Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Conditional Time Evolution increases entropy?

+ 0 like - 0 dislike
466 views

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:

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

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

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

Similarly:


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

Hence, the new density matrix is given by:

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

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 $\rho_{iso}$. 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_-) )  $

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

Using an entropy inequality :

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

Hence,

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

Since $\rho'(\pm)$ are pure states:

$$  \Delta S_R  \geq p_+ \ln p_+ + p_- \ln p_-   $$

asked Oct 13, 2020 in General Physics by Asaint (90 points) [ revision history ]
edited Oct 13, 2020 by Asaint

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOve$\varnothing$flow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...