# Quantify the probability in guessing the Hamiltonian?

## Background

Let's say I know my experimentalist friend has been measuring the eigenvalues of a physical system. I can see the $$M$$ measurements are noted in a sheet of paper and I assume the dimensionality of the Hamiltonian to be $$K$$. I also note the dimensions of eigenvalues of that of energy. I see non-unique eigenvalues and (randomly) guess an eigen-operator $$\hat O$$ was measured in between the measurements ,i.e, a measurement of operator $$\hat O$$ was done after every eigenenergy measurement. I would like to guess the Hamiltonian of the system.

## Question

Now, my question how to quantify this probability of a reasonable guess of the Hamiltonian (any strategy is allowed including the one given below) of it being particular dimension $$K$$ given $$M$$ measurements as correct ? I would prefer if one included degerate Hamiltonians in their calculations (if possible)?

## Why it's a difficult problem

This only means to show the probabilistic nature of the problem. We will think of this in terms of eigen-energies primarily other formulation will require another layer of probability. Now:

Let, us write a variable Hamiltonian $$H_j$$ where $$j$$ is a counting system adopted which avoids redundancy . From the eigenvalue equation for energy:

$$H_j |\lambda_i \rangle= \tilde \lambda_i|\lambda_i \rangle$$

From the spectral theorem I can reconstruct a particular Hamiltonian, see:

$$H_j= \sum_{i} \tilde \lambda_i |\lambda_i \rangle \langle \lambda_i|$$

Now, if we include degeneracies in the argument after measuring $$\tilde \lambda_\alpha$$ one can conclude:

$$\frac{\partial}{\partial \tilde \lambda_\alpha} H_j= \sum_{\kappa} |\lambda_\kappa \rangle \langle \lambda_\kappa|$$ How does one conclude this? When one measures a particular eigenvalue and assumes a degenerate $$H_j$$:

$$\tilde \lambda_\alpha \to \sum_{\kappa}^M |\lambda_\kappa \rangle$$

Note: all degeneracies obey:

$$H_j |\lambda_\kappa \rangle= \tilde \lambda_\alpha|\lambda_\kappa \rangle$$

Non-uniqueness: given the same eigenvalue $$\tilde \lambda_\alpha$$ cannot one distinguish between $$H_j$$ from $$H_\delta = H_j - |\tilde \lambda_\alpha \rangle \langle \tilde \lambda_\alpha |$$

This is my attempt. Let's say the list looks like:

$$\lambda_1$$, $$\lambda_2$$, $$\lambda_1$$, $$\lambda_3$$, $$\dots$$, $$\lambda_n$$

where $$\lambda_i$$ are numbers. The variables (unknowns) are the Hamiltonian and the energy eigenvectors

1. We assume $$M$$ (number of measurements) is a large number. This enables us to say the distribution observed $$\{ M \}$$ is the most probable distribution*.
2. By *most probable we mean the expectation empirically is the most probable expectation value to be measured. For example the expectation is: $$\langle H \rangle = \frac{m_1 \lambda_1 + m_2 \lambda_2 + \dots + m_n \lambda_n}{M}$$
3. Where we count the frequency of the of a particular eigenvalues with $$\lambda_k$$ with $$m_i$$. Hence, for a particular eigenvalue $$m_i$$.
4. Obviously intermediary measurements are being done otherwise the same energy eigenvalue would be on the list. We will assume the observable used (to change the state) was $$\hat O$$ with kets $$|o_i \rangle$$. Then the probability of the eigenvalue $$\lambda_k$$ arriving through that particular state (with no degeneracy)$$| \lambda_k \rangle$$ is $$P_i( \lambda_k ) = | \langle o_i | \lambda_k \rangle |^2 = \frac{m_k}{M}$$ Do note: the eigenvector is essentially not known.
5. We use the following trick (from statistical mechanics) of multinomial theorem: $$(a_1+a_2 +a_3 + \dots +a_M )^M = \sum \frac{M!}{b_1!b_2! \dots b_M!} {a_1}^{b_1}{a_2}^{b_2} \dots {a_n}^{b_n}$$
6. Now choosing $$b_j \to m_j$$ and $$a_j \to P(\lambda_j)$$. We should get a coefficient of the probability of obtaining our particular distribution which is: $$P_i(\{M \}) = \frac{M!}{m_1!m_2! \dots m_M!}$$
7. Using point ($$1$$) the probability obtained is the most probable distribution. Hence, by varying the eigenkets claim the correct eigenkets would maximize the below: $$P_i(\{ M \}) = \max P_i(\{ M' \}) \implies \frac{M!}{m_1!m_2! \dots m_M!} = \max \prod_k \frac{M!}{(M | \langle o_i | \lambda_k \rangle |^2)!}$$ Taking log and maximising will work as well.
8. It should be possible to talk about a mix of eigenkets such as $$| o_k \rangle$$, $$| o_i \rangle$$, etc (using density matrices). It should also be possible consider variations of degenerate eigenvalues (in a similar style shown in the question). Again one would have to consider which outcome maximizes the probability.