# Search results

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...

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?

Not at all.
17. ### I Commutation between covariant derivative and metric

Well, by fixing "n" I mean that you first perform the summation and then derive, what means that you get a 1-covariant tensor for each term of the sum. In each term of the sum n is definitely fixed. That's what I mean.
18. ### I Commutation between covariant derivative and metric

It must be the partial derivative of ##A^n## if you understand ##\nabla_i g_{kn}## with ##n## fixed (a 1-covariant tensor). They are equivalent ways of seeing it. Just as the covariant derivative of a contraction can be seen just as the partial derivative of it or as the contraction of the...

21. ### I Is the surface of a sphere locally flat?

I know it. I did not mean that the fact that I cannot imagine it means that it does not exist. I was just expressing my ignorance of such a coordinates system. Thanks.
22. ### I Geodesics parametrization

Let us consider a sphere of a unit radius . Therefore, by choosing the canonical spherical coordinates ##\theta## and ##\phi## we have, for the differential lenght element: $$dl = \sqrt{\dot{\theta}^2+sin^2(\theta)\dot{\phi}^2}$$ In order to find the geodesic we need to extremize the...
23. ### I Commutation between covariant derivative and metric

I already realized what happens. Following my reasoning in the first equation displayed I should have written the partial derivative of A, not the covariant one. This rescues it all. Indeed, it seems to me that it is a general rule that covariant derivative of a contraction, i.e. the covariant...

27. ### I Is the surface of a sphere locally flat?

Sure, I only mention the first set of coordinates in order to define the metric tensor.
28. ### I Is the surface of a sphere locally flat?

That definetely solves it. Many thanks!
29. ### I Is the surface of a sphere locally flat?

I can't imagine a set of coordinates that fulfills the non-zero first order correction condition. Indeed, once you impose that the metric is euclidean at p, the new coordinates become fixed and, thus, my choice was the only one possible to be made.
30. ### I Is the surface of a sphere locally flat?

Given a certain manifold in ##R^3## I've been told that at every location ##p## it is possible to encounter a reference frame from which the metric is the euclidean at zero order from that point and its first correction is of second order. This, nevertheless does not match with the following...