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02:31
so if $A$ is the matrix representation of some linear operator in the basis $\{\phi_i\}$ and $B$ is the change of basis matrix from $\{\phi_i\}$ to a different basis $\{\psi_i\}$, is the matrix representation of $A$ in $\{\psi_i\}$ just $B^\dagger AB$?
02:46
or is it inverse rather than adjoint
i think it might be inverse. idk
@Relativisticcucumber hahaha awww!
@Relativisticcucumber lol unless you're in a primary school playground, then "four fingers and a thumb!" was a common "gotcha" refrain xD
03:37
no wait it actually is adjoint
i think
!!!
04:09
user
Hey.
You doing linear algebra?
yurp!
I thought the transition matrices ought have inverses in the change of basis formula.
If the operator were hermitian, there would be an orthogonal basis such that the transition matrix be unitary and that operator would be unitarily similar to a diagonal matrix.
04:56
user image
2
@naturallyInconsistent ☝🏻😺
meow
im so confusd
bugt i need a break
@Allie 😬
@Allie get some hot beverage.
I am getting more physics related contents in my YouTube feed now.
Her PhD is developing new nuclear fuel, if that is what she meant.
 
4 hours later…
09:31
Happy new year, everyone! :)
@User1865345 you're infected
@Allie isn't this explained on e.g. Wikpedia?
In mathematics, an ordered basis of a vector space of finite dimension n allows representing uniquely any element of the vector space by a coordinate vector, which is a sequence of n scalars called coordinates. If two different bases are considered, the coordinate vector that represents a vector v on one basis is, in general, different from the coordinate vector that represents v on the other basis. A change of basis consists of converting every assertion expressed in terms of coordinates relative to one basis into an assertion expressed in terms of coordinates relative to the other basis. Such...
Hello, Tobias!
Fear not, no SC today :P
Hallo Herr Feinmann
SC? :d
09:54
aaahh superconductivity haha
sorry, I am tired hehe
10:05
@Allie the german entry is even a bit more detailed, so if you can somehow translate it... de.wikipedia.org/wiki/…
10:36
Indeed
 
2 hours later…
12:43
@User1865345 I don't want to say right away that UN as fuel is a bad idea; but it would have a few serious drawbacks that would make it impossible to use in our current reactors.
It could work in a gas-cooled fast breeder, though.
Ahh. I see. Thanks for highlighting. I have no idea about these stuffs.
@TobiasFünke seems so 😅
@TobiasFünke I do occasionally venture on German wiki articles, because most of the times they provide more perspective than their English counterparts.
Of course, those are auto translated by Google and few errors also appear thus.
12:59
hehe yeah
but luckily you can understand equations and definitions in math more easily than other stuff in any language
Indeed.
But one should know German to the extent that they may read a paper or so.
I personally want to read Mathematische Statistik – ein Einführung in Theorie und Methoden by Hermann Witting.
The book is unique for it has so many materials covered unlike any other English books. And many papers of my interest frequently cite the material.
I don't know why no one bothered to translate the work. (Because secretly they enjoy English speaking people struggling reading German books.)
 
1 hour later…
14:13
@User1865345 haha
15:13
Why isn't equation 2.5.5 on page 64 (page 78 of the pdf) in contradiction with equation 2.5.3 on page 65 (page 77 of the pdf) stat.ucla.edu/~ywu/research/documents/weinberg1.pdf
Why can he choose $C_{\sigma,\sigma'}$ to be $\delta_{\sigma,\sigma'}$
15:43
Is this theorem true?
@SillyGoose physics.stackexchange.com/a/780377/310229 you answered the question lol
@DIRAC1930 :)
@DIRAC1930 this link doesn't work for me
The link is to Weinbergs 1st QFT book
@DIRAC1930 did your question get answered by the stack question/answer?
It is still a bit weird that Weinberg assumes that there exists a certain $L(p)$ that has that property
15:57
I think it is also kind of weird (retrospectively) to assert the existence. Though that might just be my naivety. My speculation is that from physical intuition one already knows that the $U(L(p))$ should be boosts, so that is maybe why their existence is conjectured at this point.
16:07
Page 63 here maybe motivates it archive.org/details/…
I need to go through it properly though
Weinberg seems to be asserting that there is an equivalence between an orbit $\{\Psi_{p, \sigma}: p = \Lambda k, \ \Lambda \in SO^+(1,3) \}$ and the set of all (possibly) pairs of Casimir invariant values $(\eta_{\mu\nu}p^\mu p^\nu, \text{sgn}(p^0))$
@DIRAC1930 Actually, what do you mean by this?
Well isn't he assuming that the state transforms under the trivial representation of whatever $C$ is
I do not believe so. (2.5.3) is (to my eye) a concrete expression expressing a generic representation of $SO^+(1,3)$ on a representation space spanned by (WLOG) energy-momentum eigenstates.
Now looking at it, (2.5.5) is not that weird either. The casimir invariants provide a natural label of $SO^+(1,3)$ orbits. So, pick a representative of each orbit; the $\Psi_{k, \sigma}$; and then we know how to move between momenta; the $U(L(p))$, which exists because we are looking at an orbit, which by definition has any two elements connected by the group action being considered.
If the $\Psi_{p, \sigma}$ transformed under the trivial representation of $SO^+(1,3)$, then $U(\Lambda) \Psi_{p, \sigma} = \Psi_{p, \sigma}$ for all $\Lambda \in SO^+(1,3)$ since the trivial representation is defined by $U(\Lambda) = 1 \ \forall \Lambda \in SO^+(1,3)$
16:26
This is a separate question, but how does QFT resolve the issue of the spinors transforming under the non-unitary (finite dimensional) representations of the Lorentz group. I understand that the spinors thought of as field operators now transform under a unitary representation, but isn't the Hilbert space a Fock space made of one particle states which must be spinors hence it is not clear how the wavefunctions transform under a unitary representation
i.e. $U^\dagger \Psi^\alpha U = D^{\alpha}_{}\beta \Psi^\beta$ is clear
But the Hilbert space is formed of the spinor wavefunctions
i.e. $\psi^\alpha \oplus S_- (\psi^\alpha \otimes \psi^\beta) \oplus \dots$
How does one define a unitary action on these
Is it true that a sufficiently nice (need I say finite-dimensional?) quantum system governed by hamiltonian $H$ with $n_i$-fold degenerate eigenvalues $\lambda_i$ has $\sum_i n_i^2$ conserved quantities?
A sanity check is $\sigma_z$ has two distinct non-degenerate eigenvalues. And we deal with such systems using two labels $S^2, S_z$, corresponding to two conserved quantities, which is consistent with $\sum_i n_i^2 = 1 + 1$.
For $H = \sigma_z \otimes \sigma_z$ the stated result predicts $4$ conserved quantities. There are at least four by inspection $\{H, \sigma_z \otimes 1, 1 \otimes \sigma_z, S^2\}$
But then $H = \sigma_z \otimes \sigma_z + \sigma_z \otimes 1 + 1 \otimes \sigma_z$ is predicted to have $10$ conserved quantities...
Neat...there actually does seem to be 10 conserved quantities... (this is the commutator table of $H$ with the standard generators of $\mathfrak{su}(4)$ with $1 \otimes 1$ also included as a generator$)
Four "raw" conserved quantities and 6 conserved quantities comprised of linear combinations of two generators
the generators of $S_H$...
17:11
@TobiasFünke Yes that makes sense and all but im just really confused by why in this one case they do adjoint rather than inverse
and its not a unitary matrix
@Allie in which case?
The math itself makes sense to me but im just confused
I don't know
In the book they are starting with a basis set that is not orthonormal
note that a book/lecture notes etc. can contain typos
17:12
and they have the Fock matrix $F$ in that non-orthonormal basis set
ok; that is standard so far
and so they introduce a transformation $X$ that transforms the basis set into an orthonormal set
and they say that $X^\dagger FX$ is the representation of $F$ in the new orthonormal basis. I'm just confused why its adjoint here? the math makes sense when I wrote it out completely, nothing confusing there, but I'm confused why it happened to be an adjoint in this case
hmh yeah, I don't know
So the question is, why does introducing a Fock space automatically give you a unitary representation
It seems like that can't possibly be true
no, X is not unitary
they explicitly state that it isn't
17:19
Oh sorry I was talking about a different question
in QFT
oh im sorry
i think it has something to do with the orthonormality condition
well, is the basis you start with orthogonal?
17:39
Does anyone know how the issue of transforming the Fock space unitarily is resolved (which is not automatically evident since it is formed of one particle spinor states that do not transform unitarily under Lorentz transforms)
18:03
@TobiasFünke like the basis for $F$? no
only the new representation $X^\dagger FX$ is a representation of the fock operator in an orthonormal basis
can you send a screenshot or so of the relevant passage?
I suppose this is understood by the fact that if one has a unitary action on the C/A operators i.e. $U a^\dagger_p U^\dagger$, then one has a unitary action on states i.e. ${a}_p^\dagger {a}_k^\dagger \dots |0 \rangle$
what is 3.165 @Allie?
And the vacuum by definition is Poincare invariant
But then surely the same would apply to a one particle theory
18:16
$X^\dagger SX = 1$ where S is the overlap matrix in the original basis
so in that case i imagine it as "the overlap in the orthonormal basis is $\delta_{ij}$" which makes perfect sense but I again don't understand why it's an adjoint? like i do when i write out the math but i just dont get it conceptually
yeah, does not make too much sense for me either
im gonna examine how the non-orthogonality plays a role in this. afrter i work out and eat lunch
maybe also the non-linearity of the eigenvalue equation plays a role (?). I mean $F$ depends on $C$, no?
but I really don't know
no
C are the coefficients of the basis vectors that form ur molecular orbitals
F is already constructed from the C you got last iteration
ah okay
Die Symmetrische Orthogonalisierung ist ein von Per-Olov Löwdin (1916–2000) entwickeltes, in der Quantenchemie häufig eingesetztes Orthogonalisierungsverfahren. Als solches dient es dazu, aus einem gegebenen nichtorthogonalen Satz von Vektoren einen orthogonalen Satz zu erzeugen, bei dem für je zwei verschiedene Vektoren das Skalarprodukt gleich Null ist. == Beschreibung == Gegeben sei eine Basis S {\displaystyle S} für einen Untervektorraum V {\displaystyle V} eines reellen oder komplexen endlichdimensionalen…
I could only find this (in German)
18:26
in other words, C will contain the eigenvectors of the fock operator
hmm i will try to translate and see what i can glean lol
what you do is you diagonalize the overlap matrix
ye thats what the book says
well, I still don't really understand what they mean in the book :d sorry
 
2 hours later…
20:13
I am supremely shocked that the statement that the number of conserved quantities of a finite-dimensional quantum system is not stated (seemingly) anywhere. This seems like important data to have on a system.
meow
20:48
> Long ago, when shepherds wanted to see if two herds of sheep were isomorphic, they would look for an explicit isomorphism. In other words, they would line up both herds and try to match each sheep in one herd with a sheep in the other. But one day, along came a shepherd who invented decategorification. She realized one could take each herd and ‘count’ it, setting up an isomorphism between it and some set of ‘numbers’, which were nonsense words like ‘one, two, three, . . . ’ specially designed for this purpose.
21:11
iirc, that story usually is told using rocks, or a stick with notches carved in it to match the number of sheep first.
ymmv
21:31
@think_meaning_buildß I mean if you're in a barter or some other exchange, or other "matching" kind of situation you usually physically match objects in one set with another, even to this day.
22:28
Does anyone know how the issue of unitarity is resolved. The field operator transforms unitarily but it is still unclear how the explicit Fock wavefunctions transform since they are built from the one particle wavefunctions according to $\Psi = \psi^\alpha + S_-(\psi^\alpha \psi^\beta) \dots$
Fields do not transform under a unitary rep of poincare
22:49
(14) is not really that general at all, right? Generically, $\rho$ can be, say, a functional of the Hamiltonian and any other conserved observable?
@SillyGoose The text is only saying it's more general than choosing $\rho$ to be constant, which it is.
Yes that is understood; the unitary operators transform the C/A operators which kind of induces a transformation of the bispinor coeficcients.

But the states are built up of the bispinors which are required to transform unitarily however they don't
It's not clear how this is resolved
@DIRAC1930 What do you mean "the states are built up of the bispinors"?
Well one builds a Fock space out of one particle wavefunctions
symmeterised correctly
i.e. $\psi^\alpha + S_-(\psi^\alpha \otimes \psi^\beta) +\dots$ or something like that
I don't know what that notation is supposed to mean or where the problem is. Your one-particle space should already carry a unitary representation of the Poincaré group (Wigner's classification) so the Fock space inherits a unitary representation, too.
23:03
How does one show this? $\psi^\alpha$ transforms like $S^\alpha{}_\beta \psi^\beta(\Lambda x)$ where the part causing the transformation of the function argument is unitary but $S$ is not
what is $\psi^\alpha$?
The wavefunction of the Dirac equation
Where is a wavefunction coming from if we're doing relativistic QFT?
How else would one build a Fock space
Like every QFT text does: You decompose the free field into c/a operators and those create the particle states
23:06
$<x|p> = \psi^\alpha(p)$ I think
So it's built in there
What kind of non-sensical expression is that? Where's an $\lvert x\rangle$ coming from in a relativistic theory and how does the r.h.s. carry an index but the l.h.s doesn't?
I don't know
In non-rel, the field operators are $\Psi^\dagger(x) = \sum_i {a}^\dagger_i \psi^*_i(X)$
So the wavefunction is something like $<x| \hat{\Psi}^\dagger | 0>$
I think
Well the complex conjugate of the wavefunction
So $<p | \Psi^{\,\alpha \dagger }|{0}>= \psi^\alpha(p)$ maybe
In the above $\psi^\alpha(p)$ is the bispinor
The state space is the space of many particle wavefunctions
This answer by Xenomorph constructs the unitary representation on states and (momentum!) wavefunctions explicitly.
The Fock space is an algebraic construction used in quantum mechanics to construct the quantum states space of a variable or unknown number of identical particles from a single particle Hilbert space H. It is named after V. A. Fock who first introduced it in his 1932 paper "Konfigurationsraum und zweite Quantelung" ("Configuration space and second quantization"). Informally, a Fock space is the sum of a set of Hilbert spaces representing zero particle states, one particle states, two particle states, and so on. If the identical particles are bosons, the n-particle states are vectors in a symmetrized...
are my momentum wavefunctions
There's no escaping that they need to transform unitarily
Unless there is
The question is, exactly how does QFT resolve the fact that a unitary operator acting on the momentum space wavefunction doesn't exist
L&L gives a somewhat satisfactory reason "In the relativistic theory, however, the particle density is not a scalar, but is the time component of a 4-vector."
indicating that the finite part doesn't need to transform unitarily since the particle density is not invariant
23:39
I just linked you an answer that constructs the representations

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