Quantization of Free Field: Maxwell Field

Quantization of Maxwell Field

Canonical Quantization of Maxwell Field

The Hamiltonian of Maxwell Field is(in Feynman Gauge):

H=12d3x {i=13((0Ai)2+(Ai)2)(0A0)2(A0)2}

Where Aμ are vector field. We will use the canonical commutation relationship. For Ai they can be written equivalently as:

[Aμ(x),(0Aν)(x)]=iημνδ3(xx)

Proof With Feynman Gauge, we have the canonical momentum as: πμ=Fμ0ημ0νAν If we use the canonical commutation relationship as : [Aμ(x),πν(x)]=iδνμδ3(xx) Because the canonical momentum of Aμ is πμ=δLδ0Aμ which should be a covariance vector. And any other commutator vanish, it is equivalent to: [Aμ(x),νA0(x)(0Aν)(x)ην0((0A0)(x)+iiAi(x))]=iδμνδ3(xx) For that [Aμ(x),Aν(x)]=0 , we will have that x[Aμ(x),Aν(x)]=0 That is [Aμ(x),iAν(x)]=0 . Then the canonical commutation relationship should be: [A0(x),(0A0)(x)]=iδ3(xx)[Ai(x),(0Ai)(x)]=iδ3(xx) That is [Aμ(x),(0Aν)(x)]=iημνδ3(xx) q.e.d.

One can rewrite them into the plane-wave expansion. In Schrodinger Picture:

Aμ(x)=d3k(2π)3/212ω|k|λ=03(ϵμ(k,λ)eikxa(k,λ)+ϵμ(k,λ)eikxa(k,λ))

Where a is operators and ω|k|=|k| . Polarization vectors have the property:

ϵ(k,λ)ϵ(k,λ)=ηλλ ; ληλλϵμ(k,λ)ϵν(k,λ)=ημν

We can write the inverse transformation, which will represent a,a by Aμ and 0Aμ :

a(k,λ)=ληλ,λd3x(2π)3/2 eikxϵμ(k,λ)(ω|k|2Aμ(x)+i12ω|k|0Aμ(x))a(k,λ)=ληλ,λd3x(2π)3/2 eikxϵμ(k,λ)(ω|k|2Aμ(x)i12ω|k|0Aμ(x))

And their commutation relationship:

[a(k,λ),a(k,λ)]=[a(k,λ),a(k,λ)]=0 ; [a(k,λ),a(k,λ)]=ηλλδ3(kk)

Proof With that: Aμ(x)=d3k(2π)3/212ω|k|λ=03(ϵμ(k,λ)eikxa(k,λ)+ϵμ(k,λ)eikxa(k,λ))0Aμ(x)=d3k(2π)3/2iω|k|2ω|k|(ϵμ(k,λ)eikxa(k,λ)ϵμ(k,λ)eikxa(k,λ)) The inverse of Fourier transformation is: 12ω|k|λ=03(ϵμ(k,λ)a(k,λ)+ϵμ(k,λ)a(k,λ))=d3x(2π)3/2 eikxAμ(x)iω|k|2ω|k|λ=03(ϵμ(k,λ)a(k,λ)ϵμ(k,λ)a(k,λ))=d3x(2π)3/2 eikx0Aμ(x) That is: λϵμ(k,λ)a(k,λ)=d3x(2π)3/2 eikx(ω|k|2Aμ(x)+i12ω|k|0Aμ(x)) With ϵμ(k,λ)ϵμ(k,λ)=ηλλ , we have: a(k,0)=d3x(2π)3/2 eikxϵμ(k,0)(ω|k|2Aμ(x)+i12ω|k|0Aμ(x))a(k,i)=d3x(2π)3/2 eikxϵμ(k,i)(ω|k|2Aμ(x)+i12ω|k|0Aμ(x)) Where i=1,2,3 . And similarly: a(k,0)=d3x(2π)3/2 eikxϵμ(k,0)(ω|k|2Aμ(x)i12ω|k|0Aμ(x))a(k,i)=d3x(2π)3/2 eikxϵμ(k,i)(ω|k|2Aμ(x)i12ω|k|0Aμ(x)) One can also write it as: a(k,λ)=ληλ,λd3x(2π)3/2 eikxϵμ(k,λ)(ω|k|2Aμ(x)+i12ω|k|0Aμ(x))a(k,λ)=ληλ,λd3x(2π)3/2 eikxϵμ(k,λ)(ω|k|2Aμ(x)i12ω|k|0Aμ(x)) Then we need to compute their commutator: [a(k,λ),a(k,λ)]=λ1,λ2d3x1d3x2(2π)3 eikx1+ikx2ϵμ1(k,λ1)ϵμ2(k,λ2)    (iω|k|4ω|k|[Aμ1(x1),0Aμ2(x2)]+iω|k|4ω|k|[0Aμ1(x1),Aμ2(x2)])=λ1,λ2d3x1d3x2(2π)3 eikx1+ikx2ϵμ1(k,λ1)ϵμ2(k,λ2)    (ω|k|4ω|k|ω|k|4ω|k|)ημ1μ2δ3(x1x2)=0 And for the same reason [a(k,λ),a(k,λ)]=0 . Then: [a(k,λ),a(k,λ)]=λ1,λ2d3x1d3x2(2π)3 eikx1ikx2ϵμ1(k,λ1)ϵμ2(k,λ2)    (iω|k|4ω|k|[Aμ1(x1),0Aμ2(x2)]+iω|k|4ω|k|[0Aμ1(x1),Aμ2(x2)])=λ1,λ2d3x1d3x2(2π)3 eikx1ikx2ϵμ1(k,λ1)ϵμ2(k,λ2)    (ω|k|4ω|k|ω|k|4ω|k|)ημ1μ2δ3(x1x2)=λ1,λ2ϵμ(k,λ1)ϵμ(k,λ2)(1)δ3(kk)=ηλ1λ2δ3(kk) q.e.d.

With this result and the hold-order discussion in Classical theory, we can write the Hamiltonian as:

H^=d3k (λ=13a(k,λ)a(k,λ)a(k,0)a(k,0))

There is also a quantity at the right-hand-side. But like usual, we ignore it.

One can also write these operators in Heisenberg Picture :

Aμ(x)=d3k(2π)3/212ω|k|λ=03(ϵμ(k,λ)eikxa(k,λ)+ϵμ(k,λ)eikxa(k,λ))

Proof We can also check that: [a(k,λ),H^]=ω|k|a(k,λ) ; [a(k,λ),H^]=ω|k|a(k,λ) It hold for λ=0 component cause the η00=ηii . Then: aH(k,λ)=eiω|k|ta(k,λ) And: Aμ(x)=d3k(2π)3/212ω|k|λ=03(ϵμ(k,λ)eikxa(k,λ)+ϵμ(k,λ)eikxa(k,λ)) q.e.d.

Extended Hilbert Space and Physical States

Why the extended Hilbert is needed One can find that this quantization procedure leads to a Hamiltonian `without` lower bound. Different from the Dirac Field, using anti-commutation relationship has no help, because the negative sign of `scalar polarization` come from the classical theory. Moreover, in classical electromagnetic field theory, we know that `dof`of electromagnetic wave is 2. But here we have four. In classical we use gauge condition to eliminate them, but in quantum theory, for example, Lorentz gauge μAμ=0 cannot be a equality for operator. Because we have π0μAμ , let it hold will conflict to commutator relationship. At last, we find that commutator for λ=0 component is irregular: [a(k,0),a(k,0)]=δ3(kk) . That is to say if we normalize the vacuum 0|0=1 , then 0|a(k,0)a(k,0)|0=δ3(kk)<0 . Shows that our Hilbert space should cover some state with negative modulus. This is dangerous to Quantum theory especially for possibility interpretation. In general, we have to accept that there are some non-physical states in Hilbert space. And on those physical states the theory is legal.

[Definition] : The Hilbert generated from the quantization procedure above are larger than what we need. And it is not positive definite. Physical subspace is a subset of the Hilbert space, it is positive definite and describe the physical process can happen in the real world. States in Physical subspace are physical states, and the physical subspace is determined by the condition:

ψphys.|μAμ|ψphys.=0

Which is the quantum version of Lorentz gauge.

A stronger constraint(much more convenient for it is an eigen-equation) is

(a(k,3)a(k,0))|ψphys.=0 ; ψphys.|(a(k,3)a(k,0))=0

Proof If |Ψ satisfy (a(k,3)a(k,0))|Ψ=0 , then we have: Ψ|μAμ|Ψ=d3k(2π)3/212ω|k|λ(ikμϵμ(k,λ)eikxΨ|a(k,λ)|Ψ+ikμϵμ(k,λ)eikxΨ|a(k,λ)|Ψ)=d3k(2π)3/212ω|k|λ(i(δλ0δλ3)eikxΨ|a(k,λ)|Ψ+i(δλ0δλ3)eikxΨ|a(k,λ)|Ψ)=d3k(2π)3/212ω|k|(ieikxΨ|a(k,3)a(k,0)|ΨieikxΨ|a(k,3)a(k,0)|Ψ)=0 q.e.d.

And this stronger condition leads to

ψphys.|a(k,0)a(k,0)|ψphys.=ψphys.|a(k,3)a(k,3)|ψphys.

This shows that in physical subspace , the expectation of Hamiltonian is:

ψphys.|H^|ψphys.=d3k ω|k|λ=1,2ψphys.|a(k,λ)a(k,λ)|ψphys.

The right-hand-side is positive definite.

And we have the structure of physical subspace:

[Definition] : vacuum of the Maxwell Field is the ground state of it. Denoting it as |0 , it satisfies:

a(k,λ)|0=0 ; k,λ=0,1,2,3

Obviously, |0 is a physical state.

[Theorem] : Any physical states can be expressed as:

|Φc=R^c|ΦT

Where |ΦT is the state which contains only transverse photons(polarizations) , and operator R^c creates the admixture of longitudinal and scalar photons:

R^c=1+d3 c(k)L^(k)+d3kd3k c(k,k)L^(k)L^(k)+

Where c are arbitrary functions and L^(k)=a(k,3)a(k,0) . And:

Φc|Φc=ΦT|ΦT

That’s why they(added states) are also called as states-with-zero-norm .

Proof To prove that we should check: [L^(k),L^(k)]=δ3(kk)+δ3(kk)=0 Then we have: L^(k)R^c=R^cL^(k) For L^(k)|ΦT=0 by the definition, we have: L^(k)R^c|ΦT=0R^c|ΦTphysical subspace Any other states are this state with different number of scalar photons and longitudinal photons, obviously they are not physical state. This property also leads to: [R^c,R^c]=0 Then: Φc|Φc=ΦT|R^cR^c|ΦT=ΦT|(1+aL^)|ΦT=ΦT|ΦT q.e.d.

[Theorem] : The states |Φc for a given |ΦT form an equivalence class . The member of this equivalence class are related by gauge transformations. Explicitly speaking, one has:

Φc|Aμ|Φc=ΦT|Aμ|ΦT+μΛ

Where Λ is a function satisfies μμΛ=0 , it will come from gauge transformation.

Proof One has: Φc|Aμ|Φc=ΦT|R^c[Aμ,R^c]|ΦT+ΦT|[R^c,Aμ]|ΦT+ΦT|Aμ|ΦT Where we used R^c|ΦT=|ΦT . By using the fact that the commutators of Aμ with L^(k),L^(k) are c-numbers: [Aμ,L^(k)]=12ω|k|eikx(ϵμ(k,3)+ϵμ(k,0)) We obtain: Φc|Aμ|Φc=d3k c(k)[Aμ,L^(k)]+d3k c(k)[L^(k),Aμ]+ΦT|Aμ|ΦT And with ϵμ(k,3)+ϵμ(k,0)kμ , which can be easily checked by the Lorentz-covariance. We have: Φc|Aμ|Φc=d3k c(k)[Aμ,L^(k)]+d3k c(k)[L^(k),Aμ]+ΦT|Aμ|ΦT=d3k2ω|k|kμ(eikxAc(k)+eikxAc(k))+ΦT|Aμ|ΦT=μΛ+ΦT|Aμ|ΦT Where: Λ=id3k2ω|k|(eikxAc(k)eikxAc(k)) Since k0=|k| , we have Λ=0 . The gauge function Λ is consistent with the Lorentz gauge condition. q.e.d.

With these theorems, we can use |ΦT as the representation of each equivalence class. And in this meaning the Maxwell Field is totally quantized.

[Definition] : 1-particle state is defined as:

|k,λ=1,2=2ω|k|a(k,λ=1,2)|0

And the Lorentz covariance:

Λkkspecialk,λ|Λkkspecialk,λ=k,λ|k,λ=2ω|k|δλλδ3(kk)

And the Lorentz covariant integral:

d3k2ω|k|=d4k δ(kμkμ)Θ(k0)

[Definition] : Consider the coordinate transformation(special Lorentz Transformation) xx=Λ(ω)x , the corresponding operator on Hilbert space is U(Λ(ω)) , then we have:

U(Λ(ω))|k,λ=|Λ(ω),λ

The vacuum is unique, then:

U(Λ(ω))a(k,s)U(Λ(ω))1=ω|Λ(ω)k|ω|k|a(Λ(ω)k,λ)

Then one can check that the field operator will be:

Aμ(x=Λ(ω)x)U(Λ(ω))Aμ(x)U(Λ(ω))1=(Λ(ω)1)    νμAν(x=Λ(ω)x)

Proof The proof is similar to what we did before, just noting Λkkspecialϵμ(k,λ) will be independent of k ,, which means: ϵμ(Λk,λ)=Λ    νμϵν(k,λ) Then: U(Λ(ω))Aμ(x)U(Λ(ω))1=d3k(2π)3/22ω|Λk|2ω|k|λ(eikxϵμ(k,λ)a(Λk,λ)+eikxϵμ(k,λ)a(Λk,λ))=(Λ1)    νμAν(Λx) q.e.d.

Causality of Maxwell Field

Our quantization will not violent to the causality demanded by special Relativistic Theory. That is to say:

[Theorem] : In Heisenberg Picture, for any two points x,y with spacelike interval: xy2<0 , the field operators are commutable:

[Aμ(x),Aν(y)]=0 ; xy2<0

Proof With the quantization that [a(k,λ),a(k,λ)]=ηλλδ3(kk) . One has: [Aμ(x),Aν(y)]=d3k1d3k2 δ3(k1k2)(2π)34ω1ω2λ1,λ2{ηλ1λ2eik1x+ik2yϵμ(k1,λ1)ϵν(k2,λ2)+ηλ1λ2eik1xik2yϵμ(k1,λ1)ϵν(k2,λ2)}=d3k(2π)32ω|k|ημν(eik(xy)eik(xy)) With the similar argument, the causality is proved.

Propagator of Maxwell Field

[Definition] : Feynman Propagator of Maxwell Field is defined as:

iΔF(xy):=0|T(Aμ(x)Aν(y))|0

And Maxwell Field is bosonic. One can compute that:

iΔF(xy)=d4p(2π)4eip(xy)(iημνpμpμ+iϵ)

The calculation is similar to Scalar field, with m=0 , and 0|a(k,λ)a(k,λ)|0=ηλλδ3(kk) .. That is why there are a factor ημν here.

This form only work for Feynman Gauge: ξ=1 . The generic form is:

iΔF(xy)=d4p(2π)4eip(xy)(i){ημνpμpμ+iϵ+1ξξpμpν(pμpμ+iϵ)2}

Discrete Symmetry of Maxwell Field

Spatial Inversion

We hope the Spatial Inversion induces a unitary operator so as to

UPAμ(x0,x)UP1=ηγAμ(x0,x)

For Maxwell Field, ηγ=1 , which is different from the real-valued scalar field.

For creation/annihilation operators:

UPa(k,λ)UP1=a(k,λ)

Charge Conjugation

Charge conjugation for photon is:

UCAμ(x)UC1=Aμ(x)

Same to creation/annihilation operators:

UCa(k,λ)UC1=a(k,λ)

Time Reversal

Time reversal induces an anti-unitary operator. With conjugation operator K we can write it as UTK , where UT is a unitary operator. For creation/annihilation operators:

(UTK)a(k,λ)(UTK)1=a(k,λ)

For λ denotes the helicity, time reversal inverse momentum and spin, so λ should be leave invariant.

Then we can prove that:

(UTK)Aμ(x0,x)(UTK)1=ημμAμ(x0,x)

Do not sum for μ here.

Proof For ϵμ(k,λ)=P    νμϵν(k,λ) , and P    νμ=ημν , we have: (UTK)Aμ(x0,x)(UTK)1=d3k(2π)3/212ω|k|λ=03(ϵμ(k,λ)eikxa(k,λ)+ϵμ(k,λ)eikxa(k,λ))=ημμd3k(2π)3/212ω|k|λ=03(ϵμ(k,λ)eikxa(k,λ)+ϵμ(k,λ)eikxa(k,λ))=ημμAμ(x0,x) Do not sum for μ here. q.e.d.