Reading through Peskin & Schroeder, more specifically quantization of KG equation (section 2.3), I've come across the following footnote

$^\dagger$ This procedure is sometimes called second quantization, to distinguish the resulting Klein-Gordon equation (in which $\phi$ is an operator) from the old one-particle Klein-Gordon equation
I don't understand why this remark is relevant (it sounds like "I'm doing this thing but I'm not doing it")$^1$. I've encountered a similar remark in David Tong's [notes](https://www.damtp.cam.ac.uk/user/tong/qft.html) on QFT (page 9 and then section 2.8) but regarding Schrödinger equation. In that case Tong suggested that given a lagrangian that yields the correct Schrödinger equation, but
This looks very much like the Schrödinger equation. Except it isn’t! Or, at least, the interpretation of this equation is very different: the field $\psi$ is a classical field with none of the probability interpretation of the wavefunction.
Now, it is not clear to me whether Tong means that a lagrangian like $$\mathcal{L}=\frac{i}{2}[\dot{\psi}\psi^*-\psi\dot{\psi^*}]-\nabla\psi\cdot\nabla\psi^*$$ does not *intrinsically* contain the probability interpretation (so it is a complex field obeying that equation) but one may *decide* to use such interpretations, making $\psi$ an *actual* wavefunction or it *cannot* be intepreted$^2$ as the wavefunction and in such case *why* is it so? Any other reference explaining this in detail would be appreciated.

I'm aware some related questions exist, so I want to stress the question is the one a few lines above: are we saying $\psi$ cannot be the wavefunction of a system or just that without further information it is just a field?

$^1$ According to my class notes, this is just a formal procedure and what we're really doing is defining the field operators and writing down Heisenberg equation (e.g. in the Schrödinger case below). I'm not fully safisfied with that explanation, though as imposing the field CCR seems more general.

$^2$ Later, Tong quantizes this field and developing some calculations, it derives the Schrödinger equation from the quantum field equation. That is not what I'm asking, I'm asking if one can attribute the wavefunction interpretation from the beginning.

Appendix - My notes

In my notes, for the Schrödinger case, having the single-particle complete set of wavefunctions $\{\lvert\psi_\alpha\rangle\}$, we constructed Fock space and defined the annihilation operator $$\hat{a}_i\lvert n_1, n_2,... n_i,...\rangle=\sqrt{n_i}\lvert n_1, n_2,... n_i,...\rangle$$ (and the creation operator as its adjoint). After that, we defined the field operator as follows $$\hat{\Psi}(\vec{r}):=\sum_\alpha\langle\vec{r}\lvert\psi_\alpha\rangle\hat{a}_\alpha=\sum_\alpha\psi_\alpha(\vec{r})\hat{a}_\alpha$$ so $$\hat{\Psi}^\dagger(\vec{r})=\sum_\alpha\psi^*_\alpha(\vec{r})\hat{a}^\dagger_\alpha$$ which satisfy the usual commutation relations as one can easily prove. Incidentally, using the completeness of momentum basis it may be more familiar to write it in the form $$\hat{\Psi}(\vec{r})=\int\frac{d^3\vec{p}}{(2\pi)^3}e^{i\vec{p}\cdot\vec{r}}\underbrace{\sum_\alpha\langle\vec{p}\lvert\alpha\rangle\hat{a}_\alpha}_{\hat{a}_\vec{p}}.$$ Then we switched to Heisenberg picture $$\hat{\Psi}(\vec{r},t)=e^{i\hat{H}t}\hat{\Psi}(\vec{r})e^{-i\hat{H}t}$$ Finally we wrote down Heisenberg EoM $$i\frac{\partial}{\partial t}\hat{\Psi}(\vec{r},t)=[\hat{\Psi},\hat{H}].$$

  • 4
    $\begingroup$ "I don't understand why this remark is relevant." It is relevant because the authors are trying to get ahead of common misconceptions before the reader misconceives of them. They are trying to tell the reader that despite being equations composed of very similar looking symbols, the underlying meaning is very different. $\endgroup$
    – hft
    Oct 28, 2022 at 22:26
  • 1
    $\begingroup$ @Feynman_00 Remember that David Tong derived the Schrodinger equation as a limit of the Klein Gordon equation. Since the KG eqn. has no probabilistic interpretation, the Schrodinger eqn. that David Tong derived in that context also didn't have a probabilisitc interpretation $\endgroup$
    – Ryder Rude
    Oct 29, 2022 at 6:18
  • 1
    $\begingroup$ @Feynman_00 I could see it as the poster. Maybe they don't let others see it. But as I said, both approaches to QFT are considered valid, whether you start with the creation/annihilation operators and manually define the $\phi(x)$ fields, or you quantise a classical field theory. As you said, the latter seems to be more general as it's a systematic quantisation procedure, of the same nature as non relativistic QM. $\endgroup$
    – Ryder Rude
    Oct 30, 2022 at 0:37
  • 1
    $\begingroup$ This and this post also discuss similar things. Weinberg's book uses the "manually define fields approach". $\endgroup$
    – Ryder Rude
    Oct 30, 2022 at 0:42
  • 1
    $\begingroup$ The idea of the manual approach is that, in principle, you don't need field operators to calculate scattering amplitudes. You could write the interaction Hamiltonian directly in terms of creation/annihilation operators. But some properties like Locality and Lorentz invariance of interactions lets us write the interaction using field operators. The people who support the manual approach usually see particles as fundamental and fields as calculation tools. See this $\endgroup$
    – Ryder Rude
    Oct 30, 2022 at 2:39

1 Answer 1


Quantum mechanics of a single particle

Quantum mechanics can be formulated in many ways. Let's review 2.

1. The Schrodinger picture

Position $x$ and momentum $p$ are time-independent operators, and the wavefunction $\psi(x, t)$ is a complex valued function of the particle's position and of time. The dynamics is defined via the Schrodinger equation \begin{equation} i\hbar \frac{\partial \psi}{\partial t} = H(x, p=-i\hbar \nabla) \psi \end{equation}.

2. The Heisenberg picture

Position $x(t)$ and momentum $p(t)$ are time dependent operators, and the state is time independent (you can think of the state as defining the initial conditions). The dynamics are described by Heisenberg's equations \begin{eqnarray} \frac{dx}{dt} &=& \frac{i}{\hbar} [H, x] \\ \frac{dp}{dt} &=& \frac{i}{\hbar} [H, p] \end{eqnarray} This mimics Hamilton's equations from classical mechanics when written in terms of the Poisson bracket (here I'll use the subscript ${\rm cl}$ to denote that in this equation $x_{\rm cl}$ and $p_{\rm cl}$ are classical functions, not operators) \begin{eqnarray} \frac{dx_{\rm cl}}{dt} &=& \{H_{\rm cl}, x_{\rm cl}\} \\ \frac{dp_{\rm cl}}{dt} &=& \{H_{\rm cl}, p_{\rm cl}\} \end{eqnarray}

Quantum field theory

Our goal is now to describe relativistic quantum mechanics. Let's first explain an incorrect approach, then describe two correct ways to do it.

0. An incorrect approach

The Schrodinger equation looks non-relativistic; it has one time derivative on the left hand side, and two spatial derivatives on the right hand side. So let's guess that the right way to describe a relativistic quantum particle is to generalize the Schrodinger equation. This was the original idea that led to the Klein-Gordon equation (describing a free particle) \begin{equation} -\frac{1}{c^2} \frac{\partial^2 \psi}{\partial t^2} + \nabla^2 \psi = 0 \end{equation} It turns out that this equation is relevant for relativistic quantum field theory, however this equation is not a generalization of the Schrodinger equation. This is the main point that the textbook authors want you to realize.

In fact, the entire notion of creating relativistic quantum mechanics is doomed from the start, because a relativistic quantum theory is necessarily one where the particle number can change. Tong's lecture notes, for example, give a physical argument for this; due to the uncertainty principle, if you localize a particle well enough it will have relativistic momentum and there is enough energy to transition to states where particle-anti-particle pairs are created.

1. The Schrodinger picture

The Schrodinger picture is rarely used in quantum field theory, but it can be formulated.

The wavefunction becomes a wavefunctional. It is still a function of all the relevant degrees of freedom of the system. However, instead of the degrees of freedom being the position of a single particle, the degrees of freedom are the values of a field $\Phi(x)$ at every location $x$. Therefore, the wavefunctional is a complicated object, $\psi[\Phi(x), t]$, which assigns a single probability amplitude for every possible field configuration $\Phi(x)$. (This usually takes students some thinking before they fully get it).

With this properly generalized and much more complicated wavefunctional, the Schrodinger equation actually looks essentially the same: \begin{equation} i \hbar \frac{\partial \psi}{\partial t} = H \psi \end{equation} except now the Hamiltonian is a functional of the field operator $\Phi(x)$ and its conjugate momentum $\pi(x)$.

2. The Heisenberg picture

This approach is much more common, and we will finally see the Klein-Gordon equation appearing in the correct place.

The field operators obey the Heisenberg equations of motion \begin{eqnarray} \frac{\partial \Phi(x,t)}{\partial t} &=& [H, \Phi] \\ \frac{\partial \pi(x,t)}{\partial t} &=& [H, \pi] \end{eqnarray} For the case of a free field, the Hamiltonian is \begin{equation} H = \int d^3 x \frac{1}{c^2} \frac{1}{2} \pi^2(x) + \frac{1}{2} (\nabla \Phi)^2 \end{equation} You can work out all these commutators and you will find that $\Phi$ obeys the Klein-Gordon equation \begin{equation} -\frac{1}{c^2} \frac{\partial^2\Phi}{\partial t^2} + \nabla^2 \Phi = 0 \end{equation} Here we see that the same equation appears that we might have guessed by generalizing the one particle Schrodinger equation. But the correct interpretation is not that $\Phi$ is a wavefunction of a particle. The interpretation of this equation is that it describes the evolution of the operator $\Phi(x, t)$; it is properly understood as a generalization of the Heisenberg equation, not the Schrodinger equation. It is analogous to the corresponding equation of a classical field, much like how Heisenberg's equations in quantum mechanics are analogous to the classical Hamiltonian equations for a single particle.

  • $\begingroup$ In the case of Schrödinger equation my notes say that we are defining the operator $\hat{\Psi}=\sum_\nu \psi_\nu(x)a_\nu$ and then writing down Heisenberg equation. Also my notes say that the procedure here described is just formal (and suggest to ignore it) and what is really going on is the definition above (note: I'm not saying I agree with my notes). 1) What is your comment about that? 2) Can we do something similar for e.g. KG equation? $\endgroup$ Oct 29, 2022 at 5:56
  • 1
    $\begingroup$ @Feynman_00 You should check out this post. There are many approaches to QFT. Some start with creation/annihilation operators and manually define the spacetime field operators. But another approach quantises a classical field, and derives the creation/annihilation operators from there. I also can't figure out which approach is more fundamental $\endgroup$
    – Ryder Rude
    Oct 29, 2022 at 6:11
  • 1
    $\begingroup$ @Feynman_00 I don't think I can answer that without actually seeing your notes. The way I understand it is the way I wrote it in my answer. $\endgroup$
    – Andrew
    Oct 29, 2022 at 6:19
  • $\begingroup$ I've updated my question with an appendix about that. @RyderRude The link you have provided leads to a removed question. $\endgroup$ Oct 29, 2022 at 7:13
  • 2
    $\begingroup$ @Feynman_00 The way I would interpret what you've written in the appendix of your question is that you are considering a system whose Hamiltonian happens to lead to the Schrodinger equation as the classical equation of motion. This fits in the framework of my answer, but since we are "doubling up" on concepts it is a little confusing. If you quantize in the Schrodinger picture, you'd have a wavefunctional depending on this field, obeying the functional Schrodinger equation. In the Heisenberg picture, the field happens to obey the Schrodinger equation, but the field is not a wavefunction. $\endgroup$
    – Andrew
    Oct 29, 2022 at 12:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.