Recent content by Jufa

1. The integral does not converge...

I am asked to compute ##[\phi(x), \phi^\dagger(y)]## , with ##\phi = \int \frac{dp^3}{(2\pi)^3}e^{-ipx}\hat{a}(\vec{p})## and with z=x-y a spacelike vector. And show that this commutator does not vanish, which means that for this non-relativsitic field i.e. with ##p^0 = \frac{\vec{p}^2}{2m}##...
2. Linear momentum of the Klein Gordon field

Yes. it is definitely not right. There's a two at the denominator missing and also the momentum p should multiply the whole expression. Thank you very much for your answer, what you told me about the expression being odd rescues everything.
3. Linear momentum of the Klein Gordon field

The correct answer is: #P = \int \frac{dp^3}{(2\pi)^3}\frac{1}{2E_{\vec{p}} \big(a a^{\dagger} + a^{\dagger}a\big)# But I get terms which are proportional to ##aa## and ##a^{\dagger}a^{\dagger}## I hereunder display the procedure I followed: First: ##\phi = \int...
4. A Concept of wavefunction and particle within Quantum Field Theory

Many thanks for your answer, it helped me a lot! Also I checked your paper and found it really interesting.
5. A Concept of wavefunction and particle within Quantum Field Theory

-1st: Could someone give me some insight on what a ket-state refers to when dealing with a field? To my understand it tells us the probability amplitude of having each excitation at any spacetime point, but I don't know if this is accurate. Also, we solve the free field equation not for this...
6. A Why we can perform normal ordering?

Many thanks for your explanation!
7. A Why we can perform normal ordering?

Get it, thanks!
8. A Why we can perform normal ordering?

As explained in the summary, it seems that the commutators of some operators (creation and anihilation) can be ignored when quantising the hamiltonian of the Klein Gordon Field. I wonder why we are allowed to do such a thing. Is that possible because we are solely within a semiquantum...
9. I Bipartite quantum negativity

Let as consider a system ##H = A\otimes B## I've been said that quantum negativity, i.e. taking the partial transpose w.r.t A or B and summing the magnitude of the negative eigenvalues obtained, is a measure of how entangled are the parties A and B. First question: Why is it that we do not...

Many thanks!
11. I Restricted Boltzmann machine uniqueness

If some moderator can explain why my text is strikethrough I would appreciate it.
12. I Restricted Boltzmann machine uniqueness

I am dealing with restricted boltzmann machines to model distributuins in my final degree project and some question has come to my mind. A restricted boltzmann machine with v visible binary neurons and h hidden neurons models a distribution in the following manner: ## f_i= e^{ \sum_k b[k]...
13. I Quantum negativity

May be it is a good idea to give a little bit of context of the problem I am facing. In few words, I am trying to reconstruct a GHZ state of 4-qbits by means of different tomography methods and, apart from computing the fidelities of the obtained estimators, I am really interested in seeing how...
14. I Quantum negativity

When I computes the negativity (with the partial transpose) of the density matrix corresponding to the GHZ I obtain zero, no matter what is the partition I choose. I've read somewhere that this is because GHZ's distillable entanglement is zero, which I don't really understand because I haven't...
15. I Commutation between covariant derivative and metric

Is there a way of closing the thread?