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I'm trying to prove something. Sorry this post is so long but I wanted to keep things as basic as possible so people have an easier time understanding.

Let's assume we have a quantum system $\rho$ which evolves unitarily. During the process the Hamiltonian $H$ is getting externally varied, so that the evolution of $\rho$ is described by an unitary U.

Now what we do is to make a measurement of $\rho$ at the beginning. Let's say we obtain some state $\left|n\right>$ with probability $P_n$, with $P_n = \left<n|\rho|n\right> = \rho_{nn}$.

Now the process of the Unitary is implemented and our state $\left|n\right>$ evolves accordingly to U. After some time we make a second measurement and obtain $\left|m^f\right>$ where the f denotes that it is the final state.

Now we are interested in the conditional probability to find this $\left|m^f\right>$ under the condition we found state $\left|n\right>$ before. This conditional probability can be expressed as: $P_{n,m} =|\left<m^f|U|n\right>|^2$.

Additionally we have another unitary $V$ which transforms the hamiltonian basis $\left|m^f\right> = V \left|i\right>$ such that we can rewrite $P_{n,m} =|\left<m|V^{\dagger}U|n\right>|^2$.

My first question: Is there any smart way I could rewrite $P_{n,m}$ ? I'm not very good with the dirac notation and all I was able to achieve is rewrite the square:

$P_{n,m} =|\left<n|U^{\dagger}V|m\right>\left<m|V^{\dagger}U|n\right>|$

That's all.

I would need to that because ultimately what we have been doing is measuring two energy values. We are intersted in getting these states $\left|n\right>$ and $\left|m\right>$ because eventually we could substract them to obtain the work $W_{(nm)} = E_n - E_m$.

So putting all pieces together we define the Work:

$W = \sum_{nm} p_{(nm)}W_{(nm)}$

where $p_{(nm)}$ is the total probability of first getting state n and later obtainin m so:

$p_{(nm)} =\rho_{nn}P_{n,m} $

and somehow I need to take this $W = \sum_{nm} p_{(nm)}W_{(nm)}$ and rewrite it in several ways to achieve things like:

$W = \sum_{nm} p_{(nm)}W_{(nm)} = \sum_n \rho_{nn}E_n - \sum_{nm} \rho_{nn}P_{n,m}E^f_m$

which ok....I have no idea how to derive this (which I would like to in bracket notation) but I can kinda agree because it basically says "Prob. of initial state times initial energy - Prob. of final state times final energy"

But can anyone help me to understand how to obtain this result via bracket notation or at all ?

And then they go further and say:

$W = \sum_n \rho_{nn}E_n - \sum_{nm} \rho_{nn}P_{n,m}E^f_m $

$ = \sum_n \rho_{nn}E_n - \sum_{nm} \rho_{nn}E^f_m U_{kn}U^*_{ln}V_{lm}V^*_{km}$

where * is supposed to be the complex conjuagte of the unitary operators. I have no idea where k and l come from though. Any help to solve this would be greatly appreciated.

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    $\begingroup$ Use "\langle" and "\rangle" instead of "<" and ">" for your brackets $\endgroup$ – Aaron Stevens Dec 15 '18 at 15:14
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If your set of states $|n\rangle$ is complete, you can always insert an identity in betwee two operators, a bra and a ket, or an operator and a bra-ket:

$$\mathbb I=\sum_n |n\rangle\langle n|$$

In this way you can make appear the matrix elements of U and V in the decomposition you wrote:

$$\langle m|U V^\dagger|k\rangle= \langle m|U\sum_n |n\rangle\langle n| V^\dagger|k\rangle=\sum_n \langle m|U |n\rangle\langle n| V^\dagger|k\rangle\equiv \sum_n U_{mn} (V^\dagger)_{nk}$$

And we know that

$$(V^\dagger)_{nk}=V_{kn}^*$$

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  • $\begingroup$ Thanks with this I was able to solve the first part! It's obvious things like these that I just haven't learned properly yet. $\endgroup$ – Benjamin Jabl Dec 15 '18 at 16:45
  • $\begingroup$ You’re welcome bro, I didn’t go further because I don’t want to steal you all the fun, but if you need more development tell me $\endgroup$ – Francesco Bernardini Dec 15 '18 at 16:47
  • $\begingroup$ Next comment, made some error $\endgroup$ – Benjamin Jabl Dec 15 '18 at 21:40
  • $\begingroup$ 1. Well I was able to solve it and obtained in the end what I wanted! Again thank you. Now I'm kinda trying to understand the result. Reminder: I now got the results for work: 1.$\sum_i \rho_{ii} E_i - \sum \rho_{ii} E^f_j U_{ki}U^*_{li}V_{lj}V^*_{kj} $ 2. $\sum_i \rho_{ii} E_i - \rho_{im} E^f_j U_{ki}U^*_{lm}V_{lj}V^*_{kj} $ where we can see that they are identical if m=i, or in other words when $\rho$ has only diagonal elements. $\endgroup$ – Benjamin Jabl Dec 15 '18 at 21:54
  • $\begingroup$ 2.Now this is something interesting I was hoping you could tell me a bit about. $\rho$ being diagonal means that it is a classical physical system that has NO coherence. Now I was trying to understand the meaning of this which I'm still not so sure about. To confirm I tried to make the same calculation (as in initial energy - final energy = work) for a $\rho$ that HAS off-diagonal terms (thus should have coherence). $\endgroup$ – Benjamin Jabl Dec 15 '18 at 21:55

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