Background
So let's presume I have discrete "indistinguishable Hamiltonians" (*translations of Hi - the i'th Hamiltonian) as system 1. By this we mean:
⟨Hi(x1,x2,…,xn)⟩=⟨Hj(y1,y2,…,yn)⟩
Let the wavefunctions be |ϕ⟩.
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 Zj is the j'th partition function:
Zj(t−)=Tr e−βHj
where t− is before the measurement, β is the (kbT)−1 (with T as temperature) and the i'th Hamiltonian. The probability associated is:
pi−j(t−)=e−βEi−jZj
Notice, due to the cyclic trace property pi−j(t−)=pi−j′(t−) and Zj=Zj′ and therefore we will remove the j dummy index:
pi(t−)=e−βEiZ
After the measurement using the Born rule we have:
pi(t+)=|⟨ϕ|Ei⟩|2=e−βiEiZ(t+)
where
|Ei⟩ is an energy eigenket and
Z(t+) is given by:
Z(t+)=∑ie−βiEi
The measurement being an irreversible process causes the system to get out of thermal equilibrium hence the βi. To determine βi we take log:
−ln|⟨ϕ|Ei⟩|2=∑jβjEj−βiEi
Summing over i to N both sides (where ∑iE0=∑ip0i=N):
−ln∏i|⟨ϕ|Ei⟩|2=(N−1)∑iβiEi
Hence,
−(N−1)ln|⟨ϕ|Ei⟩|2=−ln∏j|⟨ϕ|Ej⟩|2−(N−1)βiEi
⟹Nln|⟨ϕ|Ei⟩|2+ln∏j≠i(1−∑k≠j|⟨ϕ|Ek⟩|2)=−(N−1)βiEi
Writing in terms of probability**:
⟹Nlnpi+ln∏j≠i(1−∑k≠jpk)=−(N−1)βiEi
Taking the derivative:
⟹Ndpipi−∑j≠i∑k≠jdpj(1−pk)=−(N−1)(Eidβi+βidEi)
Now, dpidt→0 hence,
dβdt→0
and
dEidt→0
Dividing by dpi:
⟹Npi−∑j≠i∑k≠j1(1−pk)dpjdpi=−(N−1)(Eidβidpi+βidEidpi)
Question
Is this theoretically sound? If so, is it possible to experimentally verify this equation?
Edit (My attempt to solve)
Continuing from **:
⟹Nln(1−∑j≠ipj)+∑j≠iln(1−∑k≠jpk)=−(N−1)βiEi
Taylor expanding 1 term:
⟹−N∑j≠ipj−∑j≠i∑k≠jpk≈−(N−1)βiEi
⟹∑j≠i(Npj+∑k≠jpk)≈(N−1)βiEi
⟹∑j≠i(pj(pj+∑k≠j(pj+pk))≈(N−1)βiEi
⟹∑j≠ipj(1+∑k≠jpj)≈(N−1)βiEi
Expanding:
⟹1−pi+(N−1)∑j≠i(pj)2≈(N−1)βiEi