Graduate Conditional time evolution entropy and the experimenter?

[Someone_physics]

Question
---
So I've done a calculation which seems to suggest if I combine the system of a measuring apparatus to say an experimenter who "reacts" to the outcome of the the measurement versus one who does not. Then the change in entropy in both these situations is bounded by:

$$ \Delta S_R \leq \Delta \bar S_R $$

Where $\Delta \bar S_R$ is the change in entropy where experimenter is indifferent and $\Delta S_R$ is the change in entropy where the experimenter reacts.

Are the below calculations correct? Am I interpreting it correctly if I argue the experimenters choice to "react" changes the entropy?

Summary of Calculations
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1. We isolate the experimenter, measuring apparatus and Hamiltonian system.
2. We start with a Hamiltonian coupled to an electric field whose states are in superposition of spin up and down.
3. If the experimenter measures state of the system to be up he (reacts and) halves the electric field however if he (reacts and) doubles the electric field.
4. Using the sudden approximation we can calculate change in the density matrix and also calculate the entropy change.
5. Since the system is isolated the net change in entropy is 0. Hence, we can also calculate the entropy change for the everything besides the Hamiltonian system.
6. Now we can also calculate the entropy change when the experimenter is indifferent to the outcome of the measurement and create an inequality.

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 let's 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:

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

Note: if the experimenter is indifferent to the measurement. Then one can show the change in the entropy of the rest of the environment $\Delta \bar S_R$is equal to the change in the system:

$$ \Delta \bar S_R = - \Delta S_m $$

Hence,

$$ \Delta S_R \leq \Delta \bar S_R $$

 

[PeterDonis]

Am I interpreting it correctly if I argue the experimenters choice to "react" changes the entropy?

Since by "react" you mean "change the physical conditions" (in this case the electric field), then of course "reacting" will change the entropy as compared with not reacting, since not reacting and reacting produce different physical conditions (in this case, different electric fields).

(Note that I haven't checked the details of your math; I am simply making an obvious general observation.)

I don't see what the point of analyzing this is, however.

 

[dRic2]

If you consider the experimenter as system A and your system as system B then more (stronger) interactions (the "reaction" is an interaction) between A and B, implies more constraints on the evolution of those system. More constraints means a small number of states available thus a smaller entropy change. This is not rigorous (I know too little to help you with calculations) but I think this is a well studied topic in statistical mechanics/quantum thermodynamics. I think https://arxiv.org/pdf/1305.2268.pdf page 9 could help. There is brief part explaining von Neumann entropy and the "problem" of entangled states. Maybe could give you some insights. I didn't read the paper though.

Anyway, long story short, I didn't check your calculation, but your result seems to be expected from what I know about stat mech. Hope someone else gives a feedback on this to be sure.

 

[Demystifier]

Am I interpreting it correctly if I argue the experimenters choice to "react" changes the entropy?

Try to answer the following questions. When does the change of entropy happen, is it when the experimenter makes the choice, or is it when the reaction (which finishes some time after the the experimenter's choice) finishes? What if the experimenter decides today to perform the reaction tomorrow, will the entropy change happen today or tomorrow?