Hi, @JackRod. Yes, I still have back pain. Are you still working on improving your English?

@RyderRude You might enjoy some of Greg Egan's stories involving MWI themes, eg Quarantine.

Greg isn't merely a scifi author with a pop-sci understanding of QM. His stories are always hard scifi, although he does occasionally bend or modify the laws of physics that we know for narrative purposes. He has some physics articles aimed at the general reader, but he doesn't shy away from including maths in those articles.

Some of his stories feature the Qusp a quantum AI processor.

> By isolating decision making from the outside world on a quantum level, bearers of Qusps only ever make one choice, rather than leaving copies of themselves in the future quantum multiverse (of that decision in particular) to make others.

> That way, for every choice a Qusp posthuman makes, they do not doom versions of themselves to making worse choices - but they also can't take comfort that a different version of themselves might have made a better choice, in turn. In consequence, decisions matter to Qusp posthumans far more even than for their human ancestors.

The first Qusp story is Singleton. I should warn you that it has some dark moments...

How do I generically physically interpret the fact that two hamiltonians over the same hilbert space are isospectral

To back it up a little, let's say we have a hamiltonian $\hat{H}$. Generically, it seems that hamiltonians which are isospectral to $\hat{H}$ are of the form $U H U^{-1}$ where $U$ is a unitary operator over the same hilbert space.

(there seem to be some notion of duality in that two hamiltonians can actually be isospectral without being connected via unitary conjugation)

Okay... well if we accept this language of unitary conjugation, then my natural interpretation is that a hamiltonian isospectral to $\hat{H}$ is "the same hamiltonian" written with respect to a different basis

But I do not get what it physically means to change the basis in this way, especially since $U$ is an arbitrary unitary transformation.

Moreover, say we start with $\hat{H}$ and then "transform to a new basis" $U\hat{H}U^{-1} = \hat{H}'$. What if we do not transform any of our vectors $v \in \mathcal{H}$ into the new basis? How do I interpret then $\hat{H}' \cdot v$?

I think we cannot perform that operation as you cannot mix bases.

@SillyGoose ur stack messages remind me of my favorite childhood book

its called martha blah blah

about a dog that eats soup in order to speak

then she loses her ability to speak

and after a long journey they recover the alphabet soup for her

in the interim she utters nonsense streams of consciousness, hence the name "martha blah blah"

LOL

@Relativisticcucumber honk

once more i have TPS questions XD @ACuriousMind

is this why you said that implicitly we define that isomorphism

"A Hilbert space $\otimes_i \mathcal{H}_i$" => we deal with this Hilbert space and this space only. Hence, if your Hamiltonian with respect to a TPS $\mathcal{T}$ does not end up being an operator over this Hilbert space $\otimes_i \mathcal{H}_i$ you must define an isomorphism such that the operator is made to be defined over that space

@SillyGoose Isospectral Hamiltonians connected via unitary transformations are just the same physical system viewed from different bases

Those not connected via unitary transformations - i.e. the "dual" ones - genuinely describe two at first glance different physical systems

@SillyGoose I guess so?

@PM2Ring This is very interesting. Whenever I make a godawful choice, I reason that was gonna happen anyway

I will check out this story. Thanks.

So it looks like mass renormalization in rel systems turns into the renormalization of the energy dispersion relation in non-rel systems and we also have wavefunction renormalization and I'm guessing also the renormalization of the coupling constant (however I haven't got that far yet)

And the position of the pole in the denominator gives you the renormalized dispersion relation

I am finding natural numbers EXTREMELY mysterious now

On one hand, you just know what they are, and there is no scope for ambiguity. On the other hand, any attempt to rigorously define them leads to undecidable statements

It's like you just know them as sure as you know yourself but you also don't know infinitely many things about them

Or if it even makes sense to speak of "them". If " They" even exist in a unique sense

And any attempt to mathematically define them does not pin down a unique mathematical structure. And yet, we just know that they are 1,2,3,4,5....

Hey! What about 0?! :P

Primary school told me 0 isn't natural

There is no general agreement on that, I was trying to spread discord on stupid conventions :P

For example, we just know that elements of this 1,2,3,4,5... set are either even or odd. But in robinson's arithmetic, it's an undecidable statemenr

And we accept that the 1,2,3,4,5 .... that we know satisfy Goodstein's theorem. But it's an undecidable statement in Peano arithmetic

So when we talk about the set "1, 2,3,4,5...." what mathematical structure do we even mean?

Even in ZFC, there must be something like Goodstein's theorm that's still left undecided

So how can we know which extension of ZFC is "right" for further pinning down the "1, 2,3,4,5..." that we know

Or if the "1, 2,3,4,5.." that we know are even a unique mathematical structure. Maybe our intuition is ambiguous itself

And how do we know that we picked the correct extension of Peano arithmetic when we proved Goodstein's theorem? Maybe a counter-example natural number of Goodstein's theorem can be found if we have somehow picked the "wrong" extension

@Mr.Feynman u have the more correct definition. Zero is natural according to Peano arithmetic :)

But it's still a convention i guess

Even in the same course I had to change convention :P

@RyderRude For example, in Real Analysis natural numbers $\mathbb{N}$ were defined without the zero element and the set including also zero was called $\mathbb{N}_0$. In the context of power series it was more useful to consider $\mathbb{N}_0$ as natural numbers

In rel physics, are we looking to see where the pole is and declaring this the physical mass $M_p$. Then by this, we have the condition that $\Sigma(P^2=M_p)=0$? But then how do we know that $\Sigma(P^2=M_p)=0$ is the case when we can't even calculate it?

But then if the above is the case the pole will be at the bare mass $M_B$ since in the denominator, we will require $M_P - M_B - \Sigma(P^2=M_P)=0$ meaning $M=M_B$

@DIRAC1930 Why would this imply $\Sigma(M_p) = 0$?

Apparently thats one of the renormalization conditions

Theres a second way of doing things where $M_P^2 = M_B^2 + \mathrm{Re} \Sigma(P^2=M_P^2)$

what are "the renormalization conditions"

Its in Peskin and Schroder IIRC

Also in L&L

I'm not asking this question because I don't know how renormalization works, I want to figure out where you think this condition comes from

because in general you can pick any renormalization conditions you want, they are not exactly unique

I don't have either of these books here right now but I find it hard to believe they just state that condition without explaining how it relates to solving the problem with the divergence of $\Sigma(p)$

I have no idea what's going on. My understanding is that the self energy shifts the pole of the propagator and the point at which there is a pole is the physical mass.

yes, this gives you an equation like $M_p - M_b = \Sigma(M_p)$ to first order

now, the problem is that $\Sigma(p)$ is divergent, so you need to do something to rescue the theory

because right now you have a difference between two alleged constants that diverges to $\infty$, which is bad

Do I just throw the divergent part of the self energy onto $M_B$ and the finite part onto $M_P$?

not exactly, but the idea is indeed to absorb the divergence into $M_b$

Just from complex analysis, we also have that at the pole $1- \partial \Sigma/\partial p^2=0$ I think

We regularize with some energy cutoff $\Lambda$ so that $\Sigma_\Lambda(M_p)$ remains finite at finite $\Lambda$ and then we say that that equation up there is actually the definition of $M_b(\Lambda)$ in terms of $M_p$ and $\Sigma(M_p)$

@DIRAC1930 that matters for figuring out the wavefunction renormalization $Z$ but not for the mass renormalization

$M_P - M_B(\Lambda) = \Sigma_\Lambda (M_P)$. This one?

Anyway, now we have $M_p = \Sigma_\Lambda(M_p) - M_b(\Lambda)$ and this remains actually finite as $\Lambda\to\infty$

but it just gives us $M_p = M_p$ when you do that limit

i.e. we have no predictive power for $M_p$ in this case, the value of $M_p$ becomes an input to the theory

How do we know that value remains finite in that limit?

I mean you can just plug it in

but we did a circular definition here where $M_b(\Lambda)$ is defined in terms of $\Sigma_\Lambda(M_p))$

the two divergences are just the same and cancel each other if you subtract them

I'm confused what to plug in. $M_B$ is a number from the Lagrangian. How do I get it to be dependent on $\Lambda$

anyway, the condition $\Sigma(M_p) = 0$ is what the statement "The free parameter $M_p$ should be the physical mass" means in the renormalized perturbation theory with the counterterms etc.

the whole point is that it can't stay a number, because the difference of two numbers can't be divergent

this should really all be explained in the books you're reading, but if you're looking for another account of renormalization, I'm partial to the explanation for QED's renormalization in chapter 5 of Weigand's QFT notes [pdf link]

(I'm partial to it because that was the QFT course I learned QFT in :P)

Yes I'm just getting confused by them using the propagators of QED as an example instead of something more simple and also the fact that these terms aren't divergent in cond matter

I like Schwarze's book for renormalization

So we make the claim that observable quantities are independent of the regulator. This kind of forces you to make the bare mass dependent on the regulator leading to $M_P = \Sigma_\Lambda( M_P) - M_B(\Lambda)$.

Schwartz?

Schwarz?

If I do the calculation, I will find that $M_P$ is independent of the regulator and I should just get a number out

@DIRAC1930 well, the thing is from the way you defined $M_B(\Lambda)$ you just get $M_P = M_P$

which is true, but useless, and the reason the physical mass of particles is an input parameter in QFT, not a prediction

I'm confused what I can actually predict then lol

I could have guessed $M_P=M_P$ without using any qft lol

I need to read into the fine structure constant

@DIRAC1930 everything else, in hep-th contexts mostly scattering amplitudes

Is this statement correct chat.stackexchange.com/transcript/message/63162293#63162293

So I can just throw away all terms in $\Sigma_\Lambda (M_P)$ that are dependent on the regulator?

Or equivalently, you can perhaps use the probability of some scattering process as the input and predict the masses

But im not sure

It's, in principle, not much different from fixing the value of G using experiments

U need one experiment to fix the value, and u predict every othr experiment

In general, for n free parameters, u need n experiments i guess

Cuz u need n equations

But isn't the coupling constant a free parameter?

The coupling constant has 2 b fixed using experiments @DIRAC1930

G is sort of a coupling constant for Newtonian gravity. U fix it using one experiment

Idk y people criticize renormalization for "fixing the theory" using experiments

We've been doing that since Newton

It's nothing controversial

I guess people found fixing the infinities part controversial

In layman intros to renormalization, u always read : "the original theory gives infinities. We use experimental input to fix it"

But there's no "original theory"

Using the experiments has always been the standard procedure to arrive at the theory

The confusion is that a free theory has a mass equal to the one you put it. Then just adding a tiny perturbation, that mass does something really weird instead of just correcting it slightly

it has some nice explanations like "at any energy level, the interactions provide an infinite correction to the bare mass nd charge"

Becuz of particle dressings and stuff

This may be confusing. But renormalization is perfectly sound

It does not deserve a reputation of being an ad-hoc fix

Even many working physicists still consider it ad-hoc

I'm confused about why divergent integrals are always mentioned when I have seen in texts that e.g. Weinberg that renormalization has nothing to do with infinities

@DIRAC1930 It's not just about the tiny perturbation term. also consider that we are considering the correction to the bare quantities OVER an infinite difference in energy levels. The bare quantities correspond to lattice spacing equals 0, or infinite energy. But our experiments have a finite energy level

This is y the correction from the interaction "dressing" is infinite

@DIRAC1930 many people define renormalization as connecting the parameters of the lagrangian to actual measurables

There is phySE about this. I will get it

This view of renormalization has nothing to do with infinities

It's regularization that has to do with infinities

if u consider QFT as defined on a lattice, then renormalization has nothing to do with infinites. The correction term from the interaction dressings is finite then @DIRAC1930

So the bare quantities are also finite

Renormalization merely fixes them according to experiments

I guess there seems nothing weird about making sure $M_B - \Sigma(P_M)$ is finite. Just to make sure I've understood correctly does this mean that the mass term in the Lagrangian will be $M_B(\Lambda)$ i.e. the Lagrangian will be dependent on the cutoff (or other regulator used)

I'm also pretty confused about this. This means the classical field limit usinf Ehrenfest's theorem becomes very weird becuz of the weird lagrangian

But yeah, the parameters in the original lagrangian are supposed to be infinite. We split it up into two in the "counterterm renormalization"

About the classical field limit, i just consider that we are doing lattice qft. This makes it not weird

Also consider that classical field theories sometimes need infinite parameters anyway. But that's for field-point particle theories, instead of interacting field theories

So is the initial motivation that the Dyson equations are invariant under the multiplication of those renormalization parameters

$Z_1$ etc.

Actually ignore that

I am most likely wrong but from what I can tell, the situation in hep-th is almost the reverse to the situation in cond-mat. In hep-th, we have no idea what the bare mass etc is, and can only measure the physical mass etc. In cond mat, we know the bare parameters (they are the physical mass etc. outside of the particular condensed system) and therefore the corrections from the self energy etc. are actually the modified effective parameters that you would measure in a particular cond mat system

Actually I dont know

I've got a small question. Let $A: H_1 \rightarrow H_1$ and $B: H_2 \rightarrow H_2$. Then, consider the space $H = H_1 \otimes H_2$. We can now talk about the operator $A \otimes B: H \rightarrow H$. So, without any choice of basis, $A \otimes B$ is as explicit we can write this operator, right? However, once we choose a basis, we can represent $A \otimes B$ as a dim$H_1$ * dim$H_2$ by dim$H_1$ * dim$H_2$ matrix?

@SillyGoose sure, just like you can write any operator on $H_1\otimes H_2$ as such a matrix - there's nothing special to $A\otimes B$ about this

if you choose bases of $H_1$ and $H_2$, the tensor product $A\otimes B$ is given in terms of matrices in those bases (and the induced basis of the tensor product) by the Kronecker product

By "if you choose bases of H_1 and H_2", does that mean let $\{\psi_i\}$ be a basis for H_1 and $\{\phi_i\}$ be a basis for H_2, then we choose the basis $\{\psi_i \otimes \phi_j\}$ for $H_1 \otimes H_2$?

@SillyGoose yes

ah

Why isn't equation 10.3.19 on pg 441 of Weinbergs QFT book (Vol 1) not $Z \delta m^2 = - \Pi^*_{\mathrm{LOOP}}(-m^2)$ instead of $Z \delta m^2 = - \Pi^*_{\mathrm{LOOP}}(0)$ which is what is written

my jasmine tea came in :D

I thought those were insects

they start as rolled up balls and then unfurl when brewed

i found a seemingly quite nice playlist on density matrices and using a python package to nicely perform some related manipulations :0: youtube.com/playlist?list=PLhI5X1mNN8ghnH1ckAcclxY6-CPmZvWrk

I think it may be a typo because his similar expression for a Dirac field (eqn 10.3.32) is similar to the one I had in my last message

So to connect this all to the Landau Fermi Liquid theory, there must be some sort of renormalization conditions that are physically motivated

I have seen for $\Sigma(E,p)$, $\Sigma(0,0)=0$ and $\Sigma'(0,0)=0$ (which I assume to be something physically motivated to do with the Fermi surface) however I'm not sure if these are renorm conditions or something else

I think maybe the above is the condition that the Fermi surface stays at the same place when interactions are turned on