Any quick way to see that $$A(n+1)=\frac{(n+2)A(n)-(2n+1)A(1)}{n-1}$$ admits a closed form $A(n)=cn^3+dn$ for $c,d$ constants to be determined?

I tried plugging the recurrence relation into itself iteratively but the second term is a bit nasty to deal with

By the way, this is the recurrence relation arising in the quantum anomaly of the Virasoro algebra

@naturallyInconsistent Gaussian potential but the space is limited to $[-1,1]$ say i.e. hard walls at $x=\pm 1$

@Mr.Feynman I don't recall having the $A(1)$ term. Can't it be set to zero somehow?

I read this part from Blumenhagen's CFT book which is pretty detailed in most calculations. Pg. 16

i think i am having a misunderstanding about the pauli matrices. so i have seen that the spin operator is proportional to the pauli matrices for x,y, and z. but what basis are these in (the pauli and thus sigma matrices)? if i write a "spin state" $ \vert + \rangle = \vert 0 \rangle + \vert 1 \rangle$ (ignoring normalization), then this is specifically in a basis, say the $z$ basis, so how can i act on it with $S_{x,y,z}$ / $\sigma_{x,y,z}$?

@Mr.Feynman If you set the 2nd term to zero, you do get the central extension term: $n^3-n$, and the normalization is fixed by requiring $c=1$ for free boson etc

@Relativisticcucumber I think the basis is the one in which $\sigma_z$ becomes diagonal. So if you want to act the spin operators in terms of Pauli matrices on a generic state, you have to express it in terms of eigenvectors of $\sigma_z$

@ACuriousMind Raman please

what?

Raman scattering in QFT language, do you know?

Books

In what world is pinging me with "Raman please" a remotely acceptable way to start a conversation?

Oh sorry. I thought you read all text when you come

You are the caretaker of this room

I literally got online like a minute ago

You very fast I thinked

@NairitSahoo even if these two statements are true (they are not), you are being extremely rude.

@Sanjana so if i have, say, $\sigma_x \vert + \rangle$ with $\vert + \rangle$ defined as above, then what i get is truly a value that can be interpreted as spin in the $x$ direction? i dont need to change the basis of everything?

@Relativisticcucumber What do you mean by "the spin operator is proportional to the Pauli matrices"? There's no "the spin operator", you have a family of spin operators: the "spin in direction of the unit vector $\hat{n}$" is $\hat{n}\cdot \vec \sigma = \sum_i n_i \sigma_i$, and then e.g. choosing $\hat{n}$ as the z-axis you get back that $\sigma_z$ is the spin in that direction

@Sanjana do we even have a solution to the problem when the potential is Gaußian and not restricted to a box? But this is a red Herring: If we have a suitable ground state, then it will have no nodes. If the ground state energy is smaller than the peak of the Gaußian, then it will be a symmetric state with two peaks, one on each side of the Gaußian. If the Gaußian is a tiny perturbation, then just one peak in the middle.

of course everything here is basis dependent in the sense that if you choose a different z-axis then you need to transform everything that you've expressed in terms of the components $n_i$ or $\sigma_i$, but that's how it is with vectors

@Relativisticcucumber What ACM said + If you study the Weyl and Wigner ways to classify spin states, you will realise that, to even begin, you have to align the x y z axes of every point in spacetime before you start. So then it makes sense that all the Pauli matrices are fixed and pre-determined.

@ACuriousMind oh i had seen that for instance $S_x = \frac{\hbar}{2}\sigma_x$ i think

but hm i am still confused. so this would tell me that if i choose $\hat{n}$ as the $x$ axis, then i get $\sigma_x$ i assume. but then should i rewrite $\vert + \rangle$ in the $x$ basis so that when i do $\sigma_x \vert + \rangle$ the result i get is sensible?

@Sanjana it is to be determined later at least that's what GSW do

@ACuriousMind you wouldn't say that if you were a barista and someone popped up saying "coffee please"

By the way, my first read was "Ramen please", which made more sense :P

i had in mind someone at a bar ordering a drink called "raman"

@Relativisticcucumber What do you mean by "i do $\sigma_x\lvert +\rangle$"?

measuring an operator is not applying it to the state

@Relativisticcucumber also this equality really only makes sense if you have an independent definition of what $S_x$ is supposed to be, otherwise this is just a definition

@ACuriousMind oh no but that is how i was taught to determine the $S_{x,y,x}$ matrix forms D: to see when i act on the state if i get the eigenvalues i want

unless i am remembering very wrong which is also possible

I think I'm confused about what the setup and the goal here is

what do you have given, and what question do you want to answer?

i am also confused BLEH. so i have the state $\vert + \rangle$ and would like to calculate the expectation values of $\sigma_{x,y,z}$. i did so for $\sigma_z$ but then i was unsure about $\sigma_{x,y}$ since the state is written in the $z$ basis

what's the definition of the state $\lvert +\rangle$?

$\vert 0 \rangle + \vert 1 \rangle$ with these being down and up in the $z$ direction

and normalization of $\frac{1}{\sqrt{2}}$

so what do you know about $\sigma_y$ and $\sigma_z$?

@Relativisticcucumber I just told you that you have to first align the axes everywhere in spacetime. Part of this means that the standard $\sigma_x$ and $\sigma_y$ operators are already being represented in the $z$ basis. You can just use the standard representations and be done with it.

if you knew their matrix forms in the z-spin basis this would be trivial, but I don't quite understand what information you want to deduce the values here from if you don't know them

I concur with ACM: It seems like you are either given a very abstract introduction to QM, or you are yourself imposing upon yourself a treatment of QM that is way too abstract and unhinged from how we would be doing physics, both experimentally and theoretically. If you do not have an understandable introduction to QM, it would be no wonder that you get confused by such things.

@Mr.Feynman When KPop meets metal, you get Ramenstein. :)

lol

@PM2Ring I don't know what that thing is but it's the second time in a week I read that word

I mean, the original word without pun

@ACuriousMind i have their matrix form also but this is where i get lost because i was told they are written in the "standard basis that does not have a name" so i dont really know what this means

well that's just unnecessarily confusing :P

@SillyGoose GR

if your matrices look like the ones on Wiki they're in the $\lvert 0\rangle,\lvert 1\rangle$ basis

also note that the Pauli matrices are a basis for su(2), so one needs to know all three to be able to construct an arbitrary agebra element as a linear combination

@Relativisticcucumber That is why I had to emphasise that your x y z axes are already aligned. Then you can see why it is that the Pauli matrices are the same everywhere, all expressed in the z basis

Why are generalized velocities assumed to be constant during a virtual displacement?(See hand and finch analytical mechanics where he defines virtual displacements) I'm asking this question since Goldstein nowhere uses this assumption in his derivation of Lagrange equations in his 1st chapter

By generalized velocities I mean $\dot{q_j}$

the usual derivation of Euler Lagrange equation is using the principle of least action

in this derivation, we change by a small path $\epsilon (t)$ and then set the functional derivative to zero

@RyderRude The "usual derivation" of the principle of least action is in turn derived from d'Alembert's principle of virtual displacements and Newton's laws (cf. e.g. physics.stackexchange.com/a/131392/50583), and that's clearly what Arjun is asking about. If you don't know what a question is about, you don't need to pretend to.

yes. im aware of this

i thought Goldstein might have directly assumed the principle of least action, like most texts do

Goldstein, chapter 1.4, title: "D'Alembert's Principle and Lagrange's Equations"

oh

this is why i wrote "usual derivation", in case they were using some other derivation

@Arjun The whole derivation in Goldstein makes no sense if the virtual displacements would also change the $\dot{q}_j$, for instance the expansion of $\delta r$ in terms of $\delta q$ and the appearance of $\dot{p}$ in d'Alembert's principle, etc. It's not a single step that doesn't work, but the whole idea falls apart if you try to talk about changing velocities, too (also the name "displacement" wouldn't fit anymore in such a case)

note that while Goldstein doesn't spell out the "velocities are constant" part explicitly, it's there in that he starts with a virtual displacement as being a change of the coordinates $r_i$ - he doesn't talk about changing velocities, so they don't change

@ACuriousMind GAH this makes sense.

@naturallyInconsistent i was confused because i thought i was told the pauli matrices are NOT in the $z$ basis but if they are then i see BAH

@Relativisticcucumber Consider the $\sigma_x$ Pauli matrix. If it were in the $x$ basis, then $\sigma_x=\begin{pmatrix}+1&\\&-1\end {pmatrix}$ which is what you would think $\sigma_z$ would be. Needless to say, it is extremely confusing.

i should have known

i was just confused because when i tried to look this up i couldnt find a solid answer which made me thing there was smth more to this also

that is why I made it absolutely clear that you have to first align the x y z axes. This implies an implicit agreement to measure everything in the z basis

@naturallyInconsistent bah i see now

@Mr.Feynman I can make one up. However, it is quite convoluted. I mean, there is a step in the middle that is like, if you didn't tell meow that it is of this form, miao miao would not have considered trying to search for it. Alas, miao miao is going to rush to gym in a moment and will thus be gone for hours.

@Relativisticcucumber Ah, maybe I should have explained why it has to be of this form. Remember that the spin half operator is proportional to the Pauli matrices. Now, you know that the only result of measuring spin half is that you get $\pm\frac\hslash2$ as the result. You also know that these are two-dimensional, i.e. two non-degenerate states. Thus, in its own basis, it must be a diagonal matrix of that form.

@ACuriousMind If $\dot{q}_j$ changes by an infinitesimal amount as q changes by $\delta q$ i fail to see why $\delta r$ or $\dot\dot{r}$ would be affected since r is only a function of q_j and not of $\dot{q}_j$

blebs consider a two level system, maybe an atom and 2 spin 1/2 items like electrons. i have seen that a triplet state is a state with two unpaired electrons. i have also seen that the three triplet states can be written as $\vert s = 1, s^z = 1 \rangle, \vert s = 1, s^z = 0 \rangle, \vert s = 1, s^z = -1 \rangle$. my interpretation is that, in the first case, we must have both electrons [...]

[...] be spin up, and in the third case, both are spin down. however, to satisfy the middle case of $s^z = 0$, we must have one up and one down. in this case, they can fill the same level, and we do not have a triplet state? so how is the middle case a triplet state?

@Arjun I'm not sure what you mean

@Relativisticcucumber and I'm not sure what you mean either :P

@ACuriousMind well do you agree with the three states i wrote being triplet states? or am i wrong on that

The two electron have a four-dimensional space of states with a basis of state of definite spin for the individual electrons $\lvert \uparrow \uparrow\rangle, \lvert \uparrow\downarrow\rangle, \lvert \downarrow\uparrow\rangle,\lvert \downarrow\downarrow\rangle$. The "middle" state of the triplet is $\lvert 1,0\rangle = \lvert \uparrow\downarrow\rangle + \lvert \downarrow\uparrow\rangle$. What is the question?

@ACuriousMind how is this a triplet? i thought a triplet state is a state with 2 unpaired electrons?

I don't really know what you mean by "unpaired"

@ACuriousMind I'm trying to understand why an infinitesimal change in velocity would affect the $\delta r$ expression in any way..and also it is not clear for me why a change in velocity during the displacement would lead to the crumbling of Goldstein's derivation

the 4d space of states for the two spin-1/2 objects splits into a 3d triplet and a 1d singlet subspace

@ACuriousMind im referring to the intro/image on wiki here en.wikipedia.org/wiki/Triplet_state

I have no idea what that picture is trying to depict :P

so how should i understand the difference between singlet and triplets?

the definition of a triplet state isn't "2 unpaired electrons", it's what the intro sentence of the article says: It's any of the $s=1$ states where you have three possible values for $s_z$, hence "triplet"

compare that to the $s=0$ state where you only have a single choice $s_z=0$, hence "singlet"

@ACuriousMind Should'nt the expression for $\delta r$ remain the same since r is a function of q_j 's and not the velocities..so a change in velocity shouldn't really affect $\delta r$?

OH

so their uniting characteristic is not a physical one

@Arjun but how do you "change velocity" without changing $r$? The velocity is the derivative of $r$! Again, it's not that any single step here wouldn't work, it's that the whole setup doesn't really allow you to talk about independent changes in velocity at this stage

at some point we will be allowed to treat $q$ and $\dot{q}$ as independent in some sense, but this is before that (and also many texts don't explain that part very well)

@ACuriousMind i always wondered ab this

@Relativisticcucumber I don't know what this means, either :P

@ACuriousMind i never got a satisfactory answer :,(

@ACuriousMind its ok i think i get it

triplets having angular momentum 1 and singlets having angular momentum 0 sounds pretty physical to m

@ACuriousMind i was thinking too much about the name triplet :P

but it was right in front of me

LIKE A CHILD

if there are triplets we call each a triplet

wow

damn that word was probably chosen bc it would be obvious

@Relativisticcucumber see the last paragraph of this answer of mine and also physics.stackexchange.com/q/428990/50583

i could be ashamed or amused and i choose amused

@ACuriousMind NO

i clicked on the answer

and i have upvoted it already