Quantization of the electromagnetic field

When an electromagnetic field is quantized, the field energy becomes discontinuous. After quantization, an electromagnetic (EM) field consist of discrete energy parcels, photons. Photons are massless particles of definite energy, definite momentum, and definite spin.

In order to explain the photoelectric effect, Einstein assumed heuristically in 1905 that an electromagnetic field consists of parcels of energy hν, where h is Planck's constant. In 1927 Paul A. M. Dirac was able to weave the photon concept into the fabrics of the new quantum mechanics and to describe the interaction of photons with matter. He applied a technique which is now generally called second quantization, although this term is somewhat of a misnomer for EM fields, because they are, after all, solutions of the classical Maxwell equations. In Dirac's theory the fields are quantized for the first time and it is also the first time that Planck's constant ℏ enters the expressions. In its original work, Dirac took the phases of the different EM waves ("modes") and the mode energies as dynamic variables to quantize (i.e., he reinterpreted them as operators plus commutation relations between them). At present it is more common to quantize the Fourier component of the vector potential. This is what will be done below.

Second quantization

Second quantization starts with an expansion of a scalar of 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.

Electromagnetic field and vector potential

See here for more detail.

As the term suggests, an EM field consists of two vector fields, an electric field E(r,t) and a magnetic field B(r,t). Both are time-dependent vector fields that in vacuum depend on a third vector field A(r,t) (the vector potential) through

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

The Fourier expansion of the vector potential enclosed in a finite box of volume V is

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

where the wave vector k gives the propagation direction of A(r,t); the length of the wave vector is |k| = 2πν/c = ω/c, with ν the frequency of the mode (Fourier component) and c is the speed of light. The two perpendicular unit vectors e^(μ) ("polarization vectors") are perpendicular to k. The k-component of A (a vector perpendicular to k) lies in the plane spanned by e⁽¹⁾ and e⁽⁻¹⁾. The Fourier coefficients $a_{\mathbf {k} }^{(\mu )}(t),\;{\bar {a}}_{\mathbf {k} }^{(\mu )}(t)$ can be seen as the variables defining the vector potential, in the following they will be promoted to operators.

Quantization of EM field

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

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

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 and apply the quantization rules:

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

subject to the boson commutation relations

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

The square brackets indicate a commutator, defined by

${\big [}A,B{\big ]}\equiv AB-BA$

for any two quantum mechanical operators A and B.

The quantized fields (operator fields) are the following

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

where ω = c |k| = ck.

Hamiltonian of the field

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

${\begin{aligned}H&={\frac {1}{2}}\sum _{\mathbf {k} ,\mu =-1,1}\hbar \omega {\Big (}{a^{\dagger }}^{(\mu )}(\mathbf {k} )\,a^{(\mu )}(\mathbf {k} )+a^{(\mu )}(\mathbf {k} )\,{a^{\dagger }}^{(\mu )}(\mathbf {k} ){\Big )}\\&=\sum _{\mathbf {k} ,\mu }\hbar \omega {\Big (}{a^{\dagger }}^{(\mu )}(\mathbf {k} )a^{(\mu )}(\mathbf {k} )+{\frac {1}{2}}{\Big )}\end{aligned}}$

By the use of the commutation relations the second line follows from the first. Note that ℏω = h ν = ℏ c |k|, which is the well-known Einstein expression for photon energy. 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.

Digression: harmonic oscillator

The second quantized treatment of the one-dimensional quantum harmonic oscillator is a well-known topic in quantum mechanical courses. We digress and say a few words about it. The harmonic oscillator Hamiltonian has the form

$H=\hbar \omega {\big (}a^{\dagger }a+{\tfrac {1}{2}}{\big )}$

where ω ≡ 2πν is the fundamental frequency of the oscillator. The ground state of the oscillator is designated by | 0 ⟩ and is referred to as vacuum state. It can be shown that $\scriptstyle a^{\dagger }$ is an excitation operator, it excites from an n fold excited state to an n+1 fold excited state:

$a^{\dagger }|n\rangle =|n+1\rangle {\sqrt {n+1}}\quad {\hbox{in particular}}\quad a^{\dagger }|0\rangle =|1\rangle \quad {\hbox{and}}\quad (a^{\dagger })^{n}|0\rangle \propto |n\rangle .$

Since harmonic oscillator energies are equidistant, the n-fold excited state | n⟩ can be looked upon as a single state containing n particles (sometimes called vibrons) all of energy hν. These particles are bosons. For obvious reason the excitation operator $a^{\dagger }$ is called a creation operator.

From the commutation relation follows that the Hermitian adjoint $\,a$ de-excites:

$a|n\rangle =|n-1\rangle {\sqrt {n}}\quad {\hbox{in particular}}\quad a|0\rangle \propto 0\rightarrow a|0\rangle =0,$

because a function times the number 0 is the zero function. For obvious reason the de-excitation operator $\,a$ is called an annihilation operator.

Suppose now we have a number of non-interacting (independent) one-dimensional harmonic oscillators, each with its own fundamental frequency ω_i. Because the oscillators are independent, the Hamiltonian is a simple sum:

$H=\sum _{i}\hbar \omega _{i}{\Big (}a^{\dagger }(i)a(i)+{\tfrac {1}{2}}{\Big )}.$

Making the substitution

$i\rightarrow (\mathbf {k} ,\mu )$

we see that the Hamiltonian of the EM field can be looked upon as a Hamiltonian of independent oscillators of energy ω = |k| c and oscillating along direction e^(μ) with μ=1,−1.

Photon states

The quantized EM field has a vacuum (no photons) state | 0 ⟩. The application to it of, say,

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

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

The operator

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

is the number operator. When acting on a quantum mechanical photon 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

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

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

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

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

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

A photon state is an eigenstate of the number operator. This is why the formalism described here, is often referred to as the occupation number representation.

Photon energy

Earlier the Hamiltonian,

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

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

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

The effect of H on a single-photon state is

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

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

$H{\big |}(\mathbf {k} ,\mu )^{m};\,(\mathbf {k} ',\mu ')^{n}\,{\big \rangle }=\left[m(\hbar \omega )+n(\hbar \omega ')\right]{\big |}(\mathbf {k} ,\mu )^{m};\,(\mathbf {k} ',\mu ')^{n}\,{\big \rangle },$

with

$\omega =c|\mathbf {k} |\quad {\hbox{and}}\quad \omega '=c|\mathbf {k} '|.$

Example photon density

In this article the electromagnetic energy density was computed that a 100kW radio station creates in its environment; at 5 km from the station it was estimated to be 2.1·10⁻¹⁰ J/m³. Is quantum mechanics needed to describe the broadcasting of this station?

The classical approximation to EM radiation is good when the number of photons is much larger than unity in the volume

$\left({\frac {\lambda }{2\pi }}\right)^{3},$

where λ is the length of the radio waves. In that case quantum fluctuations are negligible and cannot be heard.

Suppose the radio station broadcasts at ν = 100 MHz, then it is sending out photons with an energy content of νh = 1·10⁸× 6.6·10⁻³⁴ = 6.6·10⁻²⁶ J, where h is Planck's constant. The wavelength of the station is λ = c/ν = 3 m, so that λ/(2π) = 48 cm and the volume is 0.111 m³. The energy content of this volume element is 2.1·10⁻¹⁰ × 0.111 = 2.3 ·10⁻¹¹ J, which amounts to

3.5 ·10¹² photons per $\left({\frac {\lambda }{2\pi }}\right)^{3}$

Obviously, 3.5 ·10¹² is much larger than one and hence quantum effects do not play a role; the waves emitted by this station are well into the classical limit, even when it plays non-classical music, for instance of Led Zepplin.

Photon momentum

Introducing the operator expansions for E and B into the classical form

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

yields

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

The 1/2 that appears can be dropped because when we sum over the allowed k, k cancels with −k. The effect of P_EM on a single-photon state is

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

Apparently, the single-photon state is an eigenstate of the momentum operator, and $\hbar \mathbf {k}$ is the eigenvalue (the momentum of a single photon).

Photon mass

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,

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

From p²/E²,

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

Use

$E^{2}=\hbar ^{2}\omega ^{2}\quad \mathrm {and} \quad p^{2}=\hbar ^{2}k^{2}={\frac {\hbar ^{2}\omega ^{2}}{c^{2}}}$

and it follows that

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

so that m₀ = 0.

Photon spin

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:

$S_{z}\equiv -i\hbar {\Big (}\mathbf {e} _{x}\otimes \mathbf {e} _{y}-\mathbf {e} _{y}\otimes \mathbf {e} _{x}{\Big )}\quad {\hbox{and cyclically}}\quad x\rightarrow y\rightarrow z\rightarrow x.$

The products 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). They are related to the polarization vectors through a unitary transformation

$\mathbf {e} ^{(1)}\equiv {\frac {-1}{\sqrt {2}}}(\mathbf {e} _{x}+i\mathbf {e} _{y})\quad {\hbox{and}}\quad \mathbf {e} ^{(-1)}\equiv {\frac {1}{\sqrt {2}}}(\mathbf {e} _{x}-i\mathbf {e} _{y}).$

The spin operators satisfy the usual angular momentum commutation relations

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

Indeed, use the dyadic product property

${\big (}\mathbf {e} _{y}\otimes \mathbf {e} _{z}{\big )}\;{\big (}\mathbf {e} _{z}\otimes \mathbf {e} _{x}{\big )}=(\mathbf {e} _{y}\otimes \mathbf {e} _{x})(\mathbf {e} _{z}\cdot \mathbf {e} _{z})=\mathbf {e} _{y}\otimes \mathbf {e} _{x}$

because e_z is of unit length. In this manner,

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

Define states

$|\mathbf {k} ,\mu \rangle \equiv {a^{\dagger }}^{(\mu )}(\mathbf {k} )\,|0\rangle \leftrightarrow \mathbf {e} ^{(\mu )}e^{i\mathbf {k} \cdot \mathbf {r} }.$

By inspection it follows that

$-i\hbar {\Big (}\mathbf {e} _{x}\otimes \mathbf {e} _{y}-\mathbf {e} _{y}\otimes \mathbf {e} _{x}{\Big )}\cdot \mathbf {e^{(\mu )}} =\mu \mathbf {e} ^{(\mu )},\quad \mu =1,-1,$

and therefore μ labels the photon spin,

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

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