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9:34 AM
How do we tell the difference between 2 electrons and 1 electron in terms of the wavefunctions?
9:52 AM
Different representation in terms of operators
also they will usually be using different Hilbert spaces
One particle Hilbert space, or the antisymmetrized two particles Hilbert space
10:07 AM
$\psi(\mathbf{r}_1)$ vs $\psi(\mathbf{r}_1,\mathbf{r}_2)$
such that $$\psi(r_1, r_2) = - \psi(r_2, r_1)$$
11:02 AM
@Slereah what does $\psi(r_1, r_2) = -\psi(r_2, r_1)$ represent?
That's the anticommutation of fermions
2 hours later…
12:59 PM
@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
what is a particle according to QM ?
anything which has momentum is a particle ?
1:21 PM
It's just a particle, a particle is a particle
uhhh ? :|
2:05 PM
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.
@JohnRennie actually, how does an electron absorb a photon?
@LeakyNun You mean when we excite an electron in a hydrogen atom?
idk if spectroscopy uses that
Have you read the answer I linked above? You need to read that first to get an idea of how particles are described in quantum field theory.
they're waves in a field?
2:15 PM
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.
@JohnRennie So EM radiations are not necessarily waves ?
2:19 PM
@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.
And this description immediately explains how particles can be created and destroyed.
If you consider a water wave, then you can start with still water and add energy e.g. by throwing a stone in, and this creates a wave.
@JohnRennie what's the relationship between electrons and photons in terms of QFT?
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.
but I've heard that a moving electron produces a photon
2:24 PM
And the key thing is that the electron and photon quantum fields can exchange energy.
is there a water analogy for that?
@LeakyNun An accelerating electron produces photons.
@LeakyNun Not really, because water only has one kind of wave.
Electrons and photons are different kinfs of waves in different quantum fields.
maybe it's some water in some parallel universe
and then you have portals to exchange energy
@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)
@LeakyNun :-)
@Ishwaran Photons carry momentum. The momentum of a photon is p = h/λ
2:26 PM
@JohnRennie h = plank constant ryt
Q: If photons have no mass, how can they have momentum?

david4devAs 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...

@Ishwaran Yes, and λ is the photon wavelength.
I'll have to leave this or now as my lunch is ready.
sure have a good lunch !
2 hours later…
4:01 PM
@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?
4:51 PM
@glS yeah, though i think the original paper stuck with just doing projective measurements and unitaries?
which might make our lives easier, at the cost of being suboptimal
5:13 PM
@Semiclassical sure. That's fine. So $p(a|b)=\lvert \langle\Phi_a|(I\otimes U_b)|\Psi\rangle \rvert^2$
5:47 PM
6:02 PM
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 $b$)
do they give a clear solution?
6:24 PM
in QFT each particle has its own field?
multiple particles can belong to the same field in a given problem
if the k(2s + 2px) orbital has lower energy than the 2px orbital, why doesn't any electron live there? (k is a normalization constant)
6:47 PM
Are gauge transformation and canonical transformations related?
Both of these transformations seem to map sets of canonical coordinates..
7:03 PM
@ManasDogra Gauge transformations are canonical transformations (but not all canonical transformations are gauge)
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?
I don't know what "gauge transformations aren't coordinate transformations" is supposed to mean
Reparameterisation's of e.g. a relativistic point particle are gauge transformations
What distinguishes a "coordinate transformation" from just a "transformation" (i.e. a map from phase space to itself)?
Okay okay..now i get it..
Another question..Why is Lorenz condition a gauge fixing condition?
I read that gauge fixing conditions must completely fix the gauge..but lorenz gauge doesn't do that.
7:12 PM
It's a 'partial gauge'
Q: Why does Lorenz gauge condition $\partial_\mu A^\mu =0$ pick exactly one configuration from each gauge equivalence class?

jakFor 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 :/
That's why I was wondering..
terminology isn't universally consistent, partial gauge conditions aren't gauge conditions, manifolds with boundaries aren't manifolds, etc.
Okay so it was a simple matter of terminology.
Even a page later they say what they said on that page is of limited validity
Oh..sorry I see it now..
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 $A->A+\partial phi$..
7:25 PM
@ManasDogra They show the application to Maxwell theory explicitly at the end of the book
It's a Hamiltonian perspective on the same thing which tries to completely recognize the issues with converting between the Lagrangian and Hamiltonian formalisms
Thank you everyone

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