Is this argument really sufficient to show irreducibility? If I am processing accurately, the text is essentially saying the image of the representation acts transitively on the basis of $\mathcal{H}^{(+)}$. Intuitively, this means that no subspace of $\mathcal{H}^{(+)}$ is disconnected (in the sense of unrelated by action of some $\pi(g)$) from the rest of the space. Intuitively, this means that this representation is not reducible.
However, I have not found a clear statement relating the transitivity of the action of the image of a representation and the quality of being irreducible.
On the contrary, I found an MO post which suggests a counterexample to the claim.
Guys I'm reading about the hartree-fock equations and so it starts by saying we're trying to minimize $\langle \Psi | \hat{H} | \Psi \rangle$ subject to the normalization constraint $\langle \Psi | \Psi \rangle = 1$
So it sets up a Lagrange multiplier which makes sense, they write $L = \langle \Psi | \hat{H} | \Psi \rangle - E (\langle \Psi | \Psi \rangle - 1)$
@HerrFeinmann but I feel like the best villains are the ones with extreme restraint. I feel like nI gets too mad too quickly XD for some reason I also can’t see you being a villain. I feel like Qmechanic could be. One day someone f’s the tags up too bad and it’s the last straw.
where E is the Lagrange multiplier. I guess my confusion comes from the fact that they're taking the energy to be the Lagrange multiplier E, rather than the minimized expectation value of the Hamiltonian $\langle \Psi | \hat{H} | \Psi \rangle$. In other contexts ive seen that the Lagrange multiplier in this sort of equation is a chemical potential instead, and honestly in this situation I feel like taking the expectation value makes more sense if we're trying to solve the ground-state energy
@Allie up to now, $E$ is just a Lagrange multiplier
@SillyGoose any basis vector is cyclic; you can transform any basis vector to any other (which differ only by a finite number of occupations/spins) by applying a finite number of operators from the algebra
then you need some denseness argument, to conclude that any vector is cyclic
alternatively you could try to prove that only the commutant is proportional to the identity
@Allie ...do the derivation, and see what comes out. note that you are a) searching for the many-body energy, and b) to get that, you need to solve the HF equations, which involve a HF energy. so there are also two notions of "energy" involved here
do any of you have thoughts about the concept of "imposter syndrome"? my problem with it is that there isn't such a thing as an objective judgement of how capable a person is. it just seems like... person a's subjective judgment vs person b's subjective judgment.
yes, but the premise is that someone who is capable feels like they are incapable. it's not considered to apply to someone who is "genuinely" incapable.
Yes, I think so too. It is obvious that in order to understand what one does not know one has to know quite a few things. So completely incapable persons (for a certain skill set) won't notice too much
@HerrFeinmann actually I have an issue with a tag. It concerns the tags "Hilbert space" and "Quantum states"... I don't know why they've been marked as duplicate tags
@Allie don't underestimate the joy and pleasure from understanding a new topic/subject or whatsoever :) hehe
@TobiasFünke I shouldn't say this out loud but I'm very - uhm - casual with tag choice. I try to make them pertinent to the question but I don't really care :P
I'm a little confused in the derivation of London equation (probably this is a textbook level question): as I understand there are two relevant electron densities: $n_s$ for SC electrons and $n_n$ for normal ones
@HerrFeinmann but it is not clear where you are even getting that. Like, are you trying to obtain the appropriate equations for superconductivity? If so, shouldn't it be the case that the normal current faces resistance and will decay away, leaving just the superconducting contribution? You'd have to ask a much more answerable question
@HerrFeinmann As far as I remember, there are somewhat two derivations, no? For example, one can derive London's equation by working with the "macroscopic wave function" describing the superfluid, and its square amplitude is interpreted as $n_s$. Then nowhere $n_e$ will appear
OTOH, you can also derive London's equation by assuming the "two fluid model" (I don't know if this is the correct English term), where you then assume $n=n_s$ (and thus $n_e=0$), I think
@naturallyInconsistent mhhh, I'm just deriving London's equations, so not the BCS and I wanted to understand why they only consider the current coming from superelectrons. I don't really know what is the case because I'm totally new to SC
I have some confusion about dynamic of states. If i have a fixed lattice with some values, say spin values, and i force the temeprature to be constant, and i know the state currently to be something of the form (+.-.+.-.-.-.+.+..)
Will this state change? will it have a dynamic? what causes the change?
I thought that the system will miimize energy, so it will go for the state with least energy, which (+,+,+,+,+.... )
>For the DC case, the superconducting channel inductor will short out the normal channel resistance. Thus, the two fluid model reproduces the fact that a superconductor has no DC resistance.
Which is what I was talking about, I am working with DC. Since I really suck with circuits I don't understand that "will short out the normal channel". Does that mean that it's like there is no normal channel?
@Madder This is a contradictory setup - if you know the actual microstate (+,-,+,-,...) of the system, then it doesn't have a temperature. Temperature is a statistical notion and only a property of macrostates, i.e. when you are uncertain about the microstate. If you actually know the microstate you don't need to do statistical physics and there's no temperature.
@ACuriousMind I agree with what you say, but there is some "caveat". For example, in (classical) molecular dynamic simulations, one typically evolves a bunch of atoms according to some interaction potential, and one can introduce a temperature by coupling the system with a reservoir; the effect is e.g. that the velocities are changed. In this sense, one can describe a system coupled to a heat bath. At each time (step) however you know the microstate.
@HerrFeinmann I'd have to think about it
It is not clear to me either. Have you read the other source?
@qwerty I have no reason to doubt it describes a real phenomenon: People who have comparable objective achievements to others in their field but think the others are "actually good" at it while they themselves are "faking it". I think it's sometimes overused to spread a kind of positivity where you're not supposed to doubt yourself at all instead of combatting an actual overly negative self-image, though.
i guess my point was that I feel that "comparable objective achievement" outside of something like standardised testing (which i also think doesn't account for mitigating factors), isn't something that makes that much sense. we know that for instance not all degrees are the same, and you don't need formal metrics or awards to be good at something.
I should add to my previous comment that there are problems though, and it is not clear if the trajectories of the particels are "real". It is a complicated matter
@qwerty I think you're focusing too much on the objective achievements somehow measuring actual "ability" or whatever. That's not the point. The point is that if there are ten people with e.g. similar numbers of papers in similarly prestigious journals and one of them thinks "man, those others are actually good at this but I'm just faking it", then (unless the person actually did do some intentional deception to get the papers published or whatever) that's a person with impostor syndrome.
The claim isn't "those ten people are all objectively of equal ability" the claim is that it's generally irrational for one (or more of them) to believe that somehow the achievements of their peers demonstrate "actual ability" while their own do not
if we agree that the achievements don't actually measure anything (which is what you seem to want to argue) then it's even more irrational to see them as a reason to feel as the inferior impostor
but the "cyclic" in cyclic group refers to the fact that there exists an $n \in \mathbb{N}$ such that $n \neq 0$ such that $g^n = e$. the cycle being $g^0 = e, g, g^2, ..., g^n=e$.
@SillyGoose If I understood you well then yes, it seems so. (My initial thought regarding denseness, I think, does not suffice either). One of the criteria I mentioned above is what I've said: If every vector (except 0) must be cyclic, then the representation is irreducible. Having a "cyclic basis" is a good starting point, but I doubt it is enough. But I haven't thought about it yet
as you already said, that makes no sense because the group does certainly not act transitively on $0\in V$
furthermore, e.g. unitary representations will preserve vector norms and so cannot map vectors of different norms to each other. Representations are almost never transitive on $V$ even if you remove the $0$
I see so transitivity is very non-natural for this purpose i guess.
is there a natural weakening of transitivity that is equivalent to the standard definition of irreducible representation? i mean irreducibility (topological irreducibility in Tobias's reference) is (seemingly) about analyzing the set of stable subspaces of $V$. so perhaps transitivity is just not a useful raw concept...
@TobiasFünke by the way, do you use the c*-algebra approach to qm when thinking about condensed matter?
@TobiasFünke what is the purpose of the "finite number of operators" hypothesis? in the text it is also emphasized that each configuration (e.g. in the + or - representation space) can be reached from another by a finite number of applications of an operator.
Well. I am no expert on this, I only have a bit of knowledge which helped me to understand QM better. But I don't work with it. I know Sewell's books and find them interesting, especially regarding the thermodynamic limit
@SillyGoose if you wanted to talk about an "infinite number of operators" in any sense you would first have to prove convergence of whatever limit you're trying to imply with that
it's just the proper and rigorous way to state that you are considering something like $A^n,n\in\mathbb{N}$ or polynomials in $A$ or whatever the context here exactly is
but shouldn't it be proper and rigorous to prove that applying an operator more than a finite number of times is nonsensical. e.g., such a thing is not well-defined for a benign operator; not just to ignore the possibility because we haven't defined its meaning yet?
@ACuriousMind "to see them as a reason to feel as the inferior impostor": imo it's not achievements as reason to feel an imposter; it's being aware of your own ignorance or stupid questions... vs when it (subjectively) shows when someone else knows what they're talking about, when you think wow, they gonna realise i'm an idiot/imposter...
aka it's all vibes, and arguing to replace one set of vibes with another. i guess it's obvious this isn't a, uh, pun intended, purely academic question here. anyways it's very late here, I should sleep...
@TobiasFünke i've been finding this sewell book very interesting indeed. i feel like i have learned something conceptually new, which hasn't happened in a long time :P
separately, did you guys learn the completion of the rationals to the reals in terms of cauchy sequences? my real analysis course i think did not, but we then learned about cauchy sequences later without highlighting the connection with the rational to real completion...
For operator norm = 1 it could converge in finite dimensions by compactness of the unit sphere but since the unit sphere in infinite dimensions isn't compact that doesn't work either.
anyway i feel like now learning that the rational to real completion can be stated in terms of Cauchy completeness makes the concept of Cauchy sequences more concrete.
@SillyGoose it's not obvious to you why an operator that shortens vectors with each application could only yield zero when applied infinitely many times?
it is perhaps a good guess which then needs to be checked. but i would not describe that situation as the answer itself being obvious
@ACuriousMind I mean this is like saying it is obvious the series defined by $\sum_{n} \frac{1}{n}$ converges because intuitively the contributions to the sum should vanish at some point, which we know to be false (under standard notion of series convergence).
The proof is really simple. Convergence in norm, which is submultiplicative. If norm A<1, then norm A^n<=(norm A)^n<1. In particular, the RHS converges to 0, and thus A^n converges (in norm) to the zero element. Finally, limits are unique here, concluding the proof
@SillyGoose It's not a guess, that's already the full proof: If $v_i\to v$ then $\lvert \lvert v_i\rvert\rvert\to v$. Now assume $A^i v\to w$. Then $\lvert A^i v \rvert \to \lvert w\rvert$. But $\lvert A^i v\rvert \leq \lvert A^i \rvert\lvert v\rvert \leq \lvert A\rvert^i \lvert v\rvert$ by the properties of the operator norm, and for $\lvert A\rvert < 1$ the rightmost value goes to zero, so $\lvert w\rvert \leq 0$, i.e. $w=0$ is the only possible limit if it exists.
@SillyGoose I see what you mean and I agree. But the proof is something which can be checked immediately or is something like an easy homework exercise. What is "obvious" and what not is of course a matter of experience with the subject
@ACuriousMind Then explain to me what they mean? they pick the state of least energy (all +++...) and say "due to temperature fluctations some spins flip". And i am trying to understand, starting from the initial state, what causes the dynamic such that some spins flip? the system is already known and is at lowest energy) I do not really understad the "T fluctations". What if T=const? does it even make sense
Well it is a niche-topic that you might not be familiar with. here is the exact quote:
"Let’s consider the two-dimensional Ising model at low temperatures and suppose that it is in the state of minimal energy in which all the spins have values +1. The thermal fluctuations create domains in which there are spin flips, such as the domain in fig..."
But generally speaking, you can consider a system where you can take 1 or 0 in each point and you have a grid of NxN points. And a microstate is when you know the values. pointing up means =1
they are probably talking at a heuristic/cartoon level so as to illustrate some underlying physical mechanism of what is probably more formally some statistical mechanics computation
Maybe you can answer me this, because i might have minsconception about microstates. if i pick a microstate in this frame, it will not evolve with time? will it evolve with Temperature?
if you are talking about the ising model in isolation and you are handed an initial state, it evolves in time according to usual quantum mechanics, i.e., the time evolution operator due to the ising model hamiltonian
if you are talking about the ising model coupled to a heat reservoir (that is an idealized environment of constant temperature with which the ising model system can transfer energy with), then you are now working in the realm of equilibrium statistical mechanics in which there is no "evolution"
well the ising model given the discussion of ferromagnetic domains should make no sense in classical mechanics as you cannot have nonzero magnetization at statistical equilibrium in a classical mechanics framework :P (cf. Bohr-van Leeuwen theorem. so perhaps i am not of much help.
The logic is supposed to be that at low temperatures, any microstate in that low-temperature macrostate differs from the ground state only slightly, i.e. by small domains that are flipped. I don't think this is supposed to be a time evolution of the spins flipping, you're supposed to consider what the microstates look like as you turn up the temperature starting at absolute zero
I just want to understand something crucial for the chain of thought in my brain. if i take any snapshot of the system to be in a microstate and spesicy it to be exactly $s$ ie i know the configuration as a whole. And T = constant. Does this change with time? What happens if T is now changing, does the snapshot change ? or do these questions make no sense?
@Madder you're implicitly using the canonical ensemble, which is equilibrium statistical mechanics. there is no "time evolution" here. the system is presumed to be in "equilibrium". so microscopically there is perhaps dynamics, but at the level of statistical mechanics the distribution of states is stationary.
at least not in this context - if you have a single specified microstate why would there be a "partition function"?
there's no statistics to be done if you have a definite microstate instead of a non-trivial probability distribution (a macrostate)
The usual statistical definition of temperature is in terms of the entropy of the probability distribution (the macrostate). If you know the microstate, your probability distribution is trivial (just a point) and the entropy is zero (like in the unique ground state at absolute zero)
the partition function is there to give me the probability of the system being in a state because i do not know, right?
but when i KNOW, then Z is useless , okay! This means the system is frozen in time. correct? And there is no use of speaking about temperature, because it is statistical.
@Madder to the first part: Yes. But the system is not necessarily "frozen in time" - the microstate would still evolve by the Hamiltonian, likely into a different microstate
the whole point of "thermodynamic equilibrium" is that it is a macrostate that encompasses a bunch of microstates that all would evolve into each other, but the overall probability distribution is stationary
temperature as introduced in an undergrad course on statistical mechanics is defined to be a quantity that signifies equilibrium of energy transfer. therefore, temperature is not only a statistical mechanical quantity, but also an equilibrium statistical mechanical quantity. that is, it is only well-defined at equilibrium. there are probably extensions of the definition, however.
@ACuriousMind Ok this is very interesting because i honestly thought nothing would happen.
So there is evolution even after you specify the microstate (in the case of this model). It is somehow more visual in the gas case because i can see the particles moving due to newtonian laws, i just did not consider some dynamic due to he hamiltonian of the model.. so i guess the answer is "no" its not frozen in time even if T= cst
I mean - the state can be frozen in time, that's not forbidden, e.g. the ground state at absolute zero is certainly stationary, but it's not in general what microstates do
in principle one should be able to specify the initial state of the system, and apply the dynamic relation to evolve the system. the point of equilibrium statistical mechanics is that you cannot actually do the aforementioned in practice and we must find an alternative, statistical description of what a system might look like (with various probability weights) once that system settles down and reaches "equilibrium".
Moreover, statistical mechanics also takes into account the fact that a system interacts with an idealized environment, so it models systems as they might exist in the actual world. A cup on a table is not in isolation. it is constantly interacting with a huge number of things goings on in the air surrounding it. details completely ignored by usual first considerations of a system in isolation.
Is there any way i can understand the statement they are making in other formulation or words that is more intutive? i am still very confused about the "thermal fluctation" part.
I do not know why i cant get over this, but wikipedia defines: In statistical mechanics, thermal fluctuations are random deviations of an atomic system from its average state. and once you pick a state, you are throwing these statistical definitions out of the window.
you can think in the cartoon picture that there are thermal fluctuations in the energy of a system on the order of $k_B T$. Then, suppose your system occupies a state with energy $E_0$. At temperature $T$, your system might randomly jump up to energy $E_0 + k_B T$ or down to $E_0 - k_B T$ due to "thermal fluctuations".
i think you can formalize energy fluctuations in terms of the variance of the energy. there is a variance in the energy because indeed the system's actual energy is unknown. we only know the probability distribution of energies for the system given the temperature and the system's own hamiltonian.
anyway an okay book on statistical mechanics is the one by pathria. if you have never before, i think it is a good exercise to derive the canonical ensemble from the microcanonical ensemble.
it gives some understanding of what the canonical ensemble is exactly and requires you to know the basic "axioms" of equilibrium stat mech
At the bottom of the page, i am trying to show that norm convergence is stronger than weak$\star$ convergence. is this the right idea o.0. I am not certain about the second implication, but it seems "intuitively" true statement
as I said, I don't know whether or not your proof is valid; I just don't understand it, and find it rather complicated, given the fact that it is really just the definition of the norm which one has to use. (or I am braindead and miss something)
When one talks about the pre-second quantised field (wavefunction), what is the formal definition. Would it be $$\langle p^\mu | \Psi\rangle = \psi^\mu (p)$$
For the vector representation of the Lorentz group anyway
And then if $|\Psi\rangle$ belongs to the scalar representation, would it be $\langle p^\mu | \Psi \rangle = \phi(p)$?
In analogy to when one talks about the non-rel spin wavefunctions, one has $\langle x, s_z| \Psi \rangle = \psi_i (x)$
Where $i$ takes values $1,2$ depending on $s_z$ being $-1/2, 1/2$
@SillyGoose well that some point you would like to make contact with the usual formulation of QM, and ordinary QM is in terms of states as vectors and density matrices, no?
in particular, if the normal folium of a state is the entire set of states, then you know you don't lose any states by just doing "normal" QM in the associated GNS representation
i am confused though doesn't this algebraic formalism already fail for the free particle, which deals with unbounded operators? The text I am reading specializes to algebras of bounded operators. Is this an unnecessary assumption?
it's very much a necessary assumption for the mathematical formalism :P
and yes, it does not apply to $x$ and $p$. But it does apply to their exponentiated version, just like the Stone-von Neumann theorem only applies to the exponentiated form of the CCR (the Weyl relations)
another way to get an algebra of bounded operators is to take the algebras of spectral projections
you can talk about QM measurements and expectation values always in terms of projectors (the spectral projectors of the "real" observables) and this algebra of projectors is always bounded since projectors are bounded
and yet another way would be to do the usual trick and just put your theory in a box with an upper energy cutoff, then $x$ and $p$ become bounded by assumption
is there some sort of trade off or something for working with the projectors? it seems like if you include all the projectors corresponding to x then maybe the algebra of projectors is not compact or something, which would probably be bad for doing convenient representation theory?
I mean in the end you will find out all this "abstract" circus is not all that more general anyway because every $C^\ast$-algebra is isomorphic to the algebra of bounded operators on some Hilbert space (Gelfand-Naimark theorem)
maybe you should wait to see what's being done with the definitions instead of trying to pick them apart before you have seen them in action :P
i might leave the text for the time being...i at least gained something conceptually new out of it; though, unrelated to the algebraic formalism it seems
@SillyGoose the user Chiral Anomaly has (on their profile) a link to their private site, where they collect notes. IIRC, also about the algebraic approach, projectors, observables and so on
also the books by D. Dürr have a nice discussion about "observables", which more or less coincides with that of CA... there are 1-2 other nice posts here on PSE about that.
and many QI books have similar notions, see e.g. "mathematical language of quantum mechanics" (a relatively new book, and from the same authors there is also a nice, and very similar, text available on arxiv)
Why is spin any different than just introducing a 2 component wavefunction under the projective representation of SO(3)?
Haven't people been doing stuff like this when in classical mechanics write $A^\mu$ i.e. a 4 component function under the fundamental representation of the Lorentz group
One just has to formally declare in the first case that the Hamiltonian operator in position basis is a $2\times 2$ matrix
invariant under $SU(2)$
The question is why isn't $L=1$ classed as a vector particle
Why don't people say in classical physics that $A^\mu$ has an instrinsic angular momentum
@DIRAC1930 It's not really clear what you're exactly trying to do, but if you're trying to analyze a massless vector field in terms of 3d rotations/spin you're going to have a bad time because the massless nature of the associated particle actually means you should be thinking in terms of ISO(2) first. See also this answer of mine on the nature of "spin" for photons.
I have one question regarding QFT, the picture we operate it. Since the field operators can be expressed via mode expansion, and the states are expressed as occupation number states. Is it a direct consequence of the fact that we mostly, at least as far as I have seen, we operate in the Heisenberg picture, as long as interaction ain'\t present ?
If one had a Schrodinger equation for a hypothetical particle which had a 3 component wavefunction, 1 for each direction of space i.e. $\Psi= (\psi_x,\psi_y,\psi_z)$ how would one formally construct this
Why is this any different from the vector representation of the rotation group
but $L^2, L_x, L_y,L_z$ describe something different it seems
@DIRAC1930 "Formally construct" in what sense? Coming from QFT, you would get this as the wavefunction of a massive vector particle in the non-relativistic limit.
The fact that when the field operators act on a state, the state is given either as occupation number state $|n_1n_2...\rangle$ or as a state of the form $|\vec p_1,\vec p_2,...\rangle$ (I don't know what this is called), which show no time dependece @TobiasFünke
@DIRAC1930 but what is $\langle x\vert$ and what is $\lvert \psi\rangle$? In order to "formally construct" something you need to define all the symbols you're using.
then you are correct, of course in non-relativistic QM the wavefunction for a state $\lvert \psi\rangle$ of spin $s$ can be obtained by $\psi^i(x) = \langle s_i,x\vert \psi\rangle$ where $i = -s,\dots s$
But what about $\Psi = (\psi_x,\psi_y,\psi_z)$ i.e. a 'vector' particle in non-rel. This doesn't seem like it would be the spin 1 representation because that would give components for the spin direction in the $z$ axis and not $\psi_x,\psi_y,\psi_z$ like we require
The spin operator in the spin-1 representation transforms like a 3d vector, so the resulting $\psi^i(x)$ also transforms like an ordinary 3d vector. There's nothing wrong here
@DIRAC1930 I don't know what you mean by the subscripts "x", "y", "z"
we usually span the spin-1 space by -1,0,+1 (for some direction)
where are those indices on your wavefunction supposed to come from? What definition of $\Psi$ are you using other than the one we established above in terms of spin eigenstates that gives you those components?
I am talking about generalising the Schrodinger equation where one has a scalar wavefunction under the rotation group $\psi$. I am talking about having a vector wavefunction under the rotation group i.e. $\psi^i$
I mean, since the spin-1 spin space is three-dimensional, I think the three zero eigenvectors of $S_x,S_y,S_z$ would give you three components like that
if you think about it the axes in 3d are just each the zero eigenvectors of one of the rotation generators (since a rotation around an axis leaves that axis invariant)
