# Retrieving full Schrodinger equation (including potential) from QFT

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. 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. – Qmechanic Jun 9 '17 at 15:40
• This answer address the core of your question I think: physics.stackexchange.com/a/142172/154997 – user154997 Jun 9 '17 at 20:19
• @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? – Kvothe Jun 12 '17 at 8:36

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.