Chapter title: Direct experimental evidence for string theory

Complete contents: There is no direct experimental evidence for string theory.

there's plenty of direct evidence for string theory since string theory reduces to QFT and GR in some limit

If we have a quantum mechanical system and we consider it's Hilbert space. The basis kets are joint eigenstates of a complete set of commuting observables (Hamilton operator included). Then an arbitrary state $\Psi$ is given as $|\Psi\rangle=\sum_n c_n|\phi_n\rangle$. The prob. of measuring an eigenvalue $P(a_n)=|c_n|^2$

Now if the system occupies a state $|\Psi\rangle$ and we want to make a measurement regarding a physical quantity whose corresponding operator does not commute with the H operator and consequently is not part of the complete set of commuting observables, what would be the result of the measurement

would we be able to measure something ? If yes, with what probability ?

@imbAF The operator doesn't have to commute with $H$ to be measured, unless you also want to simultaneously measure also energy. The outcome of your measurement will be an eigenvalue of the operator with probability the squared modulus of the component along the corresponding eigenstate

the component along the corresponding eigenstate

you mean $|\langle u_n|????\rangle|^2$

If ???? is the state of your system and $u_n$ the eigenstate, yes

But the state of the system

is expressed as a linear combination of the joint eigenstates of the complete set of observables

and the physical quantity I want to measure, it's corresponding operator/observable does not belong here

se basically we make the projection of the state $\Psi$

in the basis of the Operator, whose physical quantity we want to measure and that does not commute with the H operator?

You want to express it in the eigenbasis of the observable you are trying to measure

@imbAF Yes

is there a logic here?

that we do this?

meaning we consider the inner product of the eigenstate of the observable, that corresponds to the measured eigenvalue, with that of the state of the system?

These are just the postulates of QM

Is this one?

I know the 7 postulates, but in none I have seen the following

See postulate II.b here

Together with II.a

@Mr.Feynman I will but I do have some questions, general questions. because I am having an exam this Monday, and perhaps you could help me with some clarifications, like this one right now

I'm not available right now, though. I'm about to leave

TT

Sorry for that

it's ok

@user726941 For a moment thought it's an April fools thing :D

If the set of eigenvectors of an operator A, can be considered as a basis of the Hilbert space of a system, is it possible to express an eigenstate of an operator B that does not commute with A, as a linear combination of it's eigenstates?

@imbAF I think so yeah but it may be infinite

But an eigenstate of the x component of Spin operator, is finite, expressed as a linear combination of the eigenstates of the S_z component operator

yeah when you're working in a finite dimensional space like that

position/momentum don't commute, leads to an infinite integral over the eigenstates of each other right?

Hello, i am wondering why, in the definition of charts, it is necessary that the subset and image of this subset are open? i see that this has implications for how we can cover surfaces with charts (e.g. $S^1$ and $S^2$), but i dont see why this complication is necessary?

@Relativisticcucumber It can't be Hausdorff if it ain't open

i.e. there is no $\epsilon$ sized neighborhood around the points at the edge of a closed set

@Amit so is it true that we cannot have a manifold with a boundary? i am looking at a picture where there is an oval shaped set M, and it explains that an atlas needs to meet two criteria, one of which is that the union of the open subsets of M is equal to M itself, but i am wondering how we can ever catch the boundaries of M?

You can, but I'm not sure it will be smooth / differentiable

Need to look into that

I think it may be possible to get around that by distinguishing $\partial{M}$ - the boundary points of the manifold, and $\text{int}(M)$ which is the interior

The charts of a manifold are charts of open sets of the manifold

as they are diffeomorphic they're gonna have to be mapped to open sets

@Relativisticcucumber Ah I think I see what's going on: I think we need to recall that as a topological space there's no problem having also closed sets in addition to the open ones. But it's the closed one which are allowed to be indeed, closed :) Then we can say our manifold has a boundary although it's not in the open sets of the topology

@Slereah yes, but then how does their union cover all of M -- does M not contain the boundary?

Manifolds typically have an empty boundary

although you can extend the definition to include manifolds with boundaries

Where the charts are diffeomorphic to subsets of $\mathbb{R}^n_{x \geq 0}$

the half space

okay i think i see

Can you explain it to me? Lol, you asked a good question about how the union of the open sets cover $M$ and I don't know the answer

@Amit well i do not understand the extension to manifold with a boundary yet, but i think that what @Slereah is saying maybe is that usually these manifolds to not have a boundary? in these cases that i am looking at, i think it's like that -- which is $S^1$ and $S^2$, and i just looked in my textbook and it says generally we deal with manifolds without a boundary, so this would not be a problem in those cases. so i guess i still dont have a resolution for when we do have a boundary lol

Plenty of manifolds have boundaries

oh no

But spacetimes are usually considered boundaryless

@Slereah okay maybe it's because this is a gr book so in this book the manifolds will be mostly boundaryless?

likely yes

excellent

@Slereah how does that business with demanding $x \geq 0$ help? is it that the boundary sets already contained in a larger open sets somehow?

It's just a version of $\mathbb{R}^n$ with a boundary?

Oh I see, and that doesn't force us to also put closed sets in the topology of $M$ right? So the boundary is only a consequence of this type of charts, i.e. the image set of $x=0$?

Some subsets of the half space have boundaries and some of them don't

Basically if they contain points on x = 0

Yeah... so it seems to me like the fact we can have overlap of charts is what saves us there

I start thinking about diffeomorphisms between stuff like $(0,1)$ and $[0,1]$ and go crazy. Never mind, it's probably not as complicated as I think :)

you do have some constructions of spacetimes with boundaries for ie idealized points

like singularities

naw naw it's april fools space time has nawwww boundary!!! this is flat earthers diffeomorphic! a spacetime with boundary, lol

i also have a follow up -- it says in this textbook "a chart, or coordinate system", and i am wondering if this is the usual coordinate system definition that we see? because i dont see why we wouldnt expect the canonical coordinate systems to be valid on an entire manifold

@Relativisticcucumber you may have seen that you can't cover the entire $S^2$ with a single chart for example

For any kind of non-trivial and non-flat space you'll need more than one chart, that's why all the GR books go into the trouble of talking about it, and compatibility condition between charts, etc.

yes in some cases i see that but i dont see why in principle you need a different coordinate system to cover a manifold even though i understand why you need separate charts

so i guess im trying to modify my understanding of coordinate system to match this definition of charts

oh, yes a coordinate system is definitely determined by a chart

for flat space for instance, we can use spherical or cartesian or whatever, and these systems all cover the entire manifold to my understanding

it's a way to coordinatize $\mathbb{R}^n$ ($n=4$ in GR) so as to correspond to points in "the real world"

hmm, no the spherical actually I think doesn't do it diffeomorphically, otherwise you could cover $S^2$ with one chart :)

but for a flat space in general yes you can use a single chart. if your manifold is itself just $\mathbb{R^n}$ then a single chart will do, some variation on the cartesian one (you can use oblique axes too)

however there's a difference between the manifold being $\mathbb{R}^n$ and the manifold looking locally as $\mathbb{R}^n$, for a smooth manifold we only have the latter

@Relativisticcucumber The modification is that in non-GR physics we always assume that our manifold is either $\mathbb{R}^3$ or $\mathbb{R}^4$ and the choice of coordinate system is just to exploit some symmetry of the problem, but the problem is still assumed to be "embedded" (not technical term here) in this 3 or 4 dimensional flat space. But in smooth manifolds and GR land, we can't assume that our manifold is globally like $\mathbb{R}^4$, only locally

So we look for charts that map patches of a space that locally looks like $\mathbb{R}^4$ to actual $\mathbb{R}^4$

But I guess it won't be completely clear until you get to expressing the metric tensor, which is also done via a specific chart or another

i have one more follow up -- if i can find an atlas for a set, is this enough to prove this set is a manifold? because the definition of manifold im reading says that it needs to be a set with a maximal atlas, but i think it is impossible to write down a maximal atlas in many cases

No, you also have to have the gluing condition

the whole issue of what happens for two overlapping charts

I'm studying some cool rotational dynamics stuff, some cool examples and so forth, I wanna know is this what engineers do on a daily basis?

Otherwise you just have a smooth set, a space that can be probed by charts to Rn

Like consider the shape of X

actually I think the definition of a "smooth atlas" contains the condition of chart compatibility

You can define charts from R to X

but it's not a manifold

@Amit I mean this : ncatlab.org/nlab/show/smooth+set

but it's different to define charts than to define an atlas, right? then with an atlas we know the charts overlap fine? @Slereah

ahhhh yes yes I see, sorry @Slereah, yes just this smooth structure doesn't mean it's a manifold, a smooth manifold is a manifold first @Relativisticcucumber :)

depends how you define an atlas

Sometimes it's just a collection of charts

@Slereah in my case it says that an atlas is a collection of charts that satisfies two conditions. one is about the union covering the set, and the second is about the charts being smoothly sewn together

Then yes

wait yes to what XD

A set of charts isn't the same as an atlas

You have varying levels of structures you can assign to a space

@Slereah okay so even in the case that i said it's still not sufficient proof of a manifold to find an atlas?

Set > topological space > smooth set > diffeological space > topological manifold > smooth manifold

they have different conditions for belonging

Well as I said, a good example is the shape of X

You can find charts to R that cover it completely

But that's not the shape of my heart

But you will have less good luck showing their compatibility

@Slereah but in this case they are not an atlas i thought?

It ain't Hausdorff either

not if you require the compatibility condition

@Amit It is very much Hausdorff!

X? but this point in the middle...

Hausdorff means that for two points on the topological space, there exists open sets that do not overlap

@Slereah yes so if i require this condition and i find an atlas that does satisfy compatibility then this is the case i am wondering if this proves this set is a manifold

that's true for X

Here's a quick check for hausdorffness : if your space can be continuously mapped into $\mathbb{R}^n$, it's gonna be Hausdorff :p

I thought that this branching thing violates hausdorff

No you are thinking of a different thing

Yeah I see that Hausdorff is okay with it now that I think about it

What was I thinking about?

Branching isn't restricted to non-Hausdorff spaces, but it's true that for manifold, they're the only examples that do

But X isn't a manifold

Oh, so I'm just manifold-centric

Oh I see, is it that if we take an infinite intersection of neighborhoods containing that midpoint, we get a set containing only a single point which isn't open?

yes

There isn't gonna be a transition map

not one that's a diffeomorphism certainly

There's no diffeomorphism from a point to a line

I see, thanks!

How is rewriting GR going?

slowly since I am still down with the sickness

Oh =[ Speedy recovery to you

Can anyone help me with the correct way of writing the following: Let's consider non degenerate discrete spectrum for the sake of simplicity. The prob. of measuring the eigenvalue $a_n$ of the Observable $A$ is: $P(a_n)=|\langle \phi_n|\psi\rangle |$. Here I am assuming a pure state, which can either be represented via the density matrix or the ket $|\Psi \rangle$. What if the system is in a mixed state, what do I write instead of $|\Psi \rangle$ at $P(a_n)=|\langle \phi_n|\psi\rangle |$?

I think you're missing a square

Yeah I am

adding the square

how do i find what I asked for?

which is the prob. of measuring an eigenvalue when the ssytem is in a mixed state

It's something to do with the trace of the density matrix but I don't remember exactly

If I consider a discrete non degenerate spectrum for when the system is in a pure state

I can easily from $P(a_n)=|\langle \phi_n|\psi\rangle |^2$ reach to $P(a_n)=\langle \phi_n|\rho|\phi_n\rangle$

so even when the system is in a mixed state, the same expression is valid

(seen in wikipedia)

But what if I wanted to derive this expression $P(a_n)=\langle \phi_n|\rho|\phi_n\rangle$ when the system was in a mixed state

this is what I can't do

I can easily make the following claim. If I can find the prob. with the following formula $P(a_n)=\langle \phi_n|\rho|\phi_n\rangle$, where $\rho$ is the density operator of the pure state, than when the system is in a mixed state, $\rho$ is the density matrix of the mixed state

But I started with a pure state, and I made a claim for when the system is in a mixed state. How can I do the opposite ?

Look if 16.1 here helps people.eecs.berkeley.edu/~vazirani/f04quantum/notes/mixed.pdf

It doesn't really

I thought that theorem is exactly what you're trying to prove

It does not because

The idea is not how to express the density operator for a mixed state

but rather

Starting with this: $P(a_n)=...$ arriving at this $P(a_n)=\langle \phi_n|\rho|\phi_n\rangle$ while utilizing the expression for the mixed state, which is via the density operator. Because as I said I can reach the above expression when I claim that the system is in a pure state, but not when it's in a mixed state

The theorem assumes a mixed state

$$\langle a_n | \rho | a_n \rangle$$

and how do you arrive at this conclusion

?

That's the last line of the theorem's proof!

They start by just expressing the probability as a probability weighted sum of each state

in the mixture

in the link you gave me

what is this

: Suppose we measure a mixed state $\{pj,|ψj\rangle\}$

so it's $$ \rho = \sum_j p_j|\psi_j\rangle\langle\psi_j| $$ right?

so now calculate what is $$\langle a_n | \rho | a_n \rangle$$

using the above definition of $\rho$

correct, but the $|\psi_j\rangle$ can be expressed as a linear combination of some basis kets, correct?

now I'm assuming they are already the basis you want to work with

they can be that too

If you have a mixed state in a certain basis why do you need to change it

Is it relevant to your problem

no one is changing anything

In a pure state $|\psi\rangle$, the state can be expressed as $|\psi\rangle=\sum_n c_n|\phi_n\rangle$

the same logic can be applied to the |\psi_j\rangle

if you do the above calculation you'll see you're just picking up the probability to measure $a_n$ in each basis state of the mixture and summing them

which is basically what the proof does

ok

the mixed state is a linear combination of density matrices, the pure one is a single density matrix

@imbAF so it's like taking what you've written here and giving this entire thing a probability, and adding another combination with say 1 minus this probability. So you see it is not expressible as such a single sum

Nice example of a mixed qubit state

More info here: qiskit.org/textbook/ch-quantum-hardware/…

@Amit can't you make the following claim regarding the pure states in the density matrix of a mixed state: $$ \rho = \sum_j p_j|\psi_j\rangle\langle\psi_j| $$ where $$$|\psi_j\rangle=\sum_n c_{n(j)}|u_n\rangle$$$ where $|u_n\rangle$ are the eigenstates of the observable for which we are making a measurement

Sure you can express the mixture states in some basis

Actually in the example I just linked to it's done

Unfortunately I don't get it

What's bugging you :)

I'll write it down

First of all let's consider the cases

Maybe you need to note that the states that represent a mixture are not necessarily eigenstates of any observable

@Amit I always consider that

Idk about spectrums, im barely able to recall enough about mixtures to be helpful lol

these are all the combinations for which i want to find the prob. of measuring an eigenvalue

well

Non-degenerate discreate spectrum = observable with discreate non degenerate eigenvalues

that's what it means

Now, I will derive something, and you might understand what I am trying to do

So why would it affect the probability calculation

Maybe it does I just dont recall

it would

if the spectrum is discrete but degenerate

than

$P(a_n)=\sum_i^{g_n}|c_n^i|^2$

At most it can change sums to integrals?

Okay maybe, I think you need someone who's more brushed up on this

@Amit are you there?

There as part of a mixed state lol

Like this? @imbAF

Whether the spectrum is degenerate or not doesn't change anything other than your projection operator will now be a sum of the projectors with the same measurement outcome

In more general terms: The question is: "Given an initial state (pure or mixed) $\hat{\rho}$ and observable $\hat{O}$, how do I determine the probability of measuring $o$ (an eigenvalue of $\hat{O}$) upon measurement of $\hat{O}$?"

Answer: Take the expectation value of the projection operator along the corresponding subspace: $\langle \sum_i \hat{P}_i \rangle$ with respect to $\hat{\rho}$.

Where the projection operator along the corresponding subspace is the sum of all of the projection operators of the eigenstates of $\hat{O}$ with eigenvalue $o$.

In the density matrix formalism, expectations of an operator $\hat{A}$ wrt to a state $\hat{\rho}$ are computed in a simple way: Tr$(\hat{\rho}\hat{A})$.

This trace operation is computed with respect to a certain basis of your hilbert space, but its value does not depend on a choice of basis. Hence, I choose a particularly convenient basis for my calculation above.

cf. Schlosshauer's book on Decoherence. In chapter 2, the density matrix formalism is explained. It begins by motivating their need, defines the trace operation, then talks about measures of mixedness, then talks about density matrices for composite hilbert spaces (composed of multiple subsystems)

I have fun with index free notation and all, but I will usually draw the line at never writing functions with their parameters in

Some papers will just use a pullback instead

Madness

Squack! It's a quantum mechanical duck!

@Slereah Can a pullback replace function arguments?

a possibly silly question: the non-covariant Riemann tensor $R^\alpha_{\nu\rho\sigma}$ as far as I understand has the same number of independent components as the covariant one $R_{\mu\nu\rho\sigma}$ ($20$ for example in $4$ dimensions). But the full set of symmetries are always shown via the covariant version.. is it because they are somehow "easier" to find that way? So, the non-covariant one also has them but the relation between dependent components is more complex?

what are some criticisms of the decoherence program?

also thanks to @Relativisticcucumber I know what a manifold is (I think) >:D

problem you mean

question mark

oh i am using program to refer to the fact that it seems like there is a select group of people who believe in decoherence as a good model of the quantum to classical transition and who are trying to push it further

oh i see

i.e. it doesn't seem like a generally accepted thing :0 but im not sure

I think the idea of decoherence is not very controversial because it seems that both the MWI and Copenhagen camps can find a way to incorporate it into their interpretation :)

speaking of camps, is anyone aware of Stochastic electrodynamics (SED)?

it seems to use terms that i associated with the "crazies", but apparently some of it's results are legit?

I always root for the crazies, if they only make crazy assumptions and follow through on them logically and sanely lol

well thats the thing, crazies who are rigorous and methodical i have a soft spot for

oh yeah, and are graceful in defeat

true

Yeah, we need those

hahah

(Looks interesting but never seen it before)

honestly I don't know enough to evaluate it fairly. i'm trying to round off my classical education before properly embarking into QM

There are so many fringe theories out there... yeah I wouldn't go into something like that without a very good reason to believe there's something to it apriori

yes

if someone can a) achieve sounds results, and b) acknowledge it's just an alternative perspective to (rather than a replacement for), the accepted methods/models

i have alot more time for them

but unfortunately their terms also seem to readily appear in the "free energy" whackjob crowd which makes me rather cautious

The Wiki article is very short on any technical details

never a good sign is it, heheh

:) It's surely a sign you have to dig in published papers to find anything

yeah i had a quick look at the papers, authors and journals. didn't seem terrible tbh

erm *journals the papers were in, haven't seen any journals dedicated to it [probably a good thing at this stage]

I'll wait for the anti matter hyperdrive before taking a look

lol

ahahah

true ;)

peace and good night

night mate