3:04 AM
@imbAF $$\left<\psi|\psi\right>=\left<\psi\right|\hat{\mathbb I}\left|\psi\right>=\int\left<\psi|x\right>\mathrm dx\left<x|\psi\right>=\int\psi^*(x)\psi(x)\mathrm dx=\int|\psi(x)|\mathrm dx=\int\rho(x)\mathrm dx=1$$ The other expression is just wrong.
3:27 AM
1

Simply, why is the energy density in a dielectric medium = $\frac{1}{2}K\epsilon_oE^2$ where K is the dielectric constant of the dielectric medium? Working proof Consider a capacitor that is charged to a potential difference of $V_o$. If its capacitance in air = $C_o$, its capacitance with a diel...

@nickbros123 whilst answering that, I think I've accidentally stumbled upon the thing that you were asking; yes, the energy's quadratic dependence upon the fields means that it is not obvious that the averaging scheme would keep that working, but apparently when we do that averaging shit, we alter some of the expressions and tune them to get those bits to work out too. Still unspecified, but it should be ok, since we have other ways to check.
4:26 AM
Hello Everyone...
5:24 AM
@ACuriousMind The snapshot is from some lectures on twistor theory but I thought that the naturalness of the holomorphic measure on $\mathbb{C P}^1$, here is a twistor theory independent context...
z.r.m is zero rest mass. $\mathbb{M_C}$ is complexified Minkowski space.
5:52 AM
@naturallyInconsistent I found a trick to average it, without averaging the energy itself due to its quadratic dependence. The idea is similar to yours with the maxwellian field averaging and talking about the averaged potential, But this scheme treats macroscopically small things as true infinitesimals, so as a result, these objects don't contribute to self energy, when in reality they do
does that even work?
I mean, in a sense you still want the energy to be accounted correctly in some way
I got the correct formula, and the dielectric formula for energy and the general formula for energy, if I remember correctly, differ by a factor of P times E
oh nice
I could only argue qualitatively after this, averagine blurs out everything that's happening so I think loss of information is expected
that is definitely the case
but as long as you can get 1) smoothing and 2) yet not get a wrong energy term, i.e. it protects both the 1st moment and the 2nd moment, is already a wonderful achievement
5:59 AM
Well, it all lies on a particular approximation, which may or may not be correct.... I'll post the relevant picture here
Page 178
Heck i don't have access to that book cuz I'm at home and not at the University...
Btw for diagrams GeoGebra Geometry for quick drawing and exporting ad tikz works best, and is really fast. So diagrams are sorted for live-texing
2
@nickbros123 lol
6:24 AM
@Sanjana this has to be one of the worst cases of physicists using algebro-geometric language pretentiously I've ever seen :P
"projective weight +2" just means "holomorphic" again - O(2) is the bundle of holomorphic forms
And again the "natural" metric would be Fubini-Study, but sure, if we want to pick a holomorphic form to integrate this one is simple enough to be "natural", too :P
Also your source is one of the few hits for "natural holomorphic measure" so it's not as if this is a well-established concept
@ACuriousMind But how do you get away with the thing you said earlier? $\lambda$s are homogeneous coordinates on $\mathbb{CP}^1$, how can they be even considered for building up an integral measure?
@ACuriousMind Yeah this is strange cz the text actually even introduces this twisting sheaf notation earlier
Hmm. The other people are Chern and Witten and all these are physics papers...
@Sanjana there's a correspondence between homogeneous polynomials in the homogeneous coordinates and sections of O(2), this should extend to the 1-forms in that bundle being such polynomials with one of the z variables in each summand replaced by dz
But really this is poor writing on the part of the text and it's not even 100% that this is what they mean
@ACuriousMind Oh ok. I think I had a discussion with you around similar concepts when we were discussing the sections of the dual to tautological bundle...
Thanks
@ACuriousMind btw I had a question regarding double cover of the space of generators of a cone here. Can you have a look at it?
6:54 AM
do u think the Heisenberg picture exists for quantum mechanics in imaginary time or not?
7:22 AM
@naturallyInconsistent Sorry, I had a doubt about this. Remember the problem with the two cylinders stacked on top of the planks? The frictional force provided by the lower plank is to the right and the lower cylinder rotates counterclockwise. I understood that. But since it rotates counterclockwise, the cylinder exerts a leftward force on the upper plank, no? Conversely, the upper plank should exert a rightward force on the cylinder below, right? Why does the solution draw $G_n$ to the left?
https://arxiv.org/abs/2402.05866

This came out this year. The abstract starts with "We formalize Feynman's construction of the quantum mechanical path integral" which seems interesting. The rest of it, I don't understand. Can somebody with the appropriate knowledge tell whether the issue of path integrals is actually solved here finally??!
@Sanjana the "intersects each generator twice" is the same as saying its a double cover
@Bml because the upper plank needs to be pushed to move rightwards.
i.e. the upper plank must push the cylinder leftwards, so $G$ is in the correct direction
@naturallyInconsistent Yes, but why has the upper plank to be pushed to move to the right? What I said is not right for what reason? Doesn't a cylinder moving counterclockwise feel a force to the right from the plank above?
The entire stack is moving rightwards, every level doing the same thing, so it has to be to the right
7:34 AM
@ACuriousMind But how does it relate with the usual definitions of double cover as given in wiki?
I mean... how do I see that the thing that is given in the text and in the wiki page are equivalent?
8:29 AM
@ACuriousMind And also, why cut a cone with a sphere anyway?

1 hour later…
9:45 AM
Hi

2 hours later…
11:39 AM
Hi guys , I'm having trouble understanding why using the equations of motion and conservation of energy, I get two different results
A stunt motorcyclist performs a jump from the platform, as shown. It starts from point A with zero velocity and constant acceleration $a= 4 m/s^2$ up to point B, 50 m away. Then the platform BC, of length BC= 10 m, rises with constant acceleration $a_1 = 2 m/s^2$ up to point C, where it breaks off. The angle that the platform makes with the horizontal axis is α = 30°.
Calculate the magnitude of the velocity at B and the take-off velocity at C
I used the equations of motion so to find the velocity at point B I wrote: $v^2_{B} = u^2 + 2ad, \quad v_{B} = 20$
then for the BC section, I did: $v^2_{C} = v^2_{B} + 2a_1l , \quad v_{C} = 20,97$
I tried to use energy conservation and practically for $v_{B}$ I get the same result ($20$), while for $v_{C}$ no.
I wrote: $\frac{1}{2}mv^2_{B} = mgh + \frac{1}{2}mv^2_{C}$
where $h = l \sin(\alpha) = 10 \cdot \frac{1}{2} = 5$
I get $v_{C} = 17,37$ , therefore a different result from what was found with the equations of motion. Could anyone explain to me why?
12:05 PM
24 mins ago, by Pizza
Calculate the magnitude of the velocity at B and the take-off velocity at C
I now realize that the translation is not very clear I think, however I have to calculate the speed at point B and the speed at point C.
12:16 PM
@Pizza where is the energy that is coming from the acceleration along BC?
@naturallyInconsistent $F = ma_2$?
$L = F \cdot l$
12:43 PM
@Kenshin hi
@Pizza but kinetic energy isnt conserved from A to B. did u use the work energy theorem instead?
@Pizza for the same reason, this wudnt work. energy isnt conserved from B to C either. KE1+PE1=KE2+PE2 only works when the only force on the body is gravity. here, there is the external force from the motorcycle
@Sanjana A double cover is just a cover where the fiber has two points. You have a cover $N\to PN$ whose fibers are lines, and now you have $N\cap S^5$ that contains two points from each of those lines, so $N\cap S^5 \to PN$ is a cover whose fibers are two points.
@RyderRude Okay, thanks so much for clarifying this for me!
if i understand correctly, you are assuming that energy at B = energy at C, but if the motorcyclist accelerates by burning fuel, and you don't account for chemical energy of that fuel, then you might gain energy
On the stretch from B to C they are moving not by inertia
but by motor
which puts in additional energy into the system
1:04 PM
@Pizza also, the problem doesnt specify if 2m/s^2 is the net acceleration of the climb, or just the acceleration provided by the motorcycle force (i.e. excluding gravity). ur first solution is correct in the former case
but it shud b net acceleration without further qualifiers i guess
1:18 PM
also, correction to my "the only force on the body must be gravity": other forces are fine as long as they do 0 work on the body.. i.e. it shudnt b something like friction or motorcycle force

3 hours later…
3:55 PM
Greetings everyone, currently preparing for my theoretical QM exam and I'm a bit confused about the generators of rotations. It seems to me that there's two "groups" of 3x3 matrices, the first $$F_x=\begin{pmatrix}0&0&0\\0&0&-i\\0&i&0\end{pmatrix}\dots$$ and the second $$L_x=\begin{pmatrix}0&1&0\\1&0&1\\0&1&0\end{pmatrix}\dots$$ (omitting the other two to save time).
It is my understanding that the $F_i$ just are a common set of generators for SO(3) while the $L_i$ seem to be closer related to spin - if I got it right, they kind of are like the Pauli matrices but for spin 1? Is this correct? I would greatly appreciate any input/text recommendations as the last part is still quite unclear to me (e.g. how do the "regular" Pauli matrices fit in there? Are those all generators of Rotations? If yes, are they different "kinds" of rotations?). Thanks!
ohhh or is the difference that $F_i$ are generators of SO(3) while $L_i$ and the Pauli matrices are representations of SO(3) and SO(2), respectively? (or maybe SU)
still it confuses me why the 3x3 ones are different then
4:48 PM
@naturallyInconsistent Don't you need $|\psi|^2$ ?
5:30 PM
@ACuriousMind Why sphere, why not something else... Is it just a choice that cuts a cone two times, that's why?
@Sanjana any other shape that intersected it two times would work as well, sure
5:52 PM
@imbAF that was a typo
and I cant fix it now
too late to fix
Ok
One more thing regarding this topic
when we do $\int \rho(x)dx$, we interpret that as the probability of finding the system at a certain location in time, depending from the boundaries of the integral
If we were to consider some other observable, how would you describe what that integral means or tells us ?
@imbAF No, you are thinking of the integral over finite regions, in which case your interpretation is correct. But we are talking about integration over all space
I know
But I am just thinking of just another observable, that isn't necessarily position
meaning $\rho(\alpha)$ and $\alpha$ is some continues eigenvalue of an arbitrary operator that we are considering
And there's still something that confuses me regarding the notations between bracket and in wave-terms (if one can say so the non-braket notation). I will explain
Btw thanks, once again for the lengthy explanations you gave me previously
@imbAF That is not how the expression tends to look in other cases; what would it mean for you to plot the wavefunction as a function of energy, say? How would you parcel up the spacetime into energy pockets?
I believe momentum can also have the same expression
and interpret the integral as the probability of the system having a momentum value that falls within the interval specified by the boundaries of the integral
would that be correct?
6:07 PM
Instead, we tend to have expressions that tease out the relevant expressions in terms of position, e.g. KE$=\frac{p^2}{2m}=\int_{\mathbb R^3}\psi^*(\vec r)-\frac{\hslash^2}{2m}\vec\nabla\cdot\vec\nabla\psi(\vec r)\,\mathrm d^3\vec r$
@imbAF yes
@naturallyInconsistent I don't understand this
What part do you not understand? Actually, maybe you should just continue reading some textbook because they will also cover this.
the expression you wrote
What you are saying is that, observables can be expressed in such a way, that they are a function of position, and for that reason we can use the expression $\rho(x)dx$. Am I getting it right?
Well, what do you not understand? It has $\psi^*$, it has $\psi$, so even though there is something in between that definitely will change things, it ostensibly still has all the ingredients needed to make the $\rho$ that you so much wanted.
@imbAF somewhat, yes
So the expression in the integral plays the role of the $\rho$ ?
6:15 PM
Well, if you want the integral to make sense, it has to look like $\int F(x)\rho(x)\mathrm dx$ for some $F(x)$ that is the expression of the quantity you are looking for, in terms of position.
And then you can pretend that the system had parts broken-down-able as $\rho(x)$ at a point $x$ and thus contributes $F(x)\rho(x)$ worth to the integral.
I have not encountered something of this sort in my studies until now tbh
And in the integral you only have $\rho(x)$, if I would have $x\rho(x)$ then I am taking about expectation value, but above I said probability to find the system somewhere. So they are two different things, the way I see it
It is the expectation value integral.
Yeah, but I wasn't talking about that
I was considering $\int\rho(x)dx$
But , returning to the initial problem I had with the notation:
Let's consider the following case: Non-degenerate continues spectrum of an operator A:
$A|\nu_{\alpha}\rangle=\alpha|\nu_{\alpha}\rangle$

$\rho(\alpha)=|c_{\alpha}|^2$ (This is one of the postulates of quantum mechanics, when considering the above mentioned case of the spectrum). Then:

$\rho(\alpha)=|c_{\alpha}|^2=|\langle \alpha|\psi\rangle|=\langle\psi|\alpha\rangle\langle\alpha|\psi\rangle=\psi^{*}(\alpha)\psi(\alpha)=|\psi(\alpha)|^2$.
6:36 PM
If you want your notation to make sense, you should not have $\left|\nu_\alpha\right>$, that is, just use $A\left|\alpha\right>=\alpha\left|\alpha\right>$
Apart from that type
There's still the issue in that expression I expanded via integration
I am still not sure what it is you wanted to show with the integration; you seem to still be MISTAKENLY considering that $\left|\psi\right|^2$ and $\int_\text{all space}\left|\psi\right|^2\mathrm dx$ are comparable, when they are in fact different things.
What is wrong here:
$\langle\psi|\alpha\rangle\langle\alpha|\psi\rangle=\int \psi^{*}\alpha d\alpha\int \alpha^{*}\psi d\alpha=\int\psi^{*}\psi d\alpha\int \alpha\alpha^{*}d\alpha=\int \psi^{*}\psi d\alpha$.?
And I had already typed out the full thing if you wanted it, namely $$\left<\psi|\psi\right>=\left<\psi\right|\hat{\mathbb I}\left|\psi\right>=\int_{\text{all }\alpha}\left<\psi|\alpha\right>\mathrm d\alpha\left<\alpha|\psi\right>$$
What even is $\psi^*\alpha\,\mathrm d\alpha$ and $\alpha^*\psi\,\mathrm d\alpha$ in the first place?
$\langle\psi|\alpha\rangle\langle\alpha|\psi\rangle$ how do you write this, involving the integral?
Consider each inner product
and expand it
6:44 PM
I already said so; I know I have an identity operator. I know whatever observable it is, by postulate, it has a complete "orthonormal" basis, for some normalisation acceptable in the continuous part of its spectrum, and I am choosing to decompose the identity operator in terms of this basis.
True, but for the identity operator you need the expression $\int|\alpha\ranglr\langle\alpha|$
And you have no integration sign here
$\langle\psi|\alpha\rangle\langle\alpha|\psi\rangle$ .
I do have it
I must have it
But how? Starting with this $|\langle \alpha|\psi\rangle|^2$ and expanding it. Where does the integral sign comes from?
I NEVER started from there and expanding that
symbol*
I am, and I should in the end get a meaningful expression
that reflects $\rho(x)$
6:50 PM
Anyway, if you want to go into even greater detail, you can just go deeper: $$\left<\psi|\psi\right>=\left<\psi\right|\hat{\mathbb I}\hat{\mathbb I}\hat{\mathbb I}\left|\psi\right>=\iiint\left<\psi|x\right>\mathrm dx\left<x|\alpha\right>\mathrm d\alpha\left<\alpha|y\right>\mathrm dy\left<y|\psi\right>=\iiint\psi^*(x)\alpha(x)\alpha^*(y)\psi(y)\mathrm dx\mathrm dy\mathrm d\alpha$$
@imbAF You did not start from a meaningful expression and thus you should never get anything useful from there. That is why I vehemently choose never to start from where you wanted to start from. This is science. There are dead ends.
That is a postulate of qm
$c(\alpha)=\langle \alpha|\psi\rangle$
But I figured it out
I know of postulates of QM that are very close to what you are choosing to start from, but it is not. You are just being confused.
Anyway, with what I have written down, you should be more than capable of seeing that it makes sense, and you can convert from my notation to what you want.
Also, not to mention that the postulates that you have considered are written with caveats---they are, for simplicity, written only in the finite matrix case, i.e. the discrete spectrum part, not the continous spectrum part.
$\langle \psi|\alpha\rangle=\langle \psi \mathbb I|\alpha\rangle=\int\langle\psi|\alpha'\rangle\langle\alpha'|\alpha\rangle d\alpha'=\int\psi^{*}(\alpha)$.
It is malformed at the end. Because there is an integral, you must have $\left<\alpha^\prime|\alpha\right>=\delta(\alpha^\prime-\alpha)$ and so the integral needs to disappear and so you must have $\left<\psi|\alpha\right>=\psi^*(\alpha)$ if this right hand thing even exists
@naturallyInconsistent Yeah, this is what I needed.
Thanks
I totally forgot
7:03 PM
This is why I think a first intro to QM (with maths) should just start with bra ket notation. Then you can write things down properly and not always have headaches like these
That's true, but I wanted to have a good, I wouldn't say understanding, but a clean writing
and shifting between bra-ket notation and the other notation, which one calls what exactly ?
wave notation ?
Classical notation
Most of physics is still done in classical notation. So sad
At some point, I am certain that I did ask ACM,, about the following:
If we consider the simple case of non-degenerate discrete spectrum, we know that, the probability of measuring some eigenvalue a_n is $P(a_n)\lange u_n|\psi\rangle|^2$ which is the coef. multiplying the eigenvector u_n
How were we able to deduce that, that this is the case? That squaring the coef. represents probability
was it through experimentation and observation
or is there an intuitive thinking process that led to this claim ?
7:43 PM
@imbAF Firstly, this is a postulate; it is not deduced. However, this mathematical form had to be motivated from somewhere, and if you knew about light waves, the motivation would be extremely obvious. Like, Imma put that into my textbook so that the entire set of postulates of QM would be physically motivated, not plucked from thin air
Also, the interpretation of those as probabilities is due to Born, hence "Born probabilities". It was not obvious beforehand that those are probabilities at all.
Indeed, that it could be taken as probabilities is a gigantic leap forward in the understanding of QM
so light waves is how it was postulated that those coef. , when taking their square amplitude, represent probl. of measuring the corresponding eigenvalue to the eigenvector that a considered coef. multiplies?
Light waves is the point of closest contact, yes.
Ofc I fail to see how, but ok, I'll take it at face value
can't one argue that through experimentation
one could prove the validity of the postulate
?
I will stay difficult to see for quite a while, sadly.
@imbAF No, no, that is not what you do with postulates. Postulates cannot be proven. It can be shown to be reasonable through comparison with experiment, but it cannot be proven.
Aha
7:56 PM
I think I know what I can say to help you think about it. The probability being the squaring of the coefficients is extremely motivated by requiring that the new modern quantum theory has to reproduce predictions of the classical Maxwellian theory of light. It is a correspondence principle-lite
It's something I guess...not that I have a full picture of the entire thing, but better than nothing :P
Yeah. I always find it crazy why the teaching of QM doesnt try to make this obvious.
I think,they want you to take some things at face value
Like, those postulates did not come by divine intervention into the mind of von Neumann or Dirac. They came from experiments.
But then, your foundations are weak
8:00 PM
The least they could do is try to motivate them, so that successive generations would not be wasting so much time second guessing their validity
Well, I was the only one that asked, how do we know that this is the case, and the response was : "It has been the case through rigorous observations"
Great bunch of nothing answer tbh
Yeah.
But from what I have already told you, you might discover it for your own. It is not exactly a difficult thing. It is just that it is somewhat long to talk about. However, in a quantum physics introductory lecture module, they MUST cover the experiment, and so it is extremely silly for them to already have covered the topic but not discuss that the experiment motivated the postulate.
which experiment ?
Double slit.
it was mentioned and explained, but it was meant as a way of showcasing the duality in nature of light
Nothing close to what you are implying
8:07 PM
Double slit, single slit, combine of both, are standard in physics curriculum, yet they are somehow not used to discuss this
@imbAF No, definitely they do. There must be an explanation, mathematical derivation, in fact, for the brightness of a dot within a bright spot v.s. another dot.
You know, that Cornu's spiral integral
I am saying that in our class that experiment wasn't tied to the coef. and probabilities
@imbAF exactly; it is extremely stupid that they already covered the experiment, yet they dont tell you that it is the one that best motivates the postulates.
I can't be the only person who had noticed this fact. The pioneers argued as if the postulates were obvious.
Griffith doesn't really say much either
No textbook, as far and wide as I have read, covered it.
And so I think I have to do it
and F--ing h*ll I'm so busy at work to write it
would be a good an necessary addition
Things don't fall from the sky
or are a certain way, because "they work"
8:13 PM
What I'm trying to say, is that, even if "they work" is the final answer, the reasoning as to why we would think to even guess such a form, should at least be presented, so that future generations can look at them for themselves, learn to use the same guessing method for other stuff, etc.
The superposition of wave functions, is physically reflected as interference, correct ?
Necessarily so; the light waves doing that is also why we have the $\frac1{\sqrt2}$
How so?
regarding 1/\sqrt{2}
Most of the mathematical machinery of basic QM is borrowed from Maxwellian description of light; at the time, they did not know how to deal with the electron, and so they just borrowed what they knew about light and then went on from there.
@naturallyInconsistent I think this is supposed to be obvious: When we do the double slit with electrons, the whole reason we talk about "wave-particle duality" is to show that the probability of finding the particle behaves like the classical intensity of the wave. What does "wave-particle duality" mean if not that?
8:21 PM
The energy of a light wave goes as $\vec E^2$; when you rotate $45^\circ$ the new $\vec E$ field goes as $\frac{E_x+E_y}{\sqrt2}$
@ACuriousMind Exactly, the wording, the terminology, makes it clear that the pioneers thought of these things and understood them, so why did they not point the link out explicitly?
imbAF, that 1/\sqrt2 business is just explained above, sorry didnt tag ya
how explicit do we have to get :P "The electron distribution on the screen looks like the classical light intensity" is what most presentations of the double slit present as the puzzle, so it's not as if this is obfuscated
maybe the problem is more that plenty of places try to teach QM to students whose grasp of classical mechanics and electromagnetism isn't strong enough to understand this implication immediately :P
@ACuriousMind Much more. We mathematically derive the intensity profile of the combined double-slit interference with finite single slit diffraction pattern, and so there is no good reason why they do not explicitly work out how the quantum probabilities are meant to exactly correspond to them.
This would actually be something very interesting to see
8:39 PM
Precisely. So my textbook on QM starts with the classical description of light, and since I'm doing it, I'm doing circular polarisation. I do this because, then, they would see that I am not making any approximation when I do the integral; the final answer does not at all depend upon taking cosine term or sine term, i.e. phase issues completely go away.
Is it important to have circular polarisation ?
And my first treatment is not with the 2-state system. It is a 1-state detector, either it clicks, or it does not click. (Not a 2-state despite this binary, because in QM the 2-state means spinor-like behaviour, and this will not have the equivalent of spin up in x direction, say)
@imbAF no, but most physics would make more sense that way. After all, we care a lot about the conservation of angular momentum
@ACuriousMind ok thanks. btw do you know that why should we demand that the integral on projective space I showed you earlier, should have projective weight zero?
@Sanjana it's another pretentious way to say that integrals are supposed to be scalars
8:43 PM
@ACuriousMind But why scalars under scale transformations?
Oh ok. The $\varphi$ on the LHS satisfy zero rest mass free field equations which are conformally covariant. That's why maybe

1 hour later…
10:08 PM
Can we talk about the collapse of the wavefunction to the corresponding eigenvector, for a measured eigenvalue, when the spectrum is continues ? I believe not, but I am not 100% sure. Perhaps the notion of collapse is different in the continues case
I mean does a collapse occur in this case?
collapse is an interpretation-dependent notion; the question makes no sense without specifying an interpretation
I thought I did specify what I meant with collapse, in my description in the beginning
The state of the system after a measurement is conducted
that's not exactly "collapse", but if the question is whether the result of a measurement can be an "eigenstate" from the purely continuous spectrum of the observable being measured, the answer is no, because these states are not normalizable and hence not actual states a system can take
we've been over this
Yes, I thought so and I was sure
So what happens in this case?
in what case
10:16 PM
when you make a measurement, for a system with continues spectrum
no measurement device can measure an observable with continuous spectrum to infinite precision, you always only measure the eigenvalue to be within some $[\lambda - \epsilon, \lambda + \epsilon]$, and the projector onto that subspace results in normalizable states
isn't the concept of subspaces (I assume Hilbert sub space) relevant when degeneracy is considered?
I have no idea what that question means
I keep telling you the issues with continuous spectrum are subtle and advanced, outside of the usual intro to quantum mechanics; why do you keep bashing your head against them :P
you talk about the projector onto that subspace
Well because, one of the postulates of QM tells us about the collapse of the wavefunction
@imbAF again, the proper definition of what happens would require spectral theory you do not yet possess
10:19 PM
Two statements are made, for discrete degenerate and non degenerate case
but the continues cases are not taken into consideration
So I thought why is that the case
@ACuriousMind aha ok
but, by example, if you measure the position of some state that is initially with wavefunction $\psi(x)$ to be within $[x_0 - \epsilon, x_0+\epsilon]$, then the resulting state is zero outside of that interval and equal to the original $\psi(x)$ within that interval, just with this cut-off function normalized again
You know, whenever you write something and it contains latex symbols, I open another tab I click on new thread and copy paste, so I can know what is written. Is there a more efficient way ?
@imbAF I mean a full-blown course on functional analysis that proves a spectral theorem for unbounded operators about measures corresponding to the discrete and continuous parts of the spectrum; you will not understand this just from reading Wiki articles
@imbAF the room description links meta.stackexchange.com/a/220976/263383, which has various ways to render MathJax in chat
@ACuriousMind I meant is that the topic
@ACuriousMind Actually this is a great explanation. And if I were to compare the wave function that is not zero within this region of space with the initial one, what's the difference that one can see when comparing both expressions ?
@imbAF A spectral theorem for unbounded operators would be the topic, the article you linked barely touches on that case
10:26 PM
Ok
Do we do this in physics at some point or not?
@imbAF I don't know what you mean by "comparing both expressions". The result of measuring $\psi(x)$ within $[x_0 - \epsilon, x_0 + \epsilon]$ is, as I said, a wave function that is equal to $\psi(x)$ within that interval and zero outside of it.
@imbAF what does "in physics" mean?
In physics major , master
I do not know what your particular courses offer you
Why, physicists never have to deal with "collapse" of a system with continues spectrum?
lots of observables of interest have discrete spectrum
and in plenty of cases where they do not you don't need to worry about this particular point
10:29 PM
@ACuriousMind And I believe, prior to the measurement $\psi(x)$ was not zero even outside $[x_0 - \epsilon, x_0 + \epsilon]$, is that correct?
@imbAF That obviously depends on the $\psi(x)$ you started with.
So my question really is that $\psi(x)$ before the measurement and after, in both cases is expressed, mathematically in the exact same way?
How can one differentiate that this was the wave before and this is the wave after the measurement
I really think you're worrying too much about this continuous case before you have even grasped the workings of QM in general
Ok
all this is much easier if you understand already the discrete case completely
10:31 PM
I believe I do
@imbAF IF the wavefunction was zero outside the interval there is no difference!
this is exactly the same as in the discrete case: If you already have an eigenstate and you measure the observable again, then the state doesn't change!
I see
The way I see it, having a good understanding of differential geometry, functional analysis and representation theory is extremely important to understand the mathematics
in many instances
10:55 PM
Actually, most of the questions you are bothering meow meow and ACM with, are linear algebra. All these representation, discrete, continuous stuff, are really easier if you already have the linear algebra language to talk about eigenstuff
@ACuriousMind That seems a bit wayyyy too simplistic. Like, Feynman did not just treat a sharp edged finite slit (I don't remember which particular bit, but I'm so lazy to go out of my very hot [AC needs servicing] lab and into yet hotter office to grab my copy of Feynman Hibbs), but also treated a Gaußian slit, so the attenutation is smooth. What happens if we do that here?
@naturallyInconsistent yes, the theory of real measurements - without sharp intervals - is much more subtle
but sharp measurements inside an interval present the most direct analogy to the discrete case
of course. It is almost always the case in physics that some obvious simple thing is the only thing that everybody agrees upon, and everything else is an invitation of a fist-fight.
we are blessed that almost no applications of QM require delving into this in much detail :P