**Background**

So let's presume I have discrete "indistinguishable Hamiltonians" (*translations of $H_i$ - the $i$'th Hamiltonian) as system $1$. By this we mean:

$$ \langle H_{i} (x_1,x_2, \dots,x_n) \rangle = \langle H_{j}(y_1,y_2, \dots,y_n) \rangle $$

Let the wavefunctions be $|\phi \rangle$.

These system of Hamiltonians are in thermal equilibrium. Now another system whose Hamiltonian is $H'_2$ is added which is in thermal equilibrium with the system $1$ is thermal equilibrium. The net system is now isolated. After being isolated system $2$ performs alot of measurements of energy eigenvalues. The following is the analysis of that:

The partition function of a subsystem of system $1$ be where $Z_j$ is the $j$'th partition function:

$$ Z_{j} (t_-)= \text{Tr } e^{- \beta H_j}$$

where $t_-$ is before the measurement, $\beta$ is the $(k_b T)^{-1}$ (with $T$ as temperature) and the $i$'th Hamiltonian. The probability associated is:

$$p_{i-j}(t_-) = \frac{e^{-\beta E_{i-j}}}{Z_j}$$

Notice, due to the cyclic trace property $ p_{i-j}(t_-) = p_{i-j'}(t_-)$ and $Z_{j} = Z_{j'}$ and therefore we will remove the $j$ dummy index:

$$p_{i}(t_-) = \frac{e^{-\beta E_{i}}}{Z}$$

After the measurement using the Born rule we have:

$$ p_i (t_+) = |\langle \phi | E_i \rangle|^2 = \frac{e^{-\beta_i E_i}}{Z(t_+)}$$

where $| E_i \rangle$ is an energy eigenket and $Z(t_+)$ is given by:

$$ Z(t_+) = \sum_i e^{-\beta_i E_i}$$

The measurement being an irreversible process causes the system to get out of thermal equilibrium hence the $\beta_i$. To determine $\beta_i$ we take log:

$$ - \ln |\langle \phi | E_i \rangle|^2 = \sum_j \beta_j E_j - \beta_i E_i $$

Summing over $i$ to N both sides (where $ \sum_i E^0 = \sum_i p_i^0 = N$):

$$ - \ln \prod_i |\langle \phi | E_i \rangle|^2 = (N-1)\sum_i \beta_i E_i $$

Hence,

$$ - (N-1)\ln |\langle \phi | E_i \rangle|^2 = - \ln \prod_j |\langle \phi | E_j \rangle|^2 - (N-1)\beta_i E_i $$

$$\implies N\ln |\langle \phi | E_i \rangle|^2 + \ln \prod_{j \neq i} (1- \sum_{k \neq j}|\langle \phi | E_k \rangle|^2) = - (N-1) \beta_i E_i $$

Writing in terms of probability**:

$$\implies N\ln p_i + \ln \prod_{j \neq i} (1- \sum_{k \neq j}p_k) = - (N-1) \beta_i E_i $$

Taking the derivative:

$$\implies N \frac{dp_i}{p_i} - \sum_{j \neq i} \sum_{k \neq j} \frac{ d p_j}{ (1- p_k)} = - (N-1)( E_i d\beta_i + \beta_i d E_i )$$

Now, $\frac{dp_i}{dt} \to 0$ hence,

$$ \frac{d \beta}{dt} \to 0 $$

and

$$ \frac{d E_i}{dt} \to 0 $$

Dividing by $d p_i$:

$$\implies \frac{N}{p_i} - \sum_{j \neq i} \sum_{k \neq j} \frac{ 1}{ (1- p_k)} \frac{d p_j}{dp_i} = - (N-1) ( E_i \frac{d\beta_i}{d p_i} + \beta_i \frac{d E_i}{d p_i} )$$

**Question**

Is this theoretically sound? If so, is it possible to experimentally verify this equation?

**Edit (My attempt to solve)**

Continuing from **:

$$\implies N \ln (1- \sum_{j \neq i} p_j )+ \sum_{j \neq i} \ln (1- \sum_{k \neq j}p_k) = - (N-1) \beta_i E_i $$

Taylor expanding $1$ term:

$$\implies -N \sum_{j \neq i} p_j - \sum_{j \neq i} \sum_{k \neq j}p_k \approx - (N-1) \beta_i E_i $$

$$\implies \sum_{j \neq i} (N p_j + \sum_{k \neq j}p_k) \approx (N-1) \beta_i E_i $$

$$\implies \sum_{j \neq i} ( p_j (p_j+\sum_{k \neq j} (p_j + p_k)) \approx (N-1) \beta_i E_i $$

$$\implies \sum_{j \neq i} p_j (1+\sum_{k \neq j} p_j ) \approx (N-1) \beta_i E_i $$

Expanding:

$$\implies 1- p_i + (N-1) \sum_{j \neq i} (p_j )^2 \approx (N-1) \beta_i E_i $$