Quantization of the electromagnetic field

Second quantization starts with an expansion of a scalar or vector field (or wave functions) in a basis consisting of a complete set of functions. These expansion functions depend on the coordinates of a single particle. The coefficients multiplying the basis functions are interpreted as operators and (anti)commutation relations between these new operators are imposed, commutation relations for bosons and anticommutation relations for fermions (nothing happens to the basis functions themselves). By doing this, the expanded field is converted into a fermion or boson operator field. The expansion coefficients have been promoted from ordinary numbers to operators, creation and annihilation operators. A creation operator creates a particle in the corresponding basis function and an annihilation operator annihilates a particle in this function.

In the case of EM fields the required expansion of the field is the Fourier expansion.

As the term suggests, an EM field consists of two vector fields, an electric field ${\displaystyle \mathbf {E} (\mathbf {r} ,t)}$ and a magnetic field ${\displaystyle \mathbf {B} (\mathbf {r} ,t)}$. Both are time-dependent vector fields that in vacuum depend on a third vector field ${\displaystyle \mathbf {A} (\mathbf {r} ,t)}$ (the vector potential), as well as a scalar field ${\displaystyle \phi (\mathbf {r} ,t)}$

${\displaystyle {\begin{aligned}\mathbf {B} (\mathbf {r} ,t)&={\boldsymbol {\nabla }}\times \mathbf {A} (\mathbf {r} ,t)\\\mathbf {E} (\mathbf {r} ,t)&=-{\boldsymbol {\nabla }}\phi (\mathbf {r} ,t)-{\frac {\partial \mathbf {A} (\mathbf {r} ,t)}{\partial t}},\\\end{aligned}}}$

where ∇ × A is the curl of A.

Choosing the Coulomb gauge, for which ∇⋅A = 0, makes A into a transverse field. The Fourier expansion of the vector potential enclosed in a finite cubic box of volume V = L³ is then

${\displaystyle \mathbf {A} (\mathbf {r} ,t)=\sum _{\mathbf {k} }\sum _{\mu =\pm 1}\left(\mathbf {e} ^{(\mu )}(\mathbf {k} )a_{\mathbf {k} }^{(\mu )}(t)e^{i\mathbf {k} \cdot \mathbf {r} }+{\bar {\mathbf {e} }}^{(\mu )}(\mathbf {k} ){\bar {a}}_{\mathbf {k} }^{(\mu )}(t)e^{-i\mathbf {k} \cdot \mathbf {r} }\right),}$

where ${\displaystyle {\overline {a}}}$ denotes the complex conjugate of ${\displaystyle a}$. The wave vector k gives the propagation direction of the corresponding Fourier component (a polarized monochromatic wave) of A(r,t); the length of the wave vector is

${\displaystyle |\mathbf {k} |={\frac {2\pi \nu }{c}}={\frac {\omega }{c}},}$

with ν the frequency of the mode. In this summation k runs over one side, positive or negative. (The component of Fourier basis ${\displaystyle e^{-i\mathbf {k} \cdot \mathbf {r} }}$ is complex conjugate of component of ${\displaystyle e^{i\mathbf {k} \cdot \mathbf {r} }}$ as ${\displaystyle \mathbf {A} (\mathbf {r} ,t)}$ is real.) The components of the vector k have discrete values (a consequence of the boundary condition that A has the same value on opposite walls of the box):

${\displaystyle k_{x}={\frac {2\pi n_{x}}{L}},\quad k_{y}={\frac {2\pi n_{y}}{L}},\quad k_{z}={\frac {2\pi n_{z}}{L}},\qquad n_{x},n_{y},n_{z}=0,\pm 1,\pm 2,\ldots .}$

Two e^(μ) ("polarization vectors") are conventional unit vectors for left and right hand circular polarized (LCP and RCP) EM waves (See Jones calculus or Jones vector, Jones calculus) and perpendicular to k. They are related to the orthonormal Cartesian vectors e_x and e_y through a unitary transformation,

${\displaystyle \mathbf {e} ^{(\pm 1)}\equiv {\frac {\mp 1}{\sqrt {2}}}(\mathbf {e} _{x}\pm i\mathbf {e} _{y})\qquad {\hbox{with}}\quad \mathbf {e} _{x}\cdot \mathbf {k} =\mathbf {e} _{y}\cdot \mathbf {k} =0.}$

The k-th Fourier component of A is a vector perpendicular to k and hence is a linear combination of e⁽¹⁾ and e⁽⁻¹⁾. The superscript μ indicates a component along e^(μ).

Clearly, the (discrete infinite) set of Fourier coefficients ${\displaystyle a_{\mathbf {k} }^{(\mu )}(t)}$ and ${\displaystyle {\bar {a}}_{\mathbf {k} }^{(\mu )}(t)}$ are variables defining the vector potential. In the following they will be promoted to operators.

By using field equations of ${\displaystyle \mathbf {B} }$ and ${\displaystyle \mathbf {E} }$ in terms of ${\displaystyle \mathbf {A} }$ above, electric and magnetic fields are

${\displaystyle {\begin{aligned}\mathbf {E} (\mathbf {r} ,t)&=i\sum _{\mathbf {k} }{\sum _{\mu =\pm 1}\omega {\left({\mathbf {e} ^{(\mu )}}(\mathbf {k} )a_{\mathbf {k} }^{(\mu )}(t){e^{i\mathbf {k} \cdot \mathbf {r} }}-{{\overline {\mathbf {e} }}^{(\mu )}}(\mathbf {k} ){\bar {a}}_{\mathbf {k} }^{(\mu )}(t){{e}^{-i\mathbf {k} \cdot \mathbf {r} }}\right)}}\\[6pt]\mathbf {B} (\mathbf {r} ,t)&=i\sum _{\mathbf {k} }\sum _{\mu =\pm 1}\left\{\left(\mathbf {k} \times {{\mathbf {e} }^{(\mu )}}(\mathbf {k} )\right)a_{\mathbf {k} }^{(\mu )}(t)e^{i\mathbf {k} \cdot \mathbf {r} }-\left(\mathbf {k} \times {{\overline {\mathbf {e} }}^{(\mu )}}(\mathbf {k} )\right){\bar {a}}_{\mathbf {k} }^{(\mu )}(t){{e}^{-i\mathbf {k} \cdot \mathbf {r} }}\right\}\end{aligned}}}$

By using identity ${\displaystyle \nabla \times e^{A\cdot r}=A\times e^{A\cdot r}}$ (${\displaystyle A}$ and ${\displaystyle r}$ are vectors) and ${\displaystyle a_{\mathbf {k} }^{(\mu )}(t)=a_{\mathbf {k} }^{(\mu )}{{e}^{-iwt}}}$ as each mode has single frequency dependence.

The best known example of quantization is the replacement of the time-dependent linear momentum of a particle by the rule

${\displaystyle \mathbf {p} (t)\to -i\hbar {\boldsymbol {\nabla }}.}$

Note that Planck's constant is introduced here and that the time-dependence of the classical expression is not taken over in the quantum mechanical operator (this is true in the so-called Schrödinger picture).

For the EM field we do something similar. The quantity ${\displaystyle \epsilon _{0}}$ is the electric constant, which appears here because of the use of electromagnetic SI units. The quantization rules are:

${\displaystyle {\begin{aligned}a_{\mathbf {k} }^{(\mu )}(t)&\to {\sqrt {\frac {\hbar }{2\omega V\epsilon _{0}}}}a^{(\mu )}(\mathbf {k} )\\{\bar {a}}_{\mathbf {k} }^{(\mu )}(t)&\to {\sqrt {\frac {\hbar }{2\omega V\epsilon _{0}}}}{a^{\dagger }}^{(\mu )}(\mathbf {k} )\\\end{aligned}}}$

subject to the boson commutation relations

${\displaystyle {\begin{aligned}\left[a^{(\mu )}(\mathbf {k} ),a^{(\mu ')}(\mathbf {k} ')\right]&=0\\\left[{a^{\dagger }}^{(\mu )}(\mathbf {k} ),{a^{\dagger }}^{(\mu ')}(\mathbf {k} ')\right]&=0\\\left[a^{(\mu )}(\mathbf {k} ),{a^{\dagger }}^{(\mu ')}(\mathbf {k} ')\right]&=\delta _{\mathbf {k} ,\mathbf {k} '}\delta _{\mu ,\mu '}\end{aligned}}}$

The square brackets indicate a commutator, defined by ${\displaystyle [A,B]\equiv AB-BA}$ for any two quantum mechanical operators A and B. The introduction of Planck's constant is essential in the transition from a classical to a quantum theory. The factor

${\displaystyle {\sqrt {\frac {1}{2\omega V\epsilon _{0}}}}}$

is introduced to give the Hamiltonian (energy operator) a simple form, see below.

The quantized fields (operator fields) are the following

${\displaystyle {\begin{aligned}\mathbf {A} (\mathbf {r} )&=\sum _{\mathbf {k} ,\mu }{\sqrt {\frac {\hbar }{2\omega V\epsilon _{0}}}}\left\{\mathbf {e} ^{(\mu )}a^{(\mu )}(\mathbf {k} )e^{i\mathbf {k} \cdot \mathbf {r} }+{\bar {\mathbf {e} }}^{(\mu )}{a^{\dagger }}^{(\mu )}(\mathbf {k} )e^{-i\mathbf {k} \cdot \mathbf {r} }\right\}\\\mathbf {E} (\mathbf {r} )&=i\sum _{\mathbf {k} ,\mu }{\sqrt {\frac {\hbar \omega }{2V\epsilon _{0}}}}\left\{\mathbf {e} ^{(\mu )}a^{(\mu )}(\mathbf {k} )e^{i\mathbf {k} \cdot \mathbf {r} }-{\bar {\mathbf {e} }}^{(\mu )}{a^{\dagger }}^{(\mu )}(\mathbf {k} )e^{-i\mathbf {k} \cdot \mathbf {r} }\right\}\\\mathbf {B} (\mathbf {r} )&=i\sum _{\mathbf {k} ,\mu }{\sqrt {\frac {\hbar }{2\omega V\epsilon _{0}}}}\left\{\left(\mathbf {k} \times \mathbf {e} ^{(\mu )}\right)a^{(\mu )}(\mathbf {k} )e^{i\mathbf {k} \cdot \mathbf {r} }-\left(\mathbf {k} \times {\bar {\mathbf {e} }}^{(\mu )}\right){a^{\dagger }}^{(\mu )}(\mathbf {k} )e^{-i\mathbf {k} \cdot \mathbf {r} }\right\}\end{aligned}}}$

where ω = c |k| = ck.

The classical Hamiltonian has the form

${\displaystyle H={\frac {1}{2}}\epsilon _{0}\iiint _{V}{\left({{\left|E(\mathbf {r} ,t)\right|}^{2}}+{{c}^{2}}{{\left|B(\mathbf {r} ,t)\right|}^{2}}\right)}{{\text{d}}^{3}}\mathbf {r} =V\epsilon _{0}\sum _{\mathbf {k} }\sum _{\mu =\pm 1}\omega ^{2}\left({\bar {a}}_{\mathbf {k} }^{(\mu )}(t)a_{\mathbf {k} }^{(\mu )}(t)+a_{\mathbf {k} }^{(\mu )}(t){\bar {a}}_{\mathbf {k} }^{(\mu )}(t)\right).}$

The right-hand-side is easily obtained by first using

${\displaystyle \int _{V}e^{ik\cdot r}e^{-ik'\cdot r}dr=V\delta _{k,k'}}$

(can be derived from Euler equation and trigonometric orthogonality) where k is wavenumber for wave confined within the box of V = L × L × L as described above and second, using ω = kc.

Substitution of the field operators into the classical Hamiltonian gives the Hamilton operator of the EM field,

${\displaystyle H={\frac {1}{2}}\sum _{\mathbf {k} ,\mu =\pm 1}\hbar \omega \left({a^{\dagger }}^{(\mu )}(\mathbf {k} )a^{(\mu )}(\mathbf {k} )+a^{(\mu )}(\mathbf {k} ){a^{\dagger }}^{(\mu )}(\mathbf {k} )\right)=\sum _{\mathbf {k} ,\mu }\hbar \omega \left({a^{\dagger }}^{(\mu )}(\mathbf {k} )a^{(\mu )}(\mathbf {k} )+{\frac {1}{2}}\right)}$

The second equality follows by use of the third of the boson commutation relations from above with k'' = k and μ = μ. Note again that ℏω = hν = ℏc|k| and remember that ω depends on k, even though it is not explicit in the notation. The notation ω(k) could have been introduced, but is not common as it clutters the equations.

The quantized EM field has a vacuum (no photons) state ${\displaystyle |0\rangle }$. The application to it, say,

${\displaystyle \left({a^{\dagger }}^{(\mu )}(\mathbf {k} )\right)^{m}\left({a^{\dagger }}^{(\mu ')}(\mathbf {k} ')\right)^{n}|0\rangle \propto \left|(\mathbf {k} ,\mu )^{m};(\mathbf {k} ',\mu ')^{n}\right\rangle ,}$

gives a quantum state of m photons in mode (k, μ) and n photons in mode (k′, μ′). The proportionality symbol is used because the state on the left-hand is not normalized to unity, whereas the state on the right-hand may be normalized.

The operator

${\displaystyle N^{(\mu )}(\mathbf {k} )\equiv {a^{\dagger }}^{(\mu )}(\mathbf {k} )a^{(\mu )}(\mathbf {k} )}$

is the number operator. When acting on a quantum mechanical photon number state, it returns the number of photons in mode (k, μ). This also holds when the number of photons in this mode is zero, then the number operator returns zero. To show the action of the number operator on a one-photon ket, we consider

${\displaystyle {\begin{aligned}N^{(\mu )}(\mathbf {k} )|\mathbf {k} ',\mu '\rangle &={a^{\dagger }}^{(\mu )}(\mathbf {k} )a^{(\mu )}(\mathbf {k} ){a^{\dagger }}^{(\mu ')}(\mathbf {k'} )|0\rangle \\&={a^{\dagger }}^{(\mu )}(\mathbf {k} )\left(\delta _{\mathbf {k} ,\mathbf {k'} }\delta _{\mu ,\mu '}+{a^{\dagger }}^{(\mu ')}(\mathbf {k'} )a^{(\mu )}(\mathbf {k} )\right)|0\rangle \\&=\delta _{\mathbf {k} ,\mathbf {k'} }\delta _{\mu ,\mu '}|\mathbf {k} ,\mu \rangle ,\end{aligned}}}$

i.e., a number operator of mode (k, μ) returns zero if the mode is unoccupied and returns unity if the mode is singly occupied. To consider the action of the number operator of mode (k, μ) on a n-photon ket of the same mode, we drop the indices k and μ and consider

${\displaystyle N(a^{\dagger })^{n}|0\rangle =a^{\dagger }\left([a,(a^{\dagger })^{n}]+(a^{\dagger })^{n}a\right)|0\rangle =a^{\dagger }[a,(a^{\dagger })^{n}]|0\rangle .}$

Use the "differentiation rule" introduced earlier and it follows that

${\displaystyle N(a^{\dagger })^{n}|0\rangle =n(a^{\dagger })^{n}|0\rangle .}$

A photon number state (or a Fock state) is an eigenstate of the number operator. This is why the formalism described here is often referred to as the occupation number representation.

Earlier the Hamiltonian,

${\displaystyle H=\sum _{\mathbf {k} ,\mu }\hbar \omega \left({a^{\dagger }}^{(\mu )}(\mathbf {k} )a^{(\mu )}(\mathbf {k} )+{\frac {1}{2}}\right)}$

was introduced. The zero of energy can be shifted, which leads to an expression in terms of the number operator,

${\displaystyle H=\sum _{\mathbf {k} ,\mu }\hbar \omega N^{(\mu )}(\mathbf {k} )}$

The effect of H on a single-photon state is

${\displaystyle H|\mathbf {k} ,\mu \rangle \equiv H\left({a^{\dagger }}^{(\mu )}(\mathbf {k} )|0\rangle \right)=\sum _{\mathbf {k'} ,\mu '}\hbar \omega 'N^{(\mu ')}(\mathbf {k} '){a^{\dagger }}^{(\mu )}(\mathbf {k} )|0\rangle =\hbar \omega \left({a^{\dagger }}^{(\mu )}(\mathbf {k} )|0\rangle \right)=\hbar \omega |\mathbf {k} ,\mu \rangle .}$

Apparently, the single-photon state is an eigenstate of H and ℏω = hν is the corresponding energy. In the very same way

${\displaystyle H\left|(\mathbf {k} ,\mu )^{m};(\mathbf {k} ',\mu ')^{n}\right\rangle =\left[m(\hbar \omega )+n(\hbar \omega ')\right]\left|(\mathbf {k} ,\mu )^{m};(\mathbf {k} ',\mu ')^{n}\right\rangle ,\qquad {\text{with}}\quad \omega =c|\mathbf {k} |\quad {\hbox{and}}\quad \omega '=c|\mathbf {k} '|.}$

Introducing the Fourier expansion of the electromagnetic field into the classical form

${\displaystyle \mathbf {P} _{\textrm {EM}}=\epsilon _{0}\iiint _{V}\mathbf {E} (\mathbf {r} ,t)\times \mathbf {B} (\mathbf {r} ,t){\textrm {d}}^{3}\mathbf {r} ,}$

yields

${\displaystyle \mathbf {P} _{\textrm {EM}}=V\epsilon _{0}\sum _{\mathbf {k} }\sum _{\mu =1,-1}\omega \mathbf {k} \left(a_{\mathbf {k} }^{(\mu )}(t){\bar {a}}_{\mathbf {k} }^{(\mu )}(t)+{\bar {a}}_{\mathbf {k} }^{(\mu )}(t)a_{\mathbf {k} }^{(\mu )}(t)\right).}$

Quantization gives

${\displaystyle \mathbf {P} _{\textrm {EM}}=\sum _{\mathbf {k} ,\mu }\hbar \mathbf {k} \left({a^{\dagger }}^{(\mu )}(\mathbf {k} )a^{(\mu )}(\mathbf {k} )+{\frac {1}{2}}\right)=\sum _{\mathbf {k} ,\mu }\hbar \mathbf {k} N^{(\mu )}(\mathbf {k} ).}$

The term 1/2 could be dropped, because when one sums over the allowed k, k cancels with −k. The effect of P_EM on a single-photon state is

${\displaystyle \mathbf {P} _{\textrm {EM}}|\mathbf {k} ,\mu \rangle =\mathbf {P} _{\textrm {EM}}\left({a^{\dagger }}^{(\mu )}(\mathbf {k} )|0\rangle \right)=\hbar \mathbf {k} \left({a^{\dagger }}^{(\mu )}(\mathbf {k} )|0\rangle \right)=\hbar \mathbf {k} |\mathbf {k} ,\mu \rangle .}$

Apparently, the single-photon state is an eigenstate of the momentum operator, and ℏk is the eigenvalue (the momentum of a single photon).

The photon having non-zero linear momentum, one could imagine that it has a non-vanishing rest mass m₀, which is its mass at zero speed. However, we will now show that this is not the case: m₀ = 0.

Since the photon propagates with the speed of light, special relativity is called for. The relativistic expressions for energy and momentum squared are,

${\displaystyle E^{2}={\frac {m_{0}^{2}c^{4}}{1-v^{2}/c^{2}}},\qquad p^{2}={\frac {m_{0}^{2}v^{2}}{1-v^{2}/c^{2}}}.}$

From p²/E²,

${\displaystyle {\frac {v^{2}}{c^{2}}}={\frac {c^{2}p^{2}}{E^{2}}}\quad \Longrightarrow \quad E^{2}={\frac {m_{0}^{2}c^{4}}{1-c^{2}p^{2}/E^{2}}}\quad \Longrightarrow \quad m_{0}^{2}c^{4}=E^{2}-c^{2}p^{2}.}$

Use

${\displaystyle E^{2}=\hbar ^{2}\omega ^{2}\qquad {\text{and}}\qquad p^{2}=\hbar ^{2}k^{2}={\frac {\hbar ^{2}\omega ^{2}}{c^{2}}}}$

and it follows that

${\displaystyle m_{0}^{2}c^{4}=E^{2}-c^{2}p^{2}=\hbar ^{2}\omega ^{2}-c^{2}{\frac {\hbar ^{2}\omega ^{2}}{c^{2}}}=0,}$

so that m₀ = 0.

The photon can be assigned a triplet spin with spin quantum number S = 1. This is similar to, say, the nuclear spin of the ¹⁴N isotope, but with the important difference that the state with M_S = 0 is zero, only the states with M_S = ±1 are non-zero.

Define spin operators:

${\displaystyle S_{z}\equiv -i\hbar \left(\mathbf {e} _{x}\otimes \mathbf {e} _{y}-\mathbf {e} _{y}\otimes \mathbf {e} _{x}\right)\qquad {\hbox{and cyclically}}\quad x\to y\to z\to x.}$

The two operators ${\displaystyle \otimes }$ between the two orthogonal unit vectors are dyadic products. The unit vectors are perpendicular to the propagation direction k (the direction of the z axis, which is the spin quantization axis).

The spin operators satisfy the usual angular momentum commutation relations

${\displaystyle [S_{x},S_{y}]=i\hbar S_{z}\qquad {\hbox{and cyclically}}\quad x\to y\to z\to x.}$

Indeed, use the dyadic product property

${\displaystyle \left(\mathbf {e} _{y}\otimes \mathbf {e} _{z}\right)\left(\mathbf {e} _{z}\otimes \mathbf {e} _{x}\right)=\left(\mathbf {e} _{y}\otimes \mathbf {e} _{x}\right)\left(\mathbf {e} _{z}\cdot \mathbf {e} _{z}\right)=\mathbf {e} _{y}\otimes \mathbf {e} _{x}}$

because e_z is of unit length. In this manner,

${\displaystyle {\begin{aligned}\left[S_{x},S_{y}\right]&=-\hbar ^{2}\left(\mathbf {e} _{y}\otimes \mathbf {e} _{z}-\mathbf {e} _{z}\otimes \mathbf {e} _{y}\right)\left(\mathbf {e} _{z}\otimes \mathbf {e} _{x}-\mathbf {e} _{x}\otimes \mathbf {e} _{z}\right)+\hbar ^{2}\left(\mathbf {e} _{z}\otimes \mathbf {e} _{x}-\mathbf {e} _{x}\otimes \mathbf {e} _{z}\right)\left(\mathbf {e} _{y}\otimes \mathbf {e} _{z}-\mathbf {e} _{z}\otimes \mathbf {e} _{y}\right)\\&=\hbar ^{2}\left[-\left(\mathbf {e} _{y}\otimes \mathbf {e} _{z}-\mathbf {e} _{z}\otimes \mathbf {e} _{y}\right)\left(\mathbf {e} _{z}\otimes \mathbf {e} _{x}-\mathbf {e} _{x}\otimes \mathbf {e} _{z}\right)+\left(\mathbf {e} _{z}\otimes \mathbf {e} _{x}-\mathbf {e} _{x}\otimes \mathbf {e} _{z}\right)\left(\mathbf {e} _{y}\otimes \mathbf {e} _{z}-\mathbf {e} _{z}\otimes \mathbf {e} _{y}\right)\right]\\&=i\hbar \left[-i\hbar \left(\mathbf {e} _{x}\otimes \mathbf {e} _{y}-\mathbf {e} _{y}\otimes \mathbf {e} _{x}\right)\right]\\&=i\hbar S_{z}\end{aligned}}}$

By inspection it follows that

${\displaystyle -i\hbar \left(\mathbf {e} _{x}\otimes \mathbf {e} _{y}-\mathbf {e} _{y}\otimes \mathbf {e} _{x}\right)\cdot \mathbf {e} ^{(\mu )}=\mu \hbar \mathbf {e} ^{(\mu )},\qquad \mu =\pm 1,}$

and therefore μ labels the photon spin,

${\displaystyle S_{z}|\mathbf {k} ,\mu \rangle =\mu \hbar |\mathbf {k} ,\mu \rangle ,\quad \mu =\pm 1.}$

Because the vector potential A is a transverse field, the photon has no forward (μ = 0) spin component.

• QED vacuum

This article incorporates material from the Citizendium article "Quantization of the electromagnetic field", which is licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported License but not under the GFDL.