@imbAF Skim Section 1 and Section 61 as quick background, then Section 64 for the Bosonic case. The relativistic Klein-Gordon case sort of just modifies things by making your Fock states have creation operators for both particles and anti-particles (with an equal number) so if you understand the non-relativistic case in 64 you can make sense of KG 2nd quantization pretty quick.

KG is then basically just the Schrodinger equation for your single particle state basis, once you know this basis you almost forget about KG, now its all about using those single particle states to construct multi-particle wave function for a system of identical bosons, and (as 64 explains, because of what 61 says) it is more natural to change our variables from coordinates to occupation numbers.

Once we do that we want to define operators which change the occuption numbers, and we're basically immediately led to creation/annihilation operators and quantum field operators, and 64 makes their definitions become obvious including the square root factors, it's basically showing that usual Harmonic oscillator thing is a general feature

Chapter 1 of Walecka's Many Particle Systems book (with Fetter) is very good too, sort of fills in some of the details L&L skip when it comes to actually working out the matrix elements that suggest how we should define creation operators etc

@ACuriousMind I dont see how that can be true; (R) simply asserts that magnetic fields are vector fields, and you can simply have vector fields with singularities like cusps or whatnot

I don't see why the relativistic case should be any different than the non-relativistic one (for second quantization).

@imbAF that's not what you do, as I've mentioned already in my previous answer to your comment...

hi

@naturallyInconsistent The problem is that the question is asked at a high-school level where none of the terms are defined to a degree of precision you could actually prove anything. The relevant theorem is that the integral curves of a smooth nowhere-zero vector field never intersect. The "reason" states something like "the magnetic field is a vector field". Now, does this question expect a high-schooler to go "ah, that's not the explanation because the condition about smooth and nowhere-zero is missing"?

To me, that completely depends on what exactly they discussed in their lessons about this, it's not a question you can judge in a vacuum.

@Slereah They should have called them cornifolds

I mean I guess they do all have corners, sometimes that corner is just the empty set

could relative collapse theories be called a realisation of Kant's ideas

in relative collapse theories, physics explicitly describes ur observations of the world, instead of the world in itself

as in, there is no way to even interpret a relative collapse theory as describing the world in itself, while u could interpret e.g. classical mech or objective collapse theories that way

so relative collapse theories, if true, enforce Kant's ideas

@ACuriousMind yeah especially as multiple-choice question, it's just about memorizing what was said in class

is it true that if i have a CCR preserving transformation on a set of bosonic annihilation operators $U\vec{a} = \vec{a}'$ that I can implement this as a unitary transformation $\exp(\sum_{ij} U_{ij} a^\dagger_i a_j - \sum_i a^\dagger_i a_i)$?

I think no...because the exp operator is not even unitary

@ACuriousMind no, i dont expect high schoolers to intuit that the question is asking for "nowhere-zero vector field never intersect" but that the statement that it is a vector field tells nothing about the avoidance of singularities like as in hairy ball theorem. Students are expected to intuit these things as they are to realise that the plus-one game they automatically reach in grade school implies the endlessness of positive integers.

Suppose $U = \exp(iX)$ is a unitary matrix that preserves CCR in the sense that $\tilde{a}_i : = \sum_{ij} U_{ij}a_j$ obey the CCR.

Is the implementation of this transformation on Fock space $\pi(U)$ defined by $\pi(U)a_i\pi(U)^\dagger = \tilde{a}_i$ given by $\pi(U) := \exp(-i\sum_{ij} X_{ij}a^\dagger_i a_j)$?

@SillyGoose It might be of use to move to bosonic holomorphic representation to try some of these things out. You will then IIRC just have a translation operator or something along those lines

what is the holomorphic representation?

Write the CCR in terms of a derivative operator and a complex variable i.e. $[\partial_{z_i},z_j] = \delta_{ij}$ or something

The inner product changes but maybe you don't need that for this

However in this representation, it is easier to see some expressions since they turn into translation operators

It may be of no help though

I wish Schrödinger wrote a whole textbook series like Sommerfield

@User198 This question does not really make sense if you think about how QM time evolution works. Why would there be a specific time after which the state changes?

@ACuriousMind Because the system is in contact with the environment?

I don't know how that's an answer to my question. I didn't ask you why the system would change, I asked you why there would be a specific time when that starts to happen.

@ACuriousMind Hm. In theory for the isolated systems there would be no change in measurements?

But maybe for systems that are not isolated, the environment influences the system and changes it.

@User198 I mean, rather tautologically, if a system is in an energy eigenstate of the actual, full Hamiltonian, then it will never change.

But for "more isolated" systems the time might be longer untill the influence of the environment penetrates and changes the system.

if the system is not in an eigenstate of the full Hamiltonian, it will start changing immediately after the measurement

Idk I am picturing it like heat

why? :P

that's not what the equations tell you

@ACuriousMind No, but say after the measurment the it is in the eigenstate.

@User198 in an eigenstate of what

If we measure the energy, of the system, after the measurements, we will observe the energy to be one of the eigenvalues of the $\hat H$

Yes?

If that $H$ is the full real-world Hamiltonian, then the system will remain in that state forever

the problem is that you're setting up the question wrongly

full real world hamiltonian is the hamiltonian that describes the energy of the whole universe?

Or what?

The full Hamiltonian that describes the evolution of that system

without any approximations etc.

Hm, ok. But that doesn't exist.

I don't know what "exist" means in that sentence

unless you think QM suddenly stops working at some scale, there certainly in principle is such a thing, even if it is infeasible for us to write it down or do computations with it in practice

Ok

But I am being more pragmatical now

Specifically for your example of an atom that is excited, the "energy" you're measuring is the energy of the atom. But in reality, the atom is coupled to the electromagnetic field, and the state "excited atom + no photons" is not an eigenstate of the Hamiltonian of the system "atom + EM field"

Ok I see

Thank you

there is no specific time after which the interaction with the EM field will destroy the excited state of an atom, that's not how anything in QM works - instead you get probabilities and half-lifes, which are often computed via things like Fermi's golden rule

I was thinking more about when in quantum computing, the stability of qubit

nothing what I just said is specific to atoms

Okok. I think I made a mistake

you always have that realistic systems are not 100% isolated and in reality are just part of a larger system with its own interacting Hamiltonian. If the coupling is weak enough, you can treat the system via perturbation theory and apply things like Fermi's golden rule to talk about the system in terms of eigenstates of its sub-Hamiltonian

whether that system is an atom, which is really coupled to the EM field, or a qubit, which is really coupled to its environment (depending on the realization of the qubit) or something else is immaterial

The mistake I did I think is this: The interactions with the environment, will not alter the measurment result (the eigenvalue of $\hat H$ if we measure energy) , but it might alter the whole state of the system.

@ACuriousMind Ok I see

Thanks

There are actual models of the interaction of quantum systems with the environment btw

For instance a simple one is the quantum system + a thermal bath of harmonic oscillators

It gives you a nice model to explain the decay of entanglement and such

@SillyGoose you might want to check the books by A. Arai

Things escalate quickly, sometimes, it seems

Is a wavefunction (in QM) mathematically an example of a distribution? en.wikipedia.org/wiki/Distribution_(mathematics)

I mean they're functions, so they are also distributions

Or more accurately they can be turned to distributions

In the Debye model using BVK boundary conditions, you can't have any phonons with k-vector shorter than 2pi/L but that's generally fine because those states are very low in energy and have low DOS anyways, right

@Slereah Ok thank you

@ACuriousMind What are you currently interested in?

/working on

@User198 What do you mean "akin to CCR". The SvN theorem shows the CCR have no finite-dimensional representation, what exactly is the question?

@DIRAC1930 I'm not in academia, what do you mean?