Forward Scattering - The Weblog of Nicholas Chapman

The weblog of Nicholas Chapman

Second Quantisation - a Quantisation Too Far<br>Posted 17 May 2026<br>In Sean Carroll's great book "The Biggest Ideas in the Universe, 2: Quanta and Fields", he writes "[Second quantisation] 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." (See note)

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

$$<br>\begin{equation}<br>M \frac{d^2x}{dt^2} = -K x<br>\end{equation}<br>$$<br>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

$$<br>F = -K x<br>$$<br>with<br>$$<br>a = \frac{d^2 x}{dt^2}<br>$$

The energy of the oscillator is kinetic energy plus potential energy.<br>The potential energy is a function V(x) which when differentiated with respect to position gives the negative force.

So

$$<br>V(x) = \frac{1}{2} K x^2<br>$$<br>since

$$<br>\frac{dV(x)}{dx} = Kx<br>$$

The kinetic energy is good old<br>$$<br>E_k(x, t) = \frac{1}{2} M v^2 = \frac{1}{2} M (\frac {dx} {dt})^2<br>$$

So the overall energy is

$$<br>E_t(x, t) = E_k(x, t) + V(x)<br>$$<br>$$<br>E_t(x, t) = \frac{1}{2} M (\frac {dx} {dt})^2 + \frac{1}{2} K x^2<br>$$

Ok so now the quantisation algorithm does something like:<br>Replace the classical momentum, \( p = Mv = M \frac{dx}{dt} \) with the operator (function from function to function)<br>$$<br>-i \hbar \frac{ \partial }{\partial x}<br>$$<br>Applied to \( \psi(x, t) \)

Replace the variable \( x \) with the operator \( \hat{x} \) in the potential part of the equation, giving

$$<br>x \psi(x, t)<br>$$

And finally replace \(E_t(x, t)\)

with<br>$$<br>i \hbar \frac{\partial \psi(x, t)}{\partial t}<br>$$

Writing down the classical harmonic oscillator equation again, this time using \(p\):<br>$$<br>E_t(x, t) = \frac{p^2}{2M} + \frac{1}{2} K x^2<br>$$<br>And making our replacements:<br>$$<br>i \hbar \frac{\partial \psi(x, t)}{\partial t} = \frac{(-i \hbar \frac{ \partial }{\partial x})^2 }{2M} \psi(x, t) + \frac{1}{2} K x^2 \psi(x, t)<br>$$

We get

$$<br>i \hbar \frac{\partial \psi(x, t)}{\partial t} = -\frac{\hbar^2}{2 M} \frac{ \partial^2 \psi(x, t)}{\partial x^2} + \frac{1}{2} K x^2 \psi(x, t)<br>$$

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

$$<br>E_n = \hbar \omega (n + \frac{1}{2})<br>$$<br>where<br>$$<br>\omega = \sqrt{K/M}<br>$$

See Quantum_harmonic_oscillator at wikipedia for more details on the solutions.

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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.<br>The zero...