What is the basic idea behind lattice theory?

[RedX]

What is the basic idea behind lattice theory or computer-based QFT calculations?

For example, take a scalar field, and the functional path integral:

$$W[J(x)]=\int [d\phi(x)]e^{i\int \mathcal L \mbox{ }d^4x+i\int J(x)\phi(x)d^4x}$$

W[J] is the starting point for all types of quantum-field-theoretic calculations.

Is this the quantity that's calculated by computers, summing over a grid of points instead of performing a path integral?

Usually when calculating this quantity by hand, J(x) is left arbitrary. For a computer calculation, don't you have to insert a specific J(x) for the computer to evaluate the path integral?

Do the computers evaluate this integral in position or momentum space? If it's in momentum space, it would be like this:$$W[J(k)]=\int [d\phi(k)]e^{iS+\frac{i}{(2\pi)^4}\int J(k)\phi(-k)d^4k}$$

where the action term S is a little more complicated to write for the interaction Lagrangian terms, but is still just a functional of $$\phi(k)$$. I guess the lattice spacing would be '1/a' where 'a' is the spacing in position space?

But still, what is inserted for J(k)? For a scattering process, I guess it would have to be a delta function in momentum space in order to attach a real particle, and sum each of these delta functions for all particles (if the path integral is evaluated in position space instead, you would need some sort of delta function times $$e^{\pm ikx}$$ for each attached scattering particle of 4-momentum k)?

I've seen some impressive charts that come from computer calculations of QCD: just using the Lagrangian of quarks and gluons, the computer calculations predict confinement and asymptotic freedom, and all the values of hadron masses (the only input being the renormalized coupling strength at the Z-mass)!

I'm just wondering if I can apply these computer calculations to things besides QCD, instead of learning all these tricks like drawing Feynman diagrams and applying Feynman rules and various tricks to evaluate Feynman rules and diagrams like Feynman parameterization and stuff.

 

[hamster143]

J(k) is the source. A delta function in momentum space, as you've observed.

One big caveat is that you're supposed to use Grassmann variables instead of normal floats, as long as your input/output states are fermions. (which is basically always.) Fortunately, integration over Grassmans is so easy that it can be done analytically.

No, you can't apply these calculations instead of learning Feynman diagrams. (Or, rather, you can, but you don't want to.) Feynman diagrams are extremely useful because, in most tasks, you get a high-precision answer from a small number of terms that you can compute on a napkin. QCD is special because it is not perturbatively convergent in the low energy limit, and that is the only reason people resort to lattice QCD. Therefore, people doing low energy QCD are stuck doing it over lattice, even if it means hours or even days of supercomputer time to get a single number.

 

[RedX]

J(k) is the source. A delta function in momentum space, as you've observed.

So take a 2x2 scattering with 4-momenta: a, b, c, and d (the difference between incoming and outgoing particles I guess would be determined by whether the energies are positive or negative). So $$J(k)=\delta^4(a-k)+\delta^4(b-k)+\delta^4(c-k)+\delta^4(d-k)$$ and:

$$W[J(k)]=\int [d\phi(k)]e^{iS+\frac{i}{(2\pi)^4}\int J(k)\phi(-k)d^4k} =\int [d\phi(k)]e^{iS+\frac{i}{(2\pi)^4}[\phi(-a)+\phi(-b)+\phi(-c)+\phi(-d)]}$$

It seems like if the action didn't have any kinetic terms (no derivatives at all), then this problem would be symmetrical and give the same result no matter what a, b, c, and d are.

It also seems that a, b, c, and d have to be lattice points to do this calculation. Or do you have to give each momentum some fuzziness equal to the size of the lattice spacing?

No, you can't apply these calculations instead of learning Feynman diagrams. (Or, rather, you can, but you don't want to.) Feynman diagrams are extremely useful because, in most tasks, you get a high-precision answer from a small number of terms that you can compute on a napkin. QCD is special because it is not perturbatively convergent in the low energy limit, and that is the only reason people resort to lattice QCD. Therefore, people doing low energy QCD are stuck doing it over lattice, even if it means hours or even days of supercomputer time to get a single number.

There are so many tricks to learn with Feynman diagrams that in the end, it looks far removed from the basic idea - which is just calculating the path integral.

I don't have access to (or the expertise to) a supercomputer or cluster, but I know a little bit of C++ and have a laptop, and when I go to class I can leave the computer on to do a calculation. I know that you have to learn a lot about numerical algorithms and be very good at programming and have access to a supercomputer to do a really good job, but I want to see what a computer can do for leptonic processes and match it to analytical human expressions.

 

[hamster143]

The action always has kinetic terms.

You don't really need to worry about J. J is introduced to make theorists' lives easier. It should be sufficient to set J to 0 and to do the integration subject to appropriate boundary conditions (with "wave packets" of incoming and outgoing particles at t_initial and t_final.)

The reason why a supercomputer is needed is not hard to see. If you want to do the computations over a modest 32 x 32 x 32 x 32 lattice, that's ~10^6 sites, so O(10^6) operations just to compute action for one configuration. And you need to sample at least O(10^6) configurations (in reality, much more than that) to get meaningful results.

There's one more thing that was not mentioned yet, which would greatly complicate your hopes to do lattice calcs on a laptop. Notice that your exponent is purely imaginary. Therefore your integrand always has magnitude 1 and it fluctuates wildly. In practice, people do Wick rotation into Euclidean space, where i becomes -1 and the integral is only substantially greater than zero for select trajectories. In process, physical meaning of the calculation becomes a bit blurred.

Things get more complicated (more computations, gauge fixing) if you do gauge theory, and even more complicated if you want to work with fermions. It would probably be best to get a book on the subject that covers all important points. There's a decent book by Montvay/Munster that I know of, but it's somewhat dated (1997), and a few more recent books.

 

[RedX]

I never thought about it like that. But yeah 32x32x32x32 is 10^6, and 32 points is nowhere close to infinity. And of course scalar theory with just one kind of scalar is easiest. So I won't even attempt to try a calculation with a computer. But I am still interested in how its's done in principle, on a computer.

The action always has kinetic terms.
You don't really need to worry about J. J is introduced to make theorists' lives easier. It should be sufficient to set J to 0 and to do the integration subject to appropriate boundary conditions (with "wave packets" of incoming and outgoing particles at t_initial and t_final.)

So I guess computer calculations look a lot different than theory calculations. Because setting J=0 would make it the vacuum-to-vacuum amplitude in the absence of a source, which is normalized usually to 1.

To offset this, you change the boundary conditions of the integration? Usually the integration is over the entire 4-space (Minkowski space). Which thinking about it now, is kind of strange - the integration is over the entire universe? I know why time is from -infinity to infinity, but I can't seem to remember why space is from -infinity to infinity. So now the incoming and outgoing particles are t_initial and t_final and not -infinity and infinity?

 

[hamster143]

If you recall how path integrals are introduced into QFT in the first place, it's the opposite process - you start with a path integral over a finite period of time with definite boundary conditions, then you relate it to the integration over the whole universe with an additional factor of $\phi(x_A) * \phi(x_B) * ...$ inside the integral, and then you prove that the thing is equal to the functional derivative of the integral with respect to J when J=0, and that is nice because you can do functional derivatives explicitly and that gives you Feynman diagrams. All that with a lot of handwaving, $-i\epsilon$'s and sweeping of the infinities under the rug.

Of course, on lattice you don't have the luxury (or the need) of integrating over the infinite spacetime, and you go back to basics.

 

[RedX]

I do recall that to motivate the definition of the path integral, a discrete path is considered first. This is done by putting discrete number of completeness relations between the initial position state and the final position state, and each completeness relation corresponds to a sum at that moment in time, and this sum will later be an integral.

But I do remember that you have to take the time to -infinity to infinity, in order to use the $$-i\epsilon$$ prescription to turn a path integral from initial position state to final position state, into one that is from initial ground (vacuum) state to final ground state.

So I guess with computers, you don't use an $$-i\epsilon$$ prescription. Which would make sense, since the $$-i\epsilon$$ term only shows up in a propagator, which is the perturbative Feynman diagram way of doing things - while a computer just evaluates the path integral directly.

The book I read had no justification when extending particles to fields to extend the domain of the field to infinity with respect to coordinates. So I guess you can use a length equal to the size of the lab? What about the time scale? I'm pretty sure these processes happen really fast, but can theory even predict how fast? The theory of scattering just assumes -infinity to infinity for time?