I have read the book on quantum field theory for some time, but I still do not get the physics underline those tedious calculations. The thing confused me most is how quantum mechanics relates to quantum field theory as an approximation in low energy limit. Take a free scalar field $\psi$ for example. It describes free spin 0 boson. And it satisfies Klein-Gordon equation $(\partial_{\mu}\partial^{\mu}+m^2)\psi=0$. In quantum mechanics, the wave function $\psi^{\prime}$ of a free particle without spin should also obey the Klein-Gordon equation or its classical limit Schrodinger equation. So the field operator $\psi$ and the wave function $\psi^{\prime}$ must have some relations. If we take canonical quantization into account, we may assume $\psi(x,t)=\frac{1}{\sqrt{2}}[a(x,t)+a^{\dagger}(x,t)]$ and interpret $a^{\dagger}$ as a creation operator. But I do not know the next step to deduce the Klein-Gordon equation or Schrodinger equation for particle's wave function $\psi^{\prime}$ in quantum mechanics if we start from quantum field theory and take low energy limit.

  • 3
    $\begingroup$ Just to be sure: QM is just 0+1 dimensional QFT, but you're asking if QM can be recovered generally as a low energy limit of the 3+1 QFT, right? $\endgroup$
    – ACuriousMind
    Commented Sep 23, 2015 at 14:17
  • $\begingroup$ In the view of path integral, quantum mechanics may be just 0+1 dimensional QFT. But here I mainly focus on canonical quantization, and I want to derive Schrodinger equation of particle's wave function if we assume the velocity of particles is low and neglect the creation and annihilation of particles. $\endgroup$
    – Eric Yang
    Commented Sep 23, 2015 at 23:44
  • $\begingroup$ I suggest you look at my answer on <physics.stackexchange.com/q/19775>. $\endgroup$ Commented Sep 25, 2015 at 4:28
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    $\begingroup$ 1. More on reduction from QFT to QM: physics.stackexchange.com/q/26960/2451 , physics.stackexchange.com/q/4156/2451 and links therein. 2. For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein. $\endgroup$
    – Qmechanic
    Commented Sep 25, 2015 at 13:58
  • $\begingroup$ Thanks a lot. And in one of links, I find that the notes of David Tong are very clear on this topic. $\endgroup$
    – Eric Yang
    Commented Sep 26, 2015 at 1:19

2 Answers 2


The field and the wavefunction look similar, but they don't really have much to do with each other. The main point of the field is to group the creation and annihilation operators in a convenient way, which we can use to construct observables. As usual I will start with the free theory.

If we want to find a connection to non-relativistic QM, the field equation is not the way to go. Rather, we should look at the states and the Hamiltonian, which are the basic ingredients of the Schrödinger equation. Let's look at the Hamiltonian first. The usual procedure is to start with the Lagrangian for the free scalar field, pass to the Hamiltonian, write the field in terms of $a$ and $a^\dagger$, and plug that into $H$. I will assume you know all this (it's done in every chapter on second quantization in every QFT book), and just use the result:

$$H = \int \frac{d^3 p}{(2\pi)^3}\, \omega_p\, a^\dagger_p a_p$$

where $\omega_p = \sqrt{p^2+m^2}$. There's also a momentum operator $P_i$, which turns out to be

$$P_i = \int \frac{d^3 p}{(2\pi)^3}\, p_i\, a^\dagger_p a_p$$

Using the commutation relations it is straightforward to calculate the square of the momentum, which we will need later:

$$P^2 = P_i P_i = \int \frac{d^3 p}{(2\pi)^3}\, p^2\, a^\dagger_p a_p + \text{something}$$

where $\text{something}$ gives zero when applied to one particle states, because it has two annihilation operators next to each other.

Now let's see how to take the non-relativistic limit. We will assume that we are dealing only with one-particle states. (I don't know how much loss of generality this is; the free theory doesn't change particle number so it shouldn't a big deal, and also we usually assume a fixed number of particles in regular QM.) Let's say that in the Schrödinger picture we have a state that at some point is written as $|\psi\rangle = \int \frac{d^3 k}{(2\pi)^3} f(k) |k\rangle$, where $|k\rangle$ is a state with three-momentum $\mathbf{k}$. $f(k)$ should be nonzero only for $k \ll m$. Now look what happens if we apply the Hamiltonian. Since we only have low momentum, over the range of integration we can approximate $\omega_p$ as $m+p^2/2m$ and ignore the constant rest energy $m$.

$$H|\psi\rangle = \int \frac{d^3p}{(2\pi)^3} \frac{p^2}{2m} a^\dagger_p a_p \int \frac{d^3 k}{(2\pi)^3} f(k) |k\rangle \\ = \int \frac{d^3p}{(2\pi)^3} \frac{d^3 k}{(2\pi)^3} \frac{p^2}{2m} f(k) a^\dagger_p a_p |k\rangle \\ = \int \frac{d^3p}{(2\pi)^3} \frac{d^3 k}{(2\pi)^3} \frac{p^2}{2m} f(k) (2\pi)^3 \delta(p-k) |k\rangle \\ = \int \frac{d^3 k}{(2\pi)^3} \frac{k^2}{2m} f(k) |k\rangle = \frac{P^2}{2m} |\psi\rangle $$

So if $|\psi\rangle$ is any one-particle state (which it is because the states of definite momentum form a basis), we have that $H|\psi\rangle = P^2/2m |\psi\rangle$. In other words, on the space of one-particle states, $H = P^2/2m$. The Schrödinger equation is still valid in QFT, so we can immediately write

$$\frac{P^2}{2m} |\psi\rangle = i \frac{d}{dt} |\psi\rangle$$

This is the Schrödinger equation for a free, non-relativistic particle. You will notice that I kept some concepts from QFT, particularly the creation and annihilation operators. You can do this no problem, but working with $a$ and $a^\dagger$ in QM is not particularly useful because they create and destroy particles, and we have assumed the energy is not high enough to do that.

Handling interactions is more complicated, and I fully admit I'm not sure how to include them here in a natural way. I think part of the issue is that interactions in QFT are quite limited in their form. We would have to start with the full QED Lagrangian, remove the $F_{\mu\nu}F^{\mu\nu}$ term since we aren't interested in the dynamics of the EM field itself, maybe set $A_i = 0$ if we don't care about magnetic fields, and see what happens to the Hamiltonian. Right now I'm not up to the task.

I hope I can convinced you that this newfangled formalism reduces to QM in a meaningful way. A noteworthy message is that the fields themselves don't carry a lot of physical meaning; they're just convenient tools to set up the states we want and calculate correlation functions. I learned this from reading Weinberg; if you're interested in these kinds of questions, I recommend you do so too after you've become more comfortable with QFT.

  • $\begingroup$ Thank you very much for your kind help, and I understand much better. The only thing uncomfortable to me is that "The fields themselves don't carry a lot of physical meaning". But why do we often say that the field is more fundamental than particle? $\endgroup$
    – Eric Yang
    Commented Sep 25, 2015 at 9:57
  • $\begingroup$ @EricYang: Particles appear and disappear, and the field is in a way responsible for doing that. That's why we say fields are more important than particles. But the physical interpretation of the fields is tricky, because in the canonical formalism they are operators, and usually not even observable. See, for example, this or this question. $\endgroup$
    – Javier
    Commented Sep 25, 2015 at 13:54
  • $\begingroup$ I may be 5 years late to the program, but how is $\frac{p^2}{2m}$ a good approximation for $\sqrt{p^2+m^2}$? $\endgroup$
    – Kcronix
    Commented Dec 7, 2020 at 4:29
  • $\begingroup$ @Kcronix The approximation is $m + p^2/2m$. $\endgroup$
    – Javier
    Commented Dec 7, 2020 at 14:17
  • $\begingroup$ Why do you ignore the constant rest mass? $\endgroup$
    – Kcronix
    Commented Dec 7, 2020 at 23:39

I'm not sure how you got that objective into your head: almost certainly that is not what you introductory second quantization text is telling you to do. Wikipedia has decent summaries of the bridge between QFT and QM, namely real scalar field theory ; multidimensional quantum oscillator; lattice phonon oscillators.

The 1+1 quantum field $\psi$ you wrote is an operator, resolvable to $$\psi(x)=\frac{1}{2\pi} \int dk ~e^{ikx} \phi_k= \frac{1}{2\pi} \int dk ~e^{ikx} (a_k+a_k^\dagger ) \sqrt{\frac{\hbar}{2\omega_k}}~,$$ where I use the more conventional letter, φ for the Fourier transformed field operator and the shorthand $\omega_k=\sqrt{k^2+m^2}$ from the K-G dispersion relation you posit.

So the quantum field you write is a linear combination of an infinity of normal modes $a_k$ and their conjugates for an infinity of abstract notional coupled oscillator operators on a 1-dimensional lattice.

The field operator commutation relation is equivalent to the standard commutation relation for each such oscillator labelled by k, and the eigenfrequency of each is the one given above, all different. You are done: each oscillator has a hamiltonian $H_k=\hbar \omega_k (a_k^\dagger a_k + 1/2) $, and surely you can convert it to an equivalent Schroedinger wave eqn--but why should you? Dirac already solved the problem for you in Fock space: the answers are always in matrix mechanics.

Anyway, you may simply manufacture the equivalent $H_k=\hat{p}^2_k/2m_k +m_k \omega^2_k \hat{x}^2_k/2$, which, in some abstract coordinate space, $x_k$, presents as a wave operator $-\hbar^2 \partial_{x_k}^2/2m_k +m_k\omega^2_k x^2_k/2$ acting on c-number wavefunctions $\psi_k'(x_k)$; not operators, as before, in QFT. Further note the field mass m and the absorbable mass $m_k$ of each oscillator are completely unrelated parameters and serve different purposes. No low energy limits really need be taken, but, of course, the Goldstone mode k=0 is at the bottom of the spectrum.

So it is a self-defeating idea to compare apples and oranges, operator quantum field functions of space-time and an infinite collection of wave functions defined in completely different spaces: as you now appreciate these act on wildly different spaces. The x of QFT is our space, but, the infinite $x_k$s of QM are abstract notional spaces for each oscillator, ideally never contemplated...


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