Lebesgue measure on ${\mathbb R}^n$ is defined as product measure resulting from Lebesgue measure on ${\mathbb R}$. The latter is strongly tied to the features present in ${\mathbb R}$: It gives $[0,1]$ the measure $1$, is translation invariant, and behaves as expected under scaling.
It follows...
according to wikipedia "The Haar measure [...] is equal to the restriction of Lebesgue measure to the Borel subsets of $\mathbb{R}$" en.wikipedia.org/wiki/Haar_measure#Examples
Also in more tantalizing pieces of informations, if you consider the projective line instead of the line, and look at the stabilizer group of $\infty$, then you obtain the affine group, and the cross-ratio is precisely the ratio of two lengths
@fqq The Lebesgue measure is the one giving you $\ell([0,1])=1$
But you can rescale it and it will still be a Haar measure
like, the whole reason the notion of "Haar measure" exists is because that uniqueness gives you an unambiguous way to integrate on arbitrary topological groups
I am thinking there is probably something relating the Haar measure, the pointed projective line $(P^1\mathrm{R}, \infty)$, and the resulting projective tangent bundle
Something about directions being the jet of the projective line into the manifold and projective frames defined similarly idk
and if given a set of appropriate measure we get back all the usual GR apparatus
Hard part is finding people who discuss this in differential geometry term because it seems that the projective geometry people that deal with conics don't like the differential geometry formalism much
and the CFT people throw around the cross ratio a lot but they don't seem to explain much about it
But I have that suspiscion that you can define all the Weyl structure in GR by mapping projective lines into projective space instead of $\mathbb{R}$ into $\mathbb{R}^n$, and then the metric part comes from choices of measure
Something along those lines, anyway
@ACuriousMind Is the famous "conformal boostrap" I keep hearing about a way to pull myself out of the conformal rabbit hole :p
If a bound state is a quantum state of a particle subject to a potential such that the particle has a tendency to remain localized in one or more regions of space. And if a stationary state emerges when, in the simple case, we consider a particle in a box (potential), in a confined region, then, they (bound and stationary) represent the same state for a system. If not, then what's the difference?
@imbAF "And if a stationary state emerges when, in the simple case, we consider a particle in a box (potential), in a confined region," I don't know where this comes from
a stationary state is just an eigenstate of time evolution
there are no assumptions on it being localized, in a box, bound, free, whatever, just that it's stationary in time
I find it hard to believe that you only ever looked at 1d potential wells - surely you looked also at the free particle in 1d, at some finite-dimensional system with spin and at something like the hydrogen atom
In any case, "stationary state" really just means "eigenstate of the Hamiltonian". It has a priori nothing to do with bound/unbound states or localization. Does that answer your question?
@imbAF they aren't the states "in which the system can be found", that phrase doesn't mean anything. They are the states in which the system can be found when do you an energy measurement
@ACuriousMind If you remember, we talked about the SE solutions. You told me that $\Psi(\vec r,t)$ is the general solution of the TDSE, and since linearity holds true for it, it can be expressed as a linear combination of particular solutions $\psi(\vec r,t)$, which, the particular solutions are represented with the ansatz.
And the ansatz can be considered, only for time independent Hamiltonians. The position dependent part in the ansatz, is the solution to TISE. This is what you said, if I am not mistaken. And I understand this. But I have a question:
The representation of the general solution in this way (with the help of the ansatz, and the help of the solutions of the TISE) is only possible if one can use the ansatz, meaning only if the hamiltonian is time independent,
which means that if the Hamiltonian is time dependent, there has to be another representation for the general solution of the TDSE, which is not similar to $\Psi(\vec r,t)=\sum \chi_i(t)\phi_(\vec r)$,which was what you wrote, when the ansatz for the particular solutions of the TDSE, was possible.
The representation you told me about, of the general solution, if possible in that way, doesn't care about whether the hamiltonian is time dependent or not?
we have 2 copies of the same system (2 particles), in 2 different states: One of the systems is in a state, which is the general solution of the TDSE, the other one is in a state, that is the solution of TISE. Is this scenario possible?
the system is in some state, and if you feed that state as the initial condition into the TDSE, then you discover how that state will evolve in time
@imbAF that doesn't actually mean anything - you can use any state as the initial condition for the TDSE, so every state is the $\Psi(\vec r, 0)$ of some solution to the TDSE
the whole point of deriving the $\Psi(\vec r,0) = \sum_i \chi_i(t)\phi_i(\vec r)$ expression is that it makes determining that solution easy - the $\phi_i$ are a basis of states and so you just take your initial state and expand it in that basis as $\Psi(\vec r,0) = \sum_i c_i \phi_i(\vec r)$ and the corresponding solution is just $\Psi(\vec r,t) = \sum_i c_i \chi_i(t)\phi_i(\vec r)$
i.e. solving the TDSE in the energy eigenbasis is "just adding phases" (the $\chi_i(t)$)
two things. At least, this last paragraph was quite insightful. Now I even understand this notation here $\Psi(\vec r,0)$ and the rest
2nd. If you can't make an ansatz, what does that mean, under which conditions is that not possible and how do you find an expression fo the solution of the TDSE?
@ACuriousMind I read that and I understand, in this particular case, when the separation is possible. I simply wanted to know how the solution looks in the frame of a time dependent Hamiltonian.
Since the ansatz is not possible, you can't solve the TISE equation. Or you don't even have to take it into consideration.
@ACuriousMind Adiabatic theorem? New to this. need to check it
Just to explain why I was confused until now. In our lecture when we were introduced to the ansatz method, it was made clear that this is possible for time independent H, and that would imply a time independent potential. Then we took the TISE, and we used it to solve the problem of a particle in a potential well. After solving it, we found the expression for the eigenstates of H. And it was said that the particle can be in one of t)
these states (which can be said after a measurement). So I found it weird, when above, you told me that even in the case of a time independent potential, the particle's state can be expressed as "that thing" which is solution of the TDSE ('that thing'= since you told me the solution of TDSE is not a state, so idk what to call it now
What I know: the hermitian adjoint of a matrix $A$ is defined as $A^{\dagger}=(A^{*})^{T}$. In QFT it's not obvious to me what the hermitian adjoint is, since I can't make sense of $\phi^{T}$
Context: I've trying to prove this:
and it is useful to know that for this exercise we have $\pi = i\phi^*$
the abstract definition of the adjoint is that given an inner product $\langle -,-\rangle$, the adjoint of an operator $A$ is defined by $\langle v,Aw\rangle = \langle A^\dagger v,w\rangle$ for all vectors $v,w$
the transpose is really a "bad concept" because you can't write it down abstractly
it's easy to explain to people when you have matrix representations of your operator but abstractly it only makes sense in the context of the adjoint (note that for real vector space with the default Euclidean inner product the transpose and the adjoint coincide)
that's because "really" the transpose should map an operator to an operator on the dual spaces and the usual naive treatment silently identifies that dual with the original space via the inner product and doesn't talk about it (see e.g. en.wikipedia.org/wiki/…)
where "worse" means "you'll never get a useful explicit representation of this space because Stone-von Neumann no longer applies, kiss your wavefunctions goodbye"
@Feynman_00 no, the field operators are operators in an infinite-dimensional space
For a matrix $A$, the notation $A^\dagger$ implies the transpose of the complex conjugate of $A$ i.e., $A^\dagger=(A^*)^T$.
What does the symbol $\hat{\phi}^\dagger$ mean for a quantum operator corresponding to a classical field $\phi(x)$? Is it okay to think of $\hat{\phi}(x)$ as an infinite d...
Let me phrase better what I mean: $\langle v, Aw\rangle=\langle A^\dagger v, w\rangle$ is the definition of the adjoint except some stuff about the domain, right?
I thought this $(v, Aw)=(A^T v, w)$ could be a definition of the transpose, where $(.,.)$ is a symmetric bilinear product (i.e. non hermitian inner product) but the space is complex
@ShikiRyougi because otherwise you would literally do nothing but "this stuff" for the rest of the year/life, and we don't know how to do QFT rigorously for the interesting cases anyway
causal diamonds are obtained by intersecting the past light cone and the future light cone. They are important in the study of quantum gravity. Can one study analogous diamonds for Riemannian geometry as opposed to semi-riemannian geometry?
@Slereah a causal diamond can be thought of having an input P and an output Q if I understand correctly. I was thinking about 2 causal diamonds intersecting transversally so that you start with 2 inputs of information and those inputs interact as you follow the worldines, a computation is then done where the diamonds overlap, then you measure the outputs at two different points
@imbAF I think the Bloch sphere is helpful in understanding what a (simple) quantum state is, and how it relates to what we can actually measure / observe.
@PM2Ring I mean this is not the easiest to understand : " The pure states of a quantum system correspond to the one-dimensional subspaces of the corresponding Hilbert space (and the "points" of the projective Hilbert space). For a two-dimensional Hilbert space, the space of all such states is the complex projective line {\displaystyle \mathbb {CP} ^{1}.}{\displaystyle \mathbb {CP} ^{1}.} "
I know what a pure state is. A linear combination of eigenstates of an operator
Admittedly, that Wikipedia article isn't an ideal place to learn about the Bloch sphere. But that's often true with Wikipedia science/maths articles: they can be helpful when revising, but they tend to get cluttered with technical details which make them confusing to newbies.
It is usually said that the points on the surface of the Bloch sphere represent the pure states of a single 2-level quantum system. A pure state being of the form:
$$
|\psi\rangle = a |0\rangle+b |1\rangle
$$
And typically the north and south poles of this sphere correspond to the $|0\rangle$ and...
@Feynman_00 it's the definition of the adjoint always but for unbounded operators it can happen that the domain of $A$ is not equal to the domain of $A^\dagger$ (see e.g. physics.stackexchange.com/q/644579/50583). The domain subtlety is that an operator is only self-adjoint when it's equal to its adjoint and the domains of itself and its adjoint are the same (otherwise it's only symmetric)
@ShikiRyougi again, it's worse in QFT, the field operators really are not operator-valued functions but operator-valued distributions. But as fqq said it isn't worth it to worry about all this at the stage where you're learning QFT because that's not the level of rigor at which QFT is commonly practiced
Lebesgue measure on ${\mathbb R}^n$ is defined as product measure resulting from Lebesgue measure on ${\mathbb R}$. The latter is strongly tied to the features present in ${\mathbb R}$: It gives $[0,1]$ the measure $1$, is translation invariant, and behaves as expected under scaling.
It follows...
For a matrix $A$, the notation $A^\dagger$ implies the transpose of the complex conjugate of $A$ i.e., $A^\dagger=(A^*)^T$.
What does the symbol $\hat{\phi}^\dagger$ mean for a quantum operator corresponding to a classical field $\phi(x)$? Is it okay to think of $\hat{\phi}(x)$ as an infinite d...
