Let me start with an analogy, namely a mass on a spring. If I am not mistaken,

  1. first quantization consists of replacing the dynamical variables by their operators and then solving the Schrödinger equation so as to obtain the stationary, energy eigenfunctions of the Hamiltonian, namely the various modes $\psi_k$ the wave function. Any wavefunction can then be expressed as a superposition $\Psi = \sum\limits_k c_k \psi_k$ of those modes. So far, we're only talking about waves, not particles.
  2. Second quantization consists of replacing the coefficients $c_k$ with the destruction operators [is this a operation a postulate by the way?] to then reveal that $\mid c_k\mid^2$ is an occupation number, or number of particles at the mode $k$.

How does the electromagnetic EM field quantization (i.e., photons) relate to the above? The classical EM field is already a wave, thereby making first quantization seem redundant. What I recall form my quantum optics course, however, is that the Maxwell equations were wrangled until we got the EM energy within a certain volume, then applied first quantization to basically end up with a quadratic potential in the conjugate, dynamical variables. Shuffling around the real and imaginary parts of those variables then produced the ladder operators, and that was the end of the story. That said, I can't relate that "textbook story" to the two quantizations I have outlined above. If I were to draw a (figurative) parallel with a mass on the spring, what would that mass be for the EM field? Would it be

  • a fictitious mass at every point along the path of the classical EM wave,
  • or a fictitious mass corresponding to the energy stored in the quantization volume?
  • 1
    $\begingroup$ Take a look at this question and the accepted answer. It's not an answer to your question, but it's related. $\endgroup$
    – DanielSank
    Jun 6, 2020 at 18:20
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    $\begingroup$ There's an answer here that contains: "If the phase-space variables of the classical theories are trajectories, we call the process of quantisation "first". If the phase-space variables are fields, we call it "second" quantisation. This is just a historical name, without any deep meaning." $\endgroup$ Jun 6, 2020 at 18:26
  • $\begingroup$ Like @AlfredCentauri said, the names "first" and "second" describe the type of classical model is being quantized: classical particles ("first"), or classical fields ("second"). Quantization can be used to construct a quantum theory of the electromagnetic field, which some people call "second" quantization because the classical theory we're starting with is a field theory. Many (most?) of us just call it "quantization," without the "first" or "second," when we already know what kind of classical model we're starting with. $\endgroup$ Jun 6, 2020 at 18:49
  • $\begingroup$ One of the motivations I find compelling towards the 2nd quantization is that photons (or EM field quanta) are always being created/annihilated when interacting with a detector. You can get away with 1st quant. for massive particles because at low energies they remain there after being detected (probability is conserved in this case, for photons it's not). Some books do talk about a possible wavefunction for the photon, but the concept is only useful if the photon doesn't interact $\endgroup$
    – ErickShock
    Jun 6, 2020 at 18:51
  • $\begingroup$ @AlfredCentauri I don't quite understand how fields can be phase space variables. Also, are we talking about classical fields (i.e., the potential vector) or the quantum mechanical probability amplitude wave function? $\endgroup$
    – Tfovid
    Jun 13, 2020 at 8:29

3 Answers 3


If you start out with the concept of light as made up from particles, then you can see the homogenic wave equation $$\partial_\mu \partial^\mu A^\nu = 0$$ as a massless, relativistic quantum mechanical equation.

Consider the two slit experiment for electrons and for light. In both cases an interference pattern is predicted by a wave equation, the Schrödinger and the wave equation. In both cases this pattern gives the probability distribution of particle detection.

Therefore it makes sense to say that the wave theory of light is equivalent to the first quantisation.


In some treatments, replacing coefficients with creation and annihilation operators is a postulate. I don't like to do it that way. I prefer to construct a Fock space and define the field operators to obey the Locality condition (which is required in a consistent relativistic quantum theory). This is closer to the original approach, pre-war, and achieves essentially the same result, but imv gives more physical motivation. It also means we are really talking of particles, although particles which obey quantum mechanics, not classical particles as conceived in Newtonian mechanics.

For electrons we can define position and momentum operators, but there is no position operator for a photon. We can talk about the position where a photon was annihilated, but we cannot talk about the position where it is. This does mean that first quantisation does not work. We can still construct field operators for photons in much the same way we do for electrons, but by imposing the requirement of the locality condition, we can avoid any explicit mention of second quantisation (we do end up with essentially the same mathematical theory).

Then, from the intuitively appealing "minimal" interaction between electrons and photons it is possible to derive the equations of classical electromagnetism without ever having had to assume them. I have shown how to do this in A Construction of Full QED Using Finite Dimensional Hilbert Space. Also, with more detail in The Mathematics of Gravity and Quanta.

  • $\begingroup$ -1. Of course we can talk about spacial distribution of a photon. Every beam splitter splits the photon, every interferometer is based on this principle. $\endgroup$
    – Alexander
    Jun 6, 2020 at 22:35
  • $\begingroup$ @Alexander, QM does not describe the spacial distribution of a photon. It describes the probability distribution and the probability amplitude for where the photon will be detected. $\endgroup$ Jun 7, 2020 at 5:42

One of the defining features of second quantization is that it allows variable number of electrons/photons/etc. As in first quantization you write the hamiltonian per particle (and commit to its perseverance), in second quantization the emphasis is on operators.

Any meaningful calculation regarding the electromagnetic field will describe some particle nonpreserving process, such as absorption/emission, thus the use of second quantization is required. Even the most basic scattering even of 2 photons to 2 photons cannot be described by first quantization, as the original photons are annihilated, virtual electrons created and annihilated and then two new photons are created. You can write first quantization equation for non-interacting theory. As @my2cts mentioned, classical theory of light, can be viewed as first quantized, and as a matter of fact, the paraxial approxiamtion in optics exactly reconstructs the Schrodinger equation.

In analogy with statistical mechanics, it can be viewed that first quantization resembles microcanonical ensemble, while second quantization resembles the grand canonical ensemble.


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