# How can very small lattices be sufficient for Quantum Monte Carlo simulations?

Quantum Monte Carlo simulations are often performed with very modest lattice sizes (such as [e-print], $$64$$ sites), due to their computational cost increasing exponentially with the system volume; while classic Monte Carlo simulations usually require much larger lattice sizes, even when making use of finite-size analysis.

Why are these small lattices enough for obtaining physically meaningful results for quantum systems? Is it simply a case of "we take what we can get", or is there some intrinsically quantum mechanism in action[1]?

[1] Such as a symmetry-breaking magnetic field removing degeneracies for free electrons in a tight binding approximation. Ref.: NIC Series Vol. 10 (ISBN 3-00-009057-6) (pdf 1, 2), page 131.

I think you are correct that there is an aspect of "take what you can get" to the sizes that are typically used in numerical methods. Even with finite size scaling (FSS), you usually try to go to the largest size that is practical with your computational resources. Case in point: people do finite size scaling with extremely small sizes for exact diagonalization calculations.

It's worth noting that the finite size effects are themselves interesting and often contain important physical quantities

There is one more specifically-quantum feature to consider: the finite size gap and the size-temperature tradeoff. In finite-size quantum systems, even gapless quantities have a finite-size gap $$\Delta \propto 1/L$$ (basically because $$L$$ becomes the maximum wavelength). This gap contributes directly to the finite-size effects, but also separates the "finite-size zero-temperature" and "finite-temperature large-size" regimes:

• When $$\beta \Delta \gg 1$$, then effectively the temperature is zero, because all the excited states, even the 'gapless' ones, are suppressed.
• When $$\beta \Delta \ll 1$$, the finite size effects are weaker because the system 'can't see' the finite-size gap.

The computational cost of QMC (at least the form I use, SSE) roughly scales like $$\beta L^d$$. Because of this finite-size gap, you need $$\beta \propto L$$ to maintain the same `effective' temperature as you go to larger and larger sizes. As a result, zero-temperature properties are much easier to access for small sizes (with a large gap, and not-too-low $$T$$ requirement).

If you want to avoid finite-size effects, then you want to avoid low temperatures so you aren't detecting this finite-size gap. (Or you need a large system so $$\Delta$$ is small).

• Thanks. I'm probably missing something obvious, but could you elaborate on the size-temperature trade-off? – stafusa May 19 at 12:01
• @stafusa, It wasn't obvious from the way I wrote it. I added two more paragraphs to explain. – taciteloquence May 20 at 9:54
• That's great, thanks! – stafusa May 20 at 12:56
• But, did you mean "not-too-high $\beta$ requirement" instead of "not-to-low $\beta$ requirement"? – stafusa May 20 at 12:57
• @stafusa That was ambiguous the way I wrote it. I corrected it to say "not too low T requirement". It's the temperature that would have to be low, not $\beta$. – taciteloquence May 21 at 2:25