3:03 AM
@ACuriousMind i even loathe this format of statement "we evolved xyz because abc"
ah the most effective way to get answers from stack exchange. come to hbar, type out the question, discover answer through attempts of making the question more refined. yippee.
4
3:21 AM
lol
B A H ~
if i have a hamiltonian $H_S + H_E + H_I$ and i write a master equation for this ham (lindbladian), does only $H_S$ effect the ground states of this system? or the whole Ham?
@Relativisticcucumber $H_E$ only acts on the environment degrees of freedom. $H_I$ interacts with both and so will be involved.
so if i want to know the ground state of the system, i can look at the eigenvalues of $H_S + H_I$? @naturallyInconsistent
and take the lowest to be the ground state i mean
bah this is quite confusing because how can i even interpret a ground state of a term like $\sigma_z(a^{\dagger} + a)$
3:44 AM
But $H_I$ links the system degrees of freedom to the environmental degrees of freedom, so it might be really difficult to find the ground state.
Usually, you just ignore the interaction for a bit, determine the ground state without interaction, and then determine the interaction corrections for the ground state if needed.
@naturallyInconsistent so is it likely to be a good assumption that the ground states of $H_S$ are the ground states of the system? or the corrections have a good chance at changing this?
Well, the fact that we can even separate out the system v.s. environment means that the system is interacting weakly with the environment enough for us to pretend it is so.
4:05 AM
okay i will try to go with this @naturallyInconsistent

3 hours later…
7:11 AM
i have a great approach to categorise randomness models. u all will like it
Definition a randomness model for a binary outcome experiment (H,T), is identified by a set of infinite sequences of $H$ and $T$
Trivial Randomness Model This is the set which consists of all infinite sequences of $H$ and $T$
all other randomness models are subsets of the above. this formalises the idea of "adding more structure"
A fair coin model is the set of all sequences of $H$ and $T$ such that both have equal frequency
A Memoriless probabilistic model is the set of all infinite sequences of $H$ and $T$ where the frequencies converge
A non probabilistic randomness model can be the set of all sequences of $H$ and $T$ where $H$ occurs at least twice every 10 trials
To choose a randomness model for an experiment, we have to make a guess based on a finite number of trials of the experiment
and finally, quantum mechanics is the only source of true randomness in nature, and it seems to be a probabilistic randomness... this means most kinds of randomness are not known to materialise in nature
@RyderRude no
7:47 AM
@Slereah @ACuriousMind what do u think about the above approach?
i think it can help in studying general randomness models
has this approach been utilised before
8:00 AM
@RyderRude no, because it is nowhere near well-defined enough to be considered a probability / randomness model.
8:15 AM
@RyderRude I have no interest in discussing this useless topic further with someone who has evidently no familiarity with how "memory" is usually used and modeled in the context of stochastic processes and prefers to make up random definitions instead of researching the extant literature, please don't ping me about it any more.

1 hour later…
9:25 AM
hows gaemi @naturallyInconsistent ? gonna hit here friday ;,(
for the ising ham (above), we have the first term as the nearest neighbor terms, which makes sense to me that the energy should be influenced by the coupling of nearest neighbor spins. what is the point of the second term? just a "self energy"?
9:42 AM
is there a formalization of this property of the sigma sum: $\sum_{i=0}^{2N-1}\sum_{j=0}^{2N-1} = (\sum_{i=0}^{N-1} + \sum_{i=N}^{2N-1})(\sum_{i=0}^{N-1}+\sum_{i=N}^{2N-1})$
my question is more like: what sort of mathematical structure does $\sum_{i=i_0}^{i_f}$ belong to?
10:09 AM
Hello Everyone...

1 hour later…
11:22 AM
@Relativisticcucumber my umbrella was horizontal. Wet everywhere. sexy time cancelled. very sad
@Relativisticcucumber yes, necessary for initial alignment.
@SillyGoose That's one part of measure theory. It would be the simplices part of differential geometry. It is the integration region part. In a sense, we push all the complicated parts to the integrand, and so this part can always be chosen to be simple, and so there is very little non-trivial mathematical structure there.
@SillyGoose I'm not sure what this question is relevant for, but formally you would define sum operators as being linear maps $\ell^1(\mathbb{R})\to \mathbb{R}$, where $\ell^1(\mathbb{R})$ is the space of summable sequences
11:42 AM
And oh yes, the road was littered with broken branches. Dangerous
12:00 PM
@SillyGoose it is like writing 1+2+3+4 = (1+2) + (3+4)
@123 456
12:34 PM
@naturallyInconsistent but as far as typhoons go -- bad or eh?
@naturallyInconsistent hm ok i saw some notes leave it out so i was confused
wondering if i should pick up some food at the store or if delivery will still be a viable option XD
so in this ham, we only have self energy for $\sigma_z$, so what is the difference between this and ising model for z-direction? if we want all directions (heisenberg model), shouldnt we have self energy for all directions?
@Relativisticcucumber I think it is not a "self-energy" term. it is the appropriate term to add to the Hamiltonian given application of a homogenous magnetic field in the $z$-direction on the spins
12:49 PM
What gives the direction of velocity?
Is there any example where direction of displacement (infinitesimal displacement in curvilinear motion) is different from direction of velocity (infinitesimal velocity)?
We know direction of displacement gives the direction of motion. What information we get from direction of velocity?
1:22 PM
Hello @naturallyInconsistent

2 hours later…
3:34 PM
@Relativisticcucumber dunno; am new to diz
4:25 PM
Hi, I have a question about the classical concept of torque in different scenarios:
For a magn. moment in a magnetic field the torque is of the form $\vec \tau=\vec \mu \times \vec B$. In certain cases, when a force is applied at some distance from a point, through which a rotation axis passes through, we write $\vec \tau = \vec r \times \vec F$.
Is there a way, for a magn. moment to start with the 2nd one and reach the first one, knowing that we can write $F=-\vec \Nabla(\vec \mu \vec B)$ ?
@imbAF You have to think carefully about what that torque and that force are about
In which case?
you cannot derive one from the other because they describe two different phenomena
the torque is the torque on the dipole about the center of the dipole, i.e. its tendency to rotate around its center in a magnetic field
If one would imagine the magn. or elect. moment as a finite line, where the external field pushes in one side and attracts the other, can't that be a viable case?
but the force you're talking about with $F=-\nabla(\mu \cdot B)$ is the force on the whole dipole, i.e. it's a force that is attached to the center of the dipole and produces motion of the center
4:36 PM
would this be the case for both types of moments? Magn. and electric ?
@ACuriousMind I see your point, and I kind of agree, but at the same time, thinking like this, wouldn't one argue that if this force acts on the whole dipole, the dipole should as a whole be displaced and not rotated around some axis, no?
that these two equations cannot be about the same force is obvious when you consider the case of a homogeneous magnetic field: The torque $\mu\times B$ is still non-zero if it's not parallel to the magnetic moment, but the derivative $\nabla(\mu \cdot B)$ is zero because $\mu \cdot B$ is constant.
@imbAF yes, exactly! the force $\nabla(\mu \cdot B)$ is a force that displaces the whole dipole in an inhomogeneous magnetic field; it is not the same phenomenon as the torque $\mu \times B$ that rotates the dipole even in a homogeneous field
You are right
@ACuriousMind so in the case of the S.G device, the former happens then?
yes, which is why you require an inhomogeneous field for Stern-Gerlach apparatuses
Ok, so quite the subtle thing to consider
I will simply write down, what it's written in my script and you perhaps can help me understand, why are things taken in this kind of way
The setup is, silver atoms, in ground state propagating in y positive direction and at some point they are exposed to an inho. magnetic field pointing in the pos. z direction
Because the system exhibits magn. moment, when exposed in a magn. field it will gain/lost energy in value of $E=-\vec \mu \vec B*. According to what we said, the magn. field inhom. will cause the displacement. But in the lecture we proceed with the following:$\vec \tau=\vec \mu \times \vec B$and$\vec \tau=\frac{d}{dt}\vec S$This torque causes precession, since the system, the silver atom, because of it's one electron in 1s, has spin, right? @imbAF yes 4:51 PM Ok I see, the formula of torque with radius and force, really threw me off I mean, if we really get down to it it becomes extremely questionable in what sense this is really a "torque" in quantum mechanics since an electron is a point-like object, not a classical dipole that can actually rotate in the literal sense of being affected by classical torque, but yes, the magnetic field causes the spin to precess additionally it causes the displacement of the atoms from their original path, and the creating of two points where they land in the screen The thing that I find, not rigorously clean is the following: For traceless symmetric tensorial representations, is there any guarantee that "polarization" vectors are null according to representation theory? @Sanjana what is a "'polarization' vector" and what does "null" mean in a representation-theoretic context? also, representations of what so that the word "traceless" means something? These are conformal primaries: representations of a conformal group$SO(d,1)$. I don't really have a good definition of "polarization" vector except that if I have the primary$A^{\mu_1 \mu_2 ... \mu_l}$then it is said in a paper that info about it can be encoded in terms of polynomials$A^{\mu_1 \mu_2 ... \mu_l}z_{\mu_1} z_{\mu_2}... z_{\mu_l}$where the$z$s are null w.r.t. the Minkowski metric$z^2=0$. I don't understand why the$z$s are taken to be null. 5:03 PM ACM The thing I wanted to ask, was, when we talk about the conservation of angular momentum, we always have in my the specific case, where the vector of it is perpendicular to some plane, right? @imbAF I don't understand the question - every vector is perpendicular to some plane. Let me clarify Also I thought this is connected to tracelessness which is why I said rep theoretic: but I might be wrong... so I shared a screenshot from the paper arxiv.org/abs/1107.3554 section 3.1 If the angular momentum over time has a periodic pattern, would that mean conservation in a way? 5:07 PM @imbAF I don't understand the question; the definition of a quantity being conserved is that it is constant in time @ACuriousMind The only thing I can handwavingly think of is that these must obey some massless Klein Gordon like equation componentwise, Fourier transforming which we get the polarizations to be null... but I am not sure The way I interpret the conservation of angular momentum, I mostly think the scenario in which the vector doesn't change it's direction or magnitude. But if the direction changes over time in a repeating pattern, would that constitute conservation of it? @Sanjana There is little representation theory here, the entire argument seems to be included in the picture! What part of the argument around eqs. (3.3) and (3.4) do you not understand? @imbAF no, obviously not, because conservation of a quantity means that the quantity is constant in time if the quantity changes, it's not constant I don't understand where the confusion here is Ok I see Well for now I have a good understanding of things But that little detail with the force and the torque, threw me off by a lot 5:27 PM @ACuriousMind Why$z^2=0$? Why are they considering two tensors which differ by$O(z^2)$? @ACuriousMind I mean, I do not understand why restriction to$z^2=0$is enough and why we don't require a more general$z_mu$for this? I have trouble figuring out how to write this in notation that isn't utterly confusing, give me a second 5:55 PM Direction of displacement gives us the direction of motion. What information gives direction of velocity? 2 hours later… 7:26 PM The tensor$f_{a_1\dots a_l}$is traceless iff$\eta^{a_i a_j} f_{a_1\dots a_l} = 0$for every pair$a_i\neq a_j$. Also, the traces$\mathrm{tr}_{ij}(f)_{a_i\dots \hat{a}_i\dots \hat{a}_j\dots a_{l}} = f_{a_1 \dots i \dots j \dots a_l} \eta^{ij}$are tensors of two degrees lower, i.e. they correspond to polynomials two degrees lower, where the$i,j$are such that we contract the ith and jth slot of the tensor. But since$f$is symmetric, all these traces are the same apart from re-labeling of the indices, so there's really only a single such "trace tensor". Let's call it$\mathrm{tr}(f)$(a @Sanjana ^ 7:44 PM @ACuriousMind Ok so this crucially depends on the fact that$f$is symmetric, right? yes Okay. Thank you so so much. This is very much clear. but, I mean, the trace of an antisymmetric tensor vanishes anyway, so why's that remarkable :P @ACuriousMind I mean this is dependent on the nature of representations. That's why... So basically the trace part gives$z^2=0\$
8:01 PM
@ACuriousMind I believe, a while back I asked you about the S.G device, that if you have the magn. field in the Z direction you measure the eigenvalues of the Z compoennt of the spin. But if after that magn. field, you have another one in the X direction, you'd measure the eigenvalues of this operator, you remember that?