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1:30 AM
neither, if you're in a superposition of non-degenerate energy eigenstates @DIRAC1930
2 hours later…
3:18 AM
Why is inferiority complex a bad thing in an individual?
How to deal with it?!
How can you change your identities which are decided by accident or birth?
How much of life is a choice of an indivitual
and how do you define individuality of someone?
Good Morning everyone btw!
@RewCie Because you are undervaluing yourself?
@RewCie Self conviction
@RewCie I think you will find determinism (not to be confused with fatalism) interesting
Hey all! I was wondering if any GR expert could help me?

If we assume isotropy and homogeneity then we get the FLRW metric upto a constant (cosmological constant). Is there any counter argument why this constant cannot be a property of the stress energy tensor?
4:23 AM
@MoreAnonymous How will you justify a IC Patient underevaulating himself?
@MoreAnonymous Self conviction sounds more like a Unique Selling Proposition for a business....
4:45 AM
@MoreAnonymous determinism sounds more like a belief than a proved theory...
2 hours later…
6:46 AM
@MoreAnonymous The CC can be treated either as a curvature term, in which case we write it on the left side, or as a vacuum energy, in which case we write it on the right side and add it to the SE tensor.
The two are mathematically equivalent. Traditionally it tends to be treated as a curvature, but now that we have discovered dark energy I guess that would be added to the SE tensor.
7:13 AM
@DIRAC1930 You may find the Bloch sphere helpful. "The points on the surface of the sphere correspond to the pure states of the system, whereas the interior points correspond to the mixed states".
2 hours later…
8:44 AM
@JohnRennie But like if I have a fluid stress energy tensor for hyrogen and one for helium. How do we their cosmological constant will be the same?
@RewCie It's more a consequence of having the laws of physics formulated as initial value problems
*How do we know their cc will be the same
@MoreAnonymous The SE tensor is the sum of the SE tensors from all the components present. So for example with a mixture of hydrogen and helium you can split the SE tensor into a contribution due to the hydrogen and a contribution due to the helium.
@JohnRennie So this is what I meant
If we assume isotropy and homogeneity then we get the FLRW metric up to a constant (cosmological constant). Let's say I have $2$ isotropic and homogenous universes one with hydrogen and another with helium. Will they have the same cosmological constant?
If you have dark energy as well you add this so your SE tensor now has three components.
I was typing the question :P
The CC is separate from the matter density. Hydrogen, helium or any other matter will never result in a SE tensor that contains a cosmological constant.
To get a CC contribution you need a form of energy with an unusual equation of state, i.e. you need dark energy or something like it.
8:59 AM
@JohnRennie Would u be kind enough to look at 7 latexed pages and tell me where I'm going wrong?
I think its a property of the material
I can look ...
Can you convert it to a pdf and upload it to swarchive.ratsauce.co.uk/.uploads
That's my upload server.
Oh wow! okay its already a pdf
Can u confirm u got it?
Can u delete this image^ ?
It says my name :p
Wouldnt remain anonymous
Oops :-(
9:02 AM
No worries
It's a low energy limit result
And kinda heuristic
I can't see how that's going to work.
Which part?
Or equation?
A component that acts like a CC has to have an energy density that is independent of the scale factor i.e. as the universe expands the energy density remains constant. This is equivalent to having a negative pressure.
I find it hard to see how any scattering process is going to satisfy this requirement.
Yes ... but in classical mechanics u only measure potential energy difference
@JohnRennie See equation 19 or 20
I can't help I'm afraid, I'm not familiar with approaching GR through an action principle.
9:08 AM
But I think you'd need an interaction force that was independent of distance to get a cc from the interparticle potential.
@JohnRennie Proof of claim>
You need the energy density to be independent of the particle density. Suppose you have N particles per cubic metre and take the pairwise sums to get the interaction energy. Call this V.
From the action perspective you would need a constant in the action and that's like towards the end Equation 28
Now double the size of your cube and the interaction energy has to increase to 8V to keep the energy density constant.
Hmm, I'd have to reach for a pen and paper to see what this meant for the interaction potential, and I don't have time for that right now.
9:12 AM
@JohnRennie I dont see why equation 19 wouldn't satisfy this?
Maybe it does. I can't comment.
@JohnRennie No worries
@Slereah is it coincidental you just pop up with the mention of GR in the chatroom or you got some cool alert bot setup? :P
9:43 AM
@MoreAnonymous I pop up often
I just don't comment if it's about a topic I don't know about
10:09 AM
@Slereah I see ... By the way we were talking about a more updates version of what I sent u btw
10:27 AM
@ACuriousMind what the hell is a (-1)-category
oh wait I think (-1)-truncation is for 0-categories
10:45 AM
@Slereah I have no idea
I guess it's like the $(-1)$-simplexes
3 hours later…
1:35 PM
What constraints are we maximizing the entropy for in the grand canonical ensemble?
One is that all the probabilities add to 1
But I'm confused about energy
I cant constrain energy to be a constant otherwise nothing would happen right?
As in, it would be like the reservoir and box were isolated
1:49 PM
Hi, anyone here?
Do you know the answer to my question?
No I don't sorry
2:32 PM
Can anyone help me with a question about thermodynamics?
" Don't ask about asking, just ask."
Since entropy is a state function that means that regardless how we reach from state A to state B (for a given system) entropy will be the same S_B - S_A
But if the process to reach state B from A is reversible then the entropy change is zero
and if it is irrever. then its bigger then zero
so this is a contradiction
So what's the correct thing to say or think about?
3:09 PM
Can someone tell me what the constraints are and why?
What exactly are we measuring in statistical physics?
Averages I assume?
So is the average energy of system A being conserved (where system B is the reservoir) an axiom?
3:28 PM
from wikipedia: "The thermodynamic variables of the grand canonical ensemble are chemical potential (symbol: µ) and absolute temperature (symbol: T). The ensemble is also dependent on mechanical variables such as volume (symbol: V) which influence the nature of the system's internal states. This ensemble is therefore sometimes called the µVT ensemble, as each of these three quantities are constants of the ensemble."
so fixed chemical potential, absolute temperature, and volume
in particular this means that the total number of particles and the system energy are not constrained and can fluctuate
(by contrast, the canonical ensemble assumes a fixed number of particles and the microcanonical ensemble further assumes a fixed energy.)
(Hence why wikipedia also denotes the microcanonical/canonical/grand canonical ensembles as the NVE, NVT, and µVT ensembles respectively.)
in short: read the wikipedia page
3:45 PM
That's not really an answer at all to my question
In short: read my question
you asked for the constraints. chemical potential, absolute temperature, and volume are the constraints.
But you don't use those as conserved charges of the system
sure you do.
what do you mean? if the chemical potential, absolute temperature, and volume of a state can change, you're not in the grand canonical ensemble
3:49 PM
'in particular this means that the total number of particles and the system energy are not constrained and can fluctuate'
This is an axiom right?
I mean what quantities do I put into this $\sum_i p_i Q_i$
so that I can maximise the entropy
$\mu$ will turn out to be one of the Lagrange multipliers right?
So one of the $Q_i$ will be $N_i$
but I don't see how that isn't holding particle number fixed
@DIRAC1930 pretty much, yes. the average number of particles for instance is given by $-\partial \Omega/\partial \mu$
when that violates the definition of a grand canonical ensemble
So we are constraining the average number of particles?
in system A
no, we're constraining the chemical potential
i'm not sure how you'd formulate the grand canonical ensemble using Lagrange multipliers
But that's just one of the constants of the lagrange multiplier anyway
Why not?
b/c grand canonical ensemble doesn't fix particle number
so i'm not sure what sum you'd write out
3:55 PM
Doesn't it fix the average number of particles in equilibrium though?
and isn't $\sum_i p_i N_i$ the average number of particles in region A?
$\sum_i p_i N_i = <N>$ right?
that's like saying that any ensemble (grand canonical/microcanonical/canonical) fixes the average pressure b/c one can compute that as a thermodynamic derivative
which seems dubious
May I know the subject of this discussion?
yes. scroll up.
$N_A + N_B$ is fixed right?
Yeah, I see the subject. Thanks.
3:58 PM
where $N_A$ is the number of particles in the system we are measuring and $N_B$ is the number of particles in the resevoir
is $A$ your system and $B$ the environment?
sure, that seems appropriate
Is that the constraint then?
i think you're right, in fact
different microstates have different energy and particle number, and since a variety of microstates are possible this means that they can fluctuate
but the individual microstates do have definite energy and particle number
So if $\sum_i p_i N_i^A = <N_A> $ is a constraint, does that imply that $N_A + N_B$ stays constant?
should, but i'm not sure how you'd formulate this
you'd have to set up two systems, one for the system and the other for the reservoir
and it's been a while since i did that kind of thing
4:10 PM
Would you have to though because those $p_i$'s only refer to system $A$ and they are normalized?
In statistical mechanics, is the best we can measure just averages?
depends on which ensemble you're talking about. if you have a canonical ensemble, temeprature isn't understood as being an average
if the temperature can fluctuate, you're not in the canonical ensemble. (you could imagine a scenario where the temperature changes, but it'd have to do so slowly enough to remain in equilibrium)
and in the canonical ensemble, there's no fluctations in particle number
so if you've got a system where heat can flow in/out of a reservoir but particles can't, then there's no need to talk about average particle number
I'm confused about what happens when things are quantum
Aren't fluctuations just part and parcel of everything?
up to a point. but if you're in an energy eigenstate, then energy doesn't fluctuate
4:17 PM
Okay, are all the microstates energy eigenstates?
hmm. that seems dubious as a general statement, but would be appropriate for the quantum canonical ensemble
I want to talk about something bothers me for last few days.
the quantum version of the canonical ensemble isn't too bad, happily. there's not really an easy version of the microcanonical ensemble, by contrast, owing to energy quantization. (energy in the microcanonical ensemble is supposed to be a free variable, and quantization obstructs that.)
the quantum grand canonical ensemble isn't too bad either
@M.ÇağlarTUFAN as ever: if you've got a question, ask, and if people are interested they'll respond
4:38 PM
Testing: I think I typed so long I can't post my message.
I started 3rd grade this semester and I take quantum physics course. I have finished 4 weeks today. First 2 weeks of the course was like some chat about before and after of quantum physics.
My instructor talked about the experiments that can be explained with classical physics and other experiments that can not be explained with classical physics.
Then he talked about some new ideas like quantum structure of light, photons etc. And other experiments like blackbody radiation, photoelectric effect and so on.
He derived the Planck Radiation Formula in class, I don't know why he did this especially. I couldn't grab the ideas. Later on, last 2 weeks was really difficult for me.
In 3rd week, my instructor introduced us wavefunctions, probablity densities, average positions, average squares and square averages, some about Fourier transforms.
Then 4th week he introduced operators, position and momentum operators in position space and momentum space. And lastly energy operator and a little bit about Shrodinger equations.
Finally I posted it! Yay!
Anyways, like I was saying 4th week has just finished and I'm having so much trouble understanding things. My instructor has his notes on his website I took and photocopied them. I have been trying to work on the notes this whole weekend and I couldn't got any far.
that sounds like a pretty standard approach, for better or worse
This just bothers me, I had pretty good scores and understanding in other courses this far but this one feels pretty bad for me :(
what book are you using?
I especially asked my instructor about book but he just said "search my name on google and go to the website, then download pdf notes" but he has recommended 2 books on his website
Let me find them quickly
1) Concepts of Modern Physics, Arthur Beiser, 6th ed, McGraw Hill, 2003 2) Quantum Physics, Berkeley Physics Course, Vol. 3, 1965
i'm forgetting what book our uni uses
4:49 PM
I wonder if you could take a look at my instructor's notes but they are in Turkish. Maybe you can just follow some names and understand them.
Also, first book on the recommendations seem to follow the same order of subjects my instructor explained in lectures (except 1st chapter). But the book has less calculations, mostly verbal explanations of things.
yeah, hence what i said about this being pretty standard
Hmm, so I'll get used to this within some time?
well, when i say "standard" i mean that historically it's the usual approach people use to introduce the subject
whether or not it's the 'right' approach is another question entirely
Hmm, I kinda understand.
the intention, i imagine, is to start with classical mechanics (particle moving in 1D potential) and give the QM version of that (Schrodinger equation)
which is a very wave-mechanics way of approaching it
the trouble is that it involves a good deal of technical machinery to set up
and stuff you just sorta have to take on faith that it works
my own preference nowadays is routes which start from the Stern-Gerlach experiment and show you how the linear algebra machinery emerges in that
4:57 PM
To teach the quantum physics?
Uh, I don't really have any idea about it :/
yeah, you wouldn't at this point in the 'standard' approach
once you get comfortable with the formalism it's fine
but you sorta get a lot tossed at you at the start
I don't remember working with any integral like $\int_{-\infty}^{\infty}{x\left|\psi\right|^2dx}$
you just sorta get told to take that as the definition of $\langle x\rangle$
5:01 PM
And this integrals bother me
Yeah right
and more generally $\langle \hat{A}\rangle = \int_{-\infty}^\infty \psi^*(x)\hat{A}\psi(x)\,dx$
I wanted to understand the idea behind this integral but couldn't get it
What is $\hat{A}$ here?
any operator
Like momentum, energy or position?
for instance, $\hat{x}$ acts as "multiply by $x$" in position space
whereas $\hat{p}=-(i/\hbar)d/dx$ in position space
5:03 PM
@M.ÇağlarTUFAN It's based in the interpretation of $\lvert \psi\rvert^2$ as a probability density
the case of $\langle x\rangle$ isn't too hard to understand in that context
it's just a general notion that if you want the expectation value of $f(x)$ with respect to a probability density $\rho(x)$, you do $\int f(x)\rho(x)$
the harder one is $\langle p\rangle$
there's nothing mysterious or ad hoc about that integral, it's a direct consequence of the probabilistic-ness of QM
Okay, I kinda understand it, a little.
5:05 PM
for $\langle p\rangle$ you can make the argument "do it in momentum space where it's easy" and then Fourier transform back
@Semiclassical that was exactly what I was going to say :P
I've got a question right here
i say that it works, at least. i'd have to write it out to be sure
and probably would end up needing to be careful about integration by parts etc
In the notes of my instructor, $\lange x \lange=\int{x|\psi|^2dx}
5:07 PM
What do you do in relativistic systems where particle number is not conserved at at all?
grand canonical ensemble still works there
so long as you can define operators for particle number and energy, grand canonical ensemble isn't fussed
But he took the square of x and then the average, the integral became $\langle x^2 \rangle=\int{x^2|\psi|^2dx}$ why not $|\psi|^4$?
@M.ÇağlarTUFAN suppose you want to figure out the average value of $x$, where the experiment is "pick a random integer uniformly from 0 to 10"
then you'd compute the average by noting that all outcomes are equally likely, so you'd have $0\cdot 1/11+1\cdot 1/11+\cdots+10\cdot 1/11=55/11=5$
what would you compute if you wanted to figure out the value of $x^2$?
5:13 PM
x here is the picked number from 0 to 10 right?
Then i would expect to get 5 as result again
for $x^2$?
Yeah, because 0^2, 1^2, 2^2 ... 10^2
Still same amount of numbers to be able to pick, 11
well, you'd still have an equal likelihood of getting the results, so 1/11 is correct for that
but now your outcomes are different: 0, 1, 4, 9,..., 100
so you'd be asking for the average value of these squared values
5:15 PM
Oh, right
Let me calculate it please
it's fine to just tell me what you'd compute, it's a little tedious to do the addition
That makes 385/11=35
so the random variable's value gets squared, not the probability itself
What's the relation between 5 and 35, except it is 7 times 5 :P
Oh, yeah
I understand what you mean
nothing too useful, though you do need to have 5^2 = 25 < 35
$\langle X^2\rangle \geq \langle X\rangle^2$
5:18 PM
Sorry to interrupt you, I'm listening
this is just a side point, anyways
the point is that you do the new expectation value by replacing $x\to x^2$
the probabilities don't change
Hmm then the probabilities here are $|\psi|^2$, right?
and it's the same idea with $x^2$ as an integral: If you have any $f(x)$, then the expected value is $\langle f(x)\rangle =\int_{-\infty}^\infty f(x)|\psi(x)|^2\,dx$
sorta yes sorta no. the problem is that, in the integer example, theres's a finite probability of getting each outcome
but the probability of picking 0 from a continuous distribution is strictly 0
so in that sense $|\psi|^2$ isn't the probability of getting any particular $x$, because that probability is zero
$\frac{1}{\infty}=0$ hmm
moreover, think about the units a bit. if $1=\int_{-\infty}^\infty |\psi|^2\,dx$, then what units does $|\psi|^2$ have?
5:23 PM
By units, do you mean distance, mass etc.?
Well, integral has no unit but integral square of something should outcome no unit.
Then I guess
not quite. $x$ is a position
so $\int_a^b 1 \,dx=b-a$
hence if $|\psi|^2$ was dimensionless, then the integral would have dimensions of length
Uhm, why?
Sorry, I couldn't understand this :/
take the example of an integral like $\int_a^b 1\,dx=b-a$. $1$ is dimensionless, but $b-a$ has dimension of length
more generally: what's the difference in the dimensions of $f(x)$ and $\int_a^b f(x)\,dx$?
5:29 PM
I would say both are dimensionless
Because both outcome a number, am I thinking wrong?
so is $\int_a^b 1\,dx$ dimensioness?
$x$ is a position here
Ah, I understand it now
note that $a,b$ are necessarily positions here and therefore have dimensions of length
Yes it would be distance between two positions so it should have length unit
right. so integrating w/r/t $x$ adds a dimension of length to your integrand
5:32 PM
so if $\int_{-\infty}^\infty f(x)\,dx=1$, what dimensions should $f(x)$ have?
length I guess
Because f(x) is position
other way around. integrating w/r/t adds a dimension of length
and you want to end up with something that's dimensionless
5:34 PM
for a non-probability example
take $\rho(x)$ to be a mass density. then by definition $M=\iiint \rho(x)\,dV$
what units does mass density have?
right, which makes sense b/c the integration over volume gives dimensions of mass/volume * volume = mass
Oh, yes. I got the point
5:37 PM
so if we want to integrate $|\psi|^2$ w/r/t to $x$ and get a dimensionless number, $\int |\psi|^2\,dx=1$, what dimensions must $|\psi|^2$ have?
It should have 1/length
Am I correct?
Oh, I was so afraid that you were going to say no :P
more generally, we expect that $\int_a^b |\psi|^2\,dx=\text{Pr}(a\leq x\leq b)$
so we say that $|\psi|^2$ has units of "probability"/length
hence $|\psi|^2$ is a probability density
if you want to interpret it as a probability, suppose $b=a+\Delta x$ for small $\Delta x$
probablity density means probability of something over a range, right?
5:40 PM
then you can approximate the left-hand side as $\int_a^{a+\Delta x}|\psi|^2\,dx\approx |\psi(a)|^2\Delta x$
so $\text{Pr}(a\leq x\leq a+\Delta x)\approx |\psi(a)|^2\Delta x$
it's more like: if you compute the probability to find the particle in a very small interval $\Delta x$, then this probability is approximately some multiple of $\Delta x$ itself
and that multiple is the probability density
in other words, while the probability gets smaller and smaller as you shrink the interval, the rate at which this probability vanishes occurs will not be zero
so it's more like "probability over a range" / "range" -> probability density
if this rings any bells, it's probably because this is exactly the situation you have for the Planck radiation law
Uhm what donyou mean by if this rings any bells?
You mean if it recalls something?
english idiom, sorry
if it sounds at all familiar, it's because this same matter of "density" arises when it comes to the Planck radiation law
Yeah, I'm pretty unfamiliar but I got it. Actually I couldn't understand the relation between these two :(
well, the point is that in the Planck radiation law, the energy contained by a single wavelength is zero
you only get some energy when you look at some range of energies
Why doesn't anything change when considering relativistic systems?
Is it because we are considering the system at a macroscopic level?
5:48 PM
@DIRAC1930 b/c all grand canonical ensemble cares about is if your particle number and energy operators exist. (that's the QM version of "if particle number and energy label your microstates".)
"the energy contained by single wavelength is zero" so any standalone wave has no energy?
But does particle number label your microstates? $[H,N] \neq 0$ in the relitivistic case
they don't, but that doesn't matter. all that matters for defining the grand canonical ensemble is whether the operators exist
@M.ÇağlarTUFAN hmm. i feel like i'm saying that wrong
strictly speaking, for a finite size box you only have a finite number of frequencies and each carries a certain energy
so this is probably some annoying fact about the infinite-system limit?
I really feel exactly like when I'm listening to my instructor in class. I could understand the $\langle x \rangle$ and $\langle x^2 \rangle$ and about dimensions of these integrals, but almost nothing other than these two :P
What I also couldn't understand is
@DIRAC1930 in QM, the grand partition function is the density matrix $\hat{\rho}=\exp\left[(\Omega +\mu \hat{N}-\hat{H})/kT\right]$ where $\Omega$ is the grand potential determined by demanding that $\rho$ have trace 1
and that expression doesn't care about whether $N,H$ commute
5:55 PM
Getting to that expression is the part I'm struggling with
hmm, i say that, but
"Note that for the grand ensemble, the basis states of the operators H, N are states with multiple particles in Fock space, and the density matrix is defined on the same basis. Since the energy and particle numbers are all separately conserved, these operators are mutually commuting."
from Wikipedia
For example, the $Z^{-1}$ term is determined by $\sum_i p_i=1$
If $\int{\rho dV}$ has unit of mass because we are integrating somrthing with units of mass/volume over volume, then with this respect if $\int_{-\infty}^{\infty}{|\psi|^2dx}=1$ and is unitless, then $|\psi|^2$ shouls have units of something/length here something is dimension of probablity density
I'm not sure what exactly the constraint in the form of $\sum_i p_i Q_i = c$ gives rise to the $\mu \hat{N}$ term
@M.ÇağlarTUFAN yeah.
hence why we can't interpret $|\psi|^2$ as a probability, but as a probability density
5:58 PM
What is actually a probablity density?
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