# Is there a Quantum-Mechanics analog to QFT's quartic $\lambda \phi^{4}$ interaction?

In QFT; for the quartic-interacting real scalar field $\phi$ we have the Lagrangian density: $$\mathcal{L} \ = \ \frac{1}{2} \left( \partial^{\mu} \phi \right)\left( \partial_{\mu} \phi \right) + \frac{1}{2} m^2 \phi^2 + \lambda \phi^{4}$$ For $\lambda =0$ we have a free bosonic field - when $\lambda \neq 0$ one way to interpret the interaction term is as the quanta of the field, the bosons, having a 2-body repulsive interaction.

My question is; what is the analog for the above interaction in ordinary Quantum Mechanics?

Suppose I have a particle $A$ with free Hamiltonian $H_{A} = \frac{\mathbf{p}_{A}^2}{2m}$, whose states span some Hilbert Space $\mathcal{H}$. Take another identical particle $B$ with free Hamiltonian $H_{B} = \frac{\mathbf{p}_{B}^2}{2m}$. The non-interacting free Hamiltonian for the combined system of the two particles is something like: $$H \ = \ \frac{\mathbf{p}^{2}_{A}}{2m} \otimes \mathbb{I}_{\mathcal{H}} + \mathbb{I}_{\mathcal{H}}\otimes \frac{\mathbf{p}^{2}_{B}}{2m} + V(\mathbf{x}_{A}, \mathbf{x}_{B})$$

Is there a potential $V(\mathbf{x}_{A}, \mathbf{x}_{B})$ which emulates the $\lambda \phi^{4}$ interaction from QFT? My naive guess is something along the lines of; $$V(\mathbf{x}_{A}, \mathbf{x}_{B}) = \lambda \big| \mathbf{x}_{A} \otimes \mathbb{I}_{\mathcal{H}} - \mathbb{I}_{\mathcal{H}} \otimes \mathbf{x}_{B} \big|^{4}$$

Is this correct?

• What do you mean by analog ? A QM toy-model (to understand properties of perturbation theory and so on) ? Or the first-quantization version for non-relativistic bosons ?
Jan 29 '18 at 9:31
• @Adam I agree that it's imprecise what I am asking for, but something like; In QED with Lagrangian density $\mathcal{L} = \bar{\psi} (i\gamma^{\mu}D_{\mu} -m)\psi - \frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ the analog in QM is two point particles $A$ and $B$ interacting through a potential $V = \frac{e^2}{4\pi\epsilon_0 | x_{A} - x_{B} |}$. I guess something along these lines... Jan 29 '18 at 19:12

The transition back to ordinary quantum mechanics is accomplished by working out the way probability amplitudes in the $N$-particle states evolve with time. When that is done, the particles interact with "contact terms", potentials that are proportional to $\delta(\mathbf{x}_i-\mathbf{x}_j)$. For details, see the development done in the first chapter of Brian Hatfield's "Quantum Field Theory of Point Particles and Strings".
That said, it is often repeated that ordinary quantum mechanics is just quantum field theory with zero spatial dimensions. Under the mapping implied the $\phi^4$ theory would become $$H = \frac{p^2}{2m} + \frac{m\omega^2}{2}x^2 + \lambda x^4.$$ With a countable number of particles being just a sum by adding indices to $p$ and $x$.
Transitioning back to continuum mechanics is tricky, and I don't know all of the details for certain (textbooks usually get a little hand-wavy about this path). I can say with confidence, though, that any term that mixes $x_i \rightarrow f_i \sqrt{h^3}$ with different indices, where $h$ is the spacing between adjacent sites, will either become a derivative or vanish on the continuum transition. The $\lambda$ of the discrete theory would have to be remapped by $\lambda\rightarrow \lambda h^{-3}$, for instance, in order to keep the $\lambda \phi^4$ term from vanishing in the limit $h \rightarrow 0$. All of the $m_i$ would have to be the same (say, $1$), by Lorentz invariance, and the springs that connect the $x_i$ would have to have spring constants that diverge like $h^{-1}$ in order to produce the spatial derivatives between needed sites. So, the intermediate Hamiltonian is given by \begin{align} H &= \sum_{i,j,k=-N}^N \left[\frac{1}{2}p_{i,j,k}^2 h^3 + \frac{1}{2} (f_{i+1,j,k}-f_{i,j,k})^2 h^2 + \frac{1}{2} (f_{i,j+1,k}-f_{i,j,k})^2 h^2 + \frac{1}{2} (f_{i,j,k+1}-f_{i,j,k})^2 h^2 \right. \\ & \hphantom{= \sum_{i,j,k=-N}^N}\ \ \left. + \frac{m^2}{2}f_{i,j,k}^2 h^3 + \lambda f_{i,j,k}^4 h^3\right] \end{align} and the continuum mechanics are recovered by applying $\lim_{h\rightarrow 0}\lim_{N\rightarrow \infty}$.
Note that the transformation where $x_i\rightarrow f_i\sqrt{h^3}$ and $p_i\rightarrow p_i\sqrt{h^3}$ is not a canonical change of variables. Starting from the Lagrangian it is straightforward to show from canonically conjugate momentum's definition that $p_i$ would transform as $p_i \rightarrow p_i h^{-3/2}$ if it were. What's going on here is that, in order to get a clean Hamiltonian that becomes an integral of a density, we need to modify the canonical commutation relation to read $$[f_{i,j,k},\, p_{n,\ell,m}] = i\hbar \delta_{in}\delta_{j\ell}\delta_{km} h^{-3},$$ where the right hand side now actually becomes a Dirac delta function in the continuum limit, instead of just saying that Kronecker deltas become Dirac deltas.
All of that said, there's a reason that the usual transition back to ordinary QM mentioned in the first paragraph is so convoluted. In ordinary QM $x$ is the observable position of some particle. In QFT the positions of particles are still the fundamental observables, it's just that now the particles are described as excitations in a field. The field strength is not directly observable, though, it can just be inferred on average when it is strong enough to not be significantly disturbed by the injection of multiple small test particles that we measure the positions of.
• One can show this in a much more direct way, see e.g. Chapter 1 in Srednicki's book. One looks at vectors $\Psi_t := \int dx \,\Psi(x_1, x_2, t) a^*(x_1)a^*(x_2)|\Omega\rangle$. In the limit $c\to \infty$ the "balanced" normal ordered terms $a^*a^*aa$ in the second quantized $\phi^4$ interaction Hamiltonian become dominant, and one can convince oneself, that $\Psi_t$ evolved under the second quantized Hamiltonian is equivalent to $\Psi(x_1,x_2,t)$ evolved under a pair-interaction Schrödinger Hamiltonian with a delta. Jun 16 '20 at 21:19