That's a bold and provocative statement, and it inspired me to look into second quantisation in more detail. However I feel second quantisation is a misstep, for various reasons. In this essay I will explain why.
Although there's a bunch of maths in this essay, there's also a bunch of animated diagrams that I hope will build some intuition as well.
First Quantisation
First off, what is quantisation? Quantisation (called 'canonical quantisation' in Quantum theory) is a kind of informal algorithm for taking a classical (pre-quantum) equation and converting it into a 'quantum' equation, involving some form of the Schrödinger equation.And once you look at stable solutions of the resulting Schrödinger equation, you find that only a discrete set of energy levels are found. These are the 'quanta' of energy.
First quantisation is what was done near the historical beginnings of quantum theory in the 1920s. Second quantisation is extending that 'algorithm' to field equations. More on that later.
Let's do an example of first quantisation with the classical harmonic oscillator equation (the equation for a mass on a spring):
Mass on a spring $$ \begin{equation} M \frac{d^2x}{dt^2} = -K x \end{equation} $$ Where K is the spring constant here, and M is the mass.
This equation basically says the acceleration is proportional to the negative displacement of the mass in the x direction.
From Newton's second law, (F=Ma), in this equation $$ F = -K x $$ with $$ a = \frac{d^2 x}{dt^2} $$ The energy of the oscillator is kinetic energy plus potential energy. The potential energy is a function V(x) which when differentiated with respect to position gives the negative force. So $$ V(x) = \frac{1}{2} K x^2 $$ since $$ \frac{dV(x)}{dx} = Kx $$ The kinetic energy is good old $$ E_k(x, t) = \frac{1}{2} M v^2 = \frac{1}{2} M (\frac {dx} {dt})^2 $$ So the overall energy is $$ E_t(x, t) = E_k(x, t) + V(x) $$ $$ E_t(x, t) = \frac{1}{2} M (\frac {dx} {dt})^2 + \frac{1}{2} K x^2 $$ Ok so now the quantisation algorithm does something like: Replace the classical momentum, \( p = Mv = M \frac{dx}{dt} \) with the operator (function from function to function) $$ -i \hbar \frac{ \partial }{\partial x} $$ Applied to \( \psi(x, t) \)
Replace the variable \( x \) with the operator \( \hat{x} \) in the potential part of the equation, giving $$ x \psi(x, t) $$ And finally replace \(E_t(x, t)\) with $$ i \hbar \frac{\partial \psi(x, t)}{\partial t} $$
Writing down the classical harmonic oscillator equation again, this time using \(p\): $$ E_t(x, t) = \frac{p^2}{2M} + \frac{1}{2} K x^2 $$ And making our replacements: $$ i \hbar \frac{\partial \psi(x, t)}{\partial t} = \frac{(-i \hbar \frac{ \partial }{\partial x})^2 }{2M} \psi(x, t) + \frac{1}{2} K x^2 \psi(x, t) $$ We get $$ i \hbar \frac{\partial \psi(x, t)}{\partial t} = -\frac{\hbar^2}{2 M} \frac{ \partial^2 \psi(x, t)}{\partial x^2} + \frac{1}{2} K x^2 \psi(x, t) $$ This is the time-dependent Schrödinger equation for a harmonic oscillator. Now where it gets really interesting is what the stable (stationary) states are. Stable states in this context are where \( |\psi(x, t)|^2 \) doesn't change over time. These states are interesting for two related reasons - they are the states you get when you make a measurement of the energy of the system (they are the energy eigenstates), and also, in the real world for a charged particle like an electron, these are states that don't immediately decay to a lower energy state by radiating away energy. This is because the electric field is related to \( |\psi(x, t)|^2 \), so if it doesn't change, no energy is radiated away.
Let's visualise the first few (lowest energy) stationary states. These states have energy levels $$ E_n = \hbar \omega (n + \frac{1}{2}) $$ where $$ \omega = \sqrt{K/M} $$ See Quantum_harmonic_oscillator at wikipedia for more details on the solutions.
Some things to notice: energy is basically the speed of rotation of the wavefunction in the complex plane, with \(\hbar\) being the conversion factor between angular velocity and joules.
In the visualisations above, I have set \( \hbar = 1 \) and \( \omega = 10 \). Therefore there is a gap of 10 units between energy levels.
One very important takeaway is that the \(n=0\) solution, called the ground state solution, has non-zero energy. (i.e. it is still rotating in the complex plane at a non-zero speed) This is the 'zero-point' energy.
Another note is that the ground state is the lowest energy normalisable (i.e. non-zero) state. The zero state \( \psi(x, t) = 0 \) is a solution to the Schrödinger equation, but is not usually considered a 'proper' solution.
Onwards to Field Theory
Consider the Klein-Gordon equation. This is one of the few very important quantum equations, along with the Schrödinger equation and the Dirac equation. It's the equation for the Higgs field, each component of the Dirac field obeys it, and if you set the mass to zero, it becomes a wave equation and hence can describe the electromagnetic field components, with some caveats (free, Coulomb gauge).Anyway, it's instructive to study, so let's proceed.
The Klein-Gordon equation is $$ (\frac{1}{c^2} \frac{\partial^2}{\partial t^2} - \nabla^2 + \frac{m^2 c^2}{\hbar^2}) \psi(t, x) = 0 $$ In 1 dimension that is $$ (\frac{1}{c^2} \frac{d^2}{d t^2} - \frac{d^2}{dx^2} + \frac{m^2 c^2}{\hbar^2}) \psi(t, x) = 0 $$ Consider the expansion of a function in terms of modes, e.g. the fourier transform of the equation: $$ \psi(t, x) = \int \frac{1}{2 \pi} a_k(t) e^{ikx} dk $$ Plugging that back into the Klein-Gordon equation: $$ (\frac{1}{c^2} \frac{d^2}{d t^2} - \frac{d^2}{dx^2} + \frac{m^2 c^2}{\hbar^2}) \int \frac{1}{2 \pi} a_k(t) e^{ikx} dk = 0 $$ $$ \frac{1}{c^2} \frac{d^2}{d t^2} \int \frac{1}{2 \pi} a_k(t) e^{ikx} dk - \frac{d^2}{dx^2} \int \frac{1}{2 \pi} a_k(t) e^{ikx} dk + \frac{m^2 c^2}{\hbar^2} \int \frac{1}{2 \pi} a_k(t) e^{ikx} dk = 0 $$ We can drop the \( \frac{1}{2 \pi} \)factor since it doesn't change the equation: $$ \frac{1}{c^2} \frac{d^2}{d t^2} \int a_k(t) e^{ikx} dk - \frac{d^2}{dx^2} \int a_k(t) e^{ikx} dk + \frac{m^2 c^2}{\hbar^2} \int a_k(t) e^{ikx} dk = 0 $$ $$ \frac{1}{c^2} \int \frac{d^2}{d t^2} [ a_k(t) e^{ikx}] dk - \int \frac{d^2}{dx^2}[a_k(t) e^{ikx}] dk + \frac{m^2 c^2}{\hbar^2} \int a_k(t) e^{ikx} dk = 0 $$ $$ \frac{1}{c^2} \int \frac{d^2 a_k(t)}{d t^2} e^{ikx} dk - \int (-k^2) a_k(t) e^{ikx} dk + \frac{m^2 c^2}{\hbar^2} \int a_k(t) e^{ikx} dk = 0 $$ Since fourier modes are linearly independent, for the left hand side to be zero, it requires the equation for each k to be zero: $$ \frac{1}{c^2} \frac{d^2 a_k(t)}{d t^2} +k^2 a_k(t) + \frac{m^2 c^2}{\hbar^2} a_k(t) = 0 $$ Use units in which $$ c = \hbar = 1 $$ Then $$ \frac{d^2 a_k(t)}{d t^2} + k^2 a_k(t) + m^2 a_k(t) = 0 $$ $$ \frac{d^2 a_k(t)}{d t^2} + (k^2 + m^2) a_k(t) = 0 $$ or $$ \frac{d^2 a_k(t)}{d t^2} = - (k^2 + m^2) a_k(t) $$ This is the equation of a classical simple harmonic oscillator with spring constant over mass $$ \frac{K}{M} = (k^2 + m^2) $$ Check by substituting in: $$ \frac{d^2 a_k(t)}{d t^2} = - \frac{K}{M} a_k(t) $$ $$ M \frac{d^2 a_k(t)}{d t^2} = - K a_k(t) $$
Note that \(a_k\) is complex here. For the equation to be satisfied, the equation needs to be satisfied for both the real component and imaginary components of \(a_k\) separately, that is: $$ M \frac{d^2}{d t^2} \Re[a_k(t)] = - K \Re[a_k(t)] $$ and $$ M \frac{d^2}{d t^2} \Im[a_k(t)] = - K \Im[a_k(t)] $$ So a component (real or imaginary) of a single mode of the Klein-Gordon equation is equivalent / can be mapped to the classical simple harmonic oscillator equation.
To see why that is, let's take a look at the plane-wave solution for the Klein-Gordon equation (see https://homepages.physik.uni-muenchen.de/~helling/classical_fields.pdf).
The plane-wave solution for the Klein-Gordon equation is : $$ \Psi(t, x) = \exp(i(\omega t - k.x)) $$ where $$ \omega^2 - k^2 = m^2 $$ In one dimension it's a sinusoidal curve travelling left or right. Its breakdown into even and odd components varies sinusoidally, and it's the amplitude of the even and odd parts that is \(\Re[a_k(t)]\) and \(\Im[a_k(t)]\) above.
To me it's clear that there is no actual physical, real-world harmonic oscillator here. The apparent harmonic oscillator is just sinusoidally varying decomposition into odd and even parts.
Accordingly I think it's a mistake to apply canonical quantisation to these harmonic oscillator equations.
Lets try and visualise the vacuum state according to second quantisation.
In this visualisation, we add together the oscillation of some number of Klein-Gordon modes. You can set the number of modes in the numeric entry box below the visualisation. Note that following second quantisation, the number of different modes is infinite, and so vacuum state corresponds to setting the number of modes to infinity. You can test that setting to a higher value results in more high frequency noise.
Number of modes:
Problems with second quantisation
Fields are already wave-like, no need to quantise again to get wave-like behaviour
First quantisation serves a crucial purpose of finding an equation with wave-like properties (the wavefunction) while also having the discrete energy levels observed in nature. This was essential, for example, for replacing the point-like electron orbiting around the nucleus with a wavefunction in a stable standing-wave pattern.In the second quantisation case, however, the quantisation is applied to fields, which already have wavelike behaviour. So there is no benefit of getting wavelike behaviour, since we already have it. What we do get, again, is discrete energy levels in the form of photons. In my view this is a mistake, which I will discuss more later.
Infinite energy density of the vacuum
When adopting second quantisation, in the ground state of the vacuum, every single mode is in the quantum harmonic oscillator ground state, and hence has some non-zero amount of energy. The obvious problem is that there are an infinite number of modes. So the vacuum has an infinite energy density!According to general relativity, spacetime curvature depends on not just mass, but energy also. And the observed curvature of the vacuum is very close to zero, not infinite! This discrepancy is named the Cosmological constant problem or Vacuum catastrophe and has been called 'probably the worst theoretical prediction in the history of physics'.
Terribly non-local behaviour
Breaking fields into modes and then quantising each mode (second quantisation) results in each mode having discrete energies. Each mode is a sinusoidal wave that stretches across the entire universe. Therefore transitions between mode energy levels apply simultaneously across the entire universe! This is the ultimate in non-local behaviour, which is generally considered a bad thing in physical theories. (Quantum entanglement still needs to be explained somehow without this, but this issue goes way beyond quantum entanglement).Black hole information paradox
Second quantisation results in some oscillation of the field in the vacuum state as visualised earlier. These random oscillations are thought to give rise to Hawking radiation at black hole event horizons. Ultimately this is thought to result in black holes 'evaporating'. But in this process information is lost, resulting in a paradox as quantum mechanics is thought to be reversible. This is the so called 'Black hole information paradox'. While I don't consider this particularly important some physicists consider it very important, so I think it's worth noting that without second quantisation this wouldn't be a problem.On photons
Photons are the quanta of the electromagnetic field, first hypothesised by Planck and Einstein. Second quantisation is considered by many to be the technical basis/story for the existence of photons - increasing the energy level in a mode corresponds to adding one photon with the mode's momentum to the universe.So without second quantisation, what explains photons?
My answer is that photons do not exist. The electromagnetic field is not quantised. This is known as a semiclassical approach. You still get quantised emission and absorption from atoms, but that is explained by the discrete energy levels of Schrödinger equation stationary solutions of the electron around the nucleus.
And almost all the other pieces of evidence for photons turn out to be explainable semiclassically:
The photoelectric effect was explained semiclassically by Lamb and Scully in 1969 in the paper The Photoelectric Effect without Photons.
Compton scattering can be explained semiclassically, as first done by Schrödinger himself in 1927 in the paper "Über den Compton effekt".
The Casimir effect can be explained with the Van de Waals force, apparently.
The blackbody radiation spectrum can be explained by the quantised energy levels of the cavity walls, instead of the E.M. field.
The only phenomena that can't be satisfactorily explained, in my opinion, are phenomena related to quantum entanglement, such as anti-bunching, Bell's inequality etc. I don't pretend to have a proper and intuitive physical understanding of entanglement, but I suspect a semiclassical approach without photons can eventually be found that is compatible with entanglement phenomena.The full quote is "The little calculation we just did to derive (4.10) for the energy density of a mode with wave number \(k\) is one of the most important ideas in quantum field theory, and therefore one of the most important in modern science. Maybe in human history." - Page 88, The Biggest Ideas in the Universe, 2: Quanta and Fields.