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user434058
04:55
@rob I am waiting for a question having "rigid rob" :P
05:14
morning
user434058
Morning
08:36
Can I invest in youtubers? What if I believe that some youtuber will grow a lot and I want to invest my money to help him out and get some ROI?
08:53
I get so f*cking pissed when someone says "state-of-the-art machine learning model"......
09:21
@ACuriousMind Is the point of sheaf v. presheaf that the local properties extend globally?
@Slereah sort of - not only globally but for all open subspaces and open covers of them you can extend local sections on the sets of the cover to a "global" section on the covered space and this extension is unique (this is often called "gluing" the sections)
the gluing can fail in two ways on a presheaf - there is no possible glued section, or there is more than one
09:55
@ACuriousMind Am I understanding it correctly that, if I consider the generalization, the étalé space ~ the bundle, the fiber is the stalk and the sheaf is the space of sections?
@Slereah not sure why you call the étalé space a generalization, but that sounds right to me, yes
@ACuriousMind Aren't there étalé spaces that aren't bundles?
it's not always a bundle but you can talk about fibers and sections without having a bundle
alright
10:15
What's the relation between the étalé space of continuous functions and the line bundle?
Are they isomorphic or not quite
what line bundle?
The trivial line bundle
$\mathbb{R}$
I think the étalé space of continuous function is in general pretty weird
Yeah
But I would guess it may be related to it?
or the jet extension of that bundle or something
it's certainly not a line bundle - the topology on the fiber is wrong, for one
the topology of the fiber is discrete, but in vector bundles like the line bundle the fiber has the topology of $\mathbb{R}^n$
and it's not even Hausdorff, see math.stackexchange.com/a/2724614/143136
if you're hoping to go back to nice differential geometry world by thinking about sheaves as their étalé spaces you're going to have a bad time :P
10:22
Let's just burn all the books and go back to hunting and gathering
I was just hoping that the bundle that gives us functions on the manifold would be somewhat related to the sheaf of functions on that manifold!
you mean because you can think about real-valued continuous functions as continuous sections of the trivial line bundle?
ah, well, the étalé construction doesn't construct the "best" space this is a sheaf of sections of, it just constructs one that always works :P
Damn lazy étalé spaces
the discrete topology on the fiber means you need to think about this more like a covering space than a bundle
10:32
enough sheaves for now
Let's go do the shopping
10:51
Heyyy
What is arcsin(sin(a)sin(b))?
11:09
What do you mean what is it?
@Korra
 
1 hour later…
12:29
@Charlie I mean like how we have arcsin(sinx)=x, do we have anything for something as complex as I mentioned?
I mean there may exist manipulations that put it in a simpler form but nothing comes to mind
it's not a particularly nice function: google.com/…
13:09
The reason the $\alpha' \to 0$ limit means we're discussing the massless spectrum is because of $M^2 = \frac{1}{\alpha'}(N^{\perp} - a)$ right? And then this is the low energy limit because $[M] = [E]$ so $[M] \to 0$ implies $[E] \to 0$ right?
$\hat{Q}$ is a Hermitian operator. What does it mean to take the sin of it? which is $\sin{\hat{Q}}$
maybe expand in taylor series?
idk, that's what you do with the exponential right
$$\hat{Q} \text{test}$$
hmm, ChatJax++ isn't working for some reason
$$\sin(\hat Q)\equiv \sum_{n=1}^\infty (-1)^{(n-1)}\frac{\hat Q^{2n-1}}{(2n-1)!}$$ maybe?
13:20
Is it possible to express it without $\hat{Q}$ as a linear combination in Q's eigenbasis?
uhh
not that I can think of
I mean you can write $\hat Q$'s matrix elements in terms of it's eigenbasis, but not as a linear combination
what's the mathjax syntax for bra ket notation?
\langle and \rangle
Let $\langle{i}|$ denote i'th basis vector in the eigenbasis of $\hat{Q}$. $\hat{Q}$ can be expressed in its eigenbasis as $\Sigma c_i |i\rangle\langle i|$.

$\sin{\hat{Q}} = \Sigma a_i \langle i|$
How do I find $a_i$?
your second equation has a bra equated to what I would expect to evaluate to an operator
13:31
Oops, I meant $i\rangle\langle i|$.
13:42
If we're in the eigenbasis of $\hat Q$ then $\sum_ic_i|i\rangle\langle i|=\Bbb 1$
I don't know if what you've written has an easy solution if you're in an arbitrary basis, seems like it might be complicated
or if it even makes sense
@Charlie alright then
Can someone help me find out b_1 in this equation?
Ooops
not without neck surgery
13:50
you'd need to explain what you're trying to do at least
@Charlie I know (≧▽≦)
The equation smack in the middle
@Charlie yeah
Brb
Gtg
@Charlie the thing is, I had an equation which I got by solving a differential equation. This new equation (solution to the differential equation) was cut up in three pieces due to discontinuities in the system
Now , what I wanted was to find the integration constants b and b_1
I got b by using a knowledge of the system and the property of cotangent function
user434058
14:10
I once saw (in a YT video on Tensors by eigenchris, Thanks @Charlie) that the element in the $i$-th row and the $j$-th column of a (transformation) matrix being written as $(MI)_j^i$, where $M$ is the transformation matrix and $I$ is just the corresponding identity matrix. Now, AFAIK, upper indices are used to indicate contravariance and lower indices indicate covariance, however, in this case, I see no way to justify the $^i_j$ using the $^{\rm contravariant}_{\rm covariant}$.
@Yashas @Charlie the proper formulation of functions of operators is called 'functional calculus' and does not rely on Taylor series, see physics.stackexchange.com/q/574621/50583. For operators that can be written as $A = \sum_i \lambda_i \lvert i\rangle \langle i\rvert$, you simply have $f(A) = \sum_i f(\lambda_i) \lvert i\rangle\langle i\rvert$.
user434058
So can it be explained by covariance and contravariance, or is it just overlapping notation? (I think the latter is true)
ah ok @ACuriousMind
@ACuriousMind Does that thing have a name? It's not obvious to me why that holds although the relation does lead me to the correct final answer.
@FakeMod A matrix/linear map on a vector space $V$ is equivalently an element of $V\otimes V^\ast$ and hence has one co- and one contra-variant index.
@Yashas What thing? My expression for $f(A)$ there should be taken as the definition of $f(A)$.
14:15
@ACuriousMind This: $f(A) = \sum_i f(\lambda_i) \lvert i\rangle\langle i\rvert$. I am to rigorously prove this. Maybe it's supposed to be obvious but it isn't for me. So I was asking if it had a name so that I could look up on it.
@ACuriousMind Isn't that the whatchamacallit theorem
Spectral whatever theorem
@Yashas What are you trying to prove it from?
user434058
@ACuriousMind Ah, I see. So since the row vector components are contravariant, so we place the row number as an upper index, and since the column covector components are covariant, so we place the column number as a lower index, right?
If you think there is something to prove then you must have a different definition of $f(A)$ to begin with
I am probably very bad at this but this is what I was up to:
$\hat{Q}$ is a Hermitian operator with spectral decomposition as $\hat{Q} = \Sigma c_i |i\rangle\langle i|$. How do I find the new coefficients in the same decomposition for $\sin (\hat{Q})$?
I am not very comfortable with operators so I like to assume that the operator is operating on a function $\psi$ and then reverse engineer the operator from the solution.
$\psi$ can be expressed as $\Sigma \langle\psi|i\rangle |i\rangle$. $\hat{Q}\psi = \Sigma_i c_i\langle\psi|i\rangle|i\rangle$. $\sin (\Sigma_i c_i\langle\psi|i\rangle|i\rangle) = \Sigma a_i\langle\psi|i\rangle|i\rangle$. I need to fin
14:17
It does not mean what you say at the end. You can't apply a function like $\sin$ to a vector
Ok, then my definition of $f(A)$ itself is wrong.
What does $\sin \hat{Q}$ mean then?
What I wrote above - it's the operator defined by $\sin(Q) = \sum_i \sin(a_i)\lvert i\rangle\langle i\rvert$.
for analytic functions like $\sin$, you could also define it via the Taylor series as others suggested but that then unnecessarily restricts you to only being able to apply analytic functions to operators.
user434058
@FakeMod Am I right?
(it's maybe worthwhile to convince yourself that the above is indeed the same operator as the power series definition)
it is apparently the spectral mapping theorem I had in mind
14:27
@FakeMod Higher order tensors can have both contravariant and covariant indices
user434058
@Charlie Doesn't even a rank 2 tensor (a matrix) have both of those?
Not necessarily, a rank-2 tensor has three possible index placements: $$T^{\mu\nu},\quad {T^\mu}_\nu,\quad T_{\mu\nu}.$$
That's why you usually see matrix operations like $y^\top M x$
$y^\top$ is the covector dual to $y$
user434058
Also, what is a good, mathematical way to assert whether a certain quantity is covariant or contravariant? I am not looking for any geometric or particularly intuitive method, because all of them seem to become harder to visualise as I learn more of tensors (in other words, I am not looking for some kind of "just double that thing and notice that this thing gets halved" method). I am looking for a fool-proof (if possible) method.
You just write it, usually
$v \in TM$ for a contravariant vector field, $\omega \in T^*M$ for a covariant vector field
user434058
14:33
@Charlie Ah, ok. However, as of now, I have only seen the middle example. Thanks.
Either that or it's implied by the way you use it
Covariant=index down, contravariant=index up, that's basically the syntax
user434058
@Slereah Umm... I am unfamiliar with tangent bundles. (It seems like a good time to learn about them)
user434058
@Charlie Yeah, but that's applicable when I know something is contra/covariant. How to determine that in the first place?
they're not important if you're just learning basic tensors
14:36
$V$ and $V^*$, if you prefer
user434058
@Charlie the bundles?
yeah, I wouldn't be worried about that atm
user434058
@Slereah A wee-bit more comfortable :-)
user434058
@Charlie Oh, yeah, sure. Thanks.
Afaik it's generally more interesting to know whether an object is tensorial at all rather than it's specific transformation law, which can usually be obtained afterwards
user434058
14:39
It sometimes feels that you guys (I am looking at you ACM) really put in much more math than needed, and I always end up quitting seeing the extravagant math that you use while explaining. (Jk, I am extremely thankful to you, please keep on teaching the way you do :-))
And the fact that the distinction between covariant and contravariant isn't always explicitly that important since the metric we generally use makes $V$ and $V^*$ isomorphic in finite dimensional spaces
can't do physics without math :p
user434058
@Charlie Alright, I am quite sure that I am getting ahead of myself now. I might not be able to hold on to any further conclusions from here. Thank you for your gelp and recommendations. Good <whatever part of day you are in>.
user434058
@Charlie But... but.. you can't just hurl all of those scary math at a beginner without batting an eye. TBH, I am quite surprised that folks don't quit physics by the time they encounter such devils...
if you want the basic idea, dual vectors are functions
If you have a vector $v$, a dual vector $\omega$ is a linear function $\omega(v)$
@FakeMod I often take a more mathematical approach to things than the average physics text, but that's because I believe it is very beneficial to know how mathematicians would view this stuff. There are places where lapses in rigor are excusable or even "necessary", but they're rarer than many think :P
14:44
You can write your vector in a basis, say $v = v_x e_x + v_y e_y + v_z e_z$
user434058
@Slereah I am all fine as long as you are talking about vectors, covectors (dualvectors) and transformations. Anything beyond that might be harder for me to understand.
As dual vectors are linear functions, you can rewrite it as $$\omega(v) = v_x \omega(e_x) + v_y \omega(e_y) + v_z \omega(e_z)$$
user434058
@ACuriousMind Yes, me too. I also like to throw in more math, but I do that (or in fact, I am capable of doing) that only with the stuff which I am super clear about. And in this case, I am super clear about nothing :)
Once you're happy with vectors and covectors, higher order tensors aren't that huge of a leap
...but I also find math interesting on its own so I don't consider it a waste of time e.g. to learn proper linear algebra to work with vectors, or to learn differential geometry to talk about vector fields, even if it takes a while
14:46
Each of these is a real number, so we can write each dual vector as a series of components themselves
user434058
@Slereah yeah, sure. But, with all respect, I cannot see where you are going with this. I am good enough with vectors and covectors, and I don't think that relearning them might be any helpful, however, I would be willing to do so. I just don't want you to waste yoir time teaching me something I already know. But again, it's your call, if you are feeling like teaching me that thing, I will surely be willing to learn.
What he's talking about is basically index contraction @FakeMod have you seen that yet?
user434058
Idk why people use "with all due respect". I mean there's just respect, you don't measure out the respect you left off, and then give that precise due respect.
it's more of a figure of speech
user434058
@Charlie are you talking about einstein notation? (That's the last thing I saw)
user434058
14:52
@Charlie Seems more like an old catchphrase to me :P
Not directly
For instance $v^i\vec e_i$ and $T_\mu G^\mu$ both use the summation convention but only the right hand one is an example of tensor contraction
user434058
@Charlie Nope, I don't know this stuff.
The bottom line is that vectors act on covector and covectors act on vectors
user434058
Any self sufficient literature/books on tensors which anyone would want to recommend?
user434058
@Charlie Wait, how do vectors act on covectors?
user434058
14:55
Never seen that.
@FakeMod Vectors are functions on dual vectors and vice versa
Defined by the action that $v(\omega) = \omega(v)$
Similarly every tensor can be defined as a variety of functions
Given a vector space $V$, the set of all linear functionals $w:V\rightarrow \Bbb R$ has vector space structure, this is the dual space, $V^*$. But similarly the original vector space $V$ is the set of all linear functionals $v:V^*\rightarrow \Bbb R$ that act on $V^*$
user434058
@Slereah Oh, so a covector applying on the vector is equivalent to the vector applying on the covector?
user434058
Alright, thanks!
14:59
Similarly, you can define a matrix as a function of a vector and a dual vector, or a function from vectors to vectors
$$M(\omega, v) = \omega(M(v))$$
user434058
@Charlie Oh, but that definition does not necessarily imply that $v(\omega)=\omega(v)$. I mean $v(\omega)=-\omega (v)$ will also satisfy such a definition, won't it?
This is just definitions to make things easier
user434058
@Slereah Oh, ok.
user434058
@Slereah The LHS of that equation is a matrix, however, the RHS is a real number. How is it so?
No they are both numbers
wait
15:03
On the LHS, I'm writing $M$ as a function, $$M : V \times V^* \to \mathbb{R}$$
On the RHS, I am writing it as $$M : V \to V$$
In both cases, the end product is a real number
user434058
@Slereah This is the first time I am seeing a matrix taking in a real number and giving out a real number. How?
The matrix isn't taking in a real number
$w$ is a covector and $v$ is a vector
user434058
@Charlie Oh, yeah. I see.
user434058
I misunderstood something...
Those are technically not quite the same function, you should use some mapping or whatever
But typically, physicists use matrices and vectors pretty loosely
user434058
15:06
I am sorry, but now I gotta go. Thanks a lot everyone.
afaik that's called currying right, where you've taken $M$, half filled it's arguments and treated it as a function taking a covector
bye @FakeMod
yeah, that's an instance of currying
I didn't want to open a new question on this, so I try the chat:
Could someone please recommend a good textbook or introductory paper on quantum loop-gravity? That'd be nice :D
If anyone happen to know such a thing :P
I don't think we have any loopy people here :P
Maybe this would help?
15:09
Seems like the right thread for me :P
Thank you very much, Charlie!
np :)
That gives a really good easy example of the difference between contravariant and covariant components of a vector
@kalle chapters 1 to 3 of "Covariant Loop Quantum Gravity: An elementary introduction to Quantum Gravity and Spinfoam Theory" are definitely worth reading at least
My only LQG book is Rovelli
Not sure I would recommend it
'tis a bit of a tough read
Is that the Rovelli ?
I mean he's a Rovelli
He's not Nino Rovelli, Italian bobsleder
Or Mauro Rovelli, chairman of Sacchettificio Monzese
15:18
He has I think two books on it, the other one just called Quantum Gravity
I have Quantum Gravity, yeah
I think the other one is way easier
.gnitseretni saw teneT
I'm pretty confused by the use of the Fourier transform when we quantise the harmonic oscillator in the free scalar field, what is $\phi(\vec p,t)$? Is this like momentum space in regular QM?
15:25
it's the Fourier transform of $\phi(\vec x,t)$ :P
So when we obtain $$\phi(\vec x)=\int\frac{\mathrm d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{\vec p}}}\left[ a_{\vec p}e^{i\vec p\cdot\vec x}+a^\dagger_{\vec p} e^{-i\vec p\cdot\vec x} right],$$ why do we have to do this in momentum space?
well I'm quite upset that that didn't work
@FakeMod This is not the most consistent example, but it has the idea: If $\mathbf{A} = x^1 \mathbf{e}_1$ is your vector, and you double the basis vector $\mathbf{e}_1$ to $\mathbf{e}_1' = 2 \mathbf{e}_1$, then $x^1$ must transform into $x'^1 = \frac{1}{2} x^1$ so that $\mathbf{A}' = x'^1 \mathbf{e}_1' = \frac{1}{2} x^1 (2 \mathbf{e}_1) = x^1 \mathbf{e}_1 = \mathbf{A}$ holds.
Here the components $x^1$ transform 'oppositely' to how $\mathbf{e}_1$ transforms, so you say the components transform 'contravariantly' to how the basis transforms. If $\mathbf{A} = x^1 \frac{\partial }{\partial x^1}$ is your vector, and you do a basis transformation double the basis vector,
in this case amounts to a coordinate transformation from $x^1$ to $x'^1 = \frac{1}{2} x^1$ so that $\frac{\partial }{\partial x'^1} = \frac{\partial }{\partial \frac{1}{2}x^1} = 2 \frac{\partial }{\partial x^1}$, then the coordinates must transform in the same way that the basis transforms,
so that $\mathbf{A}' = x'^1 \frac{\partial }{\partial x'^1} = \tfrac{1}{2} x^1 \frac{\partial }{\partial \tfrac{1}{2}x^1} = \frac{\tfrac{1}{2}}{\tfrac{1}{2}} x^1 \frac{\partial }{\partial x^1} = x^1 \frac{\partial }{\partial x^1} = \mathbf{A}$. In this case the components transform covariantly with the basis. The video gives an example using 2D projections which is better.
@Slereah Easy reads are always prefered :P
nooo
$\phi(\vec x)=\int\frac{\mathrm d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{\vec p}}} ( a_{\vec p}e^{i\vec p\cdot\vec x}+a^\dagger_{\vec p} e^{-i\vec p\cdot\vec x} ),$
15:30
ok I surrender
@kalle I recommend Winnie the Pooh
ah yes that haha thank you
I'll give that a try!
:P
It's a classic book!
Foundation of many good ideas!
as I heard...
15:31
For instance you shouldn't stick your head in a hole if the hole is too small
I was basically wondering why we have to work in momentum space
@Charlie Differential equations for Klein Gordon are easier to solve in momentum space
It is just the Fourier expansion of a function in momentum space eigenbasis. Note it depends on position so you first take the Fourier transform to go to momentum space via $\phi(p) = \int \phi(x) e^{-ipx}$ and then invert the Fourier transform to get back to \phi(x) = \int \phi(p) e^{ipx}$ something like that
The differential equation is pretty trivial in momentum space
But after that you have to integrate it back :p
@Charlie A free Hamiltonian is diagonal in momentum space, and it's easier to work there
15:34
Why do we call it "free"
Ok if it just makes things easier I'm on board
The non-free theory isn't enslaved
An easier way to understand it is that the solutions of the free Klein-Gordon equation are just easy plane waves $e^{\pm i p_{\mu} x^{\mu}}$, so you literally just write the general solution of the Klein-Gordon equation as a linear combination of these plane waves: $\phi(x) = \sum_p N_p (a_{+p} e^{+i p x} + a_{-p} e^{-ipx})$ for a normalization factor $N_p$, however from $p^2 = m^2$ we see not all four $p_{\mu}$'s are independent so you sum over just the $\mathbf{p}$'s.
@Slereah ...I think because non-interacting stuff just moves without any constraints?
it's not free vs. enslaved, it's free vs. constrained
Clearly should use integrals because $\mathbf{p}$ is continuous
If you didn't know how to guess that $e^{\pm i p x}$ were solutions of the Klein-Gordon equation, you could solve the equation using the method of Fourier transforms and arrive at the same result, which is what P&S assume you already know how to do in writing it that way
Tong's notes and videos do this more step-by-step, welcome to QFT where to understand one thing in one book you go to another book
15:39
yeah I'm on Tong's notes now, P&S seems like a good book to come back to once I've learned things
Schwarz is also alright to learn QFT
*Schwartz
Well they don't seem to do it as pedantically as one would like skimming it, Woit does it pedantically, see (43.1) to (43.8)
Forgot a $ above ahh
16:00
Can some of you suggest where to start, for learning quantum mechanics. I am in the final year of my high school, and I read a lot related to physics.
@ManishS How are you on calculus?
Decent. I know how to differentiate and integrate functions.
Errr which one is the basic one people recommend
Sakura?
I forget
86
Q: What is a good introductory book on quantum mechanics?

PhaDaPhunkI'm really interested in quantum theory and would like to learn all that I can about it. I've followed a few tutorials and read a few books but none satisfied me completely. I'm looking for introductions for beginners which do not depend heavily on linear algebra or calculus, or which provide a s...

You can check here
Thank you. Will check.
Also, how long am I reasonably expected to take, to learn most of it? (I generally grasp concepts well)
Depends what you mean by that, I guess
Anywhere from "A few months" to "Never"
16:17
Just go through those videos and see how you do, they try to teach you the math you need as you go along for QM
'a review of complex numbers' is video 4 for example
user434058
@bolbteppa Yup, I have seen that :-)
user434058
@Slereah Could you give me an example of such a $M$?
@FakeMod $\mathrm{diag}(1,1,1)$
user434058
@bolbteppa Yeah. That doubling the basis vectors works well as of now, but I am quite concerned by the lack of rigour in that method, that's why I was hoping for a mathematical method. Anyways, thanks :-)
Technically what I said is rigoroous and this is the way the wiki does it
In multilinear algebra and tensor analysis, covariance and contravariance describe how the quantitative description of certain geometric or physical entities changes with a change of basis. In physics, a basis is sometimes thought of as a set of reference axes. A change of scale on the reference axes corresponds to a change of units in the problem. For instance, by changing scale from meters to centimeters (that is, dividing the scale of the reference axes by 100), the components of a measured velocity vector are multiplied by 100. Vectors exhibit this behavior of changing scale inversely to...
It's easier if you use a metric so projections can be defined and no crazy partial derivative basis is needed
user434058
16:28
@Slereah Now what would happen if I input $v,\omega$ in that particular $M$ such that $v= \begin{bmatrix} a\\b\end{bmatrix}$ and $\omega=\begin{bmatrix}c & d \end{bmatrix}$?
Well, if we treat $M$ as a map from vectors to vectors
user434058
@Slereah No, I am talking about $M:V\times V^*\to \mathbb R$.
Then $$Mv =\begin{pmatrix} 1 & 0\\0 & 1\end{pmatrix} \begin{pmatrix} a\\b\end{pmatrix} = \begin{pmatrix} a\\b\end{pmatrix}$$
user434058
@bolbteppa Hmmm... I see. If that's valid and good enough, then I don't see any reason behind chasing a more abstract definition. Thanks!
If you want to do the bilinear function directly, it's gonna be $$M(v, \omega) = M^\mu_\nu v^\nu \omega_\mu$$
user434058
16:32
@Slereah Is this a definition, or is there a reason why it is this way?
@FakeMod As it is a tensor, $M$ is linear
So you can decompose it via the base
$$M(v, \omega) = M(v^\mu e_\mu, \omega_\nu e^\nu) = v^\mu \omega_\nu M(e_\mu, e^\nu) $$
user434058
@Slereah Oh, alright. I get what you're saying, thanks.
Then you define $M(e_\mu, e^\nu)$ as the components of the matrix
16:49
@Slereah and @bolbteppa thank you.
17:06
I still feel a bit stuck at the starting gate with the quantisation of the KG field. When we arrive at that big integral $\phi(x,\hat a,\hat a^\dagger)$ that I failed to type earlier, we're saying that a solution to the KG equation is a plane wave in the field, and that at each point the evolution of the scalar value at that point has simple harmonic motion.
Then, just like when we first quantise the classical harmonic oscillator and promote the position variable to an operator and define it in terms of the raising/lowering operators, we "second quantise" the oscillator and promote the point in the field to an operator which we then also express in terms of the raising/lowering operator
which leads me to ask, what exactly do the creation/annihilation operators act on in this second quantised case? Or does this come back to the fact that "state spaces" aren't as easy to define in QFT and I should think of it as just being something abstract
When you quantize your theory, you get the canonical commutation for $\phi$ and $\pi$
happy with that yeah
Using the definition of those two functions with $a$ and $a^\dagger$, you get commutation relations for $a$ and $a^\dagger$
ok sure
It turns out, o fortuna, that those commutation relations are just like the ones for the simple harmonic oscillator
You can use the proof for the harmonic oscillator to show that such commutations can give rise to such a quantum theory
Although things are slightly more complex since you have infinitely many such ladder operators
one for each momentum
17:12
So if we take the regular QHO and instead of going from 1 dimension to 3 dimensions, we go to "infinite" dimensions, that is effectively the system we have
Yes, pretty much
I'm a bit uncomfortable that the ladder operators are labelled by momentum, what is the analog in the regular qho?
for the quantum harmonic oscillator, the ladder operator can only change the energy by $\hbar \omega$
where $\omega$ is defined by your oscillator
In the case of the Klein Gordon system, you have to add particles with mass $m$ and momentum $p$, so there is a bit more freedom
Roughly speaking, every part of the Fock space is a particle Hilbert space
ah wait the $p$ on the ladder operators mean that when we for instance "raise" the state from the vacuum state to the first state, at that point we have a harmonic oscillator of frequency $\omega_p=+\sqrt{p^2+m^2}$
Your Fock space is something like $$(c, \psi(x), \psi(x) \chi(y) + \chi(x) \psi(y), \ldots)$$
if your Fock space vector is originally $$(c, 0, 0, 0, \ldots)$$
17:16
@Charlie look at the Woit notes, page 459, it's the usual non-relativistic case of a free particle for comparison. That's what you're doing. By just looking trying to solve a PDE, the usual method is Fourier transforms. It o fortuna turns out that KG, Maxwell etc turn out to be similar to the Harmonic oscillator because e.g. the fourier transformed KG equation is just that of a harmonic oscillator in momentum space [see P&S eq. (2.21)]
($c$ is the vacuum here)
Then the action of the raising operator $a_p$ will create a wavefunction with momentum $p$
ie
@Charlie if you have a 3d QHO, then you have one pair of $a_i, a_i^\dagger$ for each dimension (i.e. $i$ is $x$, $y$ or $z$). Here, you have uncountably many "dimensions" and your label for them is $p$.
$$(c, \psi_p(x), 0, 0, \ldots)$$
Ok I see that we have an uncountable collection of ladder operators labelled by $p$, so if we're in the vacuum state and I act on my (whatever it may be) state space with $a^\dagger_{p_1}$, that raising operator only excites one single harmonic oscillator. So I end up with one single point in my field oscillating
Hi everyone !
17:23
hello
@Charlie can you tell me when a string wave is produced such that one end is in our hand and other is tied at wall . Than will our hand be considered as free end or rigid end and why ?
I'm not sure sorry @ronakjain
@ACuriousMind can you help ?
@Charlie no problem
Who has watched Tenet? I wanna talk, it's really awesome
It's generally asked that you don't ping random people in chat with questions @ronakjain
17:26
Ok. Sorry
17:56
It has been many years
And I still don't know if diffeomorphisms are a gauge symmetry
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