The h Bar

Giskard42

4:00 PM

I usually think of charge in a gas as ions, but I haven't put enough energy into the plate to ionize it

G. Bergeron

@bolbteppa ...meh yea-ish

Giskard42

@G.Bergeron Say I induction furnace it, so there's no contact. I'm just boiling it slowly away entirely

G. Bergeron

@bolbteppa It's not meant to act on anything

@Giskard42 There could be enough energy to ionize

@Giskard42 but what will happen is that you will end up with charged gas

that will collect on the other plate

or more like soot particles

Giskard42

@G.Bergeron Sweet, so even though the gas isn't ionized, it'll still be charged?

The electrons will be distributed throughout the gas?

G. Bergeron

@Giskard42 Well, yes. But it will be ionized

Really mildly

bolbteppa

4:02 PM

I know, but in the classical case, when you integrate $\int \rho d^s x d^s p$ you get infinity because of the $2s - 1$ constraints, i.e. in phase space it makes perfect sense for constant energy to eliminate a variable and make this integral bad, but in the quantum case, it's analogue, $\mathrm{tr}(\rho)$, should also become infinite due to conservation of energy, so you write $\mathrm{tr}(\rho) = \sum_n <n | \rho | n>$ or something and conservation of energy should motivate infinity right?

Giskard42

@G.Bergeron Ah, okay.

@G.Bergeron Thanks a lot for your help!

Sorry to interrupt

bolbteppa

Hmm

G. Bergeron

@bolbteppa If the spectrum is continuous then it's like the classical space in that respect

bolbteppa

I think I got it

G. Bergeron

your concentrating a probability measure on a set of measure zero with respect to the previous measure... It has to be infinite

@Giskard42 np

So if it is continuous it won't be a sum but rather an integral

bolbteppa

4:06 PM

If $\rho$ was a constant number, then you have $\mathrm{tr}(\rho) = \sum_n <n|\rho|n> = C \sum_n <n|n> = \infty$, but the ability of writing $\ln(\rho_n) = \alpha + \beta E_n$ allows $\rho_n$ to be a constant since $E_n$ is constant

G. Bergeron

$Tr(\rho)=\int\limits_e \langle e | \rho | e \rangle$

@bolbteppa But that would mean constant on all the spectrum so only one energy state (modulo degenarcies)

bolbteppa

Still making sense of the number formalism tbh, is $\sum_n <n | n> = \infty$ even true?

G. Bergeron

@bolbteppa If $n$ run to $\infty$, yes but it may not

e.g. spins in magnetism

Your sum is $\sum_n 1$

The point is not that the delta appears like that

You put it in

as a constraint

bolbteppa

hmm

G. Bergeron

energy would be a dimension like in the classical case

bolbteppa

4:12 PM

So my infinite trace, if it were true, would justify the delta function, it exactly copies the classical case, the question is justifying that infinite trace, it seems like he might be waving his hands and saying, because we have so many particles, and because energy levels are so close, we can treat it as infinite

G. Bergeron

and you evaluate a non-vanishing distribution on that N-1 dimensional object. So the distribution on the whole space has to be a delta function with respect to the observables (Hamiltonian) associated to the constraint (energy)

@bolbteppa You don't do that in the classical case either

bolbteppa

Landau does do this in his book though

Let me get it

G. Bergeron

You put it in when you assume that energy is conserved and the system is closed

Maybe not explicitly but it appears nevertheless

bolbteppa

archive.org/stream/…

In between equations 4.3 and 4.4

G. Bergeron

Exactly he says "because of the constraint, we'll have to put in a delta function"

essentially

bolbteppa

4:17 PM

You seem to be saying, we add the constraint of a delta function to $\ln (\rho_n) = \alpha + \beta E_n$, but he is literally saying $\rho$ becomes a delta function without working out a Lagrange multiplier problem, it's automatic just from analyzing the integral $\int \rho d^s p d^s q$

G. Bergeron

The real notion is "concentrating a distribution on a set of measure zero"

forget about the log thing

you shove the delta in the integral not the log

bolbteppa

The log thing is literally the most important thing from this perspective :p

G. Bergeron

He says it becomes a delta BECAUSE of the reduced dimensionality of the manifold, another way of saying constraints on the manifold

bolbteppa

Ok he says the integral $\int \rho d^s p d^s q$ is zero, not infinity :\

G. Bergeron

@bolbteppa Not at all on this delta thing

@bolbteppa zero outside of the reduced manifold

where the constraint is violated

Look, it's like if I tell you "I have a point somewhere in a plane"

Now, I tell you I know it lies on a line

The distribution will become a delta function on the line as a distribution on the whole plane

it needs to be infinite because the probability if picking a point on this exact line is zero for any bounded distribution

It's a properties of continuums

The point about the log is that just by stating that your system is invariant under rotations, translations and isolated, which are all extensive quantity, then you have to have the microcanonical ensemble if there are no other conserved quantity

Translation invariance = $\rho$ is not a function of space, Rotation invariance = not much because $\rho$ is a scalar, time invariance = energy is conserved and fixed so it is a parameter of $\rho$.

bolbteppa

4:29 PM

So he seems to be saying, classically, that - you have conjugate variables $(p_i,q_i)$ okay? Because we're in the Hamiltonian formalism, we can treat one of these variables as, say, $q_2$, as the energy. But then $dq_2 = dE = 0$ because energy is conserved. However, the integral $\int \rho d^s p d^s q$ as an integral over $2s$ coordinates has to hold, and so $\int \rho d^s p d^s q = \int \rho d^{s-1} q d^s p dE = \int \rho d^{s-1} q d^s p \cdot 0 = 0$.

The only way to make this integral non-zero somewhere, is if $\rho$ is a distribution, in this case a delta function is chosen so that energy is constant on our hypersurface. That seems to be the logic of why it becomes a delta function. Maybe you can write it as a Lagrange multiplier problem, delta function seems scary but okay, but same outcome right?

The same logic, as Landau seems to indicate, carries over to the quantum case, no Lagrange multipliers necessary, though it may be equivalent. The question is the quantum case, my trace should be zero apparently :p

G. Bergeron

@bolbteppa Lagrange multiplier is not for the per se delta function, it is to obtain the distribution directly

bolbteppa

Yes I agree with what you've written

G. Bergeron

@bolbteppa This is mostly it

bolbteppa

So why is my trace zero

haha

G. Bergeron

@bolbteppa same outcome

it's not

it's only gonna be a delta function if the trace is an integral and then it is the same reasonning

@bolbteppa Anyway, I need to go

bolbteppa

4:33 PM

Alright thanks a lot for the help

G. Bergeron

np

bolbteppa

Oh so $\int_e <e |\rho | e> = C \int_e <e | e> = 0/\infty$?

Top of the page here archive.org/stream/… he seems to justify writing this integral

MetaEd

20 hours ago, by MetaEd

Do I need an add-on in chat that interprets these markups correctly?

bolbteppa

If $\mathrm{tr} (\rho) = \int_e <e | \rho |e> = C \int_e <e|e> = C \int_e 1 = \infty$ then we have a function that is infinite on the energy surface, as a delta function is, motivating writing it as a delta function maybe?

(So close!)

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