Conditional Time Evolution increases entropy?

Question

Question

Does the below calculation conclusively show the idea of conditional time evolution (if state measured is $x$ I do $y$ else I do $z$ ) increases the Von Neumann entropy? Has this already been established in the literature?

Calculation

Let's say I have an isolated system containing a measuring apparatus, an experimenter and a Hamiltonian system:

Let the Hamiltonian be:

$$ H_{sys}= H = \begin{pmatrix} E_0 + \mu \cdot E & - \Delta \\ - \Delta & E_0 - \mu \cdot E \end{pmatrix}$$

where $\mu$ is the dipole moment , $E$ is the electric field and $E_0$ is the ground state energy and $\Delta$ is the tunnelling element.

Let the energy eigenstates be represented by:

$$ H | - \rangle = (E_0 - \sqrt{(\mu \cdot E)^2 + \Delta^2 }) | - \rangle $$ $$ H | + \rangle = (E_0 + \sqrt{(\mu \cdot E)^2 + \Delta^2 }) | + \rangle $$

Now, the experimenter play the following game, if he measures the energy and finds the energy in the lower of the $2$ energy states he will double the electric field else he will half the electric field. Hence,

$$ H'(\pm) = H + \lambda_{\pm} I(\mu \cdot E) $$

where $I$ is the identity matrix and $ \lambda_+ = - 1/2$ and $\lambda_- = 1$

Let the energy eigenstates of $H'(\pm)$ be represented by:

$$ H' (-) | 0 \rangle = (E_0 - \sqrt{(1+ \lambda_{-})^2(\mu \cdot E)^2 + \Delta^2 }) | 0 \rangle $$

$$ H' (-) | 1 \rangle = (E_0 + \sqrt{(1+ \lambda_{-})^2(\mu \cdot E)^2 + \Delta^2 }) | 1 \rangle $$

$$ H' (+) | \tilde 0 \rangle = (E_0- \sqrt{(1+ \lambda_{+})^2(\mu \cdot E)^2 + \Delta^2 }) | \tilde 0 \rangle $$

$$ H' (+) | \tilde 1 \rangle = (E_0 + \sqrt{(1+ \lambda_{+})^2(\mu \cdot E)^2 + \Delta^2 }) | \tilde 1 \rangle $$

Let use the density matrix formalism for a pure state $|+\rangle + |- \rangle $:

$$ \rho = \frac{1}{2} \Big (|- \rangle \langle -| + |- \rangle \langle +| + |+ \rangle \langle -| + |+ \rangle \langle +| \Big) $$

OR:

$$ \rho = \frac{1}{2} \begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix} $$

Now lets say I measure the state and find it in $|- \rangle$ and the field is changed. Then using the sudden approximation:

$$ \rho' (-) = \Big (|0 \rangle \langle 0| \Big (|\langle -|0 \rangle |^2 \Big) + |1 \rangle \langle 0| \Big (|\langle -|0 \rangle ||\langle -|1 \rangle | \Big) + |0 \rangle \langle 1| \Big (|\langle -|0 \rangle ||\langle -|1 \rangle | \Big) + |1 \rangle \langle 1| \Big (\langle -|1 \rangle |^2 \Big) \Big) $$

OR:

$$ \rho' (-) = \begin{bmatrix} |\langle -|0 \rangle |^2 & |\langle -|0 \rangle \langle -|1 \rangle | \\ |\langle -|0 \rangle \langle -|1 \rangle | & \langle -|1 \rangle |^2 \\ \end{bmatrix} $$

Similarly:

$$ \rho' (+) = \Big (| \tilde 0 \rangle \langle \tilde 0| \Big (|\langle -| \tilde 0 \rangle |^2 \Big) + | \tilde 1 \rangle \langle \tilde 0| \Big (|\langle -|\tilde 0 \rangle ||\langle -| \tilde 1 \rangle | \Big) + | \tilde 0 \rangle \langle \tilde 1| \Big (| \langle -| \tilde 0 \rangle ||\langle -|\tilde 1 \rangle | \Big) + | \tilde 1 \rangle \langle \tilde 1| \Big (\langle -| \tilde 1 \rangle |^2 \Big) \Big) $$

OR:

$$ \rho ' (+)= \begin{bmatrix} |\langle +|0 \rangle |^2 & |\langle -|0 \rangle \langle +|1 \rangle | \\ |\langle +|0 \rangle \langle -|1 \rangle | & \langle +|1 \rangle |^2 \\ \end{bmatrix}$$

Hence, the new density matrix is given by:

$$ \rho ' = \rho '(+)\Big ( \frac{| \langle - | + \rangle + \langle + | + \rangle |^2}{2} \Big) + \rho '(-)\Big ( \frac{ | \langle - | - \rangle + \langle + | - \rangle |^2}{2} \Big) = p_+ \rho '(+) + p_- \rho '(-)$$

Now, let write down the explicit Hamiltonian of the isolated system:

$$ H_{iso} = H \otimes I_{\text{experimenter + apparatus}} + I \otimes H_{\text{experimenter + apparatus}} + H_{\text{int}}$$

Since, we suddenly change the electric field $H$ is time dependent and since this energy is supplied by the apparatus and experimenter (since he will expend energy to "calculate" how much to change the energy field by) along with the interaction between the systems. Note: $H_{iso}$ is time independent (as it is an isolated system)

$$ \rho_{iso} = \rho(t) \otimes \rho_R(t) $$

Where $R$ is the rest of the isolated system excluding the Hamiltonian system. Since the $H_{iso}$ is time independent so is $S(\rho_{iso})$. Comparing before ($t_-$) and after ($t_+$) the field is turned on:

$$ S(\rho_{iso}) = S(\rho \otimes \rho_R(t_-)) = S(\rho ' \otimes \rho_R(t_+) ) $$

Where $S$ is the Von Neumann entropy of both sides:

$$ S(\rho ' \otimes \rho_R(t_+) ) = S(\rho \otimes \rho_R(t_-) ) $$

Hence,

$$ S(\rho ') + S( \rho_R(t_+) ) = S(\rho ) + S( \rho_R(t_-) ) $$

Substituting $\rho '$ and using $\Delta S_R = S( \rho_R(t_+) ) - S( \rho_R(t_-) ) $

$$ S(p_+ \rho '(+) + p_- \rho '(-)) + \Delta S_R - S(\rho ) = 0 $$

Using an entropy bound :

$$ p_+ S( \rho '(+)) + p_- S(\rho '(-)) + \Delta S_R - p_+ \ln p_+ - p_- \ln p_- - S(\rho ) \leq 0 $$

Hence,

$$ \Delta S_R \leq p_+ \ln p_+ + p_- \ln p_- + S(\rho ) - p_+ S( \rho '(+)) - p_- S(\rho '(-)) $$

Note the $p_+ \ln p_+ + p_- \ln p_- = -S_m $ is nothing but the entropy of the Hamiltonian system when measurement is done but the electric field is not changed. We define $\Delta S_m = S_m - S(\rho )$ and $\rho '(-)$ and $\rho '(+)$ are pure states:

$$ \Delta S_R \leq - \Delta S_m $$