# 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 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 $$H_\delta = H_j - |\tilde \lambda_\alpha \rangle \langle \tilde \lambda_\alpha |$$

Thus, post the eigen-energy value one must make a different observable's measurement say which will force the $$|\lambda_{\tilde \kappa} \rangle \to |\beta_{d} \rangle$$. The probability of such an outcome is $$|\langle \lambda_{\tilde \kappa} | \beta_d \rangle|^2$$. Now, one measures energies again. Now we are forced to explicitly write time variable and use the notation:

$$|\psi \rangle \to U_{H_j}(t )|\psi \rangle$$

and

$$\frac{\partial}{\partial \tilde \lambda_\alpha} H_j= U^\dagger_{H_j}(t ) (\sum_{\kappa} |\lambda_\kappa \rangle \langle \lambda_\kappa| )U^\dagger_{H_j}(t )$$

The operator (which is constructed of wavefunctions) after measuring one gets:

$$\frac{\partial}{\partial \tilde \lambda_{\nu}} H_j= U^\dagger_{H_j}(t + \Delta t) (\sum_{\mu} |\lambda_\mu \rangle \langle \lambda_\mu | )U^\dagger_{H_j}(t+\Delta t )$$

Hence, given $$M$$ measurements (with no repeat eigenvalues) one reconstruct a Hamiltonian after $$M$$ measurements as -

First measurement:

$$H_{1-j} \approx \sum_{i} \tilde \lambda_\alpha |\lambda_\kappa \rangle \langle \lambda_\kappa|$$

$$\downarrow$$

Second measurement's probability (helpful to distinguish degeneracies and becoming state $$| \beta \rangle$$) is $$|\langle \lambda_{ \kappa} | \beta_d \rangle|^2$$.

$$\downarrow$$

The third measurement provided one does not return the value $$\tilde \lambda_{ \alpha}$$ (If one does get $$\tilde \lambda_{ \alpha}$$ then the phase factor is useful) one gets:

$$H_{3-j} \approx \sum_{\kappa} \tilde \lambda_\alpha |\lambda_\kappa \rangle \langle \lambda_\kappa| + \tilde \lambda_{\nu} (\sum_{\mu} |\lambda_\mu \rangle \langle \lambda_\mu | )$$

$$\downarrow$$

$$\vdots$$

To talk about the probability one will be forced to quantify the space of $$H_j$$. To include the the dimensionality we use the notation:

$$H_{j} \to H_{p,j}$$ where $$p= \dim H_j$$ and $$j\leq p$$

Hence, the number Hamiltonians less than an arbitrary cut off of $$p$$ are:

$$\text{Number of Hamiltonians} (\leq N) = \sum_{p=1 }^N \int_{1 }^p \text{Tr} \frac{H^{-1}_{p,j}H_{p,j}}{p} dj$$

1. We assume the limit of $$M \to \infty$$ in this limit. We count the frequency of the of a particular eigenvalues with $$\tilde \lambda_i$$ with $$m_i$$. Hence, for a particular eigenvalue $$m_i$$.

2. Obviously intermediary measurements are being done otherwise only repeated energy eigenvalues would be on the list. We will assume the observable used was $$\hat O$$ with kets $$|o_i \rangle$$. Then the probability of the eigenvalue $$\lambda_k$$ arriving through that particular state $$| \lambda_k \rangle$$ is $$P(\lambda_k ,| \lambda_k \rangle ) = \prod_i | \langle o_i | \lambda_k \rangle |^2$$

3. We will not consider the degenerate eigen-energies as they do not affect the spirit of the calculations.

4. The expectation of the eigen-energy is: $$\sum_j P(\lambda_j, | \lambda_j \rangle) \lambda_j = \sum_{j} \frac{m_j}{M} \lambda_j$$

5. Consider the following now in an arbitrary basis $$|m \rangle \langle m |$$: $$\alpha = H - \sum_i \lambda_i |i \rangle \langle i |$$ with $$H$$ being the correct Hamiltonian

Claim: Since the number of measurements are infinite the minimisation of the following function will yield the correct eigenbasis with most probable outcome as a Langrange multiplier of probability $$\kappa_j$$ and $$\beta_j$$ for normalisation:

$$\min_{\alpha \text{ with } | j \rangle} \alpha + I \times \sum_j \kappa_j \Big ( P(\lambda_j, | j \rangle) \lambda_j - \frac{m_j}{M} \lambda_j \Big ) + I \times \sum_j \beta_j \Big (| \langle j | j \rangle |^2 - 1) = \hat 0$$

with $$I$$ being the identity. The solution of the above should (non-uniquely) define the eigen-basis. It's solution will yield after varying all eigenkets of $$|i \rangle$$ will yield $$|k \rangle \to |\lambda_k \rangle$$

1. After obtaining energy eigenkets one can proceed to construct the Hamiltonian. Note, the limit $$M \to \infty$$ only enabled us to correctly claim our outcome was the most probable.