I believe that all of Quantum Mechanics should be retrievable from QFT (in 3+1 dimensions) by taking some appropriate limits and/or integrating out degrees of freedom.

David Tong shows in his Lecture Notes on QFT (David Tong) how the Schrodinger equation without potential follows from taking a free field and looking at a superposition of one particle states (pp. 43-45). How would one derive the Schrodinger equation for a general potential? What is the input on the QFT side that gives you a specific potential? (Either, or preferably both, in the Hamiltonian/Canonical Approach and/or in the Lagrangian/Path Integral approach.)

  • 1
    $\begingroup$ 1. More on reduction from QFT to QM: physics.stackexchange.com/q/26960/2451 , physics.stackexchange.com/q/4156/2451 , physics.stackexchange.com/q/208615/2451 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
    Jun 9, 2017 at 15:40
  • $\begingroup$ This answer address the core of your question I think: physics.stackexchange.com/a/142172/154997 $\endgroup$
    – user154997
    Jun 9, 2017 at 20:19
  • $\begingroup$ @Luc thanks. While this is certainly interesting and it is related, it does not answer the core of my question. It shows how we can compare quantum mechanical scattering to QFT scattering. This seems like a smart approach when you want to compare QM to QFT, as in QFT we generally focus only on scattering. But I'm more interested in where the rest of QM is hiding out. How do we derive the Schrodinger equation with a potential. Is there a simple procedure or do we have to introduce a complicated system of fields that happens to give a certain potential, such as a the 1/r potential in this case? $\endgroup$
    – Kvothe
    Jun 12, 2017 at 8:36

1 Answer 1


First I will try to address your specific interrogation about the existence of a generic case.

The potential in the Schrödinger equation for one particle models the interaction of that particle with "something else". Therefore, if it is to be the limit of some QFT, the latter shall feature at least two fields and their interaction: a field for the particles of the type modelled by the Schrödinger equation, and a field for the "something else", i.e. another type of particles. That interaction is what will give rise to the potential in non-relativistic Schrödinger equation, after integrating out all the degrees of freedom of the second field. So, I think that a reasoning valid for an arbitrary potential does not exist. A big part of the reason is that one has too much of a limited choice for the QFT Lagrangian, and especially the interaction term, as it has to be Lorentz-invariant, gauge-invariant, etc.

But now, the obvious question about the QED case. This is both the easiest and the most important variant of your question. Important because a lot of predictions in atomic physics rely on this limit. Precisely we would like to prove that QED degenerates into the following Hamiltonian for $N$ particles of charge $q$ and position $r_i$ and momentum $p_i$ and spin $\sigma_i$

$$H = \sum_{i=1}^N \frac{1}{2m}\left(p_i - q A_{\perp}(r_i)\right)^2 - \sum_{i=1}^N \frac{q\hbar}{2m}\sigma_i\cdot B(r_i) + \sum_{i\ne j} \frac{q^2}{8\pi\epsilon_0}\frac{1}{\|r_i - r_j\|} + H_\text{Photons}$$

where $A_\perp$ is the transverse vector potential and $B$ is the magnetic field. There is a very thorough treatment of this question in [1, B$_\text{V}$], a complement to chapter V titled "Justification of the nonrelativistic theory in the Coulomb gauge starting from relativistic quantum electrodynamics". This book is a translation of the original version in French, which is the one I own (and pulled from the card boxes where I had stashed it away!). I may try to summarise the argument in a later edit of this answer but it would be hard to do justice to Cohen-Tannoudji and Dupont-Roc.

[1] Cohen-Tannoudji, Dupont-Roc, Photons and Atoms: Introduction to Quantum Electrodynamics, Wiley, 1989


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.

Not the answer you're looking for? Browse other questions tagged or ask your own question.