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2:11 AM
1 hour later…
3:30 AM
ugh what
1 hour later…
4:38 AM
@naturallyInconsistent Hi, I have a question
How is a mode related to a state?
5:32 AM
@imbAF I have answered your this question rather multiple times already. The concept evolved sensibly over time. First, we had normal mode oscillations in classical mechanics, and it made sense there. Then we characterised all single oscillations by calling them modes in all kinds of scenarios, including in all the cases of EM waves that you have written down, TEM, TE, TM, etc. When it came time to quantise them all, each mode became its own QHO, and so each mode + spin = state.
6:03 AM
@naturallyInconsistent I think I have a grasp of it. The electromagnetic field can be expressed as superposition of modes, depending on the setup considered we have different modes i.e plane waves in free space, waveguide modes in waveguides etc. Now, the modes can be thought as quantum harmonic oscillators, and they can have different photon numbers. The state of a mode can be expressed as a superposition of fock states, and there's a probability to have an arbitrary nr. of photons
Is this understanding somehow correct?
6:16 AM
@imbAF I think the term mode is used differently for classical EM and QED. I think in QED a mode is a Fock state, which is typically an infinite plane wave.
How is that possible when a mode can have an arbitrary nr of photos while a fock state has a fixed one?
You can have x nr. of photos of a certain wavelength in one occasion and y in another
In classical EM, e.g. in waveguides as mentioned in the question you linked, mode means a normal mode of the oscillation.
and wavelength is one of the characteristics of a mode
Hm... I am confused right now
A Fock state can have an arbitrary number of particles. In fact I think it's an eigenstate of the number operator.
yeah but it can't have an arbitrary one than
|2\rangle can only have 2, |3\rangle can only have 3
6:21 AM
OK, I'm getting out of my depth here. You're saying the mode is a momentum eigenstate and a Fock state is the combination of mode and a number of particles?
but for a specific discrete value of $\lambda$ one could have as many photons as one wants, which would mean that the mode, is represented by a state, and the state of it can be expanded as a superposition of fock states, and $p(n)=|<n|\Psi>|^2$ would give the prob. of finding n photons in the mode. To me this makes logical sense, but maybe I am wrong
Oh, wait, ignore me.
Yes, a Fock state is a tensor product of modes with different numbers of particles.
For example @JohnRennie in my lecture we say that a mode can be thought as a quantum harmonic oscillator. Now by having that in mind, a qho can have different excitations, for different n, so the same logic can be applied to a mode. And since a fock state, an eigenstate of the nr. operator, has a fixed nr. of photos, this directly needs to imply that a mode is not a fock state
I think I am out of my depth so I should probably stop speculating :-)
I am trying to understand what "change of Coordinates" mean from a mathematical point of view. I am not finding well formualted or answered Similiar questions. The best i found is this:


I want to understand what are we treating our physical "play space" as. Is it a vector space in the sense of linear Algebra, where i am switching "Base" when i switch "coordinates", the replies to that question seem to indicate it is not.
6:36 AM
The coordinates are the charts
The change of coordinates is a transition function
So it is in the sense of manifolds, not per se Vectorspaces.
what is a good book that adress similiar type of questions
If you mean in the sense of vector spaces, the coordinates are a map from the vector space to $\mathbb{R}^n$, which is defined with respect to a basis
And a change of coordinate is a change of basis
But a Manifold does not need to be a vectorspace.
What do you mean then?. You can consider them as "both"?
6:55 AM
what is the physical interpretation of the amplitude of a coherent state?
@Mad If your manifold is flat, there are some special set of coordinates which are equivalent to the coordinates on a vector space
@Slereah Okay so this is a special case. But Generally speaking, the more general notion is to consider it as a manifold with charts.
Alright, that is really helpful. Thank you.

can you suggest a book that adresses these topics? or from which you learned these stuff. :)
I mean GR books I guess?
Or if you want a classical perspective you can try Moretti's book on classical mechanics
IIRC he has a big section on how coordinates work
7:54 AM
Alright thank you
8:15 AM
what got u into physics
do u think some past decade was objectively better to live in
i think things have improved compared to other times
1 hour later…
9:31 AM
@imbAF You are technically correct. Very pedantic. A mode fixes $\lambda$, oscillation direction, etc. The number of photons in that one mode is left for another quantum number to choose. This is in exact agreement with the case in classical mechanics; the amplitude of a normal mode is left arbitrary when we defined normal modes.
2 hours later…
11:27 AM
@Sanjana what was the outcome of this question finally?

@Arjun hey
How does reduce the 3-D Schrödinger equation to lower dimensions? For example if say i start with the 3-D equation and want to solve for a particle confined to a line or a plane,how does one reduce it to the respective dimensions?
How do u confine it to a line?
Like have an infinite potential well of 0 thickness?
Um..v=$\infty$ if y,z$\neq$ 0 and some V(x) on the line ,here x axis
11:37 AM
@Arjun Why are you trying to "reduce" it instead of just starting in 1d?
when we talk about particles confined to lower dimensions, the whole point is that we disregard any potential other dimensions
@ACuriousMind considering the 3-D one to be my og postulate and then I wanted to see the lower dimensional versions of it as sub cases instead of them being independent postulates
@Arjun I don't quite understand what the "postulate" here is supposed to be
the Schrödinger equation is just $\partial_t \psi = H\psi$, and the form of $H$ as $H = p^2 + V(x)$ for arbitrarily many momenta $p$ and positions $x$ in however many dimensions you want is something we take for granted in classical mechanics all the time, and that's where your 3d Schrödinger equation comes from
I would not grant any special status to the 3d one-particle version of this model
@ACuriousMind what I'm trying to say is that the 3-D version of schrödinger equation somehow contains the lower dimensional versions of it..and I'm trying to figure out a way to bring the lower dimensional versions of it from the 3 D one
@Arjun I don't know what you mean the 3-D version contains the 2-D version
@Arjun and what I'm trying to say is that this assertion " the 3-D version of schrödinger equation somehow contains the lower dimensional versions of it" isn't really true!
11:45 AM
@ACuriousMind ... I vaguely remember discussing with you ... chat.stackexchange.com/transcript/message/64663187#64663187
Do u have any recollection?
@ACuriousMind Oh! Fr!? Then are we supposed to consider the lower dimensional versions of it as independent axioms to the theory?
@Arjun no - the other half of what I'm saying is that there isn't really an axiom specifically about the 3d version
there's a general assumption we have about the form of Hamiltonians of physical systems, and your 3d case is just one example of it
I had some serious bad health back then ... :/
@ACuriousMind thanks for that..so if I got it correctly there is some general assumption regarding the Hamiltonian in any no.of dimensions..and all these 1d ,2d,3d cases are just subcases?
at least that would be my viewpoint
11:52 AM
@ACuriousMind you would be my best bet as of now : )
the problem is that while in classical mechanics there's no problem with just imposing $x_z = 0, p_z = 0$ to get something confined to the x-y plane, doing this in quantum mechanics in a technically correct way is a bit more involved and I see no need for that
@ACuriousMind I see
@ACuriousMind Is there some position operator defined for a particle in a circle?(Defined in some contrived way since the usual one doesen't work)
also note that becuz of the uncertainty principle, one cant confine a 3D quantum particle's wavefunction to a line
the closest u can get to ignoring the other degrees of freedom is in the Heisenberg picture
but not really
yeah one cant ignore anything in the Heisenberg picture either
in the classical case, one can have $dy/dt=0$ and $dp_y/dt=0$, if one chooses the potential and the initial conditions appropriately ($p_y (0)=0$)
but in the quantum case, there is no such thing as choosing $p_y (0)=0$
12:13 PM
@RyderRude So the 1-D and 2-D problems that we encounter in textbooks are all idealizations in the sense that zero uncertainty in position would mean infinite amounts of uncertainty in momenta(for those dimensions that are prohibited for the particle)..did I understand you correctly?
@Arjun when you look at reality there are no 1d or 2d objects, really
that 1d or 2d physical settings are idealizations is not a special property of quantum mechanics, this has always been the case
@Arjun yes
the scales at which the uncertainty principle becomes the problematic aspect of this idealization are below other scales where it breaks down anyway, e.g. you say you confine a particle "to a ring", but what's that ring made out of? what's confining the particle there with such precision that the ring is supposed to be 1-dimensional?
also note that, even if a free particle was initially sort of confined along a line, it wud not evolve in time according to the 1D Schrodinger eqn. the Kinetic terms of the other axes would still be relevant. the particle wud spread out along the other axes over time
@ACuriousMind reminds me of my electromagnetism days,which was full of idealizations like point particles,line charges,surface charges etc
12:20 PM
one way to keep a particle sort of confined in its time evolution is using chamber walls
@Arjun the presence of idealizations is a characteristic of physics in general, not of specific subfields
it's how we make problems tractable by eliminating the complications that don't impact the result we're trying to get at to a relevant degree
@ACuriousMind When you say "the scales at which the uncertainty principle becomes the problematic aspect of this idealization are below other scales where it breaks down anyway" do you mean uncertainty principle will NOT work beyond a certain scale?
what I mean is that you have to make the uncertainty in position really small to get an uncertainty in momentum that's large
but when you zoom in that closely that such uncertainty is not "effectively zero", you've already zoomed in past other problems with the idealization
1:30 PM
it is for a goddess called Nikkal
this is a 2500 year old song with lyrics
Early songs didnt use to have a fun element, it seems
all these r supposed to b enjoyed with eyes closed
@naturallyInconsistent For a coherent state, what is the physical interpretation of the amplitude and of the phase?
The coherent state is associated with the system which is the waveguide / cavity or a mode of the field operator?
2 hours later…
3:28 PM
@imbAF i dont think coherent states are parametrised by amplitude and phase?
But they are. They are given as $|\alpha\rangle=|\alpha|e^{i\theta}$
where $|\alpha|$ is the amplitude and $\theta$ the phase, which it's said that is the result of the uncertainty in nr. of photons and amplitude
which i don't understand what it means
or how that causes the accumulation of a phase
to be more precise
That is not really how it goes. The parameter is $\alpha$ itself, a complex quantity. Its real and imaginary parts has interpretation in QHO form, but if you want to generalise it to photons, it is less and less useful to be considering them as real and imaginary separately.
I am not aware of anybody considering it as magnitude and phase. Note that I am trying to avoid the word amplitude because that has too many possible other meanings here, whereas magnitude is always non-negative.
And you are correct in trying to avoid the word amplitude because it has multiple meanings
For example I know, the probability amplitude
in quantum mechanics
and the amplitude, as displacement from posiiton of eq. for a harmonic oscillator
But here
exactly why im avoiding that
@naturallyInconsistent In optical phase space, the $\alpha$ of a coherent state is indeed related to the amplitude of the electric field
that's very likely what imbAF is referring to
3:36 PM
@naturallyInconsistent I don't know if you can see the highlighted part :
and the phase of the $\alpha$ is also related to the phase of the field - for $\alpha = \lvert \alpha \rvert \mathrm{e}^{\mathrm{i}\theta}$ we have $\langle \alpha \vert E(t)\vert \alpha\rangle = \lvert \alpha\rvert \cos(t+\theta)$ (modulo a bunch of constants)
But I don't know what I am supposed to think of, when talking about amplitude in quantum mechanics, or in particular in quantum optics. For example for the QHO I don't know what I am supposed to think of or understand with amplitude. What is the amplitude of qho , the same as that of a classical one, which is a displacement in space?
@imbAF it depends on the context!
and whenever someone talks about "amplitude" you need to look at the context
And what are the contexts?
Like probability amplitude, I know that
sometimes they mean a probability amplitude, sometimes the classical idea of amplitude of a wave, etc.
3:40 PM
what else is there? or in the case of quantum optics?
Just one question
@imbAF look at what I wrote up there about the amplitude of the electric field
Note that I have preempted this: I already said that I think of it more sensibly as a general complex object. The QHO version of it differs from the photon version of it, so one might need to expend some effort to work out what is being meant in each case.
@ACuriousMind an oscillation ?
since the expectation value of $E$ comes out to be $\lvert \alpha\rvert$ times a $\cos$, this means the $\alpha$ is what you would classically call the amplitude of this wave
this is important to understand, since it shows that the coherent states are the correct quantum equivalent to the classical electromagnetic wave (and not the pure photon states where you have $\langle E\rangle = 0$!)
to a good extent, it means the quantum state $\lvert \alpha\rangle$ corresponds to a classical EM wave with amplitude $\lvert \alpha \rvert$ and phase $\theta$
But the amplitude, classically , is the disturbance of the medium/space, I mean it doesn't make sense to say medio for em waves.
But the amplitude of the electric field,
so that's what $|\alpha\rangle|$ is then
3:44 PM
The amplitude is just the number $A$ in a function of the form $A\sin(t)$ or $A\cos(t)$ :P
this is elementary classical waves/oscillators
I don't understand what you seem to find so difficult about this usage of the word
I am ok I get it
so this is the use of the amplitude in this case
But, would a coherent state have also a probability amplitude. Would that make sense ?
A "probability amplitude" is just an inner product between two states
so of course
you seem to invest these words with far more meaning than they have :P
@ACuriousMind what I find difficult is to materialize what A is physically. For a classical matte r wave, i.e waves when you throw a stone in the water, the amplitude is the highest or lowest point of the wave front. But for EM waves, I have a hard time picturing the amplitude. So I just take it as the largest value of the electric field, without giving it a physical meaning
@ACuriousMind It's because I do that, that I could tell that the use of amplitude needs to be precisely explained in a particular case.
3:48 PM
It is kinda important to be careful that "probability amplitude" and "transition amplitude" are a different kind of thing compared to what I have been calling the magnitude. Much less confusing choice of terminology
@imbAF what about "the largest value of the electric field" is not "a physical meaning"?
sounds very physical to me
how does that translates physically
Im with ACM on this one. "Largest value of the electric field" is more than enough physical meaning to meow
I don't know what you want here - the electric field is a physical thing just as much as anything else
there's nothing magical or unphysical about it - ever since we stopped believing in the aether, the electromagnetic field on its own is a basic fact of reality just like everything else
One more thing
@ACuriousMind I had a brainfart for a moment. Now I understand lmao
May I upload a picture here?
Am I allowed?
3:56 PM
uh, of course?
Depends on the picture
The description is:
Phasor diagram and temporal evolution of the electric field real part for multiple states. Upper: Coherent state with minimal uncertainty. The phase uncertainty is shown through the possibility of passing E(t) = 0, which can occur for a period of time. Middle: The vacuum state we know as the coherent state (amplitude equals 0). We have a certain noise present, which will be there all the time. Lower: Fock state. We have zero uncertainty in the
amplitude but therefore full uncertainty in the phase. We therefore know exactly the length of the red vector, but not at all its orientation. The Fock state therefore fills the yellow circle completely. We now have a noise band different from the vacuum state. Oscillations do occur here, but they occur with all possible phases at the same time.
I have a basic understanding of all three states
I don't understand the thing with the orientation of the fock state
and where does the full uncertainty in phase comes from?
Though @ACuriousMind if you were to plot this : $\langle \alpha \vert E(t)\vert \alpha\rangle = \lvert \alpha\rvert \cos(t+\theta)$ would you get that temporal profile?
What I am trying to say, just by plotting that, could you tell that there is a minimum uncertainty ?
how could you?
that's just $\langle E\rangle$, what does it have to do with uncertainty?
@imbAF the orientation of that vector is the phase. And phase and amplitude are conjugate, so they fulfill an uncertainty relation like position and momentum do.
@imbAF what what? Uncertainty of an observable $A$ is $\langle A^2\rangle - \lvert A\rangle^2$. I'm confused therefore why you think you could tell anything about the uncertainty in $E$ just by looking at $\langle E\rangle$
4:05 PM
is the phase/amplitude relation from time-energy for the E field
why would it be?
In the quantum mechanical case, the state can be seen as a finite size disk that size is depending on the uncertainty of the amplitude and phase.
I don't know what it means for a state to be a "finite disk"
So I believe that is how the uncertainty is presented
the phasor diagram is not "the state"
It's just a representation of some of its properties
4:06 PM
. In the classical case the state would be a point on the circle and oscillation would mean its movement along the circle. Here we have a sharp amplitude and phase.
the state itself is a still just a vector in Hilbert space
And what about the description of the fock state where it says:
this still just quantum mechanics
But why do we have uncertainty in the phase for a fock state? In my lecture, that wasn't pointed out or explained. It's just taken as a mere fact in the description there
Have you tried just computing the phase uncertainty for that state?
4:09 PM
he's saying it's an uncertainty relation in QM
@ACuriousMind for the fock state?
@imbAF yes?
like, why would you be vaguely wondering about "why" its uncertainty has a certain value when you could just try computing it or estimating it through the phase-amplitude uncertainty relation
I don't know how to do that. I also do not know how the phase for the coherent state came to be. In the lecture, this was the only explanation, for the case of the coherent state:
Through the number and amplitude uncertainty, we obtain a phase uncertainty.
What do you mean you don't know how to do that? You should have a definition of the Fock state and a definition of the phase operator, no?
otherwise I have no idea what you're even doing here, looking at basic quantum optics results without having any way to derive them
this looks like a grad course in modern optics, so QM should be a prereq no?
4:13 PM
I am here because that's what is provided in my lecture, which I am attending. I don't get to tell to the professor what he ought to do
The expression for the fock state is given as a function of the vacuum state and the $\dagger{\hat {a}}$ operator
Yeah, this is also why I said you have to consider the QHO as in QM; if you do, then you will see that the coherent state's real and imaginary parts refer to position and momentum, and so they have to satisfy some commutation relations, perfect for deriving whatever relations you now need.
And I don't even know that a phase operator exists. You just told me that
@imbAF if there's no phase operator, what would "uncertainty" even mean?
an uncertainty is always an uncertainty of an *operator*/observable
like momentum and position
look, maybe your lecture really is bad, but you seem to have only the weakest grasp of basic quantum mechanics or EM but you're trying quantum optics, and it seems your source assumes much more familiarity with the basics than you have
you either need to find a source more at your level (which I'm not sure exists) or go back and get a firmer understanding of quantum mechanics
4:16 PM
Yes because my lecture, is involved in QED, which I assume is related to 2nd quantization. something which I will learn the upcoming semester. But things are such that I had to attend this lecture this semester
so I am ahead in phase
what book are u guys using
im willing to bet some money that you will only be doing cavity QED, which is very different from QED proper
@naturallyInconsistent bingo
@imbAF sorry, but if you're ahead of the intended sequence of lectures then you don't get to complain that the lecture doesn't explain things - because it assumes you have learned them in the prerequisite lectures!
Not necessary since participants there are from other majors
so, it's necessary to explain stuffl, since we started from i.e fresnel equation etc
4:19 PM
what is scripts? can u tell me the book title im curious
it's a script so not a book
@Obliv "script" is a direct transliteration of German Skript, which means "lecture notes"
well, i don't have any book to recommend that isn't under your level I'm afraid
but finding a pdf online of a relevant text would help immensely I'd imagine
but I'm guessing if your professor is using notes for the class then such a textbook doesn't exist
@Obliv ????
the quantization of the EM field is an extremely common topic
of course textbooks that cover this exist
What is funny is that, I am told I don't have basic QM knowledge but for example, the way I got intruduced with uncertainty, minimal, wasn't by involving a phase operator, but simple by considering a wave packet example, and coming into the conclusion that there is a minimum uncertainty in measuring position and momentum
4:23 PM
I mean right? Like why wouldn't he at least reference books that contain the topics
even if the path isn't the same
That was the only encounter with uncertainty, in the context of qm. So how can I know about phase operator?
usually I would expect lecturers to talk about recommended reference material in the first session of the course
@imbAF 💀 that's rough
@imbAF If that's your only idea of what uncertainty is, then you really do only have very basic QM knowledge
Thats not my idea, but rather what I was told
4:24 PM
because the notion of uncertainty of an arbitrary observable (be it position, momentum, spin, phase, or anything else) should be part of QM; I'm certain every of the common intro books on QM discusses this at least once
And we did. I also gave the example of how I encountered
And for what was explained, I have an idea. But if what was explained is a particular specific example, it's pretty hard to extrapolate that to a general case
that's why I prefer to be as general as I can, in whatever I study, so that then, I can easily go to a specific case, characterized by specific conditions/ set up
just logical to proceed that way
Normally you don't do only one specific example for uncertainty; really if your lectures sucked I'm sorry for you but you need to read a proper QM book then
nvm I realized that example was from your QM course I think
Such as? I mean, I could read it once I finish this current semester, but I wouldn't be able to keep up with quantum optic lectures and reading that book
any of the common ones? Sakurai, Griffiths, Shankar, doesn't really matter all that much
4:30 PM
I just realized my dishwasher detergent package says quantum on it
do you think that I can learn QM from reading the back of the package
Having to read an introductory book in QM, while being undergraduate is actually extremely depressing xD
@Obliv everyone knows that before you open the dishwasher your dishes are in a superposition of dirty and not dirty
It's not for high school..
Cuz it implies, I didn't learn jack s... in 3 years
4:31 PM
@ACuriousMind Actually so true, I will never know if it did its job properly.
Which is kinda fair, as a bachelors feels like wikipedia intoduction
intro QM is typically taught in 3rd year of undergrad and many students take it 4th year as well
depends on school and program
Ok, that is kinda comforting
I'm really sad I have to learn programming and work at a place like SAP for physics students that didn't go to academia, because that will put a very large pause on my physics&math learning and I hope I don't forget everything by the time I pick it up again
(no offense ACM)
@imbAF the kind of theoretical quantum mechanics I'm talking about is usually a mandatory course in the 4th or 5th semester here in Germany, I'm a bit surprised you seem to never have had one
4:34 PM
I mean I should just balance it tbh, learn/keep rust off while I learn to code
@ACuriousMind I had QM for 3 semesters, all lecture notes
and in 3 semesters none of them defined abstractly the uncertainty of an operator, or taught you the general uncertainty relation?
time for a 4th
what did you do for one and a half years?
And the issue with reading a book, is that the same example for a concept which is hard to understand that was explained during the lecture, now, not only you have to understand what the book tells but also adjust it to what was taought in class
@ACuriousMind QM1, QM2, quantum statistical mechanics from the theory department
4:36 PM
Ideally you shouldn't have to adjust it to class aside from notational differences?
yeah, I don't buy a quantum statistical mechanics course not teaching you about uncertainty in any other way than by example of a wave
Or I guess the depth/context
@ACuriousMind you have to be more precise with what you mean
Heisenbergs uncertainty principle was intruduced to as, exactly the way I described
he is not convinced a QM stat mech course wouldn't go into uncertainty in any other way than by example of a wave
The physical intepretation was that one can't measure momentum and position simultaneously. That applies to all conjugate quantities
4:38 PM
@imbAF I mean that I don't buy that you've never seen the definition $\Delta A = \langle A^2\rangle - \langle A\rangle^2$ for the uncertainty of an operator $A$ in one-and-a-half years
Of course I have
@imbAF is that the physical interpretation? :O I don't know anything about QM I thought it was just math and no physical interpretation :P
@imbAF so why did you claim you have only ever seen uncertainty in the context of a wave and position and momentum???
minimal* uncertainty, as in the context of Heisenbergs uncertainty
and why were you surprised that there's a phase operator when you had been talking about "phase uncertainty" all along so clearly there must be a phase operator $\phi$ and the uncertainty is $\Delta \phi = \langle \phi^2 \rangle - \langle \phi\rangle^2$
4:40 PM
the phase uncertainty, which you can expresse it as
$\Delta \phi = \langle \phi^2 \rangle - \langle \phi\rangle^2$
In the lecture notes, the description as how it arises was:
Through the number and amplitude uncertainty, we obtain a phase uncertainty.
I understand perfectly the uncertainty of photon nr. in a coherent state
having a poisson distribution
And I have the calculations for that thing. So, that is why I am clear on this uncertainty
$(\Delta n)^2= <n^2>-<n>^2$
The amplitude uncertainty, is just said
that's it
that it exists
Do I have to take it at face value ?
the text clearly expects you, after having done the computation explicitly for $n$, to be able to perform the analogous computation for the amplitude
I did it for <n>
okay, so what problem do you have in doing it for the amplitude or phase operators?
the "amplitude operator" is just $E$
isn't it $|a|$ ?
if you understand all these technical aspects and your problem is just that you don't know how to do the computation, I don't understand why all this started with strange questions about the "physical meaning of amplitude" and all that
@imbAF how is that an operator, let along a self-adjoint one that could be an observable?
4:46 PM
@ACuriousMind Because I believe, giving a physical meaning to what you measure and calculate is important.
physical meaning comes from ordinary intuition in macroscopic world though and microscopic variables are not going to make sense "physically"
as callen said, the macroscopic variables are what survive the temporal&spatial averaging and what we "see" so QM studies the other variables
I shouldn't say it doesn't make sense physically, but I think understanding the math&results are more important than trying to imagine what words mean in QM context
If a system, a particle, behaves like a QHO, then you can think of the following: In the QHO two different eigenstates have different energies, for the particle in those two different eigenstates, what would one "observe" is that when the particle is in the eigenstate of higher energy, the frequency of oscillation around the equilibrium point, is the
same as in the case when the system is in the eigenstate of lower energy, but the difference would be that in the higher energy state, the displacement would be larger
that's how once can interpret, I guess the, different eigenstates of QHO
apply that in quantum optics, where modes can be associated with oscillators
@imbAF no, this is absolutely incorrect since the energy eigenstates are stationary states - they don't "oscillate around the equilibrium point" at all in any way you could observe
they do not involve in time, any "physical meaning" you think you have where you have to talk about something changing or oscillating is just wrong
if you think this kind of physical meaning is important, it is much more important to make sure the story you tell does not contradict the actual facts in the formalism
Maybe I should have said a superposition of those
but I guess that's not what I wanted to point out
@ACuriousMind What you mean is $\langle \alpha |E^2 |\alpha\rangle$ since the expectation, you already gave it earlier
and then, the expression for the uncertainty
right ?
is it bad I'm learning QM thru wikipedia
4:58 PM
the computations for the picture you initially posted are mostly performed e.g. here, it even draws the same kind of picture for the coherent state
from wiki, a stationary state is the eigenvector of energy operator so that leads me to think energy operator is a linear transformation (of hilbert spaces?)
@Obliv yes
conjugate pairs for uncertainty relations makes me think that hilbert spaces are complex linear vector spaces, which u said before were infinite dimensional?
because it leads to you asking questions like that that you would never have if you learned it in a pedagogical fashion (i.e. from a lecture or book) :P
@ACuriousMind I wouldn't disagree
5:00 PM
ok ok
Hopefully, If I can finish this : chrome-extension://efaidnbmnnnibpcajpcglclefindmkaj/fisica.net/mecanica-quantica/… by the next semester things will be cleared
I mean the fragmented knowledge I have, might bring a full picture
that's a good one. I'm going to do ballentine, people here seem to like it. also Landau&lifshitz vol. 3 is highly coveted
@MoreAnonymous I never asked it here. Let me say it now...
@ACuriousMind the article was the the munich university ?
I just grabbed a random Google result, you can find the same thing in the dozens if you just search for things like "quantum electromagnetic field coherent state"
5:03 PM
So hii everyone, MoreAnonymous's question is simple: In curved spacetime, different observers have different vacuua: does superposing vacuua corresponding to different observers mean anything at all or are those superpositions not allowed by some superselection rule?
And it doesn't even have to be curved. Is there any meaning to a superpostion of Rindler and Poincare invariant vacuua in flat space?
Btw, when you found the expectation value of the field operator, what expression you considered for E ?
@Obliv yay ballentine
@Sanjana Haag's theorem strikes again, making this question about as well-defined as asking whether one can superpose the free and interacting vacuum
the dirty secret of all those neat computations for Rindler vacua is that the Minkowski vacuum is not actually part of the Rindler Fock space (and vice versa)
5:45 PM
@ACuriousMind Hmm. That's true.
Btw I have always looked at superselection rules as extra ingredients in a theory. Are they derived from some more fundamental principle in any context?
I saw Wightman and Streater, but couldn't find anything related to that
a "superselection rule" is just a name for the state space of your theory being reducible as a representation of the algebra of observables
it's not an "extra ingredient" - presumably you did some physical argument to get to that state space in the first place
Yeah, I was referring to that very physical argument as the extra ingredient. These are motivated from experimental observations, right?
uh...depends on the theory?
I mean, you can find plenty of theoretical constructions that have nothing to do with experimental observations :P
@ACuriousMind Yes, I was looking exactly for one such example
I feel somehow that this might be something which I already know of but unable to relate
I mean... I want to see some theoretical argument whose concluding sentence looks something like "This is why we don't superpose these states in practice"
I'm not sure what the question is exactly
a superselection rule doesn't mean "you're not allowed to superpose these states"
it's that you cannot observe coherence between the elements of such a superposition, because no observable is able to distinguish between $\lvert a\rangle + \lvert b\rangle$ and $\lvert a\rangle + \mathrm{e}^{\mathrm{i}\alpha}\lvert b\rangle$ where $a$ and $b$ lie in different superselection sectors (=irreps of the algebra of observables), and so you can't tell the "superposition" apart from a mixed state, making the concept pointless in this case
5:55 PM
@ACuriousMind Bad wording of mine again. I meant exactly this "in practice" :p
so what are you looking for? that this (superpositions cannot be distinguished from mixed states) holds when a and b lie in different irreps is a straightforward unpacking of the definitions
When somebody says, we don't see superposition of states with different electric charge that is why it is not useful to look at such superpositions, they are using the fact that we don't see them in nature.
I am wondering whether there is some more fundamental/more theoretical reasoning to "derive" this.
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