01:00 - 23:0023:00 - 00:00

1:33 AM
@ACuriousMind i am looking for a simple way to prove the truth of the claim. the direct derivation is a little clunky because one needs to know what the $\sum$ does and such. it would be easier for me to instead understand what type of mathematical structure $\sum$ belongs to so that manipulations with it are clear.
so i do not make any mistakes :P

3 hours later…
4:04 AM
@imbAF you might want to actually state a fully-formed question so that you wont waste people's time waiting for you to continue
@SillyGoose The manipulations on the integration region side is almost always nice shapes. I don't think there is any way to help you get it any less clunky
4:37 AM
> Rosemary Fowler discovered the kaon particle during her doctoral research in 1948 but gave up PhD to have a family
4:55 AM
Hello Everyone...
Hello @naturallyInconsistent
Direction of displacement gives direction of motion. What information we get from direction of velocity?
stop tagging me like that
@123 this is just wrong.
Similarly direction of acceleration gives speed increasing/decreasing/direction change. What information we get from direction of force?
Sorry if you disturbed.
Direction of velocity is an information.
And idk wdym by "direction of motion".
If direction is not an information , why we need it?
yeah, that is a red flag, and part of why I didn't want to touch that with a ten foot pole
5:01 AM
If direction makes a physical quantity vector. Why direction is not purposeful. Magnitude and unit is information. Why not direction.
But regardless of that, 123, you should simply consider the simplest uniform circular motion. Let $$\vec r=\varrho\begin{bmatrix}\cos\varphi\\\sin\varphi\end {bmatrix}\qquad\implies\qquad\vec v=\dot{\vec r}=\varrho\dot\varphi\begin{bmatrix}-\sin\varphi\\\cos\varphi\end {bmatrix}$$
Step 1: follow the same thing as time derivative of position gives velocity, time derivative of velocity gives acceleration, show that $\vec a=-(\dot\varphi)^2\vec r$, the very well-known $a=-\omega^2r$ formula, known to the ancients, maybe as $a=-\frac{v^2}r$
Step 2: Then let $\varphi=0$ and draw out $\vec r,\ \vec v$ and $\vec a$.
Step 3: tell us where you got wrong in your original statements.
5:25 AM
Can you pls explain if phi is zero what will be position, velocity and acceleration. In my view all will become zero.
hello, $\cos0=1$
I have already chosen for everything to be as nice as they can be.
This is a circle with constant radius $\varrho$ and constant angular velocity $\dot\varphi=\omega$
5:40 AM
@SillyGoose I find this claim absurd; manipulating the sums directly is much easier than any abstraction, which, as nI hints at, would be to treat sums as a special case of integrating measures over unions of subspaces (they are integrals of measures on $\mathbb{N}$)
Since integrals are defined by limits of sums, you gain nothing by treating the sums this way.
You're over complicating the simple fact that addition is associative :P
6:01 AM
I find that particularly crazy too

2 hours later…
8:18 AM
blebs
why is magnetostatics not called electromagnetostatics?
the "canonical" magnetostatic conditions $\frac{\partial \vec{j}}{\partial t} = 0$ and $\nabla \cdot \vec{j} = 0$ precisely allow time-independent electric and magnetic fields don't they?

1 hour later…
9:25 AM
In example of uniform circular motion position vector, displacement and infinitesimal displacement are different things have different magnitudes and directions.
@SillyGoose how can electro and static be together when there is magnetism :o doesnt magnetism require moving charges?
ok back with more heisenberg model
we have this lovely little ham. i tried cases where the "self energy" term is $\sigma_z$ and also $\sigma_x$. in both cases, the ground state becomes "all up" in the direction of the self energy term. my question is -- is it sensible to have multiple directions in this "self energy" term? like is it sensible for instance to have both $\sigma_x$ and $\sigma_z$ included?
@BinkyMcSquigglebottom love the name
9:47 AM
@Relativisticcucumber how can you be looking at Heisenberg model and yet still having such a misconception?
@naturallyInconsistent i mean sillygoose is talking about classical e&m so i discard all stuff like heisenberg model
am i really that confused.....
Magnets exist in classical E&M
i thought that can only be explained by quantum so im confused
i thought the only classically explainable source of magnetism is moving charge?
No. Think of things historically. Before Faraday, electrostatics was a subject, magnetostatics was a subject, and there was no way to link them. It was only after experiments put them together that we had electromagnetism.
@Relativisticcucumber i mean that in the regime of magnetostatics, time-independent charge density $\rho(\vec{r})$ and zero current density $\vec{j} = 0$ is allowed. this is just precisely electrostatics.
9:58 AM
@SillyGoose if $\vec{j} = 0$, then how is there magnetism in the classical picture
that is what i am saying: the magnetostatic regime includes electrostatics as a special case
i.e., a situation without magnetic fields at all
so the magneto in magnetostatics is strange to me
but does it not include other cases?
it does include other cases
@SillyGoose right dont those other cases preclude being called electro...statics? or am i missing smth
yes, but magnetoelectricstatics would suggest that it is a situation which models static electric fields and static magnetic fields
which magnetostatics does
magnetostatics does not suggest that it includes electrostatics at all
10:01 AM
i think blebostatics should describe bleb being static
so imo would be weird to include cases where bleb is not static
10:14 AM
the problems in magnetostatics are usually when $\vec E=\vec0$
in the same sense that in electrostatics, $\vec B=\vec 0$
10:53 AM
a new survey on algebraic QFT arXiv link
11:23 AM
@Relativisticcucumber it's not "self energy", it's just the coupling to an external magnetic field. Sure, you can add a magnetic field in any direction. It is not true that the ground state is the spins aligned with the field though, that's not an eigenstate

3 hours later…
2:14 PM
hi

2 hours later…
3:46 PM
Hi guys, I'm having trouble with the solution to a QM exercise written by my professor
Let us suppose that the Hamiltonian of a system of two indistinguishable particles of spin zero which are subjected to a known constraint allows us to consider the following CSCO: $$\mathcal{B} = \{\mathbf{J}^2, \mathbf{J}_z,\mathbf{J}_1^2, \mathbf{J}_2^2 \}\longleftrightarrow |J,M; j_1,j_2\rangle$$ where $J,M$ are obtained via addition of the single angular momenta $l_1 \oplus l_2$ (namely the CG coefficients).
Since we're dealing with bosonic particles, we must ensure that the Symmetrization Postulate is satisfied. In order to do this, I first notice that
$$\hat{P}_{12}|J,M; j_1,j_2\rangle = (-1)^{j_1+j_2-J}|J,M;j_1,j_2\rangle$$
Then, I proceed by fixing the values of $l_1$ and $l_2$. We have to deal with the two cases: $l_1 \ne l_2$ and $l_1 = l_2$.
If $j_1 = j_2$ then we must have $j_1+j_2-J= \text{ even}$ since $|j_1,j_2\rangle$ is clearly symmetric under exchange $1 \leftrightarrow 2$.
when $j_1 \ne j_2$ I can restrict myself to the case $j_1>j_2$. And I must construct symmetric and antisymmetric combinations via $$S[A] = \frac{1}{2} \left(\mathbf{I}+[-]P_{12}\right) \longrightarrow |j_1,j_2\rangle_{S[A]}$$
Therefore, we have that $$\text{ if } j_1+j_2-J = \text{ odd } \Rightarrow |j_1,j_2\rangle_A$$ and viceversa
This is how I would proceed. My professor instead writes that when $l_1 \ne l_2$, then the eigenstate will necessarily be of the form $$|J,M;j_1,j_2\rangle_S$$ and ignores the possibility of forming eigenstates with $|J,M;j_1,j_2\rangle_A$
Is there a reason why I can only consider symmetric combinations of $|j_1,j_2\rangle$?Or maybe my professor overlooked this fact (I don't wanna sound pretentious, just asking an honest question)? Or am I missing something when imposing the Symmetrization postulate
4:16 PM
Was the exercise considering that the two initial states have the same space parts, and so the antisymmetric part has to go away? Also, there is no such thing as $\left|j_1,j_2\right>$ and even if there is such a fictional thing, just because $j_1=j_2$ does not mean that they are symmetric. The constituent $m_1$ and $m_2$ could have started out different.
@naturallyInconsistent Hi
I am having truble only with one thing regarding the SG device
If the magnetic field points in some arbitrary direction $\vec n$, then we construct (if one can say so) a spin component $S_{\vec n}$. An eigenstate of this spin operator has the form $|+\rangle_{\vec n}=cos(\frac{\theta}{2})e^{-i\frac{\phi}{2}} + sin(\frac{\theta}{2})e^{i\frac{\phi}{2}}$.
Now let's assume that $\phi=0$. The state will transform to:
$|\phi\rangle=cos(\frac{\theta}{2}) +sin(\frac{\theta}{2})$
@naturallyInconsistent It has to go away?
errm, yall can try to ask questions fully?
lol
I was asking why the antisymmetric part goes away
Well, if you have symmetric space part, then obviously you need the symmetric spin part
4:30 PM
The problem I am having is the following:
When $\theta=0$ you are having a magn. field pointing in the z direction. Therefore you are considering the classical case of S_z component.For which there's a 50% prob. to measure either eigenvalue.
But if considering the expression above, the state is $|\phi\rangle=|+\rangle$. Which means with 100% accuracy we only get the +\hbar/2 eigenvalue. How is that possible?
so you needed to check the context that your prof was going on
@Claudio I am confused what the notation here means, or what we are trying to show. What do the subscripts $S/A$ denote (don't say "(anti-)symmetric", that's obvious, but what does it mean, in technical terms)? Where do we want to go?
@imbAF When $\vartheta=0$ that is purely spin up and so 100% probability of getting $+\hslash/2$
But that's not possible though. Because when the SG device, has a magn. field pointing in the z direction meaining $\theta=0$ and still you get two dots on the screen. with 50-50 possibility
@imbAF Your notation makes no sense, $\cos(\theta/2) + \sin(\theta/2)$ is a number, not a quantum state, it cannot be equal to a quantum state $\lvert \phi\rangle$.
also it is an extraordinarily poor choice to denote the angle we're setting to zero by $\phi$ and then also call the state $\phi$
4:32 PM
@ACuriousMind $\cos(\theta/2)|+\rangle + \sin(\theta/2)|-\rangle$ *
@imbAF That's because the initial state was a mixed state. If you started with just spin up or spin down, then it would only have one dot. You need to be much more careful with stating things.
@imbAF This is even worse; you have $\left|+\right>_{\vec n}$ expressed in terms of $\left|+\right>$ and $\left|-\right>$
Hello World!
@imbAF what are $\lvert +\rangle, \lvert -\rangle$? Clarity of thought starts with clarity of notation
The eigenstates of the Z component of the spin vector operator
I believe I am stating things clear but I will try better
@imbAF Also, I'm not following you at all here - what does the value of $\theta$ - which is a property of the eigenstate have to do with any magnetic field whatsoever?
4:35 PM
@ACuriousMind I want to build well normalized eigenstates of the form $\frac{1}{\sqrt{2}}(|j_1,j_2\rangle \pm |j_2,j_1\rangle)$ which can be matched to $|J,M\rangle$ so that, overall, we eventually have $$P_{12}|j,j_z,j_1,j1_2\rangle = (+1)|j,j_z,j_1,j_2\rangle$$
@Claudio so, uh, just take the usual CG expansions and take their (anti-)symmetric parts?
@Claudio and I just told you that $\left|j_1,j_2\right>$ does not exist.
@imbAF isn't this an exercise in Sakurai? To find the eigenvectors of $S_n$? Maybe you should check that solution of that one. I think it's chapter 1 of the book
@Claudio ACM is channelling Socrates so don't interrupt.
@Claudio they're not trying to find the eigenstates, they're having some sort of confusion with these states in the context of a Stern-Gerlach experiment
4:39 PM
@naturallyInconsistent wait $|j_1,j_2\rangle , j_1 = j_2$ does exist
@Claudio No, you always start with $\left|j_1,m_1\right>$ and $\left|j_2,m_2\right>$ and then combine them to get stuff. It is never just the jays
I'm not going against that. I just use a different notation, since the Hamiltonian does not depend on $J_{iz}$
The notation is $\left<j_1,j_2,J,M|j_1,j_2,m_1,m_2\right>$
@Claudio If you dont have m then you cannot state when $m_1=m_2$ v.s. $m_1\neq m_2$ and so forth.
It gets hairy when $j_1=j_2$ but $m_1\neq m_2$
wait a second, you said before that if the spin part must be symmetric??? But spin is 0
in this case the eigenstates only live in $\mathcal{E}_{\text{space}}$ correct?
When we started studying the S.G device, the case considered was a magnetic field of the type $\vec B= B_z \hat{\vec e_z}$. A magnetic field of this sort, let us consider, for lack of better words, the Z component of the spin vector operator. What one gets in the end are the two dots on the screen. Now let's consider another, a bit more general case (considered in the lecture).
If the magnetic field points in some arbitrary direction $\vec n$, then we construct (if one can say so) a spin component $S_{\vec n}$. An eigenstate of this spin operator has the form $|+\rangle_{\vec n}=cos(\frac{\ An arbitrary state has the form:$|\psi\rangle=e^{i\chi}|+\rangle_{\vec n}$Now let's assume that$\phi=0$(which is assumed in the lecture) and$\theta\neq 0$. The state will transform to:$|\phi\rangle=cos(\frac{\theta}{2})|+\rangle +sin(\frac{\theta}{2})|-\rangle$(disregarding the global complex phase e^{i\chi}). In the lecture it was only given that we are considering this state, I, above, tried to give a reasonable explanation as to how one gets this state. Now one can play with the values of$\theta$. 4:46 PM @ACuriousMind that's exaxctly what I did, didn't I? I was just wondering why states of the form$|J,M,j_1,j_2\rangle_A$where$j_1+j_2-J = \text{ odd }$cannot exist @Claudio So your subscript$S$and$A$means "(anti-)symmetric part of this"? yep @Claudio oh yes sorry. I mistyped. I meant the principle quantum number$n$, are they the same? there's no principal quantum number the Hamiltonian reads: $$c_1L_1^2 + c_2L_2^2 + c_3L^2$$ @imbAF Sorry, I still don't really know what you're trying to say. What does "In the lecture it was only given that we are considering this state" mean? You're considering this state in what sense? The Stern-Gerlach experiment is this: You shoot a bunch of uniformly randomly oriented spin-1/2 particles through an inhomogeneous magnetic field. The result is two sharply seperated outgoing rays/dots on a screen. Where exactly do we need to "consider" this state? 4:49 PM @Claudio then you need to tell us a lot more context I wanted to make it more general since there are multiple exercises of this type. In this specific case, the constraint is the sphere$S^2$, the particles have spin 0 @ACuriousMind First of all$P(+\hbar/2)=cosn^2(\frac{\theta}{2})$. This is the probability of measuring +\hbar/2 right? Let's start with single questions, I don't know any other way of bringing you were I want @imbAF "measuring hbar/2" for what? what observable are we measuring on what state? of the S_z component First you create the spin component, so you have one magn. field in$\vec n$direction I don't know what that means 4:51 PM then you have another one in the pos. z direction how does one "create" a spin component what does any of this have to do with magnetic fields? what is this$S_{\vec n}$? It's the spin operator in the$\vec n$direction which you can gain by doing what? I don't know what "gain" means in that sentence it's just an operator it exists 4:52 PM @imbAF The SG serves the purpose of creating a "family" of states with spin in a certain direction, yours is just an exercise Well there's the difference it's not a thing, it can't be "created" or "destroyed" you can't "create" a matrix, it's just a linear operator on a vector space Again, all of this is just linear algebra In the lecture, we were told: And I guess you will understand this: Moechte man also SIlveratome im Zustand$|\psi\rangle$praeparieren so muss man die SG-aparatur so anordnen, dass, ihre Achse in Richtung \vec n zeigt @ACuriousMind that's what I was trying to say this before, it might be just an exercise. I think he's just getting confused because of the SG devices The axis of S.G is related to what, if not to the direction of the magnetic field ? I am not confused though, not one bit 4:55 PM The proper translation is very literal: If one wants to prepare silver atoms in the state$\lvert \psi\rangle$[I have no idea what$\lvert \psi\rangle$is because you only wrote down$\lvert +\rangle_{\vec n}$and$\lvert \phi\rangle$], one has to orient the magnetic field of the SG apparatus in the direction of$\vec n$. obviously the statement is describing the general case, when one wants to prepare states of some kind therefore it considers the general symbolic for a state$|\psi\rangle$I struggle to see how one was supposed to get this from your description, but the implicit idea here is that such a device produces two rays, one ray of$\lvert +\rangle_{\vec n}$states and one ray of$\lvert -\rangle_{\vec n}$states. You "prepare" stuff in the$\lvert +\rangle_{\vec n}$beam by just throwing away the lower beam. Yes of course So what's your question about this? :P 4:58 PM Ok, so we did establish that for you to get$S_{\vec n}$the magnetic field needs to point in the$\vec n$direction? I don't know what you mean by "get$S_{\vec n}$" @ACuriousMind this ? you get beams of the two eigenstates of this operator, you don't "get the operator" of course then say that! 4:59 PM you get the eigenstates You get the eigenstates of the operator$S_{\vec n}$you block the particles in the eigenstate$|-\rangle_{\vec n}$, arbitrary choice and you make a hole were$|+\rangle_{\vec n}$hit, so you let only them pass through the screen yes And we want to measure the eigenvalues of S_z. So i believe that after the particles pass the screen, a 2nd device pointing in the z direction (the magn. field that is) is there. Is that correct? I mean, you can do that, sure there's infinitely many things you could do with this beam :P Yeah, but this particular thing is where I am having truble not understanding but whether it makes sense. So I will continue. Let's backtrack for now. When the particles were passing through the first device the probabilities were either 50-50 (for a particle to land in any of the two spots) or$|cos(\frac{\theta}{2})|^2$and$|sin(\frac{\theta}{2})^2$. which would be the correct choice here? @naturallyInconsistent Sorry to tag you again. I re-read your messages and, at the light of the additional context I added before, what is the explanation behind not considering antisymmetric combinations$|j_1,j_2\rangle_A$coupled with the right$|J,M\rangle$part? I still don't get why we have to consider the eigenvalues of$\{J_{1z},J_{2z}\}$which do not appear in the Hamiltonian at all 5:06 PM @imbAF It obviously depends on what the input state to the device is I would say the latter I dont know either, because I don't see how to do quantum theory on a sphere That I don't know, and since it wasn't said or mentioned it means it doesn't play a role for the end result cuz otherwise it should have been mentioned I'm afraid that's just wrong :P usually you start with a source where the spins are uniformly randomly oriented, and then the results of any SG apparatus (regardless of$\vec n$) are 50-50 yes I believe so too 5:07 PM but if you start with a specific pure spin state, the probabilities of course are different @naturallyInconsistent I swear that exercise is exactly like this hahah. Maybe we could just throw that piece of information away and assume that$l_1 = l_2 = 0,1,2,3,....$without specifying the constraint We were given this, I will just give you the end result and have in mind the setup i described, with 2 devices, one pointing in$\vec n$and one in$\vec e_z$: After passing throught he first device, all atoms will be in the state:$|\psi\rangle=cos(\frac{\theta}{2})|+\rangle + sin(\frac{\theta}{2})|-\rangle$ultimately this doesn't matter if all you want is a beam of$\lvert +\rangle_{\vec n}$- unless you feed in pure$\lvert -\rangle_{\vec n}$, you will at least get some$\lvert +\rangle_{\vec n}$Sadly, I would still not know, because I also think your antisymmetric state ought to exist, only that I don't know what other thing we can do to produce the 2nd antisymmetry that would multiply with your antisymmetric state in order to satisfy the overall symmetry After the 2nd SG device we measure the component S_z:$P(+\hbar/2)=cos^2(\frac{\theta}{2})$and$P(-\hbar/2)=sin^2(\frac{\theta}{2})$These are the probabilities Now, I know from the postulates of QM how the probablities work, with the coef. multiplying the basis kets for a state that is expressed as a linear combination of basis kets 5:11 PM @naturallyInconsistent that's given by the CG coefficients, as I specified before. I had this discussion with ACM a long ago. the action of$P_{12}$on$|J,M;j_1,j_2\rangle$is the one I specified above I can not think and all and do the calculations$|\langle +|\psi\rangle|^2$@ACuriousMind But this doesn't explain the probabilities with the angle dependency Okay, what do you think is there to explain? @imbAF what are you asking? take$\theta=0$What that means for the mangetic field, first question it's in the z direction right? yes in that case you've just put the same SG apparatus twice in a row, congratulations :P 5:13 PM why? why twice? You put it once and you get both points But I don't want to divert Why what? Why is that z direction? what you mean, take a unit vector in spherical coordinates with \theta=\phi=0 that's what you get (0,0,1) no, this is important: Your setup contains two SG apparatuses, one oriented in the$\vec n$direction and one oriented in the$z$-direction, and the way you defined$\theta$, it's the angle between$\vec n$and the z-direction Good. The thing is that you havent stated a fully formed question. I dont know what it is you are asking. so when you choose$\theta = 0$, both your apparatuses are in the z-direction 5:15 PM Yes juuust making sure we're on the same page here; okay, what's the actual question? I think I see the problem. When we were introduced to this S.G thing, we started with 1 device which causes the particles to hit two spots with 50% probability This is how we started nothing about angles @naturallyInconsistent wait, maybe you were being sarcastic...are you saying that the exercise could be wrong from the beginning? Or are you saying that it is what I am doing that makes no sense? I'm bad at getting hints So, if you think about it, even if it wasn't mentioned,$\theta=0=\phi$, right? @Claudio No, I am not being sarcastic here. I just dont know how to do the question, not least because the problem is given in parts. However, I think that you have supplied enough parts for meow to know that it probably is not solvable as it is. You might have to attack your prof @imbAF no 5:18 PM and why not? Anyways, so @naturallyInconsistent is almost neutral but tends to my side, ACM said what I did is correct, so, counting my Professor in, are we 2.5 to 1 it depends fromt he initial states of the particles? Because yes, you started with a thing that always gives you 2 spots of 50:50, which is a mixed state, and mixed states are outside of basic introductory quantum mechanics. Aight, I'll write him about this and I'll let keep you informed @naturallyInconsistent @imbAF not really; if you only have a single device, what are measuring these angles relative to? 5:19 PM It is 2 spots of 50:50 no matter what your$\vartheta$and$\varphi$are the point of the angles in the two-device-setup is that they're the angles between the magnetic field directions of the two devices @ACuriousMind relative to the z axis @imbAF sure, but why would they mean anything in that case? you have just one device, why would the z-axis be special? @Claudio I wont be at all surprised that, since you have no spin parts, then the symmetrisation falls completely on the space part, and since you are describing the space part there, you are left with needing to enforce symmetry all the time. exactly 5:21 PM @ACuriousMind not special, just an arbitrary choice of the direction of the magn. field. You could very well take the x or y axis in the two-device-setup, the z-axis is special because the second device is fixed to be in the z-direction and your formulas for the probabilities are for the output of the second device trying to fit the one-device setup into this is just a category error Aight, I have to go now. Thanks for the feedback as per usual guys. That is how we started @ACuriousMind If I understand correctly, after you prepare the particles in some state, via the first device, you can use the 2nd one to measure the probl. of measuring the corresponding eigenvalues of a spin component? but anyway, you can say that$\theta=\phi =0$for the single device if you really want, but be warned that if you are going to plug them into any formula that is specific to the two-device setup I'll be a bit mad :P 5:24 PM I have to be a pedant again and say that I don't understand that sentence :P And if$\theta=\pi/2$, would you get 50-50 @naturallyInconsistent ? why "of a spin component"? I thought we had fixed the second device to be in the z-direction S_z, we measure the z-spin component operator at least that's how I call the S_z everyone else just calls it the spin-z operator :P "component" is technically correct but really superfluous in this case 5:25 PM and what you call$\hat{\vec S}$? the spin (vector) operator Yeah, so the spin (vector operator) has it's components. I mean maybe it's better to call it spin z operator I guess yes, again, you're technically right but this is just uncommon phrasing, let's not get hung up on it; so we have the two devices, their orientation at an angle$\theta$to each other. What's the issue? Nothing now that I realised that we must have 2 I was considering that with 1, you could get a 50-50s split disrgarding completely the state of those particles If I am being honest, when it was explained how one can get states$|\pm\rangle$, we were shown 1 S.G device only this, as you and as @naturallyInconsistent explained further, is not much of a use And one more thing, if for the state$|\psi\rangle$that we are considering, you get for the expected value of S_z$\lange S_z\rangle= 'frac{\hbar}{2}cos\theta$. Shouldn't you get only$\hbar/2 $? Why would you? 5:31 PM That's what we were told No, I mean why do you think you "should" get$\hbar/2$also, "that's what we were told" is never an acceptable answer for claims whose truth you can just check with a straightforward computation @ACuriousMind Now, why would I ? :p I have no answer for that But in all the state considered, we had no time dependency. Were we considering cases for some intial time t=0 ? I don't know what time has to do with any of this You have a state$\lvert \psi\rangle$. You can compute its expectation value for$S_z$. Nothing with this But in general the state$|\psi\rangle $can be time dependent, right? I don't know what you mean by that 5:36 PM One moment we're not considering time evolution here, just the inputs and outputs of SG devices I know, and for what we discussed I am clear thanks to you and naturallyincontrol Now I am, for additional knowledge asking about the case when time dependency is considered I don't know what you mean by that are you somehow changing the input to the device over time? That I don't know, hence why I am asking you. Give me 1 minute. At t=0 the system is at the state:$|\phi(t=0)\rangle=cos(\frac{\theta}{2})e^{-i\frac{\phi}{2}}|+\rangle + sin(\frac{\theta}{2})e^{i\frac{\phi}{2}}|-\rangle$. Now, the spectrum here is discrete and degenerate (we are considering the basis comprised of eigenstates of S_z), so for an arbitrary state one can write:$|\psi(t)\rangle=\sum_nc_ne^{-iE_n(t-t_0)}|\phi_n\rangle$In our case$|\phi\rangle_n$are$|\pm\rangle$So at some time t our state will be:$|\phi(t)\rangle=cos(\frac{\theta}{2})e^{-i(\frac{\phi}{2} +\frac{E_{+}}{\hbar})}|+\rangle + sin(\frac{\theta}{2})e^{i(\frac{\phi}{2} -
I hope everything is clear here, and would you agree?
I am making no claims about the setup
No, I don't agree, because you made a very explicit claim about the setup when you claimed the $\lvert \pm\rangle$ are the eigenstates of your Hamiltonian
We do not, at an introductory level, consider what happens inside of an SG apparatus, just like we treat any other measurement devices as black boxes
5:46 PM
I don't understand, why don't you agree?
When you say "In our case $\lvert \psi\rangle_n$ are $\lvert \pm\rangle$", you are making a claim about the Hamiltonian/time evolution
claim such as?
...that the $\lvert \pm\rangle$ are eigenstates of the Hamiltonian?
because otherwise why would they evolve with the phases $\mathrm{e}^{\mathrm{i}E_n(t-t_0)}$?
Considering that the magn. field points in the pos. z direction, would that qualify as a reason for the validity of my claim?
no, because you've set this up completely wrongly
if you want to consider the time evolution inside of the SG apparatus, you can't keep just considering the spin aspect of the particles - remember, they also "move" in space
5:50 PM
Not me really, but rather what was given in the lecture
but your lecture didn't consider the time evolution, did it?
Later on they did
and the setup was a particle with spin in a hom. magnetic field
...but a homogeneous magnetic field has nothing to do with SG apparatuses!
the whole point is that the field is inhomogeneous
I am not disagreeing
I am just trying to understand what they were trying to convey
which was
but you didn't tell me anything about what the lecture did, you started randomly talking about time evolution in the context of an SG apparatus
5:53 PM
Yes
we started
like the description I gave you about the states at t=0 and t different than zero
Wait, so the "in our case" in what you wrote above wasn't mean to refer to "our case", i.e the SG apparatuses we've been discussing, but to a completely different setup from your lecture??
how on earth was anyone meant to understand that, you really need to be much clearer about when you are completely changing the topic
Hold on, because I was clear
I told you initially what we considered
Than, I asked you about time dependency of the states. I can't possibly know how one achieves that, when it wasn't explained in the lecture. It was a jump
precisely the way I wrote everything
sorry, but I have no idea what we're talking about, you've lost me
Does this "time dependency" you want to discuss have anything to do with the SG apparatuses or not?
Yes
because I am trying for the 5th time to write the rest
xD
@ACuriousMind I got you one better solution
6:10 PM
yes, this is, as you said, the behaviour of a particle trapped (i.e. not moving) in a homogeneous magnetic field; it has nothing to do with the SG apparatuses, which have particles moving through inhomogeneous fields
yeah, but we are measuring probl. for the eigenvallues
And the state considered , is something that can be achieved in a S.G setup, or at least was mentioned and introduced when considering the S.G device
but again, both the state space and the Hamiltonian here are different from what you have in the SG apparatus
these two scenarios have almost nothing in common
Ok, and what is the point of this?
How would we measure the prob. unless we use a S.G device?
Because to me, this section, has no connection to what I did before
Like I don't understand at all, what it tries to tell me
oh, I mean, once you have evolved your state in time like this, you might release it from the trap and measure its spin with an SG device
there are other ways to measure spin in experimental setups, but you could do that
Since I know the SG. I will stick with that for the moment
Out of curiosity, you see that the Hamiltonian is proportional with S_z ?
6:19 PM
but note the crucial "release it from the trap" - then the time evolution according to this stops, you freeze the spin part at some final time $t_f$, then just use the SG device to measure the spin
But should the amount of trapping influence the prob. of measurement? Since the time is in the complex exponential and we take the square of the amplitude
I don't understand the question
Which part?
what does "amount of trapping" mean? We have this trap with a homogeneous magnetic field; we let the particle stay in it for a time $t_f$, then release it to fire it at an SG apparatus
What I mean is whether the amount of time the particles are trapped have an effect in the prob. of measuring an eigenvalue:

$P(\frac{\hbar}{2})$=|\langle +|cos(\frac{\theta}{2}e^{-i(...)})|^2
And I said, that it shouldn't, considering the fact that time is in the complex exponential, and it will vanish, when we will consider the square amplitude of the coef.
6:24 PM
the very notes you just posted contradict you, e.g. eqs (7.61) and (7.62) clearly show a time-dependence of the expectation value
Those are the expectation values
you seem to have a strange way of deriving what "should" happen that does not involve actually applying the mathematics of quantum mechanics
of the observables
I am talking about the probabilities of measuring the eigenvalues
@imbAF but a changing expectation value means precisely that also the probabilities for measuring the eigenvalues of that observable change!
$P(\frac{\hbar}{2})$=|\langle +|cos(\frac{\theta}{2}e^{-i(...)})|^2$is incorrect ? 6:26 PM I can't tell because a) you haven't specified the observable you're measuring and b) you haven't written down mathjax that compiles Ok And, since you read the notes, there it writes$H=\omega_0S_z$, this hamiltonian is the spin hamiltonian? That arises because of spin? I mean, there's a spin operator in there so sure, "it arises because of spin" I'm not sure what additional insight that particular phrase grants us :P But you also have the H= T+V . So the total hamiltonian is H_t= H +H_s I am sorry for the incorrect writing. I am not sure there is no T here note that I made it very clear that this particle is trapped it doesn't move, it doesn't have kinetic energy And if it would? move? 6:32 PM then you would have a kinetic term in the Hamiltonian and your space of states would not just be the two-dimensional spin space it would be a different system Do you know any such case? Can you give me an example I mean, the particles in the SG apparatuses are moving through magnetic fields :P I have to be more precise. What is the space of states in this case. It shouldn't be 2d. So how does an eigenstate look like it's what I am really after I am sure your notes (or a follow-up course) will come to that point when you discuss particles with spin more generally it is not something you can understand at the level where you are introduced to the two-level systems of non-moving particles this carefully In the chapter related to spin, we do not. perhaps in the one for the angular momentum we do. But not right now 6:37 PM sure, you will first discuss the pure spin part @ACuriousMind And you asked me this and then, at some point, you will merge that discussion of spin with a discussion of wavefunctions and get spinor-valued wavefunctions The fact that the expectation value of the S_z is not time dependent, doesn't that say anything about the observable considered when saying$P(\frac{\hbar}{2})$=|\langle +|cos(\frac{\theta}{2}e^{-i(...)})|^2$ ?
@ACuriousMind I do remember we do talk about the spinors, I also remember that I asked the prof. what exactly are spinors, since he just causally introduced em in the discussion, when solving the Dirac equation, and he told me, abstract constructs/entity......what does that even mean lol
well, the Dirac equation is one step further still, since it's additionally relativistic
Well, thank you very much for today, you really helped me
Thanks
7:41 PM
Hi, does anyone have any advice on when to use a PID controller? I am building a model rocket to reach a certain target altitude for a school competition and am not sure if a PID controller is needed to get the rocket on course. I looked online and it seems to fit the bill: clear model of system, error management (off course trajectory), sensors, clear goal. But I am not if something simpler will do. Any thoughts?

2 hours later…
9:18 PM
What are theories with no propagating dofs good for? E.g. I know that photon mass gets renormalized at loop level for $d=2$. I can see the calculation right infront of me, but what does it actually mean given that there are no dofs for photon in 2 dimensions?
@Sanjana They're toy models for "topological" or non-perturbative effects since every non-trivial effect in these theories must come from things that have nothing to do with our ordinary perturbative idea of particles interacting
@ACuriousMind How are these topological? I mean, the last time I saw something related to this had to do with the oft-ignored theta term which is a total derivative classically, but at quantum level there give rise to stuff like instantons, axions, etc.
But this is something different, right? I mean the Lagrangian (without the theta term) is not a total derivative in this case for $d=2$ is it?
@Sanjana physicists tend to call many "global" effects "topological", hence my scare quotes; you are right that 2d Yang-Mills is not topological in the strict sense since it depends on the metric