It is well known that one can derive wave equation by considering an infinite harmonic oscillator chain and taking the continuum limit. The wave equation can be looked upon as the massless limit of a KG-like equation. So what should be the discrete system so that we have the massive KG-equation when we take the continuum limit? I guess there should be some driving/damping term along with the oscillator. What should the functional form look like?
pretty new to quantum mechanics and stuff, and i had an interesting question
one that probably doesnt have much practicality but just out of curiosity
trying to phrase this correctly, a litle tired
for electrons in atomic orbitals, ionizing the atom is conceptualized as bringing thagt electron to a hypothetical "infinite" distance from the nucleus so it has no interaction with the proton. I'm curious though, if an electron is far away enough from a nucleus to have a very very negligible interaction, but still the particles are experiencing very small coulombic potentials from eachother, can the electron still be considered to be in some really high energy/quantum number orbital?
i guess this question came up from wondering like, at what point do we start having orbitals? is there a certain distance away an electron has to be to start being in an atom's atomic orbitals?
i assume not because like what would be the cutoff
As you increase the energy level of a hydrogen atom $n \rightarrow \infty$ I have learned that the energy of that energy level changes according to
$$ E_n =- \frac{13.6 {\rm eV}}{n^2},$$
and that there are technically an infinite set of these energy levels. As you get closer to $n=\infty$ the ...
In some ideal universe containing only one proton and one electron you could in principle have arbitrarily large orbitals. In the real universe the electron is affected by all the other matter around it and at some distance that environmental interaction becomes larger than the electron-proton interaction so we no longer have a state resembling an orbital.
@Allie An electron is always described by a wavefunction, but that wavefunction depends on the interaction with its surroundings.
In a hydrogen atom the interaction with the surroundings is dominated by the electric field from the proton so we get wavefunctions like the 1s, 2s, etc orbitals.
Far away from the proton the electron will interact with all the other matter in the vicinity and we'll get some complicated wavefunction that doesn't look anything like the hydrogen orbitals.
e.g. in a metal where electrons interact with many atoms around them we get states called Bloch waves.
@Sanjana This is actually pretty well-established in standard textbooks on field theory for condensed matter, or even just the continuum limit in classical mechanics texts. The field's time derivative bits come from KE of each individual mass points, so that part is obvious; each field point's springy connection to peers turn into the space derivative parts in the continuum limit, and should also be obvious and able to get you to the Klein-Gordon form. The mass term will just be field's
deviation from equilibrium position, the $\frac12kx^2$ with $x$ replaced by the field operator.
@Amit Having missed the deleted comments, presumably deleted by ACM, it is funny to note that the one spoofing, and somewhat implying that R is not rude, is somewhat proving the case by being ruder than R. If only irony could be used to generate electrical energy.
@Relativisticcucumber charge density waves should probably have more to do with polarons than plasmons.
@Relativisticcucumber Hydrogen is usually a covalently bonded molecular gas at RTP and is nowhere near able to conduct electricity effectively in that state.
@Relativisticcucumber It is usually also displaced to the side and left in its own limbo zone. It is so different in its properties (usually due to how low mass it is, that it can perform quantum tunnelling and other non-classical feats with impunity) that we tend to have to treat it separately.
@Relativisticcucumber The staircase is directly contradicting your viewpoint.
@SillyGoose Are you saying that you failed to be able to verify that the eigenvectors as written in Equation (94) solve the eigenvector with Hamiltonian from Equation (92) and eigenvalues in Equation (93) with angular values defined in Equation (95)?
How does one typically measure the momenta of quantum particles like say,electrons?Shankar says you hit it with low frequency photons and measure the change in momenta of the photons,use conservation laws to find the change in the momenta of the quantum particles.This only gives the change in momentum.How does one know what momentum the particle originally had?
2)Also,i feel like one determines momenta,energies etc of quantum particles using conservation laws,but how did we know that conservation laws actually worked in the first place?I.e how did we test them?I think we've used other mechanisms to find out the momenta,energies of particles involved..but am not sure what exactly we used.
@Arjun What are you talking about? Momentum conservation is baked into it. I'm saying that once you use a diffraction grating, the detector position that registers the electron automatically tells you the wavelength of the electron, which in turn tells you the momentum of the electron directly. The basic interaction would have to be elastic.
The important point is that it is not a "change in momentum"
@naturallyInconsistent So If i want to corroborate momentum conservation say in qm?can i use this experiment without going in a cyclical direction,i.e using momentum conservation to calculate momentum and using this value to show that momentum is conserved?
Cuz in the photon experiment this is what's happening
@Arjun also, do u think the photon, after collision, is in a superposition of "different momenta changes", or does it have a specific momentum with probability given by the born rule
@Arjun I think you are extremely mistaken. You need to first have a prediction, and then check if the experiment agrees with the prediction or throws the theory out. Without momentum conservation, you cannot even start making the predictions, whereas with it, you can get incredibly precise predictions that are agreed by experiments.
@RyderRude thanks for mentioning that,i also have that question in mind,i think it will go into a superposition and maybe there is some sort of decoherence,once we measure the momentum of photon,the of electron might just collapse to one of it's momentum eigenstates(making sure total momentum is conserved)
@naturallyInconsistent Okay,please help me understand this.How do we then know that momentum conservation actually holds,if our measurements of momenta are based on its conservation principle?
@RyderRude Shankar doesen't mention this in that particular example..maybe he'll talk about this stuff later
@Arjun If it does not hold, then our predictions should disagree wildly from it. For example, before we knew about neutrinos, the experimental evidence was so wildly badly in disagreement with it, that Bohr wanted to give up energy and momentum conservation.
@RyderRude Also what exactly are we measuring when we say we are trying to measure mometum,in qm afaik momentum corresponds to it's operator and when we measure it we'll get one of it's eigenvalues,but what exactly are we trying to measure in the lab when we say we are measuring momentum,measuring eigenvalues of some operator makes no sense to me
@Arjun I'm not familiar with how exactly measurement devices work... but let's say u (as a toy model of measurement) make the emitted photon absorbed by an atom, and then when that atom emits it back, u get it back as some spectral line or stuff like that. And then u look at the line
again, this is a completely made-up way to measure momentum of an electron. I'm just trying to demonstrate that, in the end, u want some macroscopic state (like a spectral like) to be correlated with the microscopic state like energy
one would say that this correlation is still a superposition...but since there's decoherence and stuff, at some point, we get just one output on the macroscopic device
@Ye that's what i thought as well,but it is causing me troubles since in the example you gave one uses momentum conservation of the particle + photon system to deduce the particle's momentum,but say i want to test whether momentum conservation itself holds ,if i use the above method,would'nt it be trivially true?
@Arjun You cannot have that. Think of the detectors that you actually know about. e.g. Geiger counters. You just detect clicks, localised at where the detectors are. You use the experimental setup to deduce which momenta, say, you are measuring when that detector clicks.
@naturallyInconsistent Do they measure the wavelength of electron's based off the pattern?And from that deduce their momentum.I don't have a problem if that's what's happening
@Arjun we just have a variety of experimental techniques, and when we use those techniques, we get some output that corresponds to what QM predicts for measurement of the momentum operator. Therefore, these techniques measure the momentum operator
how we came up with these techniques? They r built on classical intuition. I guess they just pretend that they're building a classical measurement device, and it happens to work for the quantum
like, for position detection, u just make a detector like u woudl normally do, and it works for the electron
when they first did the electron double slit experiment, they didn't know QM was a thing. They just built a detector using classical intuition
but again, i know very little about measurement devices. Take these comments with a grain of salt
Actually measuring the wavelength is a good example i think,p would simple be $\frac{\hbar}{\lambda}$,and the theory also predicts this,i think we can now use this to verify momentum conservation etc..@naturallyInconsistent @RyderRude
but with the collision thing i have some fundamental issues lol
@naturallyInconsistent Oh I'm very ignorant with regards to this experimental stuff..one more question say,the electrons were all initially in a mixture of momentum eigenfunctions,now I diffract them to see the pattern,will i get a pattern corresponding to that mixed state or will the electrons collapse into one particular momentum eigenstate?
@Arjun Again, many misconceptions baked into your question. You start with some incoming beam of electrons, say, with some distribution of momenta. That is your quantum state, and this can be a pure state or a mixed state. You scatter it off the diffraction grating, and measure at different positions, corresponding to different angles. That means that different detectors will click, one at a time, collecting statistics. Each click infers a particular momentum eigenstate.
The pattern only appears when you collect enough statistics to showcase the distribution. You cannot say that any one particular electron has so-or-so momenta, unless after you measured it by the clicking of a particular detector.
And then you have to contend with whatever interpretation you are using. If you are not using an interpretation with collapse, then you wont use the word collapse at all.
Actually, I don't know; don't ALL quantum textbooks discuss measurements, at least a bit? They might not be perfectly lucid, but they do discuss a bit of detection. I have yet to find a source that is particularly exemplary; I mean, this is the kind of stuff that you have to talk to experiment professors and then they will verbally vomit at you all the details.
Also, this suffers from the usual elitist attitude of textbook authors. They will outline the basic bits, and leave all the details to the reader to deduce. Makes for extremely difficult reading.
@naturallyInconsistent Maybe,I soould've framed my question better..I was essentially asking if we have a system of identical elctrons all originally in the same state..would the difrraction pattern correspond to this initial state?yes i do understand when we measure,it will collapse to some position eigenstate,but from the distribution we can construct the initial state of the system right?
I was wondering about this since the electrons might interact with the solid in which the diffraction is happening as as a result change to a state different from the original one,
If by the time they diffract ,they change to a completely new state?How would we be measuring what momenta the particles had before the diffraction happened?
i.e would'nt we be measuring their momenta after they've diffracted from the solid and before they hit the screen?
@naturallyInconsistent sorry ignore the above two questions lol..i got my answers..i should've thought for a bit before asking
@Arjun "all originally in the same state" requires some qualifications. For example, if there is no term in the Hamiltonian coupling spin with the resulting diffraction's position measurement, then whatever deduction you later do, will be independent of the spin degrees of freedom.
Also, if it is not a pure state, then your diffraction pattern is going to reflect the fact that it is a mixture.
But if you started with a pure state and you only care about the momentum degrees of freedom, yes, the statistical distribution you will get out of the measurement thereof will correspond to the initial state.
I mean, you can consider this as the physical analogue of doing Fourier transform on the position wavefunction to get the momentum wavefunction.
@Arjun Yes, this is also part of the assumptions; you would have to check and assume that the collision is adequately represented by elastic collision or not; if it is not, then you would have an infernally harder problem to solve.
@naturallyInconsistent thanks i was looking for this only..if it not a pure sine wave originally(the e^ikx p-eigenstate thing) the diffraction pattern would be something weird but consistent with the original wave(state)
@naturallyInconsistent So the collisions are assumed to be elastic?Hmm..why does no book clearly mention all these assumptions?
@naturallyInconsistent Can I say the states evolve just like,say some classical wave like EM wave,if both the waves(the state and the classical wave) originally had the same form?IK quantitatively no,since their evolution is governed by different equations,but can we atleast qualitatively say this?
@Arjun Only for simplicity; if you are not afraid of the extra work you have to put in to work out the inelastic cases, you can derive them all on your own.
@naturallyInconsistent @RyderRude Also it seems pretty weird to me how some interactions like the act of measuring would lead to a collapse of the state,but some other interactions like the electrons passing through a solid crystal while diffracting,don't lead to a collapse..So what are those characteristics that an interaction should have in order for it to lead to a collapse?
@Arjun That is only if the Hamiltonian and the measurement type are consistent with waves, e.g. free particle and you measuring the momentum of that. And even there, the clicking of the detector positions is strictly quantum, since waves don't make clicks. If you were measuring the particle-like properties, then you obviously cannot pretend that everything are classical waves.
@Arjun Easy; I do not use any interpretation that has collapse in it.
@Arjun there are no widely accepted theories that describe how collapse happens. The closest is probably decoherence + many worlds.
Decoherence describes the interaction of the system with its environment so I think that is what you're asking about.
But decoherence only describes loss of entanglement and you still need many worlds to select one out of the superposition of states that results from decoherence.
@JohnRennie Yes to the former, but it is not necessary for decoherence to be tied strictly to MWI. And in MWI, it is not necessary to select; the universe will just be in a superposition of all the possible measurement outcomes, and what is really necessary, is to postulate the Born rule in some form as to give us the experimentally measured probabilities.
@Amit in property dualism, every entity in the universe possesses some qualitative properties and some physical properties (these properties are what physics describes). And when atoms combine to form a brain, the physical properties of the brain interplay with the qualitative properties of the brain to paint the subjective experience that u r familiar with
this also explains y there is a correlation between subjective experience and brain activity
@Amit i think so
universe is local in nature. So whatever is subjective experience, it must arise from the brain, as the brain is locally close to subjective experience
Qualia may be just another way to invest our memories with meaning, to avoid the horrible thought we're nothing but a complex sense perception processor... =]
well i did just compute eigenvalues by hand another way and via mathematica. both of these ways disagree with the screen shotted method. but the screen shotted method is from a textbook, so I feel I am maybe misapplying the result
Is there a way to measure the momentum of a particle at a given instant?I've stackexchanged a bit and have found out these two answers that describe the process 1)physics.stackexchange.com/a/316004/304878: The problem I have here is yes we know what the magnitude of the momentum is but have no clue what it's direction at a given instant is(If we tried finding out,I think the particle's radius will completely go into oblivion)
2)physics.stackexchange.com/a/695696/304878 I suppose the only way to know if the particle had a velocity perpendicular to the fields E and B,is to measure the deflection,but once we measure the deflection,we're measuring the position and at that instant and hence we won't know what the total momentum of the particle at that instant is,so is there a way to measure the total momentum of a particle?
@Arjun If you're just starting out in QM, we do not model the measurement process in any detail at that level. "Collapse" is an interpretation-dependent word and its use in this context is more loaded than you might realize, and "measurements" in intro QM are simply by definition producing the eigenstates corresponding to the measured observable with the probabilities corresponding to the Born rule. At this level, there is no "how" to this process.
When you do want to model this process in more details within quantum mechanics, then you will have to look at concepts like von Neumann measurement schemes, decoherence and einselection
@ACuriousMind Collapse of a state while we were measuring the spatial location made some sense(the dot on the detector),but i was wondering how we can actually make the system collapse to a momentum eigenstate starting from a mixed state.
Momentum measurements are usually indirect, e.g. you measure the energy of a particle by stopping it in a calorimeter and then compute the momentum from the kinetic energy
@Arjun A linear combination of eigenstates is not a mixed state
"mixed" and "pure" refer to density matrices that cannot (resp. can) be represented as a vector in the state space
the terms have nothing to do with something being in a superposition or not
@Arjun If you zoom into that process you would have a brief time where the particle interacts with the screen material, it's not an instant either, it's just very brief
@Arjun Yes, it is an idealization; again, intro QM does not actually attempt to model what happens in a real-world measurement
(and also if I was teaching this course I would never use the word "collapse", let alone "instantaneous collapse". That just invites this kind of unnecessary confusion :P)
Didn't I say already that one way to do that is shoot it at a calorimeter?
another would be to shoot it at some very light object that will absorb it that's hanging by a thread and then measure the recoil (one can actually measure the momentum of photons reflected from a mirror that way)
you can also shoot it through a bubble chamber with a magnetic field through it and then deduce the momentum from the curve of the trace
@ACuriousMind But they say when we make a momentum measurement the particle will take up an eigenstate of the momentum right?But in the examples youv'e mentioned it doesen't seem to be the case for me
@Arjun How could you possibly tell in these examples whether the particle is in an eigenstate of momentum or not?
Also, no real particle will ever be in an eigenstate of position or momentum, since both have continuous spectrum and all measurements have intrinsic limits to their resolution
In the bubble chamber expt for example you know the particle is going on in a circle from the trajectory,this means the i wouldn't know everything about momentum,I only know it's magnitude in the bubble chamber
@ACuriousMind But how did you know what momentum the particle had initially?Maybe it really didn't have a definite momentum before it collided with the mirror
I agree .I was saying maybe it didn't have a definite momentum initially,you said after reflection the particle will have half the momentum the mirror would have,momentum conservation implies that the particle had half the momentum originally as well
if you don't like the mirror, then do a setup with electric and magnetic fields like a Wien filter - the particles that go straight through it are effectively measured to have a certain velocity, and hence momentum
@Arjun You need to formulate conservation laws in quantum mechanics appropriately; all the classical conservation laws "hold" in QM, but of course they can't mean exactly the same as in classical physics since the quantum objects don't have definite values for the conserved quantities most of the time
@Arjun suppose initial state is $(|p_1\rangle + |p_2\rangle) |0 _M\rangle$. 0_M is the initial momentum of the mirror. After collision, this comes $|p_1'\rangle |M_1\rangle +|p_2'\rangle |M_2\rangle$. Now, momentum conservation is individually satisfied : p1=p'1+M1 and p2=p'2+M2
So momentum conservation is satisfied in each of the "histories"
and after measurement, u only get one of the histories
in qft, this shows up as a delta function in the probability amplidue
so momentum conservation can still be expressed for superposition stuff
@ACuriousMind I've seen this example as well,how do you find out which particle goes in a straight line?This will correspond to some sort of position measuring and i would'nt know momentum completely(for example all three components of it)
suppose momentum conservation holds. Then we have $PH=HP$. So, time evolution sends eigenstates of total momentum to eigenstates of total momentum with the same eigenvalue
the means momentum conservation holds for eigenstates
for superpositions, it holds "term by term"
cuz u can express superposition using a sum of eigenstates
@ACuriousMind No I was content with whatever i saw for position,you get a smudged dot on a detector and stuff and you measure it without making any references to momentum,for momentum i could'nt find a satifactory way in which they measure it without making reference to the position of the particle in someway or the other they detect the particle or find the line in which it goes or the angle it makes etc
@ACuriousMind But clasically i can know position and momentum at the same time,In qm that is not the case in QM,when I wish to findout the momentum I can't simultaneously measure position
*to arbitrary accuracies
@RyderRude Thanks for the insight,makes sense to me.I like your different histories way of looking at things ; )
@Arjun that's doing a lot of heavy lifting here - if you compute the actual limit the uncertainty principle puts on the accuracies of position and momentum in the setups we've talked about, this limit will usually be far below the resolution the measurement device has anyway, i.e. it's totally irrelevant
I understand the problem you think you see, but I suggest it's not a problem in any real-world scenario, only when you want the simple idealizations in your intro QM text to correspond neatly to reality
@ACuriousMind Okay,tell me this one thing is it atleast possible to prepare particles with identical states?i.e multiple copies of the same particle each with same $\psi$ somehow?
Or it is also a textbook idealization?
ik maybe we can't do it exactly,but you get the point right?
@Arjun You cannot copy an arbitrary unknown state (no cloning theorem), but you can prepare a known state identically rather often. E.g. you can prepare a bunch of pure spin-up states simply by taking the upper of the two beams coming out of a Stern-Gerlach device.
@Oh that's sad,I thought maybe we can do it in the unkown case as well,then if we diffract the particle and realize it was initially in a momentum eigenstate,maybe infer that all the particles are also in momentum eigenstates
@ACuriousMind take the mirror example,say the particle was originally in some unknown state, measure the mirror's recoil and all I know is that total momentum will be conserved,now without knowing what momentum the particle had initially,before the collision there is no way i'll know what momentum the particle will have after the collision,So i'll also have to measure the particle's initial momentum,but how do you do that?
because we measure the mirror to have a definite momentum, and then momentum conservation dictates that the particle also has to be in a definite momentum state; that's what Ryder tried to explain
@ACuriousMind No,ryder also wrote a superposition for the momentum after the collison,I don't think he claimed the particle would have a definite momentum
He wrote "Now, momentum conservation is individually satisfied : p1=p'1+M1 and p2=p'2+M2" p1' and p2' for the two possibilities after the collision
the state before measurement is some arbitrary superposition of momentum eigenstates $\sum_i c_i \lvert p_i\rangle$. After interacting with the mirror the system mirror + particle is in an entangled state like $\sum_i c_i \lvert -p_i\rangle\otimes \lvert 2p_i\rangle_M$, where $\lvert \rangle_M$ is the mirror momentum state.
Now since the mirror is large and visible, we can measure its momentum just by looking at it (measuring the angle of the string it hangs by, whatever), so we know the mirror to be in one specific state, say $\lvert 2p_1\rangle_M$. So by the way entanglement works, we now immediately also know that the particle has to be in $\lvert -p_1\rangle$
This is essentially a von Neumann measurement scheme, the mirror states here are what are usually called the pointer states in the generic presentation, and this is not what you are supposed to worry about at all when just starting QM :P
@ACuriousMind Oh I see! So does it mean measuring of the mirror's recoil would also give us what momentum the particle had before the collision since : p1=p'1+M1 and p2=p'2+M2" ?Does it mean measuring of the recoil also in some sense fixes/makes a measurement in the past of the particle?
the particle has a definite state after measurement, it did not have one before
if we hadn't looked at the mirror, both mirror and particle would have remained in this superposition of different momenta (at least from our viewpoint) forever
the state $\sum_i c_i\lvert p_i\rangle$ is (apart from it actually having to be an integral instead of a sum because of continuous momentum) fully generic: The eigenstates of any observable form a basis for the state space, after all
@ACuriousMind Yh lol sorry,is it necessary that a single momentum eigenstate of the particle be entangled to a single momentum recoil of the mirror?I.e cant the particle have a superposition of momentum eigenstates clubbed with a given momentum state of the mirror?
Could you please clearly state the quantum version of momentum conservation?
I seem to have difficult time trying to get my head around a single momentum value of mirror giving a single final momentum value for the particle..in classical physics there are infinite no.of possible momenta values the particle could take,with the difference being equal to the mirror's momentum
@Arjun The momentum operator commutes with the Hamiltonian
@Arjun What do you mean? In classical physics, the particle has a definite incoming momentum $p$, after the reflection the particle has $-p$ and the mirror has $2p$
@ACuriousMind I mean it's momentum has a definite value that is arbitrary if i didn't know what it originally was,i.e only from mirror's initial and final momenta and no knowledge of the particle's momenta,i can have infinte no.of solutions to the conservation equation'
@Arjun I'm not following you - the $p$ in my statement is an arbitrary value. You measure the mirror momentum to be some value $p_M$, and then you know the incoming particle had $p_M/2$ and the outgoing particle has $-p_M/2$. That's one solution, not infinitely many!
@ACuriousMind Does this imply that in the mirror example for a given final momentum of the mirror,i will definitely only have a single value for the particle's final momentum?
Say I have a single particle and a single mirror,mirror's initial momentum being zero.
I bombard the particle with an initial speed that i don't know,but i measure the mirror's recoil speed
I'm saying for each value of the particle's initial momentum,I will have a corresponding value for the particles final momentum given by p2=p1-p_m
Is this clear in the classical case?
I'm trying to say if i don't know the value of particle's initial momentum p1,I can't deduce what p2 (it's final momentum is) just looking at the mirrors' recoil
for instance a photon reflecting off a mirror doesn't really "bounce" off it, it's absorbed by the mirror surface and then a photon is re-emitted, but the effect is the same as if you just think about it bouncing off
@Arjun what leads to the particle having a definite momentum is us measuring the momentum of the mirror and this, through entanglement, implying the definite momentum of the particle, not the collision itself. It's the measurement that changes the state.
@Arjun I'm afraid I don't know of many good electron mirrors :P
but regardless of how exactly you implement the measurement, this principle is always the same - there is some property of the time evolution that means certain states ("pointer states") of the measurement apparatus get entangled with the momentum eigenstates of the particle, and since the apparatus is macroscopic we can just look at it to measure its state, and then the entanglement implies we also know the state of the particle
2
@Arjun I mean the entanglement is just the result of time evolution - if you wanted to be really detailed about this you'd have to figure out the actual Hamiltonian for any specific measurement apparatus and then show its corresponding time evolution leads to these entangled states etc.
the study of such systems producing such peculiarly entangled states is what decoherence theory is about
but again this is far beyond intro QM, and the details of what goes on inside the measurement apparatus are completely irrelevant if you are just interested in its inputs and outputs, as we often are
here is a picture representing the Bragg formulation of x-ray diffraction.
how can we say that the light as pictured interferes constructively? If we take the picture literally, the "light trajectories" are parallel at every point, so will not coincide so will not have a chance to interfere. where is my understanding going wrong?
@SillyGoose those are not two light trajectories, they're the boundaries of the incoming beam (the parallel lines at the upper corners are the wavefronts)
@Arjun I mean...everything obeys momentum and energy conservation
@RyderisnotRude. I have had one buring question..If I'm not being too forward are you related to @RyderRude in any way?Or are you his fan or something lol
As you increase the energy level of a hydrogen atom $n \rightarrow \infty$ I have learned that the energy of that energy level changes according to
$$ E_n =- \frac{13.6 {\rm eV}}{n^2},$$
and that there are technically an infinite set of these energy levels. As you get closer to $n=\infty$ the ...
