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4:31 AM
@SirCumference you can try, but there's still one whole quarter left for 2020 to do it's thing :p
@Charlie each component becomes an operator, even in non-relativistic quantum mechanics the classical momentum vector becomes a momentum vector operator with three scalar operator components
2020/4 = 505
let's hope the last quarter isn't palindromic
1 hour later…
5:48 AM
@JohnRennie So can we say that the reaction given takes place in multiple steps? If yes then why is it not a $\beta$ emission too? Answer given to the question are the options (2) & (3).
@ArnavMahajan Because the reaction involves reverse beta decay i.e. positron emission not beta decay.
5 hours later…
11:19 AM
@bolbteppa That is a good point, thank you
12:17 PM
guys I could really use some help with this, or rather, understanding the finer details that lead to the confusion: physics.stackexchange.com/questions/581810/…
12:33 PM
Q: Should comments be used for providing "helpful hints"?

BioPhysicistI often see users, including moderators, leaving "helpful hints" comments on various posts. They could be posted on off-topic homework problems pointing the user in the right direction on how to solve the problem, but I also see them on ligitimate posts in the sense of "you should think about thi...

2 hours later…
2:46 PM
@ACuriousMind Isn't the Pauli equation evidence for spin in non-relativistic equations? (regarding your comment)
@MoreAnonymous What does that have to do with my comment?
The spin-statistics theorem isn't about whether things have spin or not
".... The spin-statistics theorem is a statement about relativistic quantum field theories ..."
it's about bosonic/fermionic behaviour being linked to half-integer/full-integer spin
The way I mildly remember it was
you can have theories where half-integer particles behave like fermions that aren't relativistic QFTs, but that's not a consequence of the spin-statistic theorem, it's just something you put in there
2:50 PM
based on the wave-function changes sign when the coordinates are interganged say $\psi(r_1,r_2) = - \psi(r_2,r_1)$ for fermions
I was ultimately under the impression this limits the eigen basis you can have.
Is this wrong?
eigenbasis of what operator?
The hamiltonian
I don't understand what you mean by that
One moment
For fermions, you have that the n-particle space is the antisymmetric part of the n-fold tensor product of the one-particle space, for bosons, it's the symmetric part
this has nothing to do with the spin-statistics theorem, that's just the definition of what a fermion/boson is
2:56 PM
Okay so it would it be better to just reframe the question has the effects of being a fermion or boson have on the wigner function?
@MoreAnonymous Yes. (But I don't know what effect you expect it to have)
Well your limiting the basis you can have right?
Since your only allowed to use this one
posting pic (one moment)
Also relevant
So, how are you defining the Wigner function for a system of bosons/fermions?
The above was for bosons obviously
But I was saying that your restricting the basis
@ACuriousMind The usual way
But but using the "new basis"
as shown in the pic
In fact only the new basis is allowed
(for bosons)
I don't know what "the usual way but using the new basis" would mean. Can you write down a formula for it?
3:08 PM
Sure ... I'll do for $2$ bosonic particles. One moment
Will be uploading second part (as well)
you can use MathJax in chat instead of uploading pictures
And $\psi (r_1,r_2)= \frac{1}{2^{1/2}} (u_{n_1} (r_1 ) u_{n_2}(r_2) + u_{n_2}(r_1) u_{n_1}(r_2)) $
Am I making sense?
Yes. So what do you mean by an "effect" on $W$?
hi folks I am in trouble.....
Well since I'm restricting only particular eigenbasis surely this will mean I will have certain $W$?
3:17 PM
Also, have you checked that this indeed behaves like the usual Wigner function w.r.t to taking expectation values?
There's no point to calling this a "Wigner function" unless it leads to the same equivalence between phase space computations and Hilbert space computations
@ACuriousMind Is there reason for to be suspicious?
@MoreAnonymous Probably not, but I'd make sure (or look for other literature discussing this function) before asking a question about it :P
@ACuriousMind arxiv.org/pdf/quant-ph/0702037.pdf they use the same function albeit for a different problem
Though not the same "new basis"
I was reading thermodynamics and I read about the entropy during reversible and entropy during irreversible process
in which case it is higher and why?
if both the process are same
@JackRod Plot twist there is no such thing as an irreversible process :P (jk dont take me seriously)
3:26 PM
@MoreAnonymous Yeah, I'm satisfied this works (the original work on Wigner functions for bosons/fermions is probably the 1959 paper "The Wigner distribution function for systems of bosons or fermions" by Schram and Nijboer, doi doi.org/10.1016/S0031-8914(59)97760-2)
@ACuriousMind the question got answered! :)
But again, what kind of "effect" are you asking for? All that has changed is that your $\psi$ is now (anti-)symmetric in some variables. You can deduce from that straightforwardly corresponding (anti-)symmetry properties of the $W$. What else are you looking for?
" I'm looking for a calculation where the effects of this manifestation disappear under the classical limit"
Which still isnt obvious to me despite the answer. Wonder if I'm being dense
What's the "classical limit"?
3:32 PM
$\hbar \to 0$
E.g. the (anti-)symmetry properties won't disappear if you just do $\hbar \to 0$, and there's no reason to expect them to
the naive classical limit of fermionic systems is not a classical system, it's nonsense :P
@ACuriousMind Curious and curiouser
there's a reason we have to introduce weird classical phase spaces with anti-commuting Graßmann parts if we want to get fermionic systems through quantization
If you actually say something instead of randomly posting five question marks someone might understand what you're trying to say.
3:37 PM
@JackRod They aren't the same
@Charlie ok.. why?
I'm not sure what you mean by why, if you look them up they have different definitions
sorry, but my thermodynamics knowledge is very thin all I understand is
entropy is the s=dq/T
what kind of definition you are talking about
@ACuriousMind What "enlightened" limit does one take then?
Reversible and irreversible processes have different definitions, they are not the same thing, so it's at least reasonable that entropy changes would be different depending on which type of process is being performed
A similar question was asked here a while ago (possibly by you I don't remember), I can't tell you the exact mechanism, but if your starting point is that reversible and irreversible processes are identical then that's where you're getting stuck
3:42 PM
ok taking example of adiabatic
@MoreAnonymous You need to look at your system and think about what it means for that system to approach its "classical limit". The classical limit in QM is not as simple as always taking $\hbar \to 0$ and getting something meaningful
usually it involves looking at some form of coherent states that approach the classical states as $N\to \infty$ and $\hbar \to 0$
@ACuriousMind hmmm .... interesting. Thanks. Any reference? I can read this up from?
@Charlie what I feel is a reversible process body is in equilibrium with each moving step
Yes, that's a reversible process
while in irreversible it is in equilibrium at final and initial point
but how would I relate this with entropy
it is going over my head again and again
3:46 PM
@MoreAnonymous if you search for "classical limit" you'll find a wealth of literature. One motivation of "deformation quantization" is that it makes $\hbar\to 0$ indeed a proper limit. See e.g. the answers and references in physics.stackexchange.com/q/32112/50583 and physics.stackexchange.com/q/56151/50583
@JackRod Reversible processes don't change the net entropy of the system and surroundings. Irreversible processes increase the net entropy of the system/surroundings.
@ACuriousMind thanks :))
ok now the picture I have after jmac comment is the entropy related to a particle that how they are moving
Can you imagine why a process might be called "reversible" if it doesn't change the entropy of its surroundings, given the second law of thermodynamics?
the second law of thermodynamics has no indication related to entropy
what i feel
3:51 PM
The second law of thermodynamics is entirely about entropy lol
Do you know what the second law is?
From Wikipedia: "The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time, and is constant if and only if all processes are reversible."
oops i mismatch second law with the first
second law basically that certain things are not reversible like if I rub a block on the table and heat production due to velocity can't be vice versa with a condition where heat produce the velocity in the block completely
means heat cannot give gurantee of 100 percent work
@JMac @Charlie
I'm not entirely sure what you've just said
what i said heat can't be entirely be converted into work
3:57 PM
I'm not a thermodynamics buff I'm afraid, if you're confused you should either pick up a textbook (most thermodynamics textbooks will talk about the second law in the first chapter or two) or ask a question on the main site
where work can convert into heat completely
I messed up with the word entropy some say it is very diffciult to understand untill i know statistical mechanics
Yeah I'm honestly a bit iffy on thermo specifics too. I took several courses; but that stuff barely sunk in at all.
I honestly found classical thermodynamics torturous, statistical mechanics was cool but as a non-physics undergrad student I've missed the boat on a lot of thermo, I plan to come back to it one day though
can you refer one standard and good book on this topic?
4:00 PM
If you want a free online set of notes you could try David Tong's notes here
He covers classical thermodynamics in chapter 4
No not notes just a good book
It depends what you're trying to get out of this, are you just trying to learn basic thermodynamics or is this for a university course you're taking? If it's the latter you should definitely read a textbook carefully, the second law is essential
The first thermo course I took was such a mess. It was called "Thermo-fluids" (so they tried to teach us the basics of thermodynamics and fluid mechanics), and the prof highly overestimated the classes background knowledge. The average on midterms was like 25% and they had to make the exam super easy and overwrite the midterms so people could pass.
Thermodynamics seems very hit or miss, it's either explained well and makes moderate sense the first time, or it isn't and your trust for the field is crippled for life :P
4:16 PM
I at least learned some of it eventually; but it was more applied than theoretical so I got better at like solving for like pressures and temperatures during cycles; but never really got a solid grasp on the theory.
@JMac can I say it is more theoretical like they have given working of diesel and petrol engine is the process is same
Diesel and petrol engines don't work the same way :P
We learned quite a bit about the different common processes. Quite an important part of the applied thermodynamics.
You'll learn that the first time you put diesel in a petrol car or vice versa
Diesel also works as a lubricant in diesel engines, so swapping there can be pretty bad AFAIK.
4:34 PM
@Charlie That should be more difficult now since diesel filler nozzles are too large for petrol cars and new diesel cars have a protection system.
Fortunately yeah. I still have vivid memories of my furious dad calling for assistance on the side of the road, 20m from the petrol station :P
I guess you can understand what this is about.
Q: Deleting closed questions

Gyro GearlooseI want to delete a closed question. The "closed" dialog offers me to edit question (which I did to only incur more close votes) or "Delete question", which is what I really want to do now. But when I click on that, I am fooled by, oh, different result than last time, no it is "Repeated deletion o...

5:03 PM
@ACuriousMind I was looking into a question and was wondering what was your thoughts on this?
Q: Will the black hole size increase?

drewdlesI was thinking about the following thought experiment, but wasn't sure about its outcome. Suppose there is a black-hole and I enter it with a partitioned box containing two different gases on either side. After passing, the event horizon the radius of a black-hole increases as I added mass to th...

(reply when you can)
5:20 PM
@JackRod my favourite is Callen
5 hours later…
10:27 PM
@MoreAnonymous I'm not really into GR or black holes in particular, but I think the answers there are correct - the no hair theorem, that outside observers never really see anything crossing the horizon and that the inside of the hole is causally disconnected from the outside mean that nothing you do on the inside can have any observable effect on the outside

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