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I've recently written a program in python that simulates the $\phi$ to the fourth scalar quantum field theory in a 4 dimensional euclidean spacetime. The lagrangian for this theory is that of a free scalar field plus a term equal to $\frac{g}{4!}\times \phi^4$, where $\phi$ is the scalar field and $g$ is bare coupling constant.

By the spectral decomposition of the two point correlation function in euclidean space-time we know that $E(\phi(0)\phi(T))$ (where $E$ means the expected value) will go like ~ $e^{-mT}$ for sufficiently large $T$s assuming that $T<\frac{L}{2}$, where $L$ is the size of the lattice, and where $m$ is the re-normalized mass of the scalar particle $\phi$.

Here's my question; what quantities can I compute on the lattice that will give me the mass of a potential bound state of two scalar particles? For example; the correlation function $C(T) = E(\phi(0)\phi(T))$ will give me the mass of a free scalar particle, as $C(T) \approx e^{-mT}$.

Edit: If you'd like I can post my code.

Edit2: I uploaded the code to github. https://github.com/chuckstables/Phi4 Please bear with me; I am sort of new to python and any sort of coding in general. This is actually one of the first things I've written.

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  • $\begingroup$ Maybe there is an analog to LSZ for lattice... I'm not a lattice person, but like your question. If I find the answer in a book I will tell you. $\endgroup$ – Iliado Odiseo Jul 18 '18 at 4:56
  • $\begingroup$ Hi, welcome to Physics SE! Please don't post formulae as pictures or plain text, but use MathJax instead. MathJax is easy for people on all devices to read, and can show up clearer on different screen sizes and resolutions. I've edited it here as an example. Look at this Math SE meta post for a quick tutorial. $\endgroup$ – user191954 Jul 18 '18 at 5:02
  • $\begingroup$ Hi John. I don't know the answer but the code would be useful to people who do not work on this field! $\endgroup$ – apt45 Jul 18 '18 at 6:25
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    $\begingroup$ Hey apt45. I uploaded the code to github, and I put the link in my post. If you have any questions about it let me know, or if you have any improvements please let me know. $\endgroup$ – chuckstables Jul 18 '18 at 19:40
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What you're going to need to do is build an pair of operators that correspond to creating and annihilating your proposed ground state, and then look at the two point function that operator generates.

For example, you can build a bound state hydrogen atom creation operator as $$\Psi_{H,n\ell m}(\mathbf{R}) = \int \mathrm{d}^3r\ \psi_{n\ell m}(\mathbf{r})\, \Psi_e\left(\mathbf{R} + \frac{m_p}{m_e+m_p}\mathbf{r}\right)\, \Psi_p\left(\mathbf{R} - \frac{m_e}{m_e+m_p}\mathbf{r}\right)$$ using the proton and electron creation operators, $\Psi_p$ and $\Psi_e$, respectively, and $\psi_{n\ell m}$ the standard "excited state" of the hydrogen atom.

Now, critically, when you look at the long term time behavior of the two point function of your compound operator, if it is a single bound state, you should find the same sort of $\mathrm{e}^{-mT}$ behavior you have with the simple two point function.

If you do not have the ability to build such a perfect compound operator you can always just look for the spectrum in the higher order operators. Notice how the compound operator above is built from a particular superposition of two more primitive operators (and $\Psi_p$ is, itself, a compound operator…). That tells you that the bound state behavior is to be found in the higher order correlators of the lower order theory. So, if you examine the correlator $$\left\langle\phi(0+\mathbf{r}_1,0)\,\phi(0-\mathbf{r}_1,0)\,\phi(\mathbf{R}-\mathbf{r}_0,T)\, \phi(\mathbf{R} + \mathbf{r}_0,T)\right\rangle$$ you should find a whole spectrum of comopnents that go like $\mathrm{e}^{-mT}$ for multiple values of $m$ that correspond to the net potential energy of the bound states (i.e. $m = 2m_0 c^2 - E_B$, where $E_B$ is the binding enrgy of the state).

In other words, you want to do an inverse Laplace transform on the time coordinate of the 4-point function in Euclidean space-time and examine the peaks. I don't know how important it is to rearrange the spatial coordinates into center of mass and separation the way I have done above, but since $\mathbf{R}$ is the position coordinate of any putative bound state particle, it seems like a good idea.

In Minkowski space-time, the exponential decay from the cluster decomposition theorem gets replaced with oscillations. Specifically, if you look at the real space Feynman propagator the decay is actually $e^{mr}/r$ in the spatial direction, and $e^{imT}/T$ in the time direction. Which suggests looking for peaks in the Fourier transform of $T$ times the 4-point correlator for $mT\gg1$ should yield the desired results.

Thinking further, though, the frequency spectrum won't have discrete peaks. It will instead be written as a sum of functions that all have a lower limit in their support (e.g. $\sqrt{1+x^2}$). Each of those lower limits will correspond to the net rest mass energy of each bound eigenstate.

Sorry I can't provide more details - I don't know them all, myself.

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  • $\begingroup$ So am I going to be getting different m's for different values of r1, r0 and R corresponding to different quantum numbers for the bound states? $\endgroup$ – chuckstables Jul 18 '18 at 22:27
  • $\begingroup$ Edit to my first comment: I believe I got it now. I wish I could edit comments after 5 minutes, but I can't so I have to post another comment. Thanks :) $\endgroup$ – chuckstables Jul 18 '18 at 22:36
  • $\begingroup$ @JohnDoe You shouldn't get different $m$'s, for every single choice of those parameters you should get all $m$'s - the difference should be the contribution that each $m$ provides to the overall state. Each $m$ corresponds to the energy of the bound state. $\endgroup$ – Sean E. Lake Jul 19 '18 at 0:33
  • $\begingroup$ So just to be clear, the correlator you built above is the correlation between two bound state particle? But isn't the wavefunction of the bound state still missing? $\endgroup$ – Turgon Jul 19 '18 at 7:48
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    $\begingroup$ @Turgon the 4-point correlator is more general than that - it contains contributions from both free and bound states. Because we're interested in bound states, specifically, we can narrow our focus to the center of mass coordinate and creating and annihilating the pair of particles simultaneously. $\endgroup$ – Sean E. Lake Jul 19 '18 at 14:48

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