01:00 - 18:0018:00 - 00:00

1:26 AM
Consider a quantum system $\mathbb{C}^{2^{\otimes N}}$. Let's say I have a Hamiltonian $H$. Consider the little adjoint map for $H$ defined by $\text{ad}_H := [H, X] = HX- XH$ for all $X \in \mathcal{B}(\mathcal{H})$. One set of eigenvectors of $\text{ad}_H$ is $B = \{\lvert E_n \rangle \langle E_m \lvert\}$ where $\lvert E_n \rangle$ are eigenvectors of $H$. $B$ is a complete basis for $\mathcal{B}(\mathcal{H})$.
In particular, I can write any density matrix $\rho \in \overline{\mathcal{B}}^+(\mathcal{H})$ as $\rho = \sum_{n, m} c(\omega, \omega') \lvert E_n \rangle \langle E_m \lvert$. (this form can be constrained further by applying definition of a density matrix)
For concreteness, consider $\mathcal{H} = \mathbb{C}^2 \otimes \mathbb{C}^2$. Is there a "simple" form for a product density matrix $\rho_1 \otimes \rho_2$ in the basis $B$?
1:49 AM
I made a typo: $c(\omega, \omega')$ should be $c_{n, m}$

1 hour later…
3:04 AM
@ACuriousMind it is much worse than that. there is no curriculum anywhere on this Earth that treats QFT in 2nd year of undergrad. What is most confusing, is why are you drip feeding one mutilated strand of ramen piece by piece into all these people's mouths?
@LeakyNun but then you ought to envision the confusion when you leave the rest of us still in the superposition~

2 hours later…
4:49 AM
@naturallyInconsistent People look upto you: don't show your true colors.
5:07 AM
@SillyGoose if you multiply two of them together then you get something like $\langle E_m \vert E_a \rangle$ in the middle right

1 hour later…
6:22 AM
@NairitSahoo students are much more likely to respect an instructor who shows how things work. I have no lack of credentials lifting students up, publically in class or privately out-of-class, as per student's introversion or extroversion makes what the appropriate handling be. There is no need to advertise my reputation, be it hiding or showing true colours; the students know who they can trust to give them helpful information. Though I am not sure what you want to say about hiding true colours.
Hello Everyone...
@LeakyNun what kind of multiplication?
6:39 AM
well i was using matrix multiplication but now i don't know if that was the right kind
@LeakyNun Good; think about it, even in matrix multiplications, you can have multiplication that goes smaller in dimensions, inner product and multiplication that expands in dimensions, outer product. Things can get both fun and hairy!
@LeakyNun hi
@LeakyNun SillyGoose is using the tensor product of operators. it is different from matrix multi which is the composition of operators
@naturallyInconsistent yeah; my problem is more that i don't actually know what the notations mean
as in
i come from a math background so i know what's tensor product but i don't know what is $\mathcal{B}(\mathcal{H})$ or $\overline{\mathcal{B}}^+(\mathcal{H})$
@naturallyInconsistent You don't have to be sure about everything. Life is like that. Get used to it.
@NairitSahoo I never asserted that I was.
6:47 AM
ok
@LeakyNun that is silly goose being silly and not stating things clearly, nothing to do with you
@LeakyNun im not familiar with this, but i think the B stands for Banach algebra
i think it means an algebra of operators on H
so it would be the linear functions $\mathcal{H} \to \mathcal{H}$?
yes.. and some other requirements i guess to make it Banach
@RyderRude why do you feel the need to answer questions when you are not familiar with things and have to guess?
6:56 AM
QM also uses C* algebras specifically, which have some additional requirements over Banach @LeakyNun
the element $\lvert E_n \rangle \langle E_m \lvert$ eats a "column vector" and gives another "column vector" so i think it's actually an element of $\mathcal{H}$ instead?
@LeakyNun no. the property u gave is one of a linear operator on $H$
an element of H would take a column vector and give a number using the inner product @LeakyNun
@LeakyNun If your $\mathcal H$ is the Hilbert space, then each "column vector" itself is in the Hilbert space; being able to spit out another "column vector" means it cannot be an element of $\mathcal H$ alone.
ah
is $H$ supposed to be an operator on $\mathcal{H}$?
@LeakyNun specifically, this expression corresponds to a matrix which has a 1 as its (n,m)th entry and 0s everywhere else. this is y these form a basis of operators like SillyGosse says
7:00 AM
yeah but idk if the domain is $\Bbb C^2$ or $\Bbb C^4$
SillyGoose is taking $C^4$, it seems
@LeakyNun The specific case, starts with each $\left|E_n\right>\in\mathbb C^2$, but combining them, you get the $\mathbb C^4$
i mean the hilbert space is $C^4$ which is a tensor product of two C^2
But in general, it is not $\mathbb C^2$ at all
@naturallyInconsistent this is where i'm confused; because the notation $\vert{E_n}\rangle\langle{E_m}\vert$ does not generalise well to $N$ dimensions?
so $H$ operates on $\Bbb C^2$, where we have $\vert E_1 \rangle \in \Bbb C^2$ and $\vert E_2 \rangle \in \Bbb C^2$ such that $H \vert E_1 \rangle = E_1 \vert E_1 \rangle$ and $H \vert E_2 \rangle = E_2 \vert E_2 \rangle$?
here $E_1, E_2 \in \Bbb R$ are the energies of the definite states
7:04 AM
@LeakyNun The actual notation that will end up being used, is $\left|E_k\right>\otimes\left|E_\ell\right>\left<E_m\right|\otimes\left<E_n\right|$ which is horrible looking and obviously also why silly goose did not write it. Also, this assumes a specification of left v.s. right, but I don't know of notation that does not make some assumption.
@naturallyInconsistent what is $N$ in your example?
@LeakyNun that's the nice part; I have not specified, and thus is independent of it. Can be any N you like
(as long as finite)
(or really, in real world quantum theory, countable)
what are the ranges of your indices $k, \ell, m, n$?
@naturallyInconsistent this is incorrect. The $H$ of SillyGoose is the Hamiltonian of the the full interacting theory on the tensor product Hilbert space. its eigenstates $|E_n\rangle$ are states on $C^2\otimes C^2$, and not on $C^2$
@LeakyNun in general, it can be anything. It is only in the specific instance that they have to be 0 or 1 alone.
7:14 AM
@naturallyInconsistent well so then you will only have 16 elements right
so N=4
@LeakyNun Yeah; I'm not sure if some symmetry exists to narrow it down a bit more but basically that.
are we implicitly assuming that $N$ is even
because i don't think that's reasonable
like each $\Bbb C^2$ is a qubit right
and $(\Bbb C^2)^{\otimes N}$ is just $N$ (independent) qubits right
i think Ryder is correct here, $H$ acts on the 2 qubits (N=2)
@ACuriousMind Rethinking about this: you can talk of particular field configs in classical field theory. What about QFT? Different field configurations in classical field theory correspond to different vacuua, right in QFT, right?
The field remains the same but the VEVs differ. So it must be the state which is changing
Or something else is going on?
@LeakyNun Only here. It is not justified; we can easily have $N=3$ in another case. But here silly goose said he was starting from qubits, which is obvious from the $\mathbb C^2$ everywhere
@LeakyNun no, $H$ acts on the whole Hilbert space which is the full tensor product Hilbert space
7:20 AM
@naturallyInconsistent i think you're misunderstanding what $N$ means; i was referring to the very beginning of Goose's question "Consider a quantum system $\mathbb{C}^{2^{\otimes N}}$"
@RyderRude that's what i said; the 2 qubits will have state space $\Bbb C^2 \otimes \Bbb C^2$
@LeakyNun sorry...
i confused it with a 2D hilbert space
so what was ur question again @LeakyNun
@RyderRude well i'm still trying to understand the notation
so $\rho_1$ is a linear function $\mathcal{H} \to \mathcal{H}$ right
that means $\rho_1 \otimes \rho_2$ is supposed to be a linear function $\mathcal{H} \otimes \mathcal{H} \to \mathcal{H} \otimes \mathcal{H}$?
no. $p_1$ acts on $C^2$ alone and $p_2$ acts on $C^2$
aha
see they never specified this XD
anyway
so first i need to define the indexing
i think we should use $\vert E_0 \rangle$ and $\vert E_1 \rangle$ for $\Bbb C^2$, leaning into the assumption that we're talking about qubits
then i think the states of the system of the 2 qubits should be indexed by $00, 01, 10, 11$
@LeakyNun i dont understand...
$|E_n\rangle$ r vectors on $C^2\otimes C^2$
cuz theyre eigenstates of $H$
7:28 AM
right, but
hmm
you need to also have the individual qubits right...
i was assuming that there's also an Hamiltonian on the individual qubits
yeah.. there would be a basis which would have that sort of indexing
but right now we're working with the full Hamiltonian's eigenbasis
we can, if we want, switch to a basis in which the vectors r labelled by individual particles
an example of this basis is the eigenbasis of the individual z-spins of the two particles. u have the states (1/2,1/2), (1/2, -1/2), (-1/2,1/2) and (-1/2,-1/2) where $\hbar =1$
@LeakyNun these labels r isomorphic to these 0-1 labels
yeah
@LeakyNun but in this basis we care about the individual z-spins, not the individual hamiltonians
i mean, if we don't have a hamiltonian on the individual qubits, how am i supposed to answer the question of what "is" $\rho_1 \otimes \rho_2$?
the individual hamiltonian is unobservable, so we dont care about that
7:35 AM
@RyderRude you can measure the spin of each particle right
@LeakyNun yeah.. but thats different from an individual Hamiltonian
ok
in that case
does it make sense to say that we still want to have $\vert E_0 \rangle \in \Bbb C^2$ and $\vert E_1 \rangle \in \Bbb C^2$
u could have that notion, but i dont remember exactly. in many cases, the full Hamiltonian is a sum of two individual hamiltonians plus an interaction term
but ofc, u can also pick any full Hamiltonian, and there need not be a notion of individual hamiltonians
and then does it make sense to say $\rho_a = \sum_{m=0,1 \\ n=0,1} a_{mn} (\vert E_m \rangle \langle E_n \vert)$
@naturallyInconsistent with individual hamiltonians, this notation wud work
@LeakyNun yeah
7:38 AM
and then
@ACuriousMind After asking, I stumbled upon this answer by you. This seems to agree with my thought process but I don't understand the part in brackets "Note that whether you say the vacuum state or the representation of the field as an operator is different doesn't matter, you get "different" Hilbert spaces (perturbatively) built on the vacuum state in both cases."
How does it not matter: I mean how do u know that changing the field rep. is equ I mean why do changing the field config and changing the state both correspond to changing the Hilbert spaces? Does changing the field reps change the Hilbert space? and Why do u emphasize on perturbatively?
@RyderRude it will work with the Hamiltonian for the general whole system too, just that it will be ugly
$\rho_a \otimes \rho_b = \sum_{k,\ell,m,n} a_{k\ell} b_{mn} (\vert E_k \rangle \langle E_\ell \vert) \otimes (\vert E_m \rangle \langle E_n \vert)$
@LeakyNun if you actually want to know the details of goose's question, you have to wait for goose to appear and quack. We are not goose and cannot tell you what goose really wants.
@naturallyInconsistent yeah... what i meant there is that SillyGoose is defining their $|E_n\rangle$ to be the states of the full Hamiltonian, so there is nothing to fix in their original notation
7:40 AM
@RyderRude it is too unclear to determine if it is one way or the other.
$\rho_a \otimes \rho_b = \sum_{k,\ell,m,n} a_{k\ell} b_{mn} (\vert E_{km} \rangle \langle E_{\ell n} \vert)$
is this correct?
yeah it does end up looking like this. u r just taking naturallyinconsistent's notation and omitting the two tensor products
@naturallyInconsistent this notation @LeakyNun
right
Again, we cannot tell. If RyderRude is correct, then the $E_n$ already runs over $N=4$ and there would be no need to have $E_{k\ell}$
well we need that because the individual $\rho_1$ and $\rho_2$ operate on the individual bits
7:44 AM
@LeakyNun again, u can use the individual spin-z basis, which is always well defined.
individual hamiltonians need not exist
i didn't say individual hamiltonian
oh. u r writing the eigenvalues as $E$ which gives energy vibes...
what would be a better notation
s^n _z maybe
to denote the $z$ spin of the nth particle
z is not required. we can use x or y too
but its the same notation. we r just not writing $E$
anyway @SillyGoose see if this is the answer you seek:
6 mins ago, by Leaky Nun
$\rho_a \otimes \rho_b = \sum_{k,\ell,m,n} a_{k\ell} b_{mn} (\vert E_{km} \rangle \langle E_{\ell n} \vert)$
7:48 AM
see the msg. it just says E_n r the Hamiltonian's eigenstates. this does not mean individual hamiltonians
yes i'm just using two variables to index one thing
Goose's E_n is my E_km
@LeakyNun it is not..
theyre different basis vectors
$|E_n\rangle$ of SillyGoose need not even factorise
Goose's $\vert E_n \rangle$ is an element in $\mathcal H = \Bbb C^2 \otimes \Bbb C^2$ yes?
yes..
right
yes i'm assuming that they come from individual qubits, spin or whatever
if not we'll need to wait for Goose to clarify
7:54 AM
@LeakyNun but if it is already diagonalising the Hamiltonian, then it will not be $E_{k\ell}$; and it is precisely such details that are missing that makes it impossible to argue fruitfully at this stage, and so why we should have never started on discussing this. I told yall we should not be guessing all these.
8:09 AM
alright
so i wanna ask a question that might seem very obvious
is there a unit system that would allow c=1 and h=1?
(ok i just looked it up and there's a wiki page with c=1 and h-bar=1)
i'm trying to not look at the page and try to see how i would come up with such a system
@LeakyNun yes; you can just make it. In fact, I just found that this is actually much more natural than the usual natural unit systems
so the problem is, i can set c=1 but i don't know how i woulld define length or time
@LeakyNun either you use 1 metre as the base and then the time unit is 1 lightmetre, or you use 1 second as the base and the length unit is 1 lightsecond. Of course, in even better units you have other bases, with all the connections then fixed by the choices you set to 1
right but i'm wondering if there's a more natural choice than both options
yes of course. my unit system is completely fixed except for one unit, and if I picked the mass of the electron as the final thing, then no dimensions appear anywhere
8:24 AM
but why electron
well, we were trying to do some plasma simulations, and in nuclear physics we definitely want to set the mass of proton or neutron to be 1, but we realised that a plasma can easily have electrons going relativistic yet not the nucleons. Having electron mass be 1 as opposed to those other two, makes the relativistic treatment of the electron mixing the with non-relativistic treatment of the nucleons much easier to do.
So, in a sense, Nature seems to be forcing our hand to pick the electron mass as 1
and then do you want to set G=1?
No, for the reason that experimentally, G is too difficult to measure and is known to the fewest digits of accuracy
if G=1 and electron mass = 1 then the gravity between two electrons is... 1/r^2?
ok, what constant did you choose then?
But we care much more about the electric interaction than the gravitational
8:32 AM
do we like set the ionization energy of hydrogen to 1 then
No, I did not need to set those.
I found that $k_B=1=c=h=e=m_e$ is all there is needed.
what is e?
fundamental charge unit
proton charge
And my unit system does not have $\varepsilon_0$ nor $\mu_0$ either
and what are their values?
Instead, I have the fine structure constant $\alpha$ appearing in many places
@LeakyNun they disappear entirely and wont need values. Of course, that is equivalent to saying that they are 1
I'm sorry, if $\varepsilon_0$ and $e$ are both set to 1, the system would be inconsistent. I had $\varepsilon_0=\sqrt{\frac2\alpha}$. There is no need for $\mu_0$ because $\mu_0=\frac1{\varepsilon_0c^2}$
No, $\varepsilon_0=\frac1{2\alpha}$
I had it disappear and so I never really needed it explicitly written down
8:41 AM
@LeakyNun for a finite dimensional hilbert space $\mathcal{H}$, $\mathcal{B}(\mathcal{H})$ is a fancy (condensed) notation for the space of all linear transformations $T: \mathcal{H} \to \mathcal{H}$ for which the trace of each transformation is defined. $\overline{\mathcal{B}}^+(\mathcal{H})$ is the subset of linear operators $T \in \mathcal{B}(\mathcal{H})$ such that $\text{tr}(T) = 1$ and all eigenvalues of $T$ are greater than or equal to zero.
oh here comes the wacky fowl
you have to tell us much more than that, miehehe
Hence, $\overline{\mathcal{B}}^+(\mathcal{H})$ is precisely the space of all density operators (matrices) associated with a given quantum system (hilbert space $\mathcal{H}$)
@SillyGoose and what are $\rho_1$ and $\rho_2$?
@LeakyNun each are density operators (matrices)
but on a different hilbert space right
8:43 AM
yes in my original message i meant to imply that $\mathcal{H} = \mathcal{H}_1 \otimes \mathcal{H}_2$
so the Hamiltonian $H$ is also diagonisable?
$H$ is hermitian and so (to my understanding) is always diagonalizable
and are there eigenvectors on $\mathcal H_1$ and $\mathcal H_2$?
what do you mean by that
by that i mean, when you say you want to find out $\rho_1 \otimes \rho_2$, you first need to define what $\rho_1$ and $\rho_2$ should look like right
and then from those expressions we will compute $\rho_1 \otimes \rho_2$
8:47 AM
Actually, you might not be able to decompose it as a tensor product. If you have entanglement, say
@naturallyInconsistent if you have entanglement then the full state space wouldn't be $\Bbb C^2 \otimes \Bbb C^2$ right
it would just be $\Bbb C^2$
You can always decompose it via some single tensor product, but not as a single tensor product of the unentangled stuff, because, by definition, entanglement
@LeakyNun no, I think it will be still some $\mathbb C^2\otimes\mathbb C^2$; there are some theorems about all these stuff.
entanglement means that the state of the second qubit is determined by the state of the first qubit right
if p1 has spin 1/2 then p2 has spin -1/2
It is not that easy. It can be some superposition entangled with another superposition
if p1 is in state a|1/2>+b|-1/2> then p2 is in state a|-1/2>+b|1/2> right
8:51 AM
A theorem, rather remarkably, asserts that a decomposition via some single tensor product exists. But it will not be the trivial decomposition. It tends to be quite involved.
@LeakyNun no, it can be something completely else
Entanglement can be positively correlated or negatively correlated
i feel like i need to do some cumbersome transition amplitude computations to answer my question...i was hoping there was a simpler solution. i shall leave it alone for now :P
@SillyGoose i can't tell you what $\rho_1 \otimes \rho_2$ "looks like" if you don't tell me what they individually "look like", right
@SillyGoose you needed to state if you meant $\left|E_n\right>$ as 4x1 vectors or 2x1 vectors.
and if they have entanglement in them.
$H$ is the Hamiltonian on $\mathcal{H}$, so its eigenstates (concretely) are $4 \times 1$ vectors
and they are eigenstates of an arbitrary Hamiltonian
8:54 AM
@naturallyInconsistent and then from that how do you determine the unit time?
@LeakyNun as I said, in the end there was only one thing to choose as all the units are based upon it, and when I fixed the electron mass as that choice, the electron mass would then define the unit of time too, through a lot of other things.
hmm
do you use E=mc^2?
that sounds wrong
@LeakyNun of course it is used
well what does that mean then
mc/h is Compton wavelength. It is thus a length. You have length, you know length = ct, so that sets the time unit too
@LeakyNun c=1 so E=m
8:58 AM
as in, what's the physical significance
(i'm gonna get on a train. cya in a few hours)
@LeakyNun same physical significance. rest mass is energy, just very difficult to convert
is it something like, in Compton scattering, an electron "behaves like" a photon with that wavelength?
something like adding two waves together?
9:18 AM
@naturallyInconsistent it is not vague! i interpreted it correctly from the start
@LeakyNun what subjects r u interested in
9:32 AM
@SillyGoose i dont understand... why would a factorisable density matrix have a simple form in the eigenbasis of an arbitrary operator?... since ur hamiltonian is arbitrary

1 hour later…
10:32 AM
Hello, how do you guys calculate the volume of a hollow cylinder but only calculating the volume of just its wall, excluding the hollow center
@NairitSahoo It's just Haag's theorem again, I simply try to not hide the potential complexity: "Different" vacua can only happen when the representations of the CCR are inequivalent, i.e. we cannot pretend everything happens inside a single Hilbert space. It's really both the field and the state that are different, but we usually ignore that at the physics level of rigor (since then things like the interaction picture stop working).
@NairitSahoo And no, there is in general no direct map between classical field configurations and the quantum state of a QFT (vacuum or not), just like there is no direct map from a classical state $(x,p)$ of a particle to its quantum state in ordinary QM.

2 hours later…
12:23 PM
@RyderRude QM right now

1 hour later…
1:45 PM
@RyderRude if it is not vague, you would not have had to wait until the answer is revealed before stating this.
well the intended meaning existed in a state of superposition until the measurement by goose collapsed the wavefunction
@LeakyNun well, you could make this argument even before the unit changes. Note that it is less a wavelength in space (but yes, that also works; if you try to squeeze an electron to less than a Compton wavelength, pair creation happens), but rather mostly a wavelength in time. i.e. the frequency of oscillation of an electron is usually compared to that
i genuinely don't know about compton scattering
@naturallyInconsistent how do the wavelengths in space and time relate for a photon?
(in a vacuum)
@LeakyNun by speed of light in vacuum. Doesn't matter if it is massive or massless
and also, so that i understand what you actually mean
1:53 PM
@LeakyNun that's not a particularly conceptually difficult thing to talk about. As long as you accept the wave picture, it basically follows automatically.
you're referring to $A\exp(i(kx-wt))$ right
@LeakyNun I kinda get that vibe from ya
more specifically, $k$ is the frequency(?) in space, and $w$ is the frequency(?) in time?
@LeakyNun yes; the difficulty is in accepting / entertaining that matter can behave like waves.
@LeakyNun angular versions of both, yes
what's the de-angular version?
@naturallyInconsistent sorry i'm a mathematician :P so only maths is real and the reality doesn't exist
1:55 PM
$k=\dfrac{2\pi}\lambda$ and $\omega=2\pi f=\dfrac{2\pi c}{\lambda_t}$
thanks
meowth
@naturallyInconsistent that matter behaves like waves is far less spookier than, well, the spooky action at a distance
thats ok, there is no spooky action at a distance
in other words, how a spatially symmetric wave that represents the electron somehow manages to be detected at only one location
also it seems like one quantum of electron is way easier to accept than one quantum of light?
(i guess that's just because in chemistry you deal with the quanta of electrons a lot)
is there electron lasers?
2:06 PM
well, at high energies, one quantum of light is easy to accept too, because the large numbers that usually hides the discreteness goes away.
The ease of accepting particle v.s. wave is very little to do with chemistry. It has to do with the fact that electron are fermions and photons are bosons. Bosons tend to combine into what seems like classical waves, and fermions repel that combination, and so we tend to see the particle aspects of them
what would electron lasers even mean?
@naturallyInconsistent ah
@LeakyNun that is actually rather impacted by the choice of interpretation. However, the mathematical predictions of what you would actually see in an experiment is pretty well-established, even if, in some interpretations, the prescriptions are approximations
@naturallyInconsistent does the fermion vs. boson thing also prevent "electron laser" from happening?
the thing is, what would your "electron laser" correspond to? Maybe it already exists
@naturallyInconsistent as in, laser is a lot of coherent photons
2:12 PM
Have you heard of superconductors? In them, the electrons pair up into what are called Cooper pairs, and then these pairs behave like bosons.
Also, what would "coherent electrons" mean? Would a coherent state description be enough?
(note, those are completely different meanings of "coherent")
i'm not sure what that means
good; the point is that I am trying to ask you what you meant when you described a laser, and now you understand that there is uncertainty in the definitions.
@naturallyInconsistent well coherent means same phase right
so that they interact constructively
but is that one photon's wavefunction of many photons?
This is not a trivial question: we can have an electron source giving off a highly coherent (in time and space) wave, but each of one single electron. Then the results work really well
each "individual photon"'s wavefunction is in phase, and they add up together to a huge wavefunction
2:20 PM
e.g. in the double slit experiment
so it's not possible to have a lot of electrons together?
@LeakyNun There is a sense in which this is correct; photons can do that in a way that electrons cannot, by fermion statistics blocking that. However, there is also another sense in which electrons also have a large wavefunction
@LeakyNun but, as mentioned in the Cooper pairing, you can
can you do that with hydrogen molecules (H2) which are bosons?
@naturallyInconsistent so it's like producing single electrons at a regular interval?
@LeakyNun I don't think H2 are bosons, IIRC. But Helium-4 is, and the result is extremely dramatic.
@LeakyNun You seem to be having a mixture of classical wave thinking and quantum thinking, and it is difficult to help you untangle that
@naturallyInconsistent wiki says that the spin is 1 or 0
2:31 PM
@LeakyNun It is not a fundamental particle. The spin-stats theorem may not hold for it. I'm not sure
@naturallyInconsistent well what does "large" mean here then
@LeakyNun spatial extent that is not localisable to some tiny value?
oh
for photons i thought "large" would mean high-energy
do note that low energy necessarily imply large wavefunctions, so, like, the association should be the other way around
as in a laser has a high energy ("because" it has a lot of photons)
2:35 PM
An infrared laser is high power low energy~
as in, compared to a single photon
because each photon is low energy
these things have serious experimental implications, and so there is a natural invisible hand, maybe more tangible than the invisible hand of economics, to push us to come to certain ways of speaking
oh lol I remembered that name right
and also you can spread a laser light and really see it
vs. when you measure an electron it collapses to one location
You can spread an electron wave and really see it too. See Hitachi's 1980s experiment on electron double slit
@LeakyNun when you measure one photon, it also collapses to one location
everything is quantum. Both are kooky
in other words, (photon : laser) :: (electron : ?)
2:40 PM
they don't behave as waves in the idealised sense we want, nor as particles in the idealised sense we want
laser behaves as waves right
@LeakyNun and im trying to get you to pin down what you think is the salient behaviour that defines the laser for you. depending upon your answer, there might be an electron equivalent, or there might not be.
@LeakyNun Lasers are incredibly quantum!
@ACuriousMind But din't u mean this in your answer here? You said "Now, to each different VEV there must belong a different vacuum state", and each VEV arises at leading order from a different field configuration, here a classical minima. So by transitivity we have each classical field configuration corresponds to a different state
@naturallyInconsistent as in, the double slit interference pattern is visible with "one laser" and you don't need to accumulate them
@LeakyNun actually, yes you do. Once you reduce the strong laser output down to single photons by single photons, the accumulation is again necessary.
2:43 PM
yeah but i'm not reducing it to photons :P
But I think you are trying to get at the bosonic character of lasers; namely, that the photons coming out are basically macroscopically large number of photons in the same state. Then yes, electrons cannot do that
i see
@LeakyNun but there is no way to discuss a laser without photons. They are exceedingly non-classical, even if you want to describe them as waves
well that would depend on the experiment wouldn't it
what goes wrong with the classical picture with, say, constructive interference and destructive interference
If your experiment is just wanting to detect the wave aspects, yes, you dont have to care, but then, electrons would also behave as waves that way. Otherwise, you have to consider the quantum nature.
@LeakyNun not much, except that classically constructive interference is always filled, whereas in quantum theory, those are probabilities
2:49 PM
@naturallyInconsistent right, but probabilities with a "macroscopically large number of photons" gives you the classical picture
essentially, yes, but that is a delicate matter. Not an "of course" kinda thing
well that's the law of large numbers
@naturallyInconsistent i see
the photoelectric effect is an experiment that demonstrates this :P
@LeakyNun are u interested in quantum interpretations
i think people have been arguing about them for a century and we have come no closer to "actually" figuring it out
yeah...all currently accessible experiments have no way to distinguish between interpretations
3:03 PM
@naturallyInconsistent what if instead i said, a laser has (very close to) a single well-defined frequency?
at some point, people would have to clarify what counts as a measurement
would bound electrons be the corresponding thing?
@RyderRude right, that thing as well
@LeakyNun I'm not sure what you are getting at.
Also, it is time to go and get drunk so cya
@naturallyInconsistent as in, $\sigma_p$ is very small
(uncertainty in momentum)
@LeakyNun some objective collapse interpretations give collapse models to make it all mathematical
but objective collapse hasn't been verified in experiments yet
3:06 PM
@RyderRude well we really see that the result is discrete
did you mean, the wavefunction hasn't been verified?
@LeakyNun yeah
@LeakyNun yeah
right
e.g. is Wigner's friend, did the friend collapse it or not collapse it
has there been a hidden variable model that works?
This hasn't been verified ofc
@LeakyNun no
@LeakyNun they can't handle relativity
3:08 PM
i don't think relativity is relevant?
it is... relativity, for one, allows particles to be destroyed.. so pilot wave theory goes to shit
people have proposed modifications ofc
there r other hidden variables theories like cellular automata
t'hooft is working on these
he has posted about it on PSE
well these are all very unfamiliar to me but thanks
i also only know the overview
can bell's theorem work with only two states?
u mean the whole system has only two states?
then there can be no entanglement i guese
Bell theorem uses two electrons, which combined have a 4d Hilbert space
3:14 PM
no, each particle has two states
Oh. Then usual Bell theorem is exactly that
they only model spin
what are the 4 states?
But it can be generalised. It's called the CHSH theorem
@LeakyNun (up,up), (down,down), (up,down), (down,up) is a basis
but aren't they supposed to be entangled?
i guess then (up, down) means that their spin was not vertical?
@LeakyNun the state chosen to prove the violation of the inequality is an entangled state
wiki says the state they choose is $|0\rangle \otimes |1\rangle + |1\rangle \otimes |0\rangle$ @LeakyNun
0 and 1 are the eigenstates of the spin-z operator
in this state, if u measure the one particle, u automatically know what the second particle will do @LeakyNun
like if u measured the left qubit to be spin down, the right qubit must be spin up
3:21 PM
so up,up is impossible?
yes.. the state is chosen that way
if they choose non entangled state, then the inequality wouldn't be violated i think
like 0x1+1x0+1x1+0x0. this is completely non entangled
as it factories into (1+0)x(1+0)
in objective collapse theories, measuring a state implies faster than light collapse
right
@LeakyNun u can also choose up, up + down, down...this will also violate the inequlity

1 hour later…
4:47 PM
For a free propagating electron in space, can we associate the spread of the wave-packet in space as the dimensions of the electron?
5:30 PM
Denoting by $\mathcal{R}$ a rotation in 3d Euclidean space, how can I show that this result is true: $$\delta[(\mathcal{R}^{-1}\mathbf{r})-\mathbf{r_0}]=\delta[\mathbf{r}-(\mathcal{R}\mathbf{r_0})]$$
@imbAF what does "the dimensions of the electron" mean?
diameter