How does the computational complexity of finding an equilibrium thermal state for a given Hamiltonian at a given temperature scale with system size under classical and quantum Monte Carlo? I know that it scales polynomially for CMC and for QMC with no sign problem, and exponentially for QMC with a sign problem, but what are the exponents?

I apologize for the vagueness of the question - I'm sure that the answer depends on the precise algorithm (and possibly also on whether the temperature is zero), and there may also be relevant inputs other than system size. In particular, I know that for systems with a sign problem, QMC takes exponentially long to reliably converge, but exactly how does the scaling compare with that for exact diagonalization?


The state $\rho$ for a quantum mechanical Hamiltonian $\hat{H}$ at temperature $T = \frac{1}{k_B\beta}$ is:

So let's consider the cost of a matrix exponential (what you call "exact diagonalization"). This is a very active field of research even today. The algorithm MATLAB R2016b uses comes from a 2010 thesis and 2016 has seen at least 3 new algorithms (Ruiz et al. 2016, Grebrimedhin et al. 2016, Guttel et al., 2016). The choice of algorithm, and the cost of that algorithm depends on the properties of $H$ and on how accurately you are aiming to get the answer, but $15n^3$ floating point operations (FLOPs) has been given as a crude estimate here, where $n$ is the dimension of the matrix, which for a Hamiltonian is $M^2$ where $M$ is the number of levels included in your quantum system.

  • $M=2$ for a two-level-system such as the spin of an electron,
  • $M=39$ for the vibrational levels of the ground electronic state of $^{6,6}$Li$_2$,
  • $M=2^Q$ for $Q$ qubits or $Q$ spin-1/2 particles,
  • $M$ is countably infinitely large for a quantum harmonic oscillator,
  • $M$ is uncountably infinitely large for a continuous-variable system such as position $\hat{x}$.

So a rough estimate of the deterministic algorithm cost is $15M^6$ FLOPs for an $M$-level system, which you are right, is exponential with respect to the number of spin-1/2 particles.

As for Monte Carlo methods, you will have to be more specific about what you are after. The number of macro-iterations required to obtain your answer with a precision of $\pm\epsilon$ is $\mathcal{O}\left(\sqrt{1/\epsilon}\right)$ and each of these macro-iterations will depend on $M$, but to give you the FLOP count you will have to be more specific in your question.

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