In single particle QM we had single particle states $$ \left| \psi \right> $$ and wave functions $$ \psi (\mathbf x) = \left< x | \psi \right> $$

Then we introduced the second quantization with creation/annhilation operators, Fock states, etc. $$ a_i^\dagger \, \left| ... , n_i , ... \right> \ =\ \sqrt{n_i+1} \, \left| ... , n_i +1 , ... \right> $$ and field operators $$ \hat \Psi (\mathbf{x}) = \sum_\nu \psi_\nu (\mathbf{x}) \, a_\nu $$ where $\psi_\nu (\mathbf{x})$ are single particle wave functions.

Now in QFT we have quantum fields, e. g. the real Klein-Gordon field $$ \phi (x^\mu) = \int \frac{\mathrm{d}^3k}{(2 \pi)^3 \, 2 \,k_0} \left( a_k \, e^{-ikx} + a_k^\dagger \, e^{ikx} \right) $$

But what exactly are those quantum fields? I read this post but it's still not really clear to me. A classmate told me that the field operators of the second quantization are the same objects as the quantum KG fields, whereas the answer in said post seems to mention the opposite. So my questions are:

  • are those two the same? If yes, how can I see that, if one depends only on 3d-space coordiantes and includes only one type of creation/annhilation operator, whereas the other depends on space-time corrdiantes and both types of operators.
  • if they are different, what is a quantum field? On what kind of states does it act and what's the effect of acting on these states?
  • $\begingroup$ "Quantum fields" is a formalism to describe a system in terms of occupation numbers of particles (E. Fermi). K-G is an equation in the Heisenberg picture, whereas at the beginning of your post you do not precise in which picture you compose the $\Psi$-operator. $\endgroup$ Jun 1, 2018 at 18:26
  • $\begingroup$ Relevant. $\endgroup$ Jun 1, 2018 at 19:19
  • $\begingroup$ This post may be helpful. $\endgroup$
    – DanielSank
    Jun 1, 2018 at 20:07

2 Answers 2


Nobody really knows exactly what a quantum field is now because there is no rigorous mathematical formulation which is fully general, or even widely applicable. For example, the Standard Model of particle physics lacks such a foundation. But ignoring that for the moment, I think what would clear up your confusion most easily to have a look at the "Schrodinger picture" version of quantum field theory, as described in e.g. Hatfield, "Quantum Field Theory of Point Particles and Strings".

I am not going to reproduce all of that here but in short here is how you go from QM to QFT (for a free scalar field):

The wave function $\psi(x)$ becomes a wave functional $\Psi[\phi(\vec{x}),t]$, which gives the amplitude for observing a particular configuration of the entire field $\phi(\vec{x})$ at time $t$. You can also use a fourier basis and then the wave functional looks like $\Psi[\hat{\phi}(\vec{k})]$. (Note, this is not the $\hat{\Psi}$ from your question; I discuss that below).

I am not going to write the Lagrangian etc, but the result is that for a free field one gets a simple harmonic oscillator for each $\vec{k}$ mode. In other words there are operators $a_k$ and $a^\dagger_k$ which lower and raise the oscillator level (=occupation number) of that mode. The wave functional can then be written as $\Psi[n_\vec{k},t]$, i.e., the amplitude to find occupation numbers ${n_\vec{k}}$ at time $t$.

In QM one has the $X$ operator which multiplies the wave function by $x$. The corresponding operator in QFT is exactly the quantum field $\phi(\vec{x_0})$: it acts by multiplying $\Psi[\phi(\vec{x}),t]$. I wrote $\vec{x_0}$ to emphasize that you're only multiplying by one value, the value of $\phi$ at the given point $\vec{x_0}$.

So the quantum field is like the $X$ operator of QM, except there is a different one for every point of space. Now recall how for the oscillator, $X$ can be expressed in terms of $a,a^\dagger$. It is similar with fields: if you convert to the occupation basis $n_\vec{k}$ then the operation of multiplying by $\phi(\vec{x})$ gets expressed in terms of the $a_\vec{k}$, $a^\dagger_\vec{k}$. This leads (almost) to the integral form that you cited in your question.

I say "almost" because the integral as you wrote it shows the Heisenberg picture form of the operator. In the QM oscillator the analog is $X(t) \equiv \exp(iHt)X(0)\exp(-iHt)$. If you do this for the $a,a^\dagger$ operators then they will get time dependence $\exp(\pm i\omega t)$, which is what you see in your integral expression.

The other object that you showed, $\hat{\Psi}(x)$, is just the destruction-only part of $\phi(\vec{x})$, and it is written in a generic basis of wavefunctions $\psi_\nu$ rather than the fourier $\vec{k}$ basis. It has no time dependence because it is in the Schrodinger picture: time dependence is in the wave functional. This "destruction-only" part of the field isn't very useful in a relativistic theory because it can't be used to build Lorentz-invariant expressions; however, it's still a valid operator which "destroys a particle at position $x$".

So to summarize: The two operators $\hat{\Psi}$,$\phi$ that you wrote are not the same. The first is the destruction-only piece of the field operator, written in Schrodinger picture. The second is the full field operator (as that term is used in relativistic QFT), written in Heisenberg picture.

  • $\begingroup$ QFT exists since long ago and has an exact meaning as a theory with variable number of (quasi) particles for which the occupation number formalism is most appropriate. Yes, we encounter problems of infinite corrections, but we know their origin and we subtract them (renormalizations). So in the end we have what we wanted: a formalism of occupation numbers. $\endgroup$ Jun 1, 2018 at 19:58
  • 1
    $\begingroup$ I've edited the post to reflect these concerns. To the comment of VK I would note that if you can prove that Yang-Mills theory exists in four dimensions, you are well on the way to winning $1 million from the Clay Mathematics foundation. Just calling it a theory of occupation numbers with some subtractions is not sufficient, or at least it has not satisfied mathematicians who have looked at this. $\endgroup$ Jun 2, 2018 at 0:38
  • $\begingroup$ Because some mathematicians want exact explicit solutions as a proof which is impossible to obtain for too many reasons. $\endgroup$ Jun 2, 2018 at 7:52
  • $\begingroup$ @WilliamNelson Thanks a lot for your answer! Your book recommendation also looks very promising. $\endgroup$
    – scavi
    Jun 4, 2018 at 14:18

They are conceptually the same. There are some differences particular to those examples you have given.

First of all, you (or the sources that you have used) haven't used the same notations for two expressions. The former one has three-vector notation, while the latter has four-vector notation.

In the former one you have written the time-independent states in a closed form, $\psi$, while in the latter one you have actually substituted the time-dependent plane-wave solution, $\psi (\mathbf{x}, t) = e^{i (\omega t - \mathbf{k} \cdot \mathbf{x})}$ where $\omega$ is $k_0$.

Again, in the former one you assumed the states are countable via $\nu$, therefore you superposed the states with a summation. While in the latter, your states are continuous, due to the plane wave form, and therefore they are superposed with an integral measure.

The former one is time-independent and isn't covariant (and presumably not relativistic), however the latter one is covariant thanks to the $\omega$ in the donominator (you can work it out by first taking the integral in 3+1 dimensions as $d^4x$ and see it indeed becomes the form you have written, see any textbook on QFT).

About the ladder operators appearing single or both: the latter one is a KG field with no electrical charge since complex fields are not invariant under charge conjugation but real fields are (cf. U(1) symmetry).


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