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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:

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


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This is my attempt. We start with the following tricks:

  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.
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