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?


1 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.

  • $\begingroup$ Nice answer, thanks! Just a doubt, you mean "$P$ represents the projection onto the subspace orthogonal to $|V_1\rangle,|V_N\rangle$", right? $\endgroup$
    – user2669
    Nov 16, 2019 at 7:22
  • $\begingroup$ Also I wanted to ask you this, while simulating the system, why do you keep the original Hamiltonian? Shouldn't the original Hamiltonian be changed in the simulated system so as to get the same eigenvalues when expectation value of the Hamiltonian on a state is measured? Say if you are evaluating the expectation value of the Hamiltonian on the state $|V_4\rangle$, this will have a contribution from the other overlapping states. One should take that into account. $\endgroup$
    – user2669
    Nov 16, 2019 at 7:29
  • $\begingroup$ It’s not clear to me that one should although you certainly could without to much effect I think. $\endgroup$
    – DaftWullie
    Nov 16, 2019 at 7:35
  • $\begingroup$ The expectation value of the Hamiltonian on a state $|V_n\rangle$, $n \neq 1,N$ will give the expectation value $1 + N \epsilon^2$. In the limit that $\sqrt{N}\epsilon$ is comparable or larger than 1, $N \epsilon^2$ will contribute. This will cause a mismatch with the Hamiltonian's spectrum. $\endgroup$
    – user2669
    Nov 16, 2019 at 8:20
  • $\begingroup$ I agree, so you can fix that if you wish. What i’m claiming is that I don’t think that change will have much of an effect on the argument I’m making. $\endgroup$
    – DaftWullie
    Nov 16, 2019 at 8:30

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