Ok after sitting on the discussion of the dirac equation from a few days ago, i have realized that my question is actually "how do we calculate gamma matrices" @Mr.Feynman -- i have since looked into this, and i always see the anticommutation relation with the metric as well as the gammas for basic cases like minkowski metric, but i have yet to find anything regarding a general approach. i think the anticommutation relation is not sufficient? maybe i am wrong tho
@Mr.Feynman omg that really is annoying... I suspect this kind of thing goes along also with an insufficiently geometric approach. Since if it isn't seen that $g$ is different in different coordinate charts anyway it may feel "wrong" to put the prime on it
I mean, if you're just talking about the same metric viewed from two different coordinate systems then putting the prime on the indices seems perfectly reasonable (if unusual) to me :P
As per a change of basis matrix $A$, it's just a different way to write $A^i{}_j$ as $A^i_{j'}$. The prime marks the rightmost index, so it's not that bad
@ACuriousMind I would argue that $g'_{\mu\nu}$ is more common than $g_{\mu'\nu'}$
As I was saying, for the change of basis it makes not a big difference, while for fully covariant/contravariant vectors I'll need time to get used to the notation and be efficient :P
@Mr.Feynman my point is that if you care about the active/passive distinction then you'd write $g'_{\mu\nu}$ for the result of an active transformation and $g_{\mu'\nu'}$ for the result of a passive transformation (i.e. change of coordinates)
that's the usual way of doing it but if you really think about it it doesn't make a lot of sense because $v'^\mu$ looks like it should be the components of some object $v'$, which presumably is not the same as $v$
In Wald there is a very nice convention he chooses. He uses english letters for abstract index notation and greek ones for the coordinates index notation. Which may be confusing if you skip the 3 pages he takes to explain it and go on to try and learn some GR lol
No, basically the idea is that he wants to use indices even for the geometric versions of the equations, ones which are true in every basis. So in order to do that, he uses english letters for those. And when he actually does work in a basis, it's always with greek letters
Any suggestion on how the electric field term should disappear in the nR limit here? I think there is a way to understand - in natural units - what happens or why I should neglect it (if that's the case)
@ACuriousMind $c\to\infty$ is the galilean limit, so we're interested in the second one you mentioned, i.e. $v\ll c$. Working on the wave equations, the only nR limit I've seen is that one, that working on time indepedent equations amounts to $E\approx m$
Also because I don't know how to implement $v\ll c$ on something like e.g. the KG equation
But anyways, this is a minor issue I found playing with the equations, I'll think of it later and get back to my beloved Dirac fields
@bolbteppa I forgot to reply to this, sorry. Apparently I can't recognize the result I'm talking about there. In the case of KG, the inner product was of the form $(\phi_1,\phi_2)=i\int d^3\vec{x}\phi_1^\ast\overset{\leftrightarrow}{\partial_0}\phi_2$. I'll write the final result in the case of 4-momentum to make the notation obvious.
The 4-momentum $P^\mu$ is the Noether charge associated to space-time translation invariance and it turns out that for a given $\phi$ be written as $P^\mu=(\phi, \hat{\Pi}^\mu\phi)$, where $\hat{\Pi}$ represents the action of the translation algebra on the infinite dimensional space of KG solutions.
ok so in this paper they claim that the zweibein fields are defined by the relation between the minkowski metric and a curved space metric (if im understanding correctly?) then towards the end when they are calculating the hamiltonian, they list these fields. i am wondering how these are generally obtained? does one calculate them usually? journals.aps.org/prd/pdf/10.1103/PhysRevD.27.2893
hm on second thought mb they can indeed be calculated from their definition.
Hm yes i must be missing smth bc im my calculations i only get the term inside the brackets -- not the trig component
in the particle in a 2 dimensional box, we write down $\psi$ as multiplication of $\psi _{x} \cdot \psi _{y} $, and we write the hamiltonian eigen equation $ H \psi=E \psi = (H_{x}+H_{y})\psi = E\psi $ with the potential=zero. all fine. what is then the motivation for us to write down$ E=E_{x}+E_{y} $? E is just the eigen value of the total hamiltonian right, that necessarily need not give us the sum of $ E_{x} and E_{y} $ right?
@nickbros123 The $x$ and $y$ problem are independent. The total Hamiltonian is the sum of the Hamiltonians (each acting respectively on $x$ and $y$ in the coordinate representation) and the wavefunction is the product of the the wavefunctions
In such case the eigenvalues are additive
When you write $H_1+H_2$ you really mean $H_1\otimes 1+1\otimes H_2$, $1$ being the identity operator on the respective space
if $H_1\lvert\psi_1\rangle=E_1\lvert\psi_1\rangle$ and $H_2\lvert\psi_2\rangle=E_2\lvert\psi_2\rangle$, then $(H_1\otimes 1+1\otimes H_2)(\lvert\psi_1\rangle\otimes\lvert\psi_2\rangle)=(E_1+E_2)(\lvert\psi_1\rangle\otimes\lvert\psi_2\rangle)$
Projecting this onto $\langle x_1, x_2\lvert:=\langle x_1\lvert\otimes\langle x_2\lvert$ yields $(\mathcal{H}_1+\mathcal{H}_2)\psi_1\psi_2=(E_1+E_2)\psi_1\psi_2$, where $\mathcal{H}$ is the Hamiltonian in the coordinate representation of each particle (which you called resp. $H_x$ or $H_y$)
Is the interpretation of a cloud of virtual particles coming from standard QM perturbation theory expansion of the state $|k \rangle_{\mathrm{full}} = \hat{A}_k |0\rangle + \lambda \dots$ where the term proportional to $\lambda$ will have the creation operators of the virtual particles?
@Relativisticcucumber I didn't learn proper QFT other than the EM field canonical quantization yet. The rest of my first course was about either applications theoreof (radiation-matter interaction using ordinary perturbation theory), relativistic one particle QM, and the Poincaré group. In other words, that was a prequel to QFT. As per Schwartz's book, I acknowledge it exists but right now I have Maggiore, Peskin and possibly Weinberg to cover
@DIRAC1930 virtual particles are not associated with actual states created by c/a operators, they're internal lines in a Feynman diagram corresponding to a sum over intermediate states
What do you get if you just use standard QM perturbation theory to find what a free state $|k\rangle$ will be once the interactions are adiabatically switched on?
@imbAF the usual position or momentum operators in 1d
@DIRAC1930 I'm not really sure what you're trying to do
the kind of perturbation theory you're talking about looks for eigenstates of the Hamiltonian to figure out energy levels, but that's not really useful in a QFT context
all energies above 0 obviously can occur just as the state of a single particle with arbitrary momentum
Say if I have the free vacuum $|0>$ I should be able to calculate the interacting vacuum $|\Omega>$ just from the above perturbation theory. I'm not trying to calculate the energies, just the states. But I think I can have discrete energies in a many body system
the z-spin angular momentum of a spin-1/2 particle is non-degenerate
the z-component of orbital angular momentum of a particle in 3d is degenerate
@DIRAC1930 Haag's theorem says it doesn't work like that; the interaction picture in QFT is a lie
if you're trying to understand what people mean when they say that the vacuum is filled with virtual particles they mean that the Feynman diagram for the vacuum energy are "vacuum bubbles" of diagrams that have only internal lines, see e.g. physics.stackexchange.com/a/695424/50583
@ACuriousMind Two more questions, if you don't mind: What are the implications of the 2nd Law of thermodynamics in Physics? And also, I am asked to prove via graph/plot/ sketch that a system can never reach the 0 temperature in Kelvin. Is there such a sketch? I myself, used the uncertainty principle as to why a system can never reac 0K temperature. But I am asked for a sketch, and I have no idea
Perhaps something that has to do with Temperature and Entropy ?
@ACuriousMind One more thing. We say that two fermions cannot be in the same state (Pauli Principle). But does the distance among them in space plays a role ? For example in 2 H-atoms, the electrons are in the same state. Right?
hm but i guess when i think of states. i think of them as a property of a particular system. hence, the energy levels (or energy eigenstates) of one hydrogen atom are utterly distinct from the energy levels in another hydrogen atom
to me they are because the energy observable is "different" between systems in that it is a map between hilbert spaces. hence, if you are mapping between different hilbert spaces the observable is distinct
e.g. if H-atom 1 is $\mathcal{H}_1$ and H-atom 2 is $\mathcal{H}_2$, then $\hat{H}: \mathcal{H}_1 \rightarrow \mathcal{H}_1$ should be identified with being the energy operator for H-atom 1, definitely not H-atom 2 since it cannot talk with that system other than trivially
hm though i guess im maybe implicitly assuming stuff. given 2 H-atoms, I assume by this language that we are dealing with two distinct subsystems, one for each H-atom. Hence, an energy eigenstate of H-atom 1 belongs to the hilbert space representing the system for H-atom 1. In particular, an energy eigenstate of H-atom 1 is not a state in the hilbert space representation the system H-atom 2. So, these states are not the same.
because by "state" i mean a (ray) vector in a "system". and by "system" I mean a Hilbert space (perhaps with some other structure like accessible observables and so on)
@ACuriousMind The interaction picture is just the Heisenberg picture if we have $H_0$. I am not dealing with asymptotic states here i.e. I am not adiabatically adding the interaction. I am just saying that $|k>$ is an eigenstate of the free Hamiltonian and peturbatively calcualting the corrections to get the eigenstate of the full Hamiltonian.
I find it quite weird how we shift our perspective as to which the quantum system is, i.e in the H-atom. The eigenfuction $\psi(n,l,ml,ms)$ is the mathematical representation of the eigenstate $|n,l,ml,ms\rangle$. This should be an eigenstate of an electron. Because only electrons are characterized by the four quantum numbers, not the atom containing all the electrons.
@SillyGoose
So the system is the electron and it's eigenvector is $|n,l,ml,ms\rangle$. There is no degeneracy of the energy for an electron. It only has one value of energy. But if we shift our perspective to the atom, then the spectrum of the Hamiltonian of the atom is 2n^2 times degenerate. As you can see, the eigenstates give information about the electron (system), but the degeneracy exist only when considering the atom (system).
To me this shift of the perspective as to which is the system is weird. It would make more sense if the eigenvectors would be such that include all the building particles of an atom. In that case the eigenstate would belong and describe the entire atom, and not only an electron which is the case for |n.l.ml.ms\rangle
Also, a question about interpretation. Once we've calculated the corrections to the Greens function through the self energy, we are left with a propagator that has the exact same form as a free propagator just with a shifted mass. Doesn't this mean that we are describing some free theory?
