# Some basics about Bracket Notation

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 $$P_n = \left = \rho_{nn}$$.

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: $$P_{n,m} =|\left|^2$$.

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

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:

$$P_{n,m} =|\left\left|$$

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.

• Use "\langle" and "\rangle" instead of "<" and ">" for your brackets – Aaron Stevens 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

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

• 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. – Benjamin Jabl Dec 15 '18 at 16:45
• 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 – Francesco Bernardini Dec 15 '18 at 16:47
• Next comment, made some error – Benjamin Jabl Dec 15 '18 at 21:40
• 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. – Benjamin Jabl Dec 15 '18 at 21:54
• 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). – Benjamin Jabl Dec 15 '18 at 21:55