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.

  • 1
    $\begingroup$ Use "\langle" and "\rangle" instead of "<" and ">" for your brackets $\endgroup$ Dec 15 '18 at 15:14

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


  • $\begingroup$ 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. $\endgroup$ 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 $\endgroup$ Dec 15 '18 at 16:47
  • $\begingroup$ Next comment, made some error $\endgroup$ 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. $\endgroup$ 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). $\endgroup$ Dec 15 '18 at 21:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.