What goes wrong if I try to simulate a system with a larger Hilbert space with a smaller Hilbert space?

Question

System 1: This has a Hilbert space of dimension $N$.

System 2: This has a Hilbert space of dimension $N'$, with the condition that $N' \ll N$. We want to simulate system 1 using the system 2, and so we use a "nearly orthogonal" basis of $N$ vectors to "span" the Hilbert space, with the vectors satisfying the following relations:

$$ \langle V_i | V_i \rangle = 1, \quad |\langle V_i | V_j \rangle| \leq \epsilon \, \forall \, i \neq j.$$

Note that $\epsilon$ is a small number and can be adjusted so as to create the basis of $N$ vectors in the System 2's space.

Now we know that if we probe these two systems using $N'$ number of operators then one can easily distinguish between these two systems, e.g. using $N'$ number of operations. However say if $N = 10^{23}$ and $N' = 10^5$, it isn't feasible to use $10^5$- point correlation functions. So we look for simpler observables to distinguish between the two systems. Which observable(s) / series of operations can demonstrate whether System 2 has a smaller Hilbert space?

As an example, let's say the system we want to understand is a gas at a temperature $T$ and is our System 1. What exactly goes wrong if we try to simulate it using a smaller Hilbert space, i.e. System 2? Do laws of thermodynamics hold?

Answer 1

This is a neat idea. However, having individual overlaps being small isn't sufficient in a quantum system. For example, imagine the set of overlaps $$ \langle V_1|V_N\rangle=0,\qquad \langle V_1|V_n\rangle=\langle V_N|V_n\rangle=\epsilon \quad \forall n=2,\ldots,N-1, \langle V_n|V_m\rangle=0 \quad \forall n,m=2,\ldots,N-2 $$ Remember that these are supposed to be simulating orthogonal states, so starting from $|V_1\rangle$, you should never be able to get to $|V_N\rangle$. Now consider the Hamiltonian $$ H=\sum_{n=2}^{N-1}|V_n\rangle\langle V_n|, $$ evolving the state $$ e^{-iHt}|V_1\rangle. $$ The evolution occurs in a 3-dimensional subspace spanned by $$ |V_1\rangle, |\Psi\rangle=\frac{1}{\sqrt{N-2}}P\sum_{n=2}^{N-2}|V_n\rangle, |V_N\rangle $$ where $P$ represents the projection onto the subspace orthogonal to $|V_1\rangle,|V_N\rangle$ and normalisation.

Now, $\langle \Psi|H|V_1\rangle=\langle \Psi|H|V_N\rangle\sim\epsilon\sqrt{N}$ (I'm not calculating this carefully here), and we can calculate other similar terms so that the Hamiltonian in this 3-dimensional subspace becomes $$ \approx\left(\begin{array}{ccc} N\epsilon^2 & \sqrt{N}\epsilon & N\epsilon^2 \\ \sqrt{N}\epsilon & 1 & \sqrt{N}\epsilon \\ N\epsilon^2 & \sqrt{N}\epsilon & \sqrt{N}\epsilon \end{array}\right), $$ where the matrix it's supposed to be simulating is $$ \left(\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right). $$ For large $N$, these are very different. For example, the first matrix can almost perfectly transfer the state $|V_1\rangle$ to $|V_N\rangle$ in time $\pi/(1+2N\epsilon^2)$, and this is something you could easily measure for reasonable choices of, say $|V_1\rangle=|0\rangle^{\otimes N'}$ and $|V_N\rangle=|1\rangle^{\otimes N'}$ because to detect that the state is no longer in $|V_1\rangle$, it would be sufficient to perform a simple magnetisation measurement, or similar.