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10:10 AM
If one considers particles as excitations of the light field, if we were to have or consider 2 or more particles with different energies, could one argue that these excitations belong to the same mode?
10:47 AM
Goldstein in his derivation of Euler equations in chapter 2 of his book assumes that y(x) is twice differentiable..it looks as if it is a sufficient condition..but is it necessary? Here he is trying to extremize the integral $\int_{x_1}^{x_2} f(y,\dot y,x)dx$,and he simply assumes y(x) is twice continuously differentiable,that made me wonder if it was necessary..
 
1 hour later…
12:02 PM
@imbAF different energies means different $\omega$ which then cannot be the same mode
But a mode state can be expressed as superposition of fock states. And how can you differ between fock states, unless you are considering diffierent photons?
Otherwise you'd differentiate between different photon nr. of the same wavelength. How can you make that differentiation ?
You can superpose almost anything, IIRC
But the problem remains, how do you differentiate between the different fock states, if the photons are of the same energy.
I don't know how to put it into words
but basically the $|1\rangle$ fock state and $|23\rangle$ fock state
have different photon numbers
Having same wavelength is not sufficient to fix a mode. For example, it could be vertically polarised v.s. horizontally polarised. Then those are different modes and you would not get to assert that those are indistinguishable photons. The whole point of talking about modes down to the specifics is that we can then say that the photons in one mode are indistinguishable from each other
@imbAF aren't these obviously different and you wrote it as different? What is your problem?
They have different photon numbers; that should be good enough
But what I am trying to say is
why stop counting at 1 for $|1\rangle$ and at 23 for $|23\rangle$
There must be like some sort of timespan or interval or something similar
that is the deciding factor, for the classification of the fock state
12:10 PM
I dont know what it is you are trying to draw attention to
I will give a very silly example. I don't know how true it holds
much prefer silly examples than weird roundabouts to nowhere
just shoot
You are considering an arbitrary mode. Whose state can be decomposed to fock states. Let's take the 5th and 123th Fock states from all the possible ones. One has 5 photons and the other one has 123 photons. The photons are of same energy, wavelength etc. So if the photons are identical in all regards, for the mode considered, then what is the deciding factor for
assigning 5 photons to one and 123 to the other. Like, there must be some condition that is fulfilled that you assign a certain nr. of photons. And as a result, you represent the state with that number. The index of the fock state is because for some reason you grouped an arbitrary nr. of photons together, and not the other way around.
This is the best I can do, in explaining what I can't understand
Which is the criteria that makes us group 5 photons together, and not more
I believe that has to do with the excitation
If 5 photons are created, all at the same time, you are having the 5th fock state
I think you are grossly misunderstanding what is happening.
Please explain
12:18 PM
It is not 5th Fock state and 123th Fock state
It is that when you have a cavity, say, and let us only consider only TEM modes with E field aligned along x axis only, etc, so that now we only have one single degree of freedom left.
Then the photons are expressed in states like $\left|\omega=\frac32\text{ has } n=3,\omega=\frac72\text{ has }n=8,\text{ others zero}\right>$
what?
That is, energy eigenstate $\omega$ that defines the mode (and wavelength and so forth), is being expressed as having $n$ photons in that mode, and then you state how many other modes have how many other photons in them, so on and so forth, until no more is stated, or in the case of coherent states, you specify how that filling scheme goes to infinity.
The labelling of which Fock state is being considered, is by $\omega$ and polarisation vector $\varepsilon$ and so forth.
This is why I have been trying to get you to consider normal modes as in classical mechanics, because this is literally that
@naturallyInconsistent if this is true, that the labeling of the fock state considered, is by $\omega$, when you decompose the mode state in fock states, aren't you considering different $\omega$'s ?
@imbAF I wrote down $\omega=\frac32$ and $\omega=\frac72$
@imbAF This might be tangential to what you are worried about, but a source that I came across recently that might be helpful to you is The Quantum Theory of Radiation by W. Heitler. Make sure you pick a later edition however. It explains the physical interpretation of the quantized electromagnetic field etc, the differences between how to interpret the quantized Dirac field vs electromagnetic field etc.
 
1 hour later…
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1:38 PM
@imbAF you're not "assigning" some photons to one state or the other, you just have states with different number of photons. Their superposition does not have a definite number. All of this is not specific to photons, when you restrict to one mode it's basically the same as the simple harmonic oscillator
I understand the part of the mode being the analogue to a qho, and the eigenstates of the qho are the fock states for a mode. But that doesn't really help me understand, the criteria for grouping a certain nr. n of photons together and not n+1, like in the case of QHO were the different eigenstates, differ from the different probabilities distribution, but I'll take it as it stands
what criteria are you talking about? What are the equivalent "criteria" by which you take a state of the QHO as $\lvert n\rangle$ and not $\lvert n+1\rangle$?
they're both just different states, I really don't understand what you think is missing here
In the qho if you would integrate the wavefunction of the 3rd excited state over the confined region, you'd get a standing wave, and that would represent the spatial probability distribution of finding the particle right?
Is what I am saying correct?
basically $|\psi|^2 x dx$
is your problem that there's no position here?
Here you can look at e.g. $\langle E(x)\rangle$ or $\Delta E(x)$ in those states, and you get different results for different states - the thing that is being described here is the EM field, not a particle with position
Ok let me put it in another way. Perhaps I can show you my problem. And I am confident that I am not overcomplicating things
when you consider two fock states $|n\rangle$ and $|n+1\rangle$. One, pretty noticeable difference is in the photon nr. clearly
And just to keep in mind, we are in a fixed mode, with fixed, $\omega$, $\lambda$ and polarization
Out of curiosity is there a difference in the way of how the photons are created? If that makes sense to ask? Meaning, $|n\ranlge$ means that n photons are created simultaneously, while for $|n+1\rangle$ n+1 are created simultaneously.
Is the way I am expressing myself correct or it doesn't make sense?
That's all I need to know
1:51 PM
it doesn't make sense to me
there is nothing "created" here - $\lvert n\rangle$ is just a state of n photons being there
and if you were to make a measurement, you'd find an arbitrary nr. of photons for the mode
would this make sense to say?
I don't know what you mean by "arbitrary"
Assuming that the mode state is expressed as a superposition of fock states
not a fixed one
if you have the state $\lvert n\rangle$ and you measure the number (somehow), you get $n$
since $P(n)=|<n\Psi>|^2$
@ACuriousMind No, I don't have that state, I have an arbitrary state $\psi$, which can be a superposition of fock states.
That's what I am having
1:55 PM
I don't know why we are not considering an arbitrary state :P
your question started being about the difference between $\lvert n\rangle$ and $\lvert n+1\rangle$
in the context of them being used when expressing an arbitrary state, as a superpositon of fock states
then you didn't really explain the problem you're having with that and now we're talking about an arbitrary state
and sure, the Fock states form a basis for the spaces of states, so you can express any state as a superposition of them
But I already got it, after this much time. Sometimes I get lost , trying to equalize every aspect of the QHO to the mode of the quantized EM field
just like any other basis of the space of states anywhere else in QM
Just to be certain, the Hilbert space of the EM field, which can be considered a superpositon of modes, would be the tensor product of the hilbert spaces of each mode. is it correct to say this?
2:00 PM
@Arjun hi. i think it depends on the Lagrangian. if the Lagrangian has at least the second degree of speed, then the term $d/dt(dL/dv)$ ends up having the second derivative of $y(t)$
@imbAF yes
Ok, now I understand pretty much the general picture
ACM could you help me with one other thing
The annihilation and creation operators describe single photons added and subtracted from our system. Therefore, we should be able to observe single photons which are in exactly one oscillation state. In a cavity this would lead to exactly one cavity mode being excited. In general this system is stationary and can therefore be described by the stationary Schrödinger equation, enabling us to easily obtain the systems energy as
What does this mean : "Therefore, we should be able to observe single photons which are in exactly one oscillation state. "
How is this implied ?
I don't know, you said it, not me :P
Because it says therefore, I assume that it is implied
But if it is not the case, ok then
also note that the Fock states are not normalizable, so there is no such thing as a photon with a definite frequency mode
it is always superpositions
2:08 PM
But why it says : " In a cavity this would lead to exactly one cavity mode being excited"?
Because we change fock states?
I think you're trying to overanalyze this passage. Assuming that's a quote, all it's trying to say is that since we have single photon states $a^\dagger \lvert 0\rangle$ we should be able to observe such states.
@RyderRude aren't they a basis?
I don't think there's anything deeper here
@RyderRude given the amount of confusion imbAF has displayed so far, do you really think there is any value in getting into that nitpick now, or did you just want to show off that you know something?
@ACuriousMind I would assume in different modes, right? Multiple single photon states, for multplie modes considered
@imbAF I don't know what you mean
2:12 PM
@ACuriousMind sorry. their confusion was about "photons that r in exactly one oscillation state", which is y i mentioned it
isn't with oscillation state meant a mode?
yes, assuming u mean the labels $\vec {p}$ of the creation operator
@RyderRude Hello! Say I solve the Euler equations ,After reading Goldstein,I feel like he is trying to imply that the obtained y(x) after solving Euler's equations will always be twice differentiable..Is that true?
@ACuriousMind Maybe I am overthinking it but you said "single photon states" and not "single photon state"
if you are considering mutplie $|1\rangle$
@imbAF oh, yes, I mean for all the different mode operators $a^\dagger_\lambda$ that there might be
if you have more than one creation operator, you have more than one single-photon state
@Arjun in physics, all functions are implicitly assumed smooth until it turns out they aren't
3
2:14 PM
@Arjun no, e.g. for L = $x\dot {x}$, the EL equation is a tautology. hence every differentiable trajectory is a solution
if you're trying to read a physics text with a mathematician's mind where you always have to show or explicitly assume how smooth functions are, you're going to have a bad time :P
@Arjun so twice differentiability is not a property of the EL equations, but of physical Lagrangians
@RyderRude But it could turnout that there is no single y(x) that would solve the given lagrangian right?
@ACuriousMind Hmm..if I understand it right, different modes will have different ladder operators $\hat{a}_i$ $\hat{a}_j$ ?
@imbAF yes, of course
2:16 PM
@ACuriousMind Ok so then, why was this not understood ?
@Arjun ive never seen example Lagrangians like that, but maybe it's possible mathematically
@imbAF because I didn't understand what you were asking about until you pointed to the plural in my "single photon states"
Ah
See, I think paying attention to small details is important in physics
given that you had asked about the total space being the product of the individual mode states I considered the existence of different single-photon states obvious
@RyderRude @ACuriousMind So all trajectories in physics are nice and would be differentiable as many times as I would need lol?
2:18 PM
Maybe I overdo it sometimes though xP
@Arjun yes
@Arjun yes. God likes it smooth
@ACuriousMind ofc it is, but hey, this is physics, and I have fragemented understanding
until I get my hands on the books you recommended
@RyderRude Dang that's a profound statement..🙏
2:21 PM
@ACuriousMind But could you help me with one last thing. I understood a lot today. We speak about how the ladder operators change the photon number. This is a very straightforward explanation but
how does that translate physically
for the mode, for the em field itself, in various setups i.e free space, cavity or what else
I don't know if I am conveying properly what I am seeking
to understand
perhaps you could point out how the c/a operators of the QHO "translate physically"?
c/a ?
creation/annihilation
it's a common abbreviation, I mean the ladder operators
If I would consider the QHO in an eigenstate, the c/a operators acting on this state, would translate physically in a change of the expected regions in which the probability of finding the system would be higher
if one were to perform $|\psi(x)|xdx$ over some interval of x
I don't know why you are so fixated on the position wavefunction
that's the one thing you cannot transfer to the EM field because the notion of position is tricky for photons
2:27 PM
But if I were to consider multiple QHO (the equivalent of multiple modes for the em field), and each one's state is a superposition of eigenstates (the equivalent of the mode state being a superposition of fock states), I can't tell you what the physical intepretation of the c/a operators acting on the state of the em. field
@ACuriousMind that's what I know from the QHO
position is just one of many observables, I don't know why you're consider that more "physical" than talking about any other observable
ok, energy
how would that translate physically? larger amplitude?
energy is already something physical
why do you need to translate it further?
cuz energy is exhibited in a certain way
also, larger amplitude of what
2:29 PM
in a certain form
Hello! I’m writing an article on how students and researchers keep up with the latest advancements in their areas. Would anyone mind answering a few brief questions about the tools and strategies they use?
@ACuriousMind of it's displacement form the equilibrium? But I am not sure, I think, you pointed out that this was wrong
previously
@imbAF if you're thinking about the HO, yes, that's wrong
again, these are stationary states, there is no notion of anything moving/changing here
Ok, and you told me not to think of position
I suggest you drop your insistence on these "physical translations" that are simply factually incorrect
2:30 PM
but of energy
@ACuriousMind you shouldn't have physical interpretation of things?
oh, you should, I just think your idea of what's "physical" is distorted
But, I don't want to go away from my point
Or what I am trying to understand
you asked me about the c/a acting on the QHO, I gave an explanation, intepretation of them
But it wasn't correct?
But my goal is this:
But if I were to consider multiple QHO (the equivalent of multiple modes for the em field), and each one's state is a superposition of eigenstates (the equivalent of the mode state being a superposition of fock states), I can't tell you what the physical intepretation of the c/a operators acting on the state of the em. field
@ACuriousMind how so?
@imbAF as in you keep talking about e.g. the amplitude of the stationary states of the QHO as their "physical interpretation", but there is no such amplitude because they're stationary. Your interpretation isn't "physical", it's just wrong
probability amplitude, you mean?
Yeah that is wrong
no, the "displacement from the equilibrium" you talked about above!
2:34 PM
I remember you pointing it out
ok I remember you pointing it out
@Freechoiceguy Top two methods: You read what's published on arXiv in the relevant tags and you go to conferences
But I avoided considering the energy, because I would be forced, to give a physical interpretation of it. Just saying different energy values, is not satisfactory to me
by "mode state", do u mean annihilation operator eigenstate or energy eigenstate
I mean there's more photons in there, if you let them out and count them (e.g. shooting them at a screen) you see different counts of blips on the screen for different numbers of photons
Explaining how the different energies are exhibited when the system is observed, would be my end goal. Hence why I considered the position because I knew that my understanding of it was accurate, or hoped it was
2:37 PM
Did you look in that book Quantum Theory of Radiation by Heitler?
Somewhere around here might be helful
Chapter 2 section 7 starting on page 54 might be helpful
title ?
of the chapter I mean or section
I really struggle to interpret c/a acting on state, whether that being a fock state, a mode one, or the field one, which would be a tensor product,
On pg 61, there is a section 'The State Vector of the Radiation Field' that may be helpful
However that is jumping too far ahead I think
@DIRAC1930 I found it
2:45 PM
@imbAF Why do you need any more interpretation than "it adds/removes a photon"?
to which mode? Or that's not important to know? And I am assuming the general case of the c/a operators acting on the state of the em field
@imbAF You have one pair of c/a operators $a_\lambda, a^\dagger_\lambda$ for each mode $\lambda$!
Ok
so is it correct to say
twice
there's 100% chance of never having a single photon state
detecting *
I don't know where that statement came from or what it has to do with the preceding discussion
Nothing really
2:48 PM
???
It's a follow up question
@ACuriousMind but I will stop with this
sorry, I'm done with this conversation where you just say "Ok" and then randomly with no indication of what even your original problem was ask a completely different question, this is very frustrating
I didn't say ok or anything. You answered my question with this
@ACuriousMind .
I would flip through the first 3 chapters of that book
@imbAF u shud say "also" here instead of "so"
2:50 PM
and skip the irrelevant parts
And after understanding it, I continued with a new, follow up question
3:06 PM
@imbAF all the states of the form $\psi (p)$ are single particle states
so there is a 100% chance of detecting a single particle
@RyderRude If $|\psi\rangle=\sum_{n=0}^{\infty} c_{n_i} |n_i\rangle$, where $|\psi\rangle$ is some state of a mode and the index "i" is used for different modes of the em. field, then:

$(a^{\dagger})^2|\psi\rangle=\sum_{n=0}^{\infty} c_{n_i}\sqrt{n_i}\sqrt{n_i+1} |n_i+2\rangle$. For $n_i \ge 0$ the lowerst fock state is $|2\rangle$. So when I say, there's 100% chance of NOT measuring 1 photon, what I wrote above is what I meant
@imbAF in the latter state u hav written down, there's a minimum of two particles in every state of the superposition, so one cant measure just a single particle here, yeah
Yeah, that's what I was trying to ask with :
"so is it correct to say
twice
there's 100% chance of never having a single photon state
detecting *"
yeah, but u didnt mention ur state there
"
I really struggle to interpret c/a acting on state, whether that being a fock state, a mode one, or the field one, which would be a tensor product,"
3:13 PM
in states of the form $\psi (p)$, there's a 100% chance of always measuring a single particle
@RyderRude well clearly not if the creation operator acts on that mode acts twice on it. that's what you just confirmed a moment ago
yes... but y wud u act the creation operator twice on it?
why wouldn't I?
why cant a system simply have the state $\psi (p)$?
It can
But my whole discussion was about the c/a operators acting on a state and trying to understand the physical meaning of that thing
3:16 PM
oh
I mean I made it pretty clear, several times
so this is what u meant by a follow up question
Ahaa yes
yeah, if u do that operator twice, then no state in the one-particle sector remains
but u never mentioned before that u were doing that operator twice?
I know, I am not the most proficient when it comes to English, but I believe I properly showed what I wanted to know
Take it once
It means that theres 100% chance of measuring at least one photon
or mor
more*
but never 0
3:18 PM
yes
@imbAF also note that u hav not mentioned the mode of the creation operator that u r applying
the entire point of the discussion
but ACM said :
"You have one pair of c/a operators $a_\lambda, a^\dagger_\lambda$ for each mode $\lambda$!"
Which is why I told him, that he already answered my question
And what I asked after that was a follow up one
How hard is to understand the process: Ask about theory ---> Generalize it ---> go to a concrete case to check your understanding
that's how things should be
logical way to test your own understanding of things
yeah. but i think that question did not have complete context
The context was already made from the different previous questions that led to the last one. But ok
I will be more precise the next time
3:24 PM
@imbAF also note that all the previous particle numbers do not simply get increased by 2. the number that gets increased is determined by the mode of the operator
what?
@imbAF yes.. but u need to provide it again at the time of asking the question
@imbAF e.g. $a^{\dagger} _i |n_j\rangle\neq c |n_j +2\rangle$, when $i\neq j$
the rhs shud b $c|n_j, 2_i\rangle$
Ofc
yeah.. then it's good :)
But one needs to consider the most general case, which in this case, would be a field state, expressed as a tensor product of mode states. When the c/a operators act on this state, they clearly belong to one of these modes, so they will act on that mode
3:29 PM
yes
Yeah, so I don't think, from as far as I know, that the c/a operators of a mode can act on another one. That's why I totally disregard that
@imbAF sorry. u r just summing over a fixed $i$ here and varying the $n$. i mis-read it
i thought u were varying both the $i$ and $n_i$ in the sum
You said, previously that $\langle n| n\rangle \neq 1$ ?
yes. these r not normalizable
it's the same thing as in QM
the momentum states normalise to delta
3:46 PM
the multi particle basis states r like the multi particle symmetrised hilbert space basis states from qm @imbAF
Yeah I understand it now
But it took time to get a clear picture of everything
@Obliv only that time itself has progressed
has anyone an opinion on murphy's c*-algebras and operator theory
4:36 PM
I just came across a neat result which might not be that well known...
Often in calculations one can integrate an expression by parts and throw off the boundary term. This can be done by applying divergence theorem valid for covariant derivatives, but what about Lie derivatives? That is, when can we write the "extra term" after integrating by parts as a boundary integral? The answer is neat: When the Lie derivative is taken w.r.t. a solenoidal vector field!
If anyone is interested, it is discussed here.
4:57 PM
I don't think I'll ever be able to learn GR
 
4 hours later…
9:03 PM
@imbAF Did you like that book?
9:20 PM
In that video with Dirac and Hund, Dirac says that he believes that the correct equation unifying relativity and quantum mechanics has most likely not been found. I always assumed he meant special relativity and thought Dirac was just giving being falsely humble, but he probably meant general relativity
You have to be a complete genius to even come up with this stuff
The book after this (called Fundametal Theory) aimed to derive all fundamental constants
"A somewhat damaging statement in his defence of these concepts involved the fine-structure constant, α. At the time it was measured to be very close to 1/136, and he argued that the value should in fact be exactly 1/136 for epistemological reasons. Later measurements placed the value much closer to 1/137, at which point he switched his line of reasoning to argue that one more should be added to the degrees of freedom, so that the value should in fact be exactly 1/137, the Eddington number."
lol
Wags at the time started calling him "Arthur Adding-one"
I don't know what Wags means in this context. I just copied and pasted that last line from wikipedia
Thanks
10:21 PM
This is a very elementary thing to ask perhaps, but does Hermitian conjugation turn a contravariant index to covariant and vice versa?
Is there even a notion of "Hermitian conjugate" of mixed tensors?
@Sanjana I'm not sure in what situation exactly you have to ask yourself that question
@ACuriousMind Do u mean, that this context does not arise usually?
I'm just asking in what context you are asking that; in most cases where I'd bother about "covariance" and "contravariance", everything is real, not complex
@ACuriousMind Okay... let's say that the tensor elements are complex numbers
that's not enough to have Hermitian conjugation
where's the complex inner product w.r.t. which stuff is Hermitian coming from?
I could contort myself to argue one way or the other in the abstract here, but really I don't see how this question ever might come up in practice
or, rather, in the situations where it does come up, the answer for that situation should be obvious
so I'm asking you what you're doing that this question is relevant for
10:29 PM
@ACuriousMind I will just tell you the story. I was reading a book where a Hermitian conjugation operation is defined and it does interchange the indices. But, then I thought... I haven't been doing this previously, say in some QFT calculation---so have I been doing things wrong? But the conversation with you seems to point out that such a situation does not occur often so I might be misremembering things
So, are you sure that Hermitian conjugation of tensors of arbitrary rank don't usually come up in physics?
The snapshot is from "Predrag Cvitanovic - Group theory_ birdtracks, Lie's, and exceptional groups-Princeton University Press (2008)"
@ACuriousMind For example, consider the "spinor indices" of gamma matrices. Do they go up if I take Hermitian conjugate of the gamma matrices? I have always simply interchanged the indices (coming from the transpose) and complex conjugated the elements... Did I do anything wrong?
see, that's why I asked for the context: Yes, if you have a complex tensor $T^{ijk}_{lm}$ acting on a complex vector space and its indices are defined w.r.t. a basis of that space, then indeed Hermitian conjugation will interchange the indices because it is transposition followed by complex conjugation
@ACuriousMind I get the interchange of the order of indices part, but why would I also have to pull the upper indices downstairs and put the covariant indices upstairs?
but in physics we also often have operator-valued tensors whose indices are from the target space of the field, not from the complex vector space, e.g. for a vector field $A_\mu$ that becomes an operator during quantization - there $(A_\mu)^\dagger$ does not change the index position since the conjugation acts on a different space (the quantum space of states) than the space from which the index is drawn (the classical target space of the field)
@Sanjana because that's what the transpose does
the transpose of a linear map $V\to W$ is a map $W^\ast \to V^\ast$ - switching the position of $V$ and $W$ is switching the index order, switching to the dual is switching the index position
10:51 PM
I spent over 4 hours working on this post before posting it, including demonstrating the research effort behind it, as well as making it as clear and useful as possible: physics.stackexchange.com/questions/816931/…
I went through this over a year ago on here
I just don't understand... why is physics stack exchange a place that so discourages asking questions?
I am literally just asking questions and doing everything in my power to comply with the posting guidelines
instantly downvoted when I cannot think of how I could've improved this post any further
if you require someone to already have a PhD in physics to post here, then say so in the guidelines
11:09 PM
and now it's closed
I really wish there were some place on the internet where lay people could go to ask physics questions
@Asklepian Had you tried to comply with the guidelines, you would not have asked three separate questions in one post.
you're right; I spent over 4 hours doing nothing
I quit
this community is rude, incurious, petty, and officious

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