Work definition does not make any sense for coherent states? TPM (two projective measurement) scheme I'm talking about the relation that if we are in a isolated, quantum system that only allows for work exchange with the surrounding system we know from the first law of thermodynamics that $\Delta E = W$ where $E$ is the systems internal energy.
There is the so called TPM (two projective measurement) scheme in which one considers an initially equilibrium system $\rho$. This system gets measured at the beginning giving us some initial energy $E_i = Tr(\rho H)$ after which it will get evolved according to some unitary $U$. Now after the evolution another measurement is taken of the system $\rho_f$ which gives us the final energy $E_f = Tr(\rho_f H)$
We know (can be proven) that as long as the state $\rho$ has been classical (only diagonal entries) the relation of the first law of thermodynamic holds for $\Delta E = E_i - E_f = W$ where W is the work that we used to drive the system with the unitary $U$, holds.
But apparently this is not the case if $\rho$ shows coherence (has off-diag.) entries. 
But the problem is I calculated $E_i$ and $E_f$ for such a coherent $\rho$ but it seems as if these off diagonal terms have no effect on the systems energy.
Bonus: it is implied that the first measurement destroys any coherence in the system making it impossible to calc. $W$ for systems that carry coherence. 
Anyone knows anything about this stuff/what am I missing or misinterpreting?
 A: I will calculate throughout in the energy basis of the free Hamiltonian of a 2 state system.
Notation : $ H_0 \lvert 1 \rangle = \hbar \omega_1 \lvert 1 \rangle , H_0 \lvert 2 \rangle = \hbar \omega_2 \lvert 2 \rangle , \omega = \omega_2 - \omega_1, H_0 = \begin{bmatrix} \hbar \omega_1 & 0 \\ 0 & \hbar \omega_2\end{bmatrix} $
Case 1: Free Time Evolution
Here I will show there is no mixing of the energy eigenstates. As a result the average energy is constant and coherence terms in the density matrix have no effect: you can't distinguish between a pure state and a mixture.
$ \lvert \psi (0) \rangle = c_1  \lvert 1 \rangle +  c_2 \lvert 2 \rangle $ 
$ \rho (0) = \begin{bmatrix} p & \alpha \\ \alpha^* & 1-p \end{bmatrix} $ 
Do time evolution under free Schrodinger equation:
$ \lvert \psi (t) \rangle = c_1 e^{i \omega_1 t}  \lvert 1 \rangle +  c_2 e^{i \omega_2 t} \lvert 2 \rangle $ 
$ \rho (t) = \begin{bmatrix} p & \alpha e^{-i \omega t} \\ \alpha^* e^{i \omega t} & 1-p \end{bmatrix}   $ 
So $ \langle H_0 (t)\rangle = Tr ( \rho(t) H_0 ) = p \hbar \omega_1 + (1 - p) \hbar \omega_2 $ ,  a constant independent of the coherence term $\alpha$.  
Case 2: Time Dependent Perturbation causing Rabi Oscillation Here I will show that if a unitary transformation is applied to the two state system so as to mix the free energy eigenstates then the coherence terms in the density matrix become important. I will defer the derivation of the Schrodinger evolution but note for now that a suitable choice of time-dependent interaction Hamiltonian between measurements can apply a unitary transformation to the state vector. Let the Rabi oscillation frequency be $\Omega$.
$\psi (t) \rangle =\cos {(\Omega t)} e^{i \omega_1 t} \lvert 1 \rangle + \sin {(\Omega t)} e^{i \omega_2 t} \lvert 2 \rangle $
$ \rho (t) = \begin{bmatrix} \cos^2(\Omega t) & \sin (\Omega t) \cos (\Omega t) e^{-i \omega t} \\ \sin (\Omega t) \cos (\Omega t) e^{i \omega t} & \sin^2(\Omega t) \end{bmatrix}   $ 
So $ \langle H_0 (t)\rangle = Tr ( \rho(t) H_0 ) = \cos^2(\Omega t) \hbar \omega_1 +  \sin^2(\Omega t) \hbar \omega_2 $
Note the average energy calculation still does not  explicitly depend on the coherence terms. However the coherence terms play a role in the time evolution. This means the result will depend on whether a measurement is performed midway through the evolution as we will now demonstrate.
We assume initial state is pure ground state.
$ \rho (0) = \begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix}   $ 
Time evolve through 45-degree oscillation:
$ \rho (45^o) = \begin{bmatrix} \frac {1}{2} & \frac{1}{2} e^{-i \omega t} \\ \frac{1}{2} e^{i \omega t} & \frac{1}{2}\end{bmatrix}   $
Case 2a. Repeat 45-degree pulse without intermediate measurement
$ \rho (90^o) = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}   $ 
Average energy = $\hbar \omega_2$
Case 2b. Repeat 45-degree pulse after intermediate measurement
The energy measurement after the first pulse results in decoherence in the energy basis resulting in the pure state being replaced by a mixed state (following Born's rule). We can discuss this step in more detail later.
$ \rho (45^o) = \begin{bmatrix} \frac {1}{2} & 0 \\ 0 & \frac{1}{2}\end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix} + \frac{1}{2} \begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix}   $ 
$ \rho (90^o) = \frac{1}{2} \begin{bmatrix} \frac {1}{2} & \frac{1}{2} e^{-i \omega t} \\ \frac{1}{2}  e^{i \omega t} & \frac{1}{2}\end{bmatrix} + \frac{1}{2} \begin{bmatrix} \frac {1}{2} & \frac{1}{2} e^{-i \omega t} \\ \frac{1}{2} e^{i \omega t} & \frac{1}{2}\end{bmatrix} = \begin{bmatrix} \frac {1}{2} & \frac{1}{2} e^{-i \omega t} \\ \frac{1}{2} e^{i \omega t} & \frac{1}{2}\end{bmatrix}   $
So average energy = $\frac {1}{2} (\hbar \omega_1 + \hbar \omega_2 ) $.
A: As an interesting example consider Rabi oscillation.Reference Sakurai, Modern Quantum Mechanics p.320.
We have a two state system with two energy levels $E_1$ and $E_2$.
$$ H_0 = E_1 \lvert 1 \rangle \langle 1 \rvert + E_2 \lvert 2 \rangle \langle 2 \rvert $$ 
We apply a time dependent perturbation at the resonant frequency $ \omega = (E_2 -E_1)/\hbar$
$$ V(t) = e^{i \omega t} \lvert 1 \rangle \langle 2 \rvert + e^{-i \omega t} \lvert 2 \rangle \langle 1 \rvert  $$
It can be shown that a pure state oscillates between the two energy eigenstates at frequency $\omega/2 $.
So the mean energy of the system oscillates between $E_1$ and $E_2$.
$$ \langle E \rangle = E_1 \sin^2 \frac{\omega t}{2} +  E_2 \cos^2 \frac{\omega t}{2} $$
Note at $45 ^o $ through a cycle the pure state will be in a equal coherent superposition of the two energy eigenstates.
Now suppose at this point an energy measurement is performed and the measurement result discarded. Now the state should be described as a equal statistical mixture of each of the two eigenstates.
Evolving the mixture from the point of measurement each eigenstate independently performs Rabi oscillation, but out of phase with each other by $90^o$.
So the average energy of the mixed state is 
$$ \langle E \rangle = \frac{1}{2} ( E_1 \sin^2 \frac{\omega t}{2} +  E_2 \cos^2 \frac{\omega t}{2} )+ \frac{1}{2} ( E_1 \cos^2 \frac{\omega t}{2} +  E_2 \sin^2 \frac{\omega t}{2} )   $$
$$ = \frac{1}{2} (E_1 + E_2) $$
So we see in the pure state the perturbation is causing a transfer of energy to and from the system (absorbtion and emission). In the mixed state there is no net energy transfer. 
