@Slereah do you understand the linear combination of atomic orbitals? I don't understand why two 1s orbitals combine to form only two molecular orbitals; like if I have one electron in each of the 1s orbitals, then I can combine them 90 deg out of phase right
the two molecular orbitals form by having the 1s orbitals superimpose either in phase or out of phase
The answer is about photons, but the explanation applies to all particles like electrons and quarks as well as photons. Particles turn out to be rather strange things.
Yes. This isn't as strange as it seems. Consider a wave on water, and specifically a tidal wave. These carry momentum and energy as that's how they smash everything in their path. Yes?
Waves on water can be any size, but suppose they were quantised so you could only have no wave or a wave of a certain size. Then they'd behave very like particles. They'd have a momentum and an energy, and they'd scatter off things like particles do.
And this is essentially how we describe a particle. It's a wave in a quantum field.
I should note that no-one knows what a quantum field really is, or even if they really exist. All we know is that maths works and successfully decsribes how particle interact.
@Ishwaran Photons are excited states of a quantum field. They can behave like waves and they can behave like particles, but they are not either of those two things.
There is an electron quantum field and a photon quantum field. Adding energy to the electron field creates an electron and taking energy out destroys an electron. There is a separate photon quantum field and adding/removing energy creates/destroys photons in te same way.
@JohnRennie Definitely the water waves carries some energy as well as momentum, But we describe Photons as a massless particle, how can we say it has some momentum. (I assume Momentum is ability of an object to push something and doesn't have a different definition)
As an explanation of why a large gravitational field (such as a black hole) can bend light, I have heard that light has momentum. This is given as a solution to the problem of only massive objects being affected by gravity. However, momentum is the product of mass and velocity, so, by this defini...
@Semiclassical sorry, forgot to answer before. So we have a bipartite shared $\rho\equiv\rho_{AB}$. Bob does a local operation, somehow selecting unitaries depending on the message he wants to send, call these $U_b$, and we thus have $(I\otimes U_b)\rho(I\otimes U_b^\dagger)$. Now Alice does a global measurement $\mu$, obtaining results $a$ with probs $\langle\mu(a),(I\otimes U_b)\rho(I\otimes U_b^\dagger)\rangle$
so the conditional probabilities are $p(a|b)=\langle\mu(a),(I\otimes U_b)\rho(I\otimes U_b^\dagger)\rangle$. The task is to find $b\mapsto U_b$ and POVM $\{\mu(a)\}$ that maximise the mutual information, right?
that's not how I remembered the problem there, though. The task was encoding info only in the choice of measurement basis, right? So Bob performs some PVM $\{\Pi^b_\alpha\}_\alpha$, and passes along the resulting state (but not the outcome). Alice never gets the measurement outcome, so she gets $\sum_\alpha (I\otimes \Pi^b_\alpha)\rho(I\otimes \Pi^b_\alpha)$, from which she wants to estimate $b$. So the task is to find $b$ from the probability distribution $p(a|b)=\sum_\alpha \langle \mu(a), (I\otimes \Pi^b_\alpha)\rho(I\otimes \Pi^b_\alpha)\rangle$ maybe?
and now the question is what's the optimal choice of $\mu$ to maximise $\sum_b p(b|b)$, right? That is, we want to maximise the measurement giving the answer "$b$" when the measurement basis was indeed the $b$ one
(assuming unbiased prior on the distribution over $b$)
Thought so, but I also know that gauge transformations aren't coordinate transformations but canonical transformations are transformation of phase-space coordinates...so how do I reconcile these two facts?
For a vector field $A_\mu$, there are infinitely many configurations that describe the same physical situation. This is a result of our gauge freedom
$$ A_\mu (x_\mu) \to A'_\mu \equiv A_\mu (x_\mu) + \partial_\mu \eta(x_\mu ),$$
where $\eta (x_\mu)$ is an arbitrary scalar function.
Therefore, ...
Yes I understand that more or less from wikipedia..In page 27 of Quantization of gauge systems by Henneaux, complete fixing of gauge is given as a condition for calling it a "gauge fixing" condition :/
One thing which I did not understand was $\delta F= \delta v^a [F,\phi_a]$ is called to be a gauge transformation generated by the first class primary constraints. I don't see the resemblance with the gauge transformation in electrodynamics or in the QFT gauge theories...things of the form $A->A+\partial phi$..
It's a Hamiltonian perspective on the same thing which tries to completely recognize the issues with converting between the Lagrangian and Hamiltonian formalisms
For a vector field $A_\mu$, there are infinitely many configurations that describe the same physical situation. This is a result of our gauge freedom
$$ A_\mu (x_\mu) \to A'_\mu \equiv A_\mu (x_\mu) + \partial_\mu \eta(x_\mu ),$$
where $\eta (x_\mu)$ is an arbitrary scalar function.
Therefore, ...
