01:00 - 18:0018:00 - 23:00

1:01 AM
@Relativisticcucumber quite a tremendous amount of variational mechanics will be much less confusing if instead of focusing upon Lagrangian or Hamiltonian, we spoke instead of action. When the action integral differs by a total derivative, we will be using the vanishing at the boundaries to insist that the action integral is numerically invariant, and that means that the physics will be the same. But we keep saying Lagrangian rather than action for this idea.
Also, the 2D area under curve $y=f(x)$ integral that we want to interpret geometrically, is actually 2 applications of the generalised Stokes's theorem in a rather non-trivial way, before it spits out numbers. What ACM said there is the important last step.

3 hours later…
3:49 AM
right i see i think that makes sense now @naturallyInconsistent
i keep getting that 2.6 should be $\phi(\vec{x},t)$. am i missing something? im also trying to understand why this shows us that the degrees of freedom are separate in the momentum case. my understanding is obtaining 2.6 should show us that the degrees of freedom are independent. and this is nice because it allows us to treat the field as a bunch of independent oscillators. [...]
[...] is it that we show that for each p, this equation is a harmonic oscillator equation, so we have a spectrum associated with every $\vec{p}$
im getting through one sentence of tong a day fantastic

@Relativisticcucumber One way to get what is happening is that you are kinda supposed to throw Equation (2.5) into Equation (2.4) and see that the resulting equation is precisely Equation (2.6). They are correct, it is $\phi(\vec p,t)$, not position. Note that in Equation (2.6) the $\vec p{}^2$ is not an operator.

@naturallyInconsistent i tried that but its not working :/ i am not able to get rid of the exponential and field in the integral so i have something like $(\partial_0\partial^0 + \vec{p} + m^2)(\int(~~~)) = 0$ where the integral is the integral 2.5

As for why the degrees of freedom have been decoupled into independent degrees of freedom, compare and contrast how in Equation (2.4), the field at each position depends upon its neighbouring field values through the derivatives, whereas if you look at Equation (2.6), that is "time derivative and one single $\vec p$ dependence, not two different $\vec p$ dependence", i.e. if you ignore the time derivative is twice, it kinda looks like Schrödinger equation for one wavefunction.
@Relativisticcucumber You can proceed in 2 ways at this point. One is the more standard way, which is to apply an inverse Fourier transform to obtain the Dirac delta distribution needed to kill the integrals. The other way is to argue that, since the overall integral is zero, and each $\vec p$ has a contribution to this integral that is completely independent of the others, it is thus necessary for each single component to be zero, in order for the entire integral to also be zero.

4:12 AM
@naturallyInconsistent since the integral encompasses both of the terms (the exponential and the field), that means we need to do some kind of integration by parts to do the first method, right?
and isolate the exponential in the integral to be able to do the inverse fourier transform?

I have to rush away, bye

3 hours later…
7:05 AM
hi

1 hour later…
8:06 AM
@Relativisticcucumber It's much easier if you just Fourier transform the entire equation instead of trying to "plug in". Do you understand the general property of the Fourier transform where the Fourier transform of $\partial_x f(x)$ is $\mathrm{i}pf(p)$?

do you happen to have a post or explanation or resource which explains how differential forms rigorously allow "dividing" by differentials?
I was looking for a stack question and answer with such explanation but i could not find one...

@Relativisticcucumber the IbP (integration by parts) is more for the method that ACM is suggesting, which is to Fourier transform the entire Equation (2.4)

@SillyGoose Do they

@SillyGoose Consider $\mathrm d\omega=f(x)\,\mathrm dx$; that is more of mapping to $\mathrm d\omega=\frac{\mathrm d\omega}{\mathrm dx}\mathrm dx$ thus $\frac{\mathrm d\omega}{\mathrm dx}=f(x)$. Note that this is really a Jacobian rather than anything else. This, however, is more an identification than a definition or anything. Just trying to say that older notation would have been expressed in a new way.

8:26 AM
Note also that this is not really a division by differentials. Consider $\mathrm df=\frac{\partial f}{\partial x}\mathrm dx+\frac{\partial f}{\partial y}\mathrm dy$; this can only be written down as so if there exists a valid $f$. This matters because this literally asserts that $\mathrm df$ is an exact form. Instead, it is common to deal with $\omega=g(x,y)\,\mathrm dx+h(x,y)\,\mathrm dy$ that is a non-exact form, and then we won't have any division concepts for this.
Rather, with non-exact forms, we are often interested in finding an integrating factor $I(x,y)$ so that $I(x,y)\omega=I(x,y)g(x,y)\,\mathrm dx+I(x,y)h(x,y)\,\mathrm dy=\mathrm df$ makes the non-exact form into an exact one.

hm okay, so $dH(q) = \frac{dH}{dq}dq$ by definition of a differential. then, if we have $dH = -\dot{p}dq$ we have $\frac{dH}{dq}dq = -\dot{p}dq$, which means that $\frac{\partial H}{\partial q} = -\dot{p}$?
in the last step, the one thing I am not sure about is identifying the "integrands" with each other and what makes it valid to do this, as you do in your $d\omega$ example

8:52 AM
@SillyGoose It is that you asserted that $\mathrm dH=$ stuff, i.e., you asserted that there is a $H$ in which the RHS is an exterior derivative thereof, that makes $\dot p=-\frac{\partial H}{\partial q}$
@SillyGoose The necessary ingredient is that $\mathrm dx$ and $\mathrm dy$ are independent variables. If they are interlinked via some constraint, then the identification might fail. Independent variables means that you can just pick one of them to be zero, and then you are obviously left with just one term to do the identification by.

@naturallyInconsistent hm sorry I am not following this

@SillyGoose Imagine for a moment that you don't know Hamiltonian mechanics. Faced with $\omega=\dot q\,\mathrm dp-\dot p\,\mathrm dq$ as a randomly defined form, you might not know that this is an exact form. Then you cannot say anything much about this form. It is only when you know that $\omega=\mathrm dH$, that there exists a well-defined Hamiltonian $H$ from which $\omega$ is the exterior derivative of, that $\omega$ is an exact form, that you can get $\dot q=\frac{\partial H}{\partial p}$ etc
$\mathrm d$ is the exterior derivative operator. $\mathrm dH=\frac{\partial H}{\partial p}\mathrm dp+\frac{\partial H}{\partial q}\mathrm dq$
Note that a lot of these maths is tautological redefinitions. We first worked out a set of maths getting the Hamiltonian mechanics to work, and then we worked out differential forms, and then we brute-force chose definitions so that the pre-existing Hamiltonian mechanics maths gets absorbed by the new differential forms notation.

hm i think i am missing something between the fact that $\omega$ is exact and the wanted result
I know the definition: $\omega$ is exact when $d\omega = 0$, but am I missing a result about exact forms perhaps? Or misunderstanding elsewise

@SillyGoose That's closedness, not exactness
Exactness is that there exists some $H$ such that $\omega = \mathrm{d}H$

oh... exact implies closed
okay i see hehe

9:07 AM
on a contractible space, all closed forms are exact (Poincaré lemma)

Trying to remember some weird mathematical paradox in the philosophy of mathematics but just googling it gives me all the usual boring paradoxes
It is some paradox about whether a reader can decide from context if a given mathematical notation is what the author intended
The argument rested on some weird function that acts just like addition up to a given number
nvm found it

@naturallyInconsistent Hm okay I am on board with the fact that writing $dH$ makes $dH$ an exact form. But I am still not understanding why $dH$ being an exact form allows the wanted identification. (i do not mean to say $H$ is a hamiltonian necessarily here)

@SillyGoose Remember that the $\mathrm{d}x^i$ are a basis of forms
if you have $a_i \mathrm{d}x^i = b_i \mathrm{d}x^i$ you have $a_i = b_i$ by the defining property of a basis

@SillyGoose Remember how I defined $\omega$ with the dotted stuff. The dotted stuff are really not at all expressible on the manifold; those are being insisted upon from on high. It is a tautology.

So if a form $\omega = \omega_i \mathrm{d}x^i$ is exact as $\omega = \mathrm{d}f$, then it follows that $\omega_i = \frac{\partial f}{\partial x^i}$
there's really nothing more going on
at least for the "differential forms make clear how differentiating is "dividing by $\mathrm{d}x^i$" part

9:19 AM
i see...how illuminating

for the specific unfortunate example @naturallyInconsistent tried to pick there's a bunch of other subtleties going on because $\dot{q}$ or $\dot{p}$ are not functions on phase space :P

I did not pick it!!!
hmmpf!

oh, right

anyway, carsick, gotta lay down for a bit

@ACuriousMind oh no ._. and here i thought i understood

9:29 AM
if you didn't worry about it before I said it you don't need to worry about it now :P

you need one of those memory erasers from MIB

9:45 AM
what r some coolest math tools

@SillyGoose yeah so he can take the evil genius path
That's not a safe idea

Kant says that ethics is duty
I kind of agree with Kant's viewpoint that whatever we observe is heavily mind dependent
sets and logic may be mind dependent
the concept of distinct things and multiple things may not exist in a mind independent way
this disagrees with Dennett's viewpoint where the universe is objectively describable as a computation
there is evidence for Kant's viewpoint in special relativity. we know that the psychological arrow of time is a mind dependent thing. the universe does not exist like that
who's to say that the universe exists as a manifold either
everything we know is ofc mind dependent in the trivial sense even in Dennett's viewpoint, as u cant know what ur mind cant detect
but this mind dependence can be more extreme than that, as in, even logic and sets can be mind dependent

2 hours later…
12:18 PM
@RyderRude hey Ryder. hope you doing good. do you remember how I arrived at Ron's answer and rabbit holes of dirac and feynman ? I also proved and realized what x + a does :p
but without dirac and feynman
Feynman's logic still helped though :P

12:31 PM
@GiorgiLagidze oh. what did u realise

imagine you got 2 particles. you assume that x1(t) and x2(t) are true paths. now write an action for them(i.e in the action integral, x1(t) and x2(t) are assumed to be true path). What you now calculate by that is action value(scalar) on true paths). Now, imagine that particles had moved on $x1(t)+\epsilon$, and $x2(t) + \epsilon$. assume/imagine, but not in reality. 2 cases could happen(either action on new paths matches the real one or don't). if they don't, you don't care about that scenario.
If the actions match, basically, what it means is L didnâ€™t change as well. $\epsilon(\frac{\partial L}{\partial x_1} + \frac{\partial L}{\partial x_2}) = 0$. Note that, even though in the action, we got true paths inserted(x1(t) and x2(t)), the Euler Lagrange still holds true even for that action, so you go, write euler lagrange for each particle, then sum it up, and right side ends up 0. i.e momentum is conserved.
@RyderRude note that this is what feynman's doing as well. hypothetically imagines if particles had used different path other than original path :P
x+a in the lagrangian basically means that you're doing x(t) + a in the action
you're not transforming a single coordinate/loocation, you assume what if particle had moved on x(t) + a. hope that makes sense
anyways, i lost 2 weeks for fucking dirac :) and still didn't understand the way i wanted to

@GiorgiLagidze r u considering the conservation of total momentum of a two particle system?
i couldnt guess y u r considering two true paths

yes, sorry. it's 2 particles in 1D

oh
yes, the EL does sum up to zero on the rhs

12:46 PM
the logic all adds up and finally makes sense what x+a means.

u can derive this by considering a potential of the form V(x_1-x_2)

yes, but my logic is general
I just don't get why Susskind(when I first watched it) only gave 2 minues explanation on this.

but note that in Feynman's method, we wud draw just one path even if it's a two particle system

yes, that's for simplicity

cuz we r considering paths in configuration slace

12:48 PM
yeah, in reality, you can have 2 paths. that doesn't change the behind the hood logic

@GiorgiLagidze no, the argument is not simplied. it considers paths in a higher dimensional space than physical space

yeah, i don't know what that means, but even if I consider it in physical space, it still works

the state of a two particle system is represeted by a single point (x1,x2). Feynman's proof considers the path traced by this point
and the variation is also of this single path
in general, u can have any number of particles and points of the form (x1,x2....x_n)
Feynman's proof doesnt care about the number of particles cuz it considers the single path

yes, but the main point is he considers x(t) + a as a hypothetical path what if system had moved on that
so all good

yes

12:51 PM
finally closing this chapter :P

ok :)
conservation of momentum is indeed obvious from.the EL eqn itself

yes, i just wanted to understand it from more intuitive point of view

u can also derive conservation of angular momentum from the EL eqn by considering polar co ordinates and rotational invariance of the lagrangian
Noether's theorem is needed for more complicated situations like in field theory
and also for energy conservation

yes. going through all this and moving on to the next topic from saturday(Hamiltonian)

cuz there is no canonical momentum for time cuz time is not an observable
@GiorgiLagidze ok :)

2 hours later…
2:51 PM
Derek just posted about that circle problem

3:08 PM
@PM2Ring u might like it

3:21 PM
it is about two types of definition of a day depending on the two types of rotations that earth makes while in its orbit
i mean u can define differently "what counts as a 360 degree rotation". the other definition givee 366.24 days in a orbit where one day is 23 something hours
this other definition has applications in studying distant stars and satellites and stuff
it is called a Sidereal day

4:08 PM
Is anyone around to discuss the extension of statistical entropy to continuous probability distributions

for a discrete probability distribution its statistical entropy is defined as $-\sum_{i \in \mathbb{N}} p_i \log p_i$. When this definition is extended to continuous probability distributions the statistical entropy is infinite, see this extract from R.Balian Microphysics to Macrophysics: Methods and Applications of Statistical Physics. Volume II
I cant convince myself it should be infinite in the continuous case and not infinite in the discrete (but still infinite) case

@Monty Do you mean you don't understand the math (i.e. how the term is infinite) or you don't understand the attempted explanation after "This can be understood"?
in any case, what exactly is the problem?

the explanation
it seems like in both cases an infinite amount of information is gained

oh, wait - what do you mean by "discrete (but still infinite)" case?
I just don't think your text means the case of discrete-but-infinite when it says that

4:24 PM
A probability distribution which takes discrete values, but an infinite number of them

oh, I understand what the phrase means, I don't understand why it would be relevant in this instance

say we are doing an experiment

you can let all of this take place on a finite interval $[x_0,x_N]$ with your $x_m$ being finitely many points $x_1,\dots,x_{N-1}$ inside of that interval
I'm pretty sure that's the situation your text is talking about, not a case where you have infinitely many $x_m$

and there are an infinite number of possible outcomes, then we are told the outcome of the experiment, in my head it sounds like we have gained an infinite amount of information...

note that each outcome is assigned a finite probability in that experiment. in the continuous case, the individual points r assigned infinitesimal probabilities

4:27 PM
@Monty yes, that's what the text is saying

this is y it's incomparable

note that it talks about a segment, so it's really talking about the finite interval case here

@ACuriousMind no the text is just talking about continuous distributions

for finite $\Delta x_m$, there are not infinitely many outcomes here
but when you send $\Delta x_m\to 0$, you turn the finitely many bins of size $\Delta x_m$ on the segment into infinitely many possible outcomes (the individual points)
you understood the argument perfectly fine, you just didn't understand correctly what the text means by the discrete case (it really means finitely many outcomes)

u cud consider an infinite discrete space but the gain in information upon observing an outcome wud not be infinite becuz each outcome is assigned a finite probability here
the probability function is just a convergent infinite series here

4:30 PM

You are correct that the same idea would show the naive discrete entropy isn't a good idea on discrete-but-infinite cases either, but probability theory doesn't like discrete-but-infinite spaces to begin with (e.g. there's no uniform probability on an infinite discrete space)

@ACuriousMind but the naive entropy is the one used in the discrete-infinite case and it works fine

@Monty well, the expression doesn't diverge
doesn't mean it still reflects the same thing

fair point

actually I don't even think its guaranteed it doesn't diverge
in the countable case you just have the constraint that $\sum_i p_i = 1$. This does not ensure that $\sum_i p_i \mathrm{ln}(p_i)$ always converges

4:33 PM
note that the gain in information is finite in the discrete infintie case becuz each outcome is assigned a finite probability, despite the number of outcomes being infinite
in the continuous case, each point is assigned an infinitesimal probability. this is y the gain in information is incomparable

@Monty and indeed there are counterexamples, see math.stackexchange.com/q/2938622/143136, so the claim that this can be "generalized without difficulties" is not strictly speaking true

24

Consider the independent variables $X_1, X_2 , X_3 \cdots$, where $X_k$ have non overlapping discrete uniform distributions over $2^k$ values: $X_1 \sim U[2\ldotp\ldotp3]$, $X_2\sim U[4\ldotp\ldotp7]$, $X_3 \sim U[8\ldotp\ldotp15]$, etc. The respective entropies are $H(X_k) = k$ bits. Now, we ...

yeah here is more example^

i think it will always diverge in the continuous case. it only sometimes diverges in the discrete case.

yup

@RyderRude What on earth are you talking about, why would it "diverge" in the discrete case? A finite sum cannot "diverge".

4:40 PM
this is because "observing particle at a point" is categorically a different kind of observation than "observing particle in a region". the former has infinitesimal probability

he means countable discrete
I see there is a difference @RyderRude

@Monty i mean countably infinite discrete

but cant justify the effect of that difference

i mean this difference does show that there is always an infinite gain of information in the continuous case

@Monty What do you think there is to justify?

4:43 PM
for the discrete infinite case, the probability of each outcome is finite. so u cud expect finite gains of information
but it turns out, some cases diverge there too
i guess something to do with the nature of those probability distributions that cud b interpreted as infinite information gains after measurement

The reason the discrete-but-infinite case is unintuitive that, as I said, there is no uniform probability on discrete-but-infinite spaces, but all of our intuition is such that when someone says "random" they usually mean "uniformly random"

yes, probbaility theory is awful in those spaces in the first space

The argument from your book is, I think, extremely clear in the case of uniform probability: If you have $N$ bins and uniform probability, then the information you gain from knowing the particle is in a certain bin is finite. If you have $\aleph_1$ bins and uniform probability, the information you gain is infinite
The comparable statement for countably many bins is meaningless to state because there's no uniform probability on them

yes, the book only addresses the finite discrete case

(I don't want to hear any comments about me assuming the continuum hypothesis :P)

4:48 PM
but we shud note that the book does not assume uniform probbaility to make this argument
the argument has to do with infinitesimal probbailities which r absent in the countably infinite discrete case
@ACuriousMind ooh u said aleph _1. i got the reference :P

Note that the argument really does depend on the shape of the probability, even if it is not spelled out explicitly: For instance, if your probability for something to be in a bin/at a point was zero to begin with, you gain no information by learning it's not there.

yeah
i also think that the information gain shud go to zero anyway for learning the particle is not at a point even if the density at that point was non zero @ACuriousMind
this is becuz of "not" in "not at a point"
the probability of point is infinitesimal. the probability of "not at a point" is 1

@RyderRude Monty is clearly just starting to learn about continuous probabilities. I see no point in confusing them further with something their text will no doubt explain in due time.

@ACuriousMind i can handle continuous probabilities

oh, in that case you probably didn't need to be reminded that "probability at a point" doesn't really mean anything, either :P

4:55 PM
yhyh all good

learning that a particle is at a point wud lead to a delta distribution. learning that it's not at a point wud create a discontinuous hole in the continuous probbaility distribution (if the probability was initially non zero there)
both these distributions r not physical

I guess the thing to understand first is how can a countably infinite prob.dist. have finite entropy (If i'm understanding entropy as the amount of information gained when told the 'position of the particle')
i guess it is what you were saying @RyderRude

u r thinking of an infinite information gain as something that happens when we narrow the information down from infinitely many outcomes to one outcome

that there was a non-zero probability of the particle being at that point in the infinite-countable case, so in that sense we havent gained infinite amount of information

but turns out, the above is not always an infinite informatiom gain

5:00 PM
yes i think we have got there

yes. the real reason it's always infinite in the continuous case is becuz u have observed something whose probability was infinitesimal

yup
and then the hard question of why do some infinite-countable probability distributions have finite entropy and some infinite...

so this is a sufficient but not necessary condition for infinite information gain, as demonstrated by ACM's and ur counter examples in the countably discrete case

beyond - the series diverges in some cases but not in others....
@RyderRude agreed

@Monty i guess it's something to do with the nature of some particular distributions in the countably infinite case

5:04 PM
@Monty I wouldn't obsess over stuff that diverges all too much trying to figure out "why". There's a bunch of distributions (even "realistic" ones that we use to describe real-world stuff) where things like the expected value or higher momenta just straight-up don't exist
there isn't a lot of useful "why" to that that I know, either

@ACuriousMind probably a good idea. Thanks
thanks aswell @RyderRude

(I'm thinking of the canonical "bad distribution", the Cauchy distribution)

@ACuriousMind it looks so pretty. cud confuse it for normal distribution

guys, I was trying to get to: $$<x|2> = \frac{1}{\sqrt{2!}}(\frac{1}{\sqrt{2x_0} })^2 (x-x_0^2\frac{d}{dx})^2 <x|0> , x_0 = \sqrt{\frac{\hbar}{m \omega}}$$

5:09 PM
the next few paragraphs in the text are interesting, introducing relative entropy (KL divergence), so I will probably have more questions another time:-)

Just found out there is an approximation known as Villain approximation
Another case where an upper case changes everything

@Mr.Feynman ah, the enemy of Hero's formula

Lmao never throught of it using the Latin name

But when I tried developing the expression $$(1/\sqrt{2})<x|(a^{\dagger})^2|0>$$ I got to: $$\frac{1}{2\sqrt{2}} \frac{1}{x_0^2}[x^2\psi_0(x)+x_0^4\frac{d^2}{dx^2}\psi_0(x)-x_0^2(x\partial_x\psi_0(x)-\partial_x(x\psi_0(x))]$$
which leads to: $$\frac{1}{\sqrt{2}} (\frac{1}{\sqrt{2x_0}})^2(x^2+x_0^4\frac{d^2}{dx^2}+x_0^2)\psi_0(x)$$, if I havent made any mistake

tip: \langle and \rangle are the right symbols for bra-kets, not < and >

5:17 PM
@ACuriousMind you know I usually use them, but its much quicker
why do I get a different result

you haven't really setup the context here - is this supposed to be the QHO?
and what's $x_0$

I'd like to get a general expression for $\langle x|n\rangle$
which should involve Hermite polynomials

sure but what's $\lvert n\rangle$? A number eigenstate of the QHO?

the energy eigenkets
for the simple harmonic oscillator

...of the QHO?
ah, yes

5:19 PM
My bad I completely forgot to mention this hahaha
QHO stands for that, good to know
I was not getting it tbh

it stands for Quantum Harmonic Oscillator

Yes I never encountered it though
I was thinking about quantum hamiltonian theory lol

so where does your $x_0^4$ come from? Isn't it just $a^\dagger = x-x_0 - \mathrm{i}p$?

it comes from this: $$1/(m^2\omega^2) (-i\hbar)^2\partial_x^2\langle x |0\rangle$$
where I used the expression for $\langle x|p^n|\alpha \rangle = (-ih)^n\partial_x^n\langle x|\alpha \rangle$

oh, wait, your $x_0$ is a bunch of annoying constants, not the midpoint of the oscillator, why call it $x_0$ :P

5:25 PM
sakurai says it sets the length scale for the HO
so I use it too

yeah, it's a natural length scale, I just find it confusing to call it $x_0$ because we usually use $x_0$ to denote some kind of "offset" or "initial value" for $x$
but it's more like the natural unit for $x$ here
anyway, not the point

I'd like to point out that I computed the mixed products in the same way $$\langle x |px|0 \rangle = \langle x|p|\beta \rangle, |\beta \rangle = x|x_0\rangle$$

In any case, schematically you have $a^\dagger = x-\mathrm{i}p$, which means $\langle x\vert 2\rangle = \langle x\vert (x-\mathrm{i}p)^2\vert 0\rangle = (x-\partial_x) \langle x\vert (x-\mathrm{i}p)\lvert 0\rangle = (x-\partial_x)^2\langle x\vert 0\rangle$, I'm not sure what you're doing

I used inner products and the properties of the position and momentum operator

there isn't really any expansion of anything necessary here, just know that the operator $x$ pulled through a $\langle x\vert$ becomes the variable $x$ and $p$ becomes $-\partial_x$

5:30 PM
Im not sure what you did in the first equality
second sorry

i.e. $\langle x\vert \hat{x} = x\langle x\vert$ and $\langle x\vert p = -\mathrm{i}\partial_x\langle x\vert$
guess I should've gotten $(x+\partial_x)$ above, but, eh, signs :P

wait, the problem remains, we both should get the same result, shouldnt we?

@ClaudioMenchinelli I don't understand what you tried to do so I don't know if you should get the same result :P

I did what you wrote but then I expanded the $(x-ip)^2$ term
and simply computed the inner products
wait when you write $(x-\partial_x)^2(\cdot)$ do you mean applying the derivative twice

ACM, your original choice of sign is correct. It is $\hat a^\dagger=\frac1{\sqrt2}(\frac{\hat x}{x_0}-ix_0\hat p)$

5:37 PM
@ClaudioMenchinelli I don't understand why you're trying to expand the term
29 mins ago, by Claudio Menchinelli
guys, I was trying to get to: $$<x|2> = \frac{1}{\sqrt{2!}}(\frac{1}{\sqrt{2x_0} })^2 (x-x_0^2\frac{d}{dx})^2 <x|0> , x_0 = \sqrt{\frac{\hbar}{m \omega}}$$
and there's that $(x-\partial_x)^2$ term, clearly unexpanded
if you want to get to something unexpanded, why expand?

because I wanted to check if that formula is correct, and now I'm doubting it, although this last statement sounds incredibly inappropriate coming from the likes of me hahah

I'm willing to bet you're just making a mistake with the (non-)commutativity of the operators somewhere
which is why I didn't try to expand the term because I'm sure I'd make one, too :P

no no no, that's unfair, I paid much attention to it, in fact I obtained: $$<x|xp|0>+<x|px|0>$$
sorry for the rangles and langles again

You might even have computed it correctly; the fact of the matter is that the result that is quoted is the one that did not expand it, so why did you expand it?
Like, maybe if you compute the final form of the wavefunction, you would actually realise that you got everything correct.

to check its correctness, I'm not sure how it does not get any second derivative of $\psi_0(x)$
like there must be a second derivative because of $p^2$, am I not right?

5:44 PM
@ClaudioMenchinelli ...if you look at this, $(x-\partial_x)^2 = x^2 + \partial_x^2 + [x,\partial_x] = x^2 + \partial_x^2 + 1$, so your expression is the same (if you put $x_0=1$ like me because you don't want to write constants)

$(x-x_0^2\frac{\mathrm d\ }{\mathrm dx})^2$ has a 2nd derivative

now that's shameful of me
the moral is: I must not underestimate commutators

did you?
I thought we all agreed that you did it correctly?

@ACuriousMind you can completely erase the last 50 messages from the record. I did not compute that binomial correctly hahahha

@ACuriousMind but you're a projective being

5:48 PM
I was thinking about $(\partial_x (\cdot))^2$

@Mr.Feynman I am?

@Mr.Feynman what you mean by projective?

Nov 14 at 10:05, by ACuriousMind
everything I say is modulo signs ;P (you're right, of course)
Then that "sign" was $i$, so I concluded ACM is projective
@ClaudioMenchinelli basically modulo a phase

yep now I get it
guys, I have a theoretical question, which arises from the fact that I don't have enought time to study all Cohen-Tann. chapters

ACM is now looking up worldwide literature to find an instance where "projective" is used as an offense to have a shot at banning me :P

5:53 PM
no way man, ACM is our benevolent overlord.

@Mr.Feynman now you're projecting what you would do onto me :P

the book says: If I have an arbitrary even observable $B_+$, with an eigenvector $|\phi_b\rangle$ and eigenvalue $b$