velocity.

Why wave packets spread

Because$$\omega \propto k^2$$ω∝k2

different Fourier components travel at different group velocities:$$v_g=\frac{\hbar k}{m}$$vg​=mℏk​

Large-$k$k components move faster than small-$k$k components.

As time passes, the packet spreads out.

This is called dispersion.

For light in vacuum,$$\omega=ck$$ω=ck

and$$\frac{d\omega}{dk}=c$$dkdω​=c

for every $k$k, so no spreading occurs.

For Schrödinger waves,$$\frac{d^2\omega}{dk^2} = \frac{\hbar}{m} \neq 0$$dk2d2ω​=mℏ​=0

and the packet inevitably disperses.

Why Lorentz invariance fails

The Schrödinger equation assumes$$E=\frac{p^2}{2m}$$E=2mp2​

which leads directly to$$\omega=\frac{\hbar k^2}{2m}.$$ω=2mℏk2​.

A Lorentz transformation mixes energy and momentum:$$E’=\gamma(E-vp)$$E′=γ(E−vp) $$p’=\gamma\left(p-\frac{vE}{c^2}\right)$$p′=γ(p−c2vE​)

and the transformed quantities no longer satisfy$$E’=\frac{p’^2}{2m}.$$E′=2mp′2​.

Thus the Schrödinger dispersion relation is preserved only under Galilean transformations, not Lorentz transformations.

That is why relativistic quantum theory replaces the Schrödinger relation$$E=\frac{p^2}{2m}$$E=2mp2​

with$$E^2=p^2c^2+m^2c^4,$$E2=p2c2+m2c4,

leading to the relativistic wave equations such as the Klein-Gordon and Dirac equations.

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Start with the free-particle Schrödinger plane wave:$$\psi(x,t)=A e^{i(kx-\omega t)}$$ψ(x,t)=Aei(kx−ωt)

Using de Broglie relations,$$p=\hbar k,\qquad E=\hbar \omega$$p=ℏk,E=ℏω

so$$\psi(x,t)=A e^{\frac{i}{\hbar}(px-Et)}$$ψ(x,t)=Aeℏi​(px−Et)

For the nonrelativistic Schrödinger equation,$$E=\frac{p^2}{2m}$$E=2mp2​

so$$\omega=\frac{\hbar k^2}{2m}$$ω=2mℏk2​

Now apply a Lorentz transformation along the $x$x-axis:$$x=\gamma(x’+vt’)$$x=γ(x′+vt′)$$t=\gamma\left(t’+\frac{v x’}{c^2}\right)$$t=γ(t′+c2vx′​)

where$$\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$$γ=1−v2/c2​1​

Substitute these into the phase:$$px-Et = p\gamma(x’+vt’) – E\gamma\left(t’+\frac{vx’}{c^2}\right)$$px−Et=pγ(x′+vt′)−Eγ(t′+c2vx′​)

Collect terms in $x’$x′ and $t’$t′:$$px-Et = \gamma\left(p-\frac{vE}{c^2}\right)x’ + \gamma(pv-E)t’$$px−Et=γ(p−c2vE​)x′+γ(pv−E)t′

Rewrite it as$$px-Et = p’x’-E’t’$$px−Et=p′x′−E′t′

so we identify$$\boxed{ p’=\gamma\left(p-\frac{vE}{c^2}\right) }$$p′=γ(p−c2vE​)​

and$$\boxed{ E’=\gamma(E-vp) }$$E′=γ(E−vp)​

Therefore the transformed wave is$$\boxed{ \psi'(x’,t’) = A e^{\frac{i}{\hbar}(p’x’-E’t’)} }$$ψ′(x′,t′)=Aeℏi​(p′x′−E′t′)​

or$$\boxed{ \psi'(x’,t’) = A e^{i(k’x’-\omega’t’)} }$$ψ′(x′,t′)=Aei(k′x′−ω′t′)​

with$$\boxed{ k’=\gamma\left(k-\frac{v\omega}{c^2}\right) }$$k′=γ(k−c2vω​)​

and$$\boxed{ \omega’=\gamma(\omega-vk) }$$ω′=γ(ω−vk)​

So the phase transforms nicely:$$kx-\omega t = k’x’-\omega’t’$$kx−ωt=k′x′−ω′t′

However, here is the important point.

For the Schrödinger wave,$$\omega=\frac{\hbar k^2}{2m}$$ω=2mℏk2​

But after Lorentz transformation,$$\omega’=\gamma(\omega-vk)$$ω′=γ(ω−vk)$$k’=\gamma\left(k-\frac{v\omega}{c^2}\right)$$k′=γ(k−c2vω​)

In general,$$\boxed{ \omega’ \neq \frac{\hbar k’^2}{2m} }$$ω′=2mℏk′2​​

So the transformed wave is not generally another valid Schrödinger plane wave with the same nonrelativistic dispersion relation.

That is the core result:$$\boxed{ \text{The phase } px-Et \text{ can be Lorentz transformed, but the Schrödinger equation itself is not Lorentz invariant.} }$$The phase px−Et can be Lorentz transformed, but the Schro¨dinger equation itself is not Lorentz invariant.​

For Lorentz invariance, the energy relation must be relativistic:$$\boxed{ E^2=p^2c^2+m^2c^4 }$$E2=p2c2+m2c4​

which leads to the Klein-Gordon equation for scalar particles or the Dirac equation for spin-$\frac12$21​ particles.

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showing that the Klein–Gordon scalar field transforms consistently under Lorentz transformations and that the theory is Lorentz invariant.

The key idea is:$$\boxed{ \text{Physics must look identical in all inertial frames.} }$$Physics must look identical in all inertial frames.​

For scalar fields, this turns out to be beautifully simple.

1. Start with the Klein–Gordon Equation

The scalar field satisfies:$$(\partial_\mu\partial^\mu + m^2)\phi(x)=0$$(∂μ​∂μ+m2)ϕ(x)=0

or equivalently$$(\Box + m^2)\phi(x)=0$$(□+m2)ϕ(x)=0

where$$\Box \equiv \partial_\mu\partial^\mu$$□≡∂μ​∂μ

is the d’Alembertian operator.

Using metric signature $(+,-,-,-)$(+,−,−,−),$$\Box = \frac{\partial^2}{\partial t^2} – \nabla^2$$□=∂t2∂2​−∇2

2. Lorentz Transformations

A Lorentz transformation changes coordinates:$$x^\mu \rightarrow x’^\mu$$xμ→x′μ

with$$x’^\mu = \Lambda^\mu_{\ \nu}x^\nu$$x′μ=Λ νμ​xν

The matrix $\Lambda$Λ satisfies$$\Lambda^T \eta \Lambda = \eta$$ΛTηΛ=η

which preserves the spacetime interval:$$x_\mu x^\mu = x’_\mu x’^\mu$$xμ​xμ=xμ′​x′μ

3. What Is a Scalar Field?

A scalar field is defined by:$$\boxed{ \phi'(x’)=\phi(x) }$$ϕ′(x′)=ϕ(x)​

This means:

• the numerical value of the field is unchanged
• only the coordinates labeling spacetime points change

This is the defining property of a Lorentz scalar.

4. Transforming Derivatives

Now derive how derivatives transform.

Using the chain rule:$$\frac{\partial}{\partial x’^\mu} = \frac{\partial x^\nu}{\partial x’^\mu} \frac{\partial}{\partial x^\nu}$$∂x′μ∂​=∂x′μ∂xν​∂xν∂​

Since$$x’^\mu = \Lambda^\mu_{\ \nu}x^\nu$$x′μ=Λ νμ​xν

the inverse transformation is$$x^\nu = (\Lambda^{-1})^\nu_{\ \mu}x’^\mu$$xν=(Λ−1) μν​x′μ

Therefore:$$\partial’_\mu = (\Lambda^{-1})^\nu_{\ \mu}\partial_\nu$$∂μ′​=(Λ−1) μν​∂ν​

or equivalently$$\partial’^\mu = \Lambda^\mu_{\ \nu}\partial^\nu$$∂′μ=Λ νμ​∂ν

So derivatives transform like four-vectors.

5. Transforming the d’Alembertian

Now examine$$\Box’ = \partial’_\mu\partial’^\mu$$□′=∂μ′​∂′μ

Substitute the transformed derivatives:$$\Box’ = (\Lambda^{-1})^\nu_{\ \mu}\partial_\nu \Lambda^\mu_{\ \rho}\partial^\rho$$□′=(Λ−1) μν​∂ν​Λ ρμ​∂ρ

Using$$(\Lambda^{-1})^\nu_{\ \mu}\Lambda^\mu_{\ \rho} = \delta^\nu_{\rho}$$(Λ−1) μν​Λ ρμ​=δρν​

we get$$\Box’ = \partial_\nu\partial^\nu = \Box$$□′=∂ν​∂ν=□

Thus:$$\boxed{ \Box \text{ is Lorentz invariant} }$$□ is Lorentz invariant​

This is the central result.

6. Lorentz Invariance of the KG Equation

Now apply this to the field equation:$$(\Box + m^2)\phi(x)=0$$(□+m2)ϕ(x)=0

Under Lorentz transformation:$$(\Box’ + m^2)\phi'(x’) = (\Box + m^2)\phi(x)$$(□′+m2)ϕ′(x′)=(□+m2)ϕ(x)

Since the RHS is zero,$$(\Box’ + m^2)\phi'(x’)=0$$(□′+m2)ϕ′(x′)=0

Therefore the Klein–Gordon equation has exactly the same form in every inertial frame.

So:$$\boxed{ \text{The KG equation is Lorentz invariant} }$$The KG equation is Lorentz invariant​

7. Lorentz Invariance of the Action

The action is$$S = \int d^4x\, \mathcal L$$S=∫d4xL

with$$\mathcal L = \frac12\partial_\mu\phi\partial^\mu\phi – \frac12m^2\phi^2$$L=21​∂μ​ϕ∂μϕ−21​m2ϕ2

Now check each term.

Kinetic Term

$$\partial_\mu\phi\partial^\mu\phi$$∂μ​ϕ∂μϕ

is a Lorentz scalar because it contracts two four-vectors.

Like:$$A_\mu A^\mu$$Aμ​Aμ

Mass Term

$$\phi^2$$ϕ2

is also scalar since $\phi$ϕ itself is scalar.

Volume Element

Lorentz transformations preserve spacetime volume:$$d^4x’ = d^4x$$d4x′=d4x

for proper Lorentz transformations.

Therefore:$$S’ = S$$S′=S

Thus the action is Lorentz invariant.

8. Physical Interpretation

The scalar field has:

• no direction in spacetime
• no spin index
• no vector structure

It behaves like temperature distributed through spacetime.

All observers agree on the field value at the same spacetime event.

9. Contrast with Spinors and Vectors

Scalar:$$\phi'(x’)=\phi(x)$$ϕ′(x′)=ϕ(x)

Vector:$$A’^\mu(x’) = \Lambda^\mu_{\ \nu}A^\nu(x)$$A′μ(x′)=Λ νμ​Aν(x)

Spinor:$$\psi'(x’) = S(\Lambda)\psi(x)$$ψ′(x′)=S(Λ)ψ(x)

where $S(\Lambda)$S(Λ) is the spinor representation of the Lorentz group.

Spinors transform much more subtly.

10. The Deep Idea

The entire structure of relativistic QFT comes from demanding:$$\boxed{ \text{the action be Lorentz invariant} }$$the action be Lorentz invariant​

That requirement strongly constrains:

• allowed fields
• allowed interactions
• allowed dynamics

For the Klein–Gordon field, Lorentz invariance emerges because:$$\boxed{ \partial_\mu\partial^\mu }$$∂μ​∂μ​

is a spacetime scalar operator.

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unless you are in a special approximation. The relativistic treatment changes the framework.

The main issue is this:$$\boxed{ \text{Relativity treats space and time together, but ordinary Hamiltonians single out time.} }$$Relativity treats space and time together, but ordinary Hamiltonians single out time.​

So the more natural object is the action or the field theory Lagrangian, not a nonrelativistic two-particle Hamiltonian.

For two relativistic particles/fields, one writes something like$$S = \int d^4x \, \mathcal L_A + \int d^4x \, \mathcal L_B + \int d^4x \, \mathcal L_{\text{int}}$$S=∫d4xLA​+∫d4xLB​+∫d4xLint​

or, schematically,$$\mathcal L = \mathcal L_A + \mathcal L_B + \mathcal L_{\text{int}}$$L=LA​+LB​+Lint​

For two scalar fields,$$\mathcal L_A = \frac12 \partial_\mu \phi_A \partial^\mu \phi_A – \frac12 m_A^2\phi_A^2$$LA​=21​∂μ​ϕA​∂μϕA​−21​mA2​ϕA2​ $$\mathcal L_B = \frac12 \partial_\mu \phi_B \partial^\mu \phi_B – \frac12 m_B^2\phi_B^2$$LB​=21​∂μ​ϕB​∂μϕB​−21​mB2​ϕB2​

and an interaction might be$$\mathcal L_{\text{int}} = -g \phi_A^2 \phi_B^2$$Lint​=−gϕA2​ϕB2​

Then the relativistic dynamics comes from$$\delta S = 0$$δS=0

rather than from a simple two-body Schrödinger Hamiltonian.

For quantum theory, the state lives in a tensor product of field Hilbert spaces:$$\mathcal H = \mathcal H_A \otimes \mathcal H_B$$H=HA​⊗HB​

but the particles are excitations of fields, not little objects with fixed wavefunctions.

A bipartite relativistic state might look like$$|\Psi\rangle = \sum_{p,q} C(p,q) \,a_A^\dagger(p) a_B^\dagger(q) |0\rangle$$∣Ψ⟩=p,q∑​C(p,q)aA†​(p)aB†​(q)∣0⟩

where $a_A^\dagger(p)$aA†​(p) creates particle $A$A with four-momentum $p$p, and $a_B^\dagger(q)$aB†​(q) creates particle $B$B with four-momentum $q$q.

The relativistic energy is$$E_{\mathbf p} = \sqrt{\mathbf p^2 + m^2}$$Ep​=p2+m2​

so the free Hamiltonian is roughly$$H_0 = \int \frac{d^3p}{(2\pi)^3} E_{\mathbf p} \,a^\dagger_{\mathbf p}a_{\mathbf p}$$H0​=∫(2π)3d3p​Ep​ap†​ap​

For two species,$$H_0 = \int \frac{d^3p}{(2\pi)^3} E^A_{\mathbf p} a_A^\dagger(\mathbf p)a_A(\mathbf p) + \int \frac{d^3q}{(2\pi)^3} E^B_{\mathbf q} a_B^\dagger(\mathbf q)a_B(\mathbf q)$$H0​=∫(2π)3d3p​EpA​aA†​(p)aA​(p)+∫(2π)3d3q​EqB​aB†​(q)aB​(q)

Then add an interaction Hamiltonian:$$H = H_A + H_B + H_{\text{int}}$$H=HA​+HB​+Hint​

But now $H_{\text{int}}$Hint​ comes from a Lorentz-invariant field interaction.

So the relativistic version of the bipartite Hamiltonian idea is:$$\boxed{ H_A \otimes I + I \otimes H_B + H_{\text{int}} }$$HA​⊗I+I⊗HB​+Hint​​

but interpreted in QFT, where $H_A$HA​, $H_B$HB​, and $H_{\text{int}}$Hint​ act on field Fock spaces.

The clean hierarchy is:$$\boxed{ \text{Nonrelativistic QM: two-particle Hamiltonian} }$$Nonrelativistic QM: two-particle Hamiltonian​ $$\boxed{ \text{Relativistic QM: constrained and delicate} }$$Relativistic QM: constrained and delicate​ $$\boxed{ \text{Relativistic QFT: fields, Fock space, Lorentz-invariant action} }$$Relativistic QFT: fields, Fock space, Lorentz-invariant action​

So the safest answer is:$$\boxed{ \text{Treat the bipartite system as two interacting quantum fields.} }$$Treat the bipartite system as two interacting quantum fields.​

Not as two particles with an ordinary potential $V(\mathbf x_A-\mathbf x_B)$V(xA​−xB​), because instantaneous potentials generally conflict with relativity.

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A spinor field is a field describing spin-$\frac12$21​ particles:

• electrons
• quarks
• neutrinos

The Dirac field is the standard example:$$\psi(x)$$ψ(x)

Unlike scalar or vector fields, spinors transform differently under rotations and Lorentz transformations.

2. Classical Macroscopic Fields Exist for Bosons

Examples:

• electromagnetic field $A_\mu(x)$Aμ​(x)
• classical light waves
• laser beams
• superfluids
• Bose condensates

Why?

Because bosons can pile into the same quantum state.

The occupation number can become enormous:$$N \gg 1$$N≫1

This allows the quantum field operator to behave approximately like a classical field.

For example:$$\hat A_\mu(x) \rightarrow A_\mu^{\text{classical}}(x)$$A^μ​(x)→Aμclassical​(x)

3. Fermions Cannot Do This

Spinor fields describe fermions.

Fermions obey:$$\boxed{ \text{Pauli exclusion principle} }$$Pauli exclusion principle​

No two identical fermions can occupy the same quantum state.

Mathematically:$$\{a_p,a_q^\dagger\}=\delta_{pq}$${ap​,aq†​}=δpq​

instead of bosonic commutators:$$[a_p,a_q^\dagger]=\delta_{pq}$$[ap​,aq†​]=δpq​

This changes everything.

4. Grassmann Nature of Fermionic Fields

In QFT, fermionic fields are not ordinary numbers.

They are Grassmann-valued:$$\psi_1\psi_2=-\psi_2\psi_1$$ψ1​ψ2​=−ψ2​ψ1​

In particular:$$\boxed{ \psi^2=0 }$$ψ2=0​

This nilpotent property means you cannot build arbitrarily large coherent amplitudes from a fermion field.

That prevents a classical macroscopic field interpretation.

5. Why Electromagnetic Waves Exist but Electron Waves Do Not

For photons:

many photons can occupy one mode:$$|N\rangle$$∣N⟩

with arbitrarily large $N$N.

This produces classical EM waves.

But for electrons:

occupation is only:$$0 \text{ or } 1$$0 or 1

per quantum state.

So there is no analog of a huge coherent classical electron field.

6. Macroscopic Matter Is NOT a Macroscopic Spinor Field

A metal contains enormous numbers of electrons.

But:$$\boxed{ \text{the electron field itself is still quantum} }$$the electron field itself is still quantum​

The matter behaves macroscopically because:

• enormous statistical averages emerge
• collective variables become classical
• densities and currents become classical observables

NOT because the spinor field becomes classical.

7. Contrast with Bose–Einstein Condensates

Bosonic field:$$\phi(x)$$ϕ(x)

can acquire a macroscopic expectation value:$$\langle \phi(x)\rangle \neq 0$$⟨ϕ(x)⟩=0

This is a true classical order parameter.

For fermions:$$\langle \psi(x)\rangle =0$$⟨ψ(x)⟩=0

in ordinary macroscopic systems.

Instead, fermions form composite bosons:

Examples:

• Cooper pairs
• helium-4 atoms
• mesons

Then THOSE bosonic composites can condense macroscopically.

8. Superconductivity Is the Key Example

Electrons themselves cannot form a classical spinor field.

But paired electrons:$$(e^- e^-)$$(e−e−)

form Cooper pairs with integer spin.

Those pairs behave bosonically.

Then you get a macroscopic condensate:$$\langle \Psi_{\text{Cooper pair}}\rangle \neq 0$$⟨ΨCooper pair​⟩=0

leading to superconductivity.

9. Deep Geometric View

Bosonic classical fields correspond to smooth amplitudes over spacetime:$$\phi(x)\in \mathbb R \text{ or } \mathbb C$$ϕ(x)∈R or C

Fermionic fields instead live in Grassmann algebra.

Grassmann quantities do not have ordinary macroscopic classical limits.

So:$$\boxed{ \text{fermionic fields are fundamentally quantum objects} }$$fermionic fields are fundamentally quantum objects​

while bosonic fields can emerge into classicality.

10. Important Caveat

People sometimes write “classical Dirac fields.”

This means:

• using the Dirac equation classically
• before quantization

But these are mathematical tools.

Physical macroscopic spinor fields are not observed in nature the way classical EM fields are.

The short version:$$\boxed{ \text{Bosons stack. Fermions exclude.} }$$Bosons stack. Fermions exclude.​

and therefore:$$\boxed{ \text{bosonic fields can become classical macroscopic fields,} }$$bosonic fields can become classical macroscopic fields,​

while$$\boxed{ \text{fermionic spinor fields cannot.} }$$fermionic spinor fields cannot.​

The post Why there are no macroscopic Spinor Fields? appeared first on Time Travel, Quantum Entanglement and Quantum Computing.

This is one of the most important conceptual distinctions in field theory.

The symbols $\partial$∂ and $\delta$δ represent completely different operations.

1. The Partial Derivative $\partial$∂

$\partial$∂ means:

“How does a function change as spacetime coordinates change?”

For a field$$\phi(x) = \phi(t,\mathbf x)$$ϕ(x)=ϕ(t,x)

the derivative$$\partial_\mu \phi$$∂μ​ϕ

describes how the field changes from point to point in spacetime.

Examples:$$\partial_0 \phi = \frac{\partial \phi}{\partial t}$$∂0​ϕ=∂t∂ϕ​

(time variation)

and$$\partial_i \phi = \frac{\partial \phi}{\partial x^i}$$∂i​ϕ=∂xi∂ϕ​

(spatial variation)

So:$$\boxed{ \partial = \text{ordinary spacetime derivative} }$$∂=ordinary spacetime derivative​

It acts inside spacetime.

2. The Variation $\delta$δ

$\delta$δ means:

“Suppose I slightly deform the entire field configuration.”

This is not motion through spacetime.

Instead:$$\phi(x) \rightarrow \phi(x)+\delta\phi(x)$$ϕ(x)→ϕ(x)+δϕ(x)

means:

• keep spacetime point $x$x fixed
• slightly change the value of the field there

So $\delta\phi$δϕ is an infinitesimal “test deformation” of the whole function.

Visual Intuition

Think of the field as a rubber sheet over spacetime.

• $\partial\phi$∂ϕ:
  measures the slope of the sheet from point to point
• $\delta\phi$δϕ:
  slightly lifts or perturbs the sheet itself

Why Both Appear Together

The Lagrangian depends on:$$\mathcal L(\phi,\partial_\mu\phi)$$L(ϕ,∂μ​ϕ)

meaning:

• the field value
• and its spacetime gradients

When varying the action:$$\delta S = \delta \int d^4x\,\mathcal L$$δS=δ∫d4xL

you must vary BOTH:$$\delta\phi$$δϕ

and$$\delta(\partial_\mu\phi)$$δ(∂μ​ϕ)

Since differentiation and variation commute:$$\boxed{ \delta(\partial_\mu\phi) = \partial_\mu(\delta\phi) }$$δ(∂μ​ϕ)=∂μ​(δϕ)​

This relation is central to the derivation.

Deep Geometric Interpretation

$\partial$∂ lives within spacetime.

$\delta$δ lives in the infinite-dimensional space of all possible field configurations.

So:

Analogy with Classical Mechanics

For a particle trajectory:$$x(t)$$x(t)

we distinguish:

Velocity:$$\frac{dx}{dt}$$dtdx​

vs variation of the path:$$x(t)\rightarrow x(t)+\delta x(t)$$x(t)→x(t)+δx(t)

The first measures motion along the path.

The second compares neighboring possible paths.

Field theory is exactly the same idea, but with fields instead of trajectories.

The Key Philosophical Idea

The action principle says:

Nature compares all nearby possible field configurations and selects the one where$$\boxed{ \delta S=0 }$$δS=0​

That does NOT mean the action is zero.

It means:$$\text{first-order change in action vanishes}$$first-order change in action vanishes

which is analogous to:$$\frac{df}{dx}=0$$dxdf​=0

for extrema in ordinary calculus.

The field equation emerges because the true field configuration is a stationary point in the space of all possible field configurations.

The post What is the difference between δ and ∂ in this derivation appeared first on Time Travel, Quantum Entanglement and Quantum Computing.

The physical field configuration is the one for which a small variation of the field,$$\phi(x)\rightarrow \phi(x)+\delta\phi(x)$$ϕ(x)→ϕ(x)+δϕ(x)

does not change the action to first order:$$\delta S=0$$δS=0

Now vary the action:$$\delta S = \int d^4x \left[ \frac{\partial \mathcal{L}}{\partial \phi}\delta\phi + \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} \delta(\partial_\mu\phi) \right]$$δS=∫d4x[∂ϕ∂L​δϕ+∂(∂μ​ϕ)∂L​δ(∂μ​ϕ)]

Since variation and differentiation commute,$$\delta(\partial_\mu\phi) = \partial_\mu(\delta\phi)$$δ(∂μ​ϕ)=∂μ​(δϕ)

so$$\delta S = \int d^4x \left[ \frac{\partial \mathcal{L}}{\partial \phi}\delta\phi + \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} \partial_\mu(\delta\phi) \right]$$δS=∫d4x[∂ϕ∂L​δϕ+∂(∂μ​ϕ)∂L​∂μ​(δϕ)]

Now integrate the second term by parts:$$\int d^4x\, \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} \partial_\mu(\delta\phi) = – \int d^4x\, \partial_\mu \left[ \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} \right] \delta\phi$$∫d4x∂(∂μ​ϕ)∂L​∂μ​(δϕ)=−∫d4x∂μ​[∂(∂μ​ϕ)∂L​]δϕ

The boundary term vanishes because we assume$$\delta\phi=0$$δϕ=0

on the boundary of spacetime.

Therefore,$$\delta S = \int d^4x \left[ \frac{\partial \mathcal{L}}{\partial \phi} – \partial_\mu \left( \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} \right) \right]\delta\phi$$δS=∫d4x[∂ϕ∂L​−∂μ​(∂(∂μ​ϕ)∂L​)]δϕ

For the action to be stationary for arbitrary $\delta\phi$δϕ, the bracket must vanish:$$\boxed{ \frac{\partial \mathcal{L}}{\partial \phi} – \partial_\mu \left( \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} \right) =0 }$$∂ϕ∂L​−∂μ​(∂(∂μ​ϕ)∂L​)=0​

This is the Euler–Lagrange equation for a field.

For the scalar field,$$\mathcal{L} = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi – \frac{1}{2}m^2\phi^2$$L=21​∂μ​ϕ∂μϕ−21​m2ϕ2

we get$$\frac{\partial \mathcal{L}}{\partial \phi} = -m^2\phi$$∂ϕ∂L​=−m2ϕ

and$$\frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} = \partial^\mu\phi$$∂(∂μ​ϕ)∂L​=∂μϕ

So the Euler–Lagrange equation becomes$$-m^2\phi-\partial_\mu\partial^\mu\phi=0$$−m2ϕ−∂μ​∂μϕ=0

or$$\boxed{ (\partial_\mu\partial^\mu+m^2)\phi=0 }$$(∂μ​∂μ+m2)ϕ=0​

That is the Klein–Gordon equation.

In plain English:$$\boxed{ \delta S=0 \quad\Rightarrow\quad \text{field chooses stationary action} \quad\Rightarrow\quad \text{Euler–Lagrange equation} \quad\Rightarrow\quad \text{Klein–Gordon equation} }$$δS=0⇒field chooses stationary action⇒Euler–Lagrange equation⇒Klein–Gordon equation

The post The principle of stationary action derivation appeared first on Time Travel, Quantum Entanglement and Quantum Computing.

In quantum mechanics, observables (like position, momentum, energy) are represented by operators. Requiring those operators to be Hermitian (more precisely, self-adjoint) is not arbitrary—it follows from a few fundamental physical requirements.

1. Measurement outcomes must be real numbers

A physical measurement always gives a real value.

If an operator $\hat{A}$A^ represents an observable, its possible measurement outcomes are its eigenvalues.

A Hermitian operator guarantees:$$\text{All eigenvalues of } \hat{A} \text{ are real}$$All eigenvalues of A^ are real

$\hat{A} = \hat{A}^\dagger$A^=A^†

If the operator were not Hermitian, you could get complex eigenvalues like $3 + 2i$3+2i, which have no physical meaning as measurement results.

2. Expectation values must be real

Even before measurement, we often compute the expectation value:$$\langle A \rangle = \langle \psi | \hat{A} | \psi \rangle$$⟨A⟩=⟨ψ∣A^∣ψ⟩

For a Hermitian operator:$$\langle \psi | \hat{A} | \psi \rangle \in \mathbb{R}$$⟨ψ∣A^∣ψ⟩∈R

If $\hat{A}$A^ were not Hermitian, the expectation value could be complex—which would make no physical sense as an “average measurement.”

3. Orthogonality of eigenstates (clean measurement structure)

Hermitian operators have a powerful property:

• Eigenstates corresponding to different eigenvalues are orthogonal

This gives us a clean decomposition:$$|\psi\rangle = \sum_i c_i |a_i\rangle$$∣ψ⟩=i∑​ci​∣ai​⟩

Where:

• $|a_i\rangle$∣ai​⟩ are eigenstates of the observable
• $|c_i|^2$∣ci​∣2 are probabilities

Without Hermiticity, this orthogonal structure breaks down → probabilities become ambiguous.

4. Probability interpretation requires it

Quantum mechanics relies on:$$P(a_i) = |\langle a_i | \psi \rangle|^2$$P(ai​)=∣⟨ai​∣ψ⟩∣2

This only works cleanly if:

• Eigenstates form an orthonormal basis
• The operator is Hermitian

Otherwise, you lose a consistent probability framework.

5. Connection to unitary time evolution

Hermitian operators also generate unitary transformations.

Example: the Hamiltonian $\hat{H}$H^$$U(t) = e^{-i \hat{H} t / \hbar}$$U(t)=e−iH^t/ℏ

If $\hat{H}$H^ is Hermitian:

• $U(t)$U(t) is unitary
• Total probability is conserved

If not:

• Probability could grow or decay → physically unacceptable

6. Deeper insight (physics intuition)

You can think of Hermitian operators as enforcing:

• Reality → measurements are real
• Stability → probabilities don’t explode
• Consistency → repeatable measurements give structured outcomes

In a deeper sense:

Hermiticity ensures that the mathematical structure of quantum mechanics aligns with the physical requirement that observations are real, probabilistic, and consistent.

The post Why are the observable operators in QM required to be Hermitian? appeared first on Time Travel, Quantum Entanglement and Quantum Computing.

This is a deep and beautiful distinction—one that sits right at the boundary between pure geometry and physics-ready geometry. Let’s build it cleanly and intuitively, then connect it to densities, derivatives, integrals, and invariant derivatives.

1. Unconnected Manifold (Bare Manifold)

What it is

An unconnected manifold (more precisely: a manifold without a connection) is just:

• A smooth space ( M )
• With coordinate charts
• And smooth transition functions

? But no notion of how to compare vectors at different points

Key limitation

You can define:

• Scalars ( f(x) )
• Tensors at a point

But NOT:

• How a vector at ( x ) relates to a vector at ( x + dx )

Derivatives here

You only have partial derivatives:

[
\frac{\partial f}{\partial x^\mu}
]

These are:

• Coordinate-dependent
• Not geometric objects (for tensors beyond scalars)

Integrals here

Integration is not automatically well-defined globally unless you introduce:

• A density or
• A volume form

Densities (critical here)

A density is something that transforms like:

[
\rho'(x’) = \left| \det \left( \frac{\partial x}{\partial x’} \right) \right| \rho(x)
]

This allows:

[
\int_M \rho(x), d^n x
]

to be coordinate invariant

? On a bare manifold, densities are what make integration possible

2. Affine Connected Manifold

Now we add structure:

What is added?

An affine connection (typically denoted ( \Gamma^\lambda_{\mu\nu} ))

This gives:

• A rule for comparing vectors at nearby points
• A notion of parallel transport
• A way to define covariant derivatives

Covariant derivative

Instead of partial derivatives, we now define:

[
\nabla_\mu V^\nu = \partial_\mu V^\nu + \Gamma^\nu_{\mu\lambda} V^\lambda
]

This is:

• Coordinate invariant
• Tensorial

Why this matters

Without a connection:

• Derivatives of tensors are NOT tensors

With a connection:

• Derivatives become geometrically meaningful

3. Densities in Connected Manifolds

Now something powerful happens.

If you also have a metric ( g_{\mu\nu} ), you get:

[
\sqrt{|g|}
]

This is a natural density.

So integrals become:

[
\int_M f(x), \sqrt{|g|}, d^n x
]

? No need to manually define densities anymore—the geometry gives you one.

4. Integrals: Bare vs Connected

Bare manifold

You must supply:

• A density ( \rho(x) )

Affine connected (with metric)

You get:

• Natural volume element ( \sqrt{|g|} d^n x )

5. Invariant Derivatives (The Core Idea)

This is where the distinction becomes profound.

Problem

Partial derivatives:
[
\partial_\mu V^\nu
]

DO NOT transform as tensors.

Solution: Covariant derivative

[
\nabla_\mu V^\nu
]

This does transform properly.

Why?

Because the connection compensates for:

• Curvature
• Coordinate distortion

6. Invariant Divergence (Important Bridge to Densities)

A beautiful identity:

[
\nabla_\mu V^\mu = \frac{1}{\sqrt{|g|}} \partial_\mu \left( \sqrt{|g|} V^\mu \right)
]

? This connects:

• Covariant derivative
• Density structure

Interpretation

• ( \sqrt{|g|} V^\mu ) behaves like a density
• Its ordinary divergence becomes invariant

This is the bridge between densities and connections

7. Big Picture Comparison

8. Physical Interpretation (Your Style)

Think of it this way:

Unconnected manifold

• A map with no notion of direction consistency
• Each point is isolated
• You can measure locally, but not compare globally

Affine connected manifold

• A map with a compass field
• You can move vectors around
• You can define:
  • Curvature
  • Geodesics
  • Conservation laws

9. The Deep Insight

The connection does something subtle but profound:

? It replaces coordinate dependence with geometric structure

And densities do something equally profound:

? They replace coordinate-dependent volume with invariant measure

10. Executive-Level Summary (your style)

• A bare manifold gives you location
• A connection gives you movement
• A density gives you measure
• A covariant derivative gives you truth independent of coordinates

If you want, we can take this further into:

• Levi-Civita connection vs general affine connection
• Torsion vs curvature
• Or tie directly into Dirac’s formulation and invariant operators (which aligns with your current reading)

The post The unconnected manifold versus the affine connected manifold – describe what these are. Densities, Derivatives, Integrals, Invariant Derivatives in particular appeared first on Time Travel, Quantum Entanglement and Quantum Computing.