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.


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


1 Answer 1


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

We start with the following tricks:

  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.
  9. Once, one has the eigenkets one can use the spectral theorem to write the Hamiltonian.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.