Homework Statement:: I need to draw circuits involving quantum gates and quantum states
I would like a platform that allowed me to draw quantum computation circuits. That is, that it provides me with the schemes and allows me to write in latex.
Many thanks in advance.
I maybe should have mentioned that both ket vectors belong to ##\mathcal{C}^2## and thus the perpendicular vectors are well defined up to a global phase. My attempt is the following:
In the basis ## \ket{\psi}, \ket{\varphi}## the matrix looks like a diagonal one, namely:
##M = Diag\Big(...
We mainly have to prove that this quantity
## \bra{\varphi} A^{\otimes n } \ket{\varphi} \pm \bra{\varphi} B^{\otimes n } \ket{\varphi} ##
is greater or equal than zero for all ##\ket{\varphi}##.
Being ##\ket{\varphi}## a product state it is straightforward to demonstrate such inequality. I...
I guess when you say ##\ket{\phi}## you mean ##\ket{\varphi}##. I don't think it makes sense to compute the entries of the matrix in a non-orthonormal basis. I don't think the associated characteristic equation has nothing to do with the one you get when an orthonormal basis is considered.
We have a matrix ##M = \ket{\psi^{\perp}}\bra{\psi^{\perp}} + \ket{\varphi^{\perp}}\bra{\varphi^{\perp}}##
The claim is that the eigenvalues of such a matrix are ##\lambda_{\pm}= 1\pm |\bra{\psi}\ket{\varphi}|##
Can someone proof this claim? I have been told it is self-evident but I've been...
Summary:: Looking for articles/books to prepare myself for the course: Quantum computation with superconducting qubits
Hello everyone. I am about to take a course in Quantum computation with superconducting qubits and I am searching for material to prepare it. I took a first course on that...
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}##...
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.
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...
-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...
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...
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...
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]...
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...
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...
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.
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...
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.
Let us consider a sphere of a unit radius . Therefore, by choosing the canonical spherical coordinates ##\theta## and ##\phi## we have, for the differential length element:
$$dl = \sqrt{\dot{\theta}^2+sin^2(\theta)\dot{\phi}^2} $$
In order to find the geodesic we need to extremize the...
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...
First, we shall mention that it is known that the covariant derivative of the metric vanishes, i.e ##\nabla_i g_{mn} = 0##.
Now I want tro prove the following:
$$ \nabla_i A_k = g_{kn}\nabla_i A^n$$
The demonstration I encounter takes advantage of the Leibniz rule:
$$ \nabla_i A_k = \nabla_i...
The metric tensor in an inertial frame is ## \eta = diag(-1, 1)##. Where I amb dealing with only 1-D space. The metric tranformation rule after a crtain coordinate chane is the following:
$$ g_{\mu \nu} = \frac{\partial x^\alpha}{\partial x'^{\mu }} \frac{\partial x^\beta}{\partial x'\nu }...
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.
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...
I don't think it is a needed that the vectors are unitary. Indeed the only effect of choosing a non Unit vector in the basis is a change in the metric.
It is just a way of giving a physical sense to these basis vectors. The three components are never used during during the derivacions. Actually we have ##e_\theta=(1, 0) ## and ##e_\phi =(0, 1) ## and the problem you point out is solved but the inconsestency remains.
As Romsofia suggested I tried to picture a simple example to see what's going on and I encounter something that I don't understand. Put this example: The surface of a sphere considereing the basis vectors as ## \vec{e_\theta} = \big(cos(\theta)cos(\phi), cos(\theta)sin(\phi)...
I think the definition of the Christoffel symbols can be made in terms of the metric following this reasoning:
$$ \partial_j g_{mn} = \partial_j \big( \vec{e_m} \cdot \vec{e_n} \big) =\nabla_j \big( \vec{e_m} \cdot \vec{e_n} \big) =
\nabla_j \big( \vec{e_m} \big)\cdot \vec{e_n} +...
I think I have encountered where my misconception comes from. I think it comes from the definition of the Christoffel symbols ## \Gamma^k_{ij} ##. These are definied in such a way that:
$$ \nabla_j \vec{e_i} = \Gamma^k_{ji}\vec{e_k} $$
What I need to know is what ##\nabla_j \vec{e_i}##...