@TobiasFünke I've tried several times but I am bad at actually publishing posts - I usually produce like a dozen unpushblished drafts that I'm not quite happy with and then lose motivation :P
In the free Klein gordon case, can a antiparticle and particle change into 2 particles and 2 antiparticles as long as all other conservation laws are satisfied?
What conservation law causes the above not to be allowed?
If the particles of the Klein Gordon theory are non interacting and the number of them are conserved, then surely the whole theory can be properly interpreted as a 1 particle theory. One knows that what was thought as probability density is actually charge density so that leaves the question regarding what the expression for probability density is
Underlying everything, there is secretely normal qm going on just not using the expressions that one naively assumes
@TobiasFünke I am reading's Baxter's book on exactly solvable models in stat mech. But I am looking at this old review (journals.aps.org/rmp/abstract/10.1103/RevModPhys.71.S358) and i am assuming their definitions of universality and scaling (section III)
The Hamiltonian $$H=\sqrt{p^2+m^2}$$ defines a one-particle quantum mechanics in the usual way. Let us call this theory RQM for short. Peskin and Schroeder claim that RQM violates causality because the (quantum-mechanical) propagator has support outside the light cone (Sec. 2.1). I do not beli...
Okay but then why not have a momentum probability density
The point is the Klein-Gordon theory should be able to be formulated properly as a single particle theory if one can formulate a non-interacting many particle theory where the number of particles and antiparticles is conserved. The question is how.
@DIRAC1930 There is no problem in formulating the theory on a formal level. It is just unphysical because the naive one-particle Hamiltonian is unbounded from below.
@DIRAC1930 I just wrote it down, it is the usual formula in terms of eigenstates of the momentum operator since the momentum operator exists as a self-adjoint operator with no problems (since translations are represented unitarily). What's unclear about that?
I have no idea what you mean, but if you think there "should" be such that thing, just go ahead and write it down. If you can say it should exist, what problem do you have in finding it?
You can't just assert that random things should exist without presenting a derivation of them.
@ACuriousMind maybe im getting it wrong but doesnt this just say that when accounting for PEP there should be an energy decrease when discussing fermions? it's not clear to me why this energy difference would correspond specifically to the energy difference between $T_0$ and $S$ tho
@DIRAC1930 You can split the space of wavefunctions $H$ into $H_+ \oplus H_0 \oplus H_-$ as the eigenspace decomposition of the KG-Hamiltonian. What's your point?
Well when you split the many particle Hilbert space into $H_+ \oplus H_-$, you have $a,a^\dagger$ operators only acting on $H_-$ and $b,b^\dagger$ operators only acting on $H^+$.
So this is 2 particles essentially
Which is claimed to be consistent
Which it probably is
But my point is that in the non interacting regime, one can consider this as just being 2 particles, one with a wavefunction in $H_+$ and another particle with a wavefunction in $H_-$
And the one should have no inconcictencies if the many body QFT is correct
Ah, this is the difference between the field Klein-Gordon Hamiltonian and the one-particle one
@TobiasFünke you're thinking of the field Hamiltonian, which is bounded below with that spectrum (if the infinite zero point energy is somehow taken care of)
but if you just look at the KG equation on one-particle wavefunctions, it's not bounded below since the "negative frequency" solutions have $-\sqrt{p^2+m^2}$ as their energy
@DIRAC1930 But that doesn't change that the Hamiltonian for one of your two "particles" is unbounded below.
I really don't understand what you're trying to do here
But that's not what you're doing, QFT certainly does not interpret the negative frequency solutions as a different particle in the case of the real scalar field.
mhm, I am looking at one of Arai's books, where he introduces the single-particle operator $\omega_m(k):=\sqrt{k^2+m^2}$, and from that constructs the second quantized Hamiltonian $\mathrm d\Gamma(\omega_m)$. The time evolution of the fields is governed by this latter SQ Hamiltonian, which I think you refer to the "field" Hamiltonian (?)
Lets get down right to basics. If I have a real Klein-gordon particle infront of me, does the notion of having a momentum probability density make sense?
ultimately of course the "correct" Hamiltonian on one-particle states is also bounded below, but it is not equal to the Hamiltonian of the classical KG equation because we throw away that infinite piece in the field Hamiltonian (or switch the role of the c/a operators or whatever phrasing you want for what makes the field Hamiltonian well-defined)
@DIRAC1930 I just wrote it down, it is the usual formula in terms of eigenstates of the momentum operator since the momentum operator exists as a self-adjoint operator with no problems (since translations are represented unitarily). What's unclear about that?
@DIRAC1930 You need to take three steps back and first define the Hilbert space (again, complete with inner product) we're taking as our space of states here
@DIRAC1930 If you think you can build a relativistic theory without making the Poincaré group unitary as your very first step then you will never build a relativistic theory.
otherwise you could tell which Lorentz frame you are in by performing an experiment where those probabilities differ in different frames. That such probability-preserving transformations must be (anti-)unitary operators is Wigner's theorem.
I have already made exactly the same point and am rapidly reaching the conclusion that this conversation is pointless
One just needs to ensure $\int \mathrm{d}^3x\psi^\dagger \psi$ transforms as the time component of the 4-vector
Otherwise you run into silly statements like 'qft is needed because the spinor matrices are non-unitary'
Hence some silly statement regarding people saying its resolved by the matricies transforming the operators which aren't part of the Hilbert space or something
@Relativisticcucumber In the singlet, the spins are anti-parallel and so there is no "Pauli exclusion effect" at work. In the triplet, the spins are parallel and Pauli exclusion forbids them to occupy the same state in the non-spin part of the space.
(1) in condensed matter and qft, the cluster decomposition principle supposes that long-range spatial correlations should vanish (essentially). (2) in textbook qft especially only isolated systems are considered. (3) then how can the CDP possibly be true if entanglement is a generic fact of life in an isolated system?
@SillyGoose The cluster decomposition principle does not state that no systems possess long-range correlations.
It merely states a factorization property e.g. of the S-matrix when the states considered are seperated from each other in a specific way (related to causality)
(there are many equivalent or almost-equivalent ways to state this causality property of the scattering operator which is why I use 'e.g.')
@DIRAC1930 Several people have extremely patiently tried to engage with your questions yet you continue to ignore all the pointers you are given to the correct solutions. You can do that, but you have no business getting upset about people disengaging from you when their input is continuously ignored.
@Arjun don't worry about it, chat messages don't disappear if you don't read them fast enough ;P The explanation I gave was designed to be brief; if the usage of the topological properties of connectedness doesn't sit well with you there should be various other derivations in the literature that discusses the structure of the Lorentz group (that it has four connected components is a result many places should show)
@ACuriousMind But they aren't the correct solutions. I am literally quoting from the methods in Dirac's book, Pauli's book, Landau & Lifshitz, some other early texts on quantum electrodynamics that build things up properly etc. Noone has offered a solution regarding the exact structure of what I am asking because it is not as simple as it seems
@Relativisticcucumber sorry, I said it the wrong way around. Here's a formal argument: If you example the spin states, then the triplet states are symmetric under exchange but the singlet state is anti-symmetric. It follows that, for the overall state to be antisymmetric, the spatial part of the wavefunction is anti-symmetric for the triplet but symmetric for the singlet. So in the singlet, the wavefunctions are allowed to be "the same" - no exclusion effect - while in the triplet, they may not
more concretely, I am thinking that one defines a correlation as $\langle A B \rangle - \langle A \rangle \langle B \rangle$. So that the above screenshot would then imply that this vanishes.
More particularly, I am working through Baxter's exposition of the general Ising model. He defines a correlation function $g_{ij} := \langle \sigma_i \sigma_j \rangle - \langle \sigma_i \rangle \langle \sigma_j \rangle$, which if the system is translationally invariant only depends on the lattice distant between sites $g_{ij} = g(\vec{r}_{ij})$.
Then he writes that it is expected that $\lim_{\lvert r \lvert \to \infty} g(\vec{r}) \to x^{-\tau} e^{-x/\xi}$ as $x \to \infty$, which according to a discussion I had yesterday with a prof is the CM-implementation of cluster decomposistion
The one caveat stated by Baxter is that the "cluster decomposition" behavior of $g$ should hold true away from a "critical point/temperature"
@ACuriousMind A (classical) canonical ensemble $\rho \sim Z^{-1}(B, \beta) \exp(-\beta H(\sigma, B))$ where $B$ is an external field, $\sigma$ is a spin configuration, and $\beta$ is inverse temperature.
@ACuriousMind erm i am still not getting it. for one, any argument based off of symmetry and antisymmetry wont be able to single out $T_0$, it would apply to all triplets, no?
@Relativisticcucumber Sure. But you already observed correctly that the spins are parallel in the other two triplet states, and flipping a spin in general might also cost a bunch of energy, so if you took those to define the "exchange energy" you might get more than you want
@SillyGoose Now we're getting to CM-specifics I am rather weak on but I believe it's indeed either a theorem or an assumption (it might depend on $H$ which) that such thermal states also satisfy cluster decomposition
But there is no claim that all states (i.e. those particular entangled states with long-range correlations you're thinking of) satisfy cluster decomposition
also, about my qualm from yesterday. so if i have a situation, where there are two regions, left and right, one electron, and some tunneling $g$ between these two regions, then i can write a hamiltonian for this system reminiscent of the normal 2-level system. if i diagonalize this and get energies $E_{\pm} = \varepsilon \pm \frac{1}{2}\sqrt{\delta^2 + 4 g^2}$, [...]
[...] then i see that the tunneling being nonzero "breaks a degeneracy". what i mean by this is that with zero tunneling, there is a $\delta$ for which $E_- = E_+$. however, the text im reading says "the tunneling removes the degeneracy, and symmetric and antisymmetric states are formed". is there a way to understand why removing the degeneracy leads to symmetric and antisymmetric states? i mean in the case of, say, a magnetic field, [...]
[...] this would make sense because zeeman splitting, but i see no rationale to this here.
kind of a reiteration of a q yesterday but tried to provide more info :PPP
@Relativisticcucumber As long as your Hamiltonian remains symmetric under exchange of left and right, its new non-degenerate eigenstates need to be eigenvectors of the exchange operators (if this is not obvious, try to think about why that is). The eigenvectors of the exchange operator are symmetric and anti-symmetric combinations of the "left" and "right" state.
by exchange operator, you mean smth that acts like $\hat{O}(\vert L \rangle \otimes \vert R \rangle) = \vert R \rangle \otimes \vert L \rangle$, right?
The Hamiltonian $$H=\sqrt{p^2+m^2}$$ defines a one-particle quantum mechanics in the usual way. Let us call this theory RQM for short. Peskin and Schroeder claim that RQM violates causality because the (quantum-mechanical) propagator has support outside the light cone (Sec. 2.1). I do not beli...
