Grand canonical system, relativistic cts Hamiltonian

[binbagsss]

Homework Statement

Question attached:

Hi,

To me this looks like a classical, continuous system, as a pose to a quantum, discrete system, so I am confused as to how to work the system in the grand canonical ensemble since , in my notes it has only been introduced as a quantum thing,(consistent with what I found only after a quick google to check I hadn't missed anything here: https://ps.uci.edu/~cyu/p115A/LectureNotes/Lecture9/html_version/lecture9.html) .

Homework Equations

I do have the relation that:

$Z(T,\mu,V)= \sum\limits_{N=0}^{N=\infty}e^{\beta \mu N} z(\beta,N,V)$ [2]
(where I have used a capital Z to denote the grand ensemble and a small z to denote the canonical ensemble)

However the derivation of this is from a quantum canonical ensemble as a pose to the classical canonical ensemble.
But even so, this seems like my best bet at attempted the question, although I will now be substituing in the classical canonical ensemble and not the quantum canonical ensemble, which from the derivation I think the formula should really be used for.

The Attempt at a Solution

So my plan is to integrate over phase-space with this Hamiltonian in the canonial ensemble and apply the formula above.

Doing the integral I get:

$z_1 = \frac{\sqrt{2\pi}}{h\beta c}V\frac{\Gamma(d)}{\Gamma(\frac{d}{2})}$

(Please ask me for details of my working on this if needed, but from what I see, that's not the issue here)

I then use Gibb's formula that $z_N=\frac{(z_1)^N}{N!}$

$\implies Z=\sum\limits_{N=0}^{N=\infty}\frac{e^{\beta \mu N}}{N!}( \frac{\sqrt{2\pi}}{h\beta c}V\frac{\Gamma(d)}{\Gamma(\frac{d}{2})})^N$ [1]

I know that $\Phi(V,T,\mu)=\frac{1}{\beta} In(Z) = -P(T,\mu) V$

and so from this I need the log of [1], and using the above its easy enough to compute $p$, however I can't see how to simplify the sum.

Is the use of [2] wrong here?

Is there a quantum way to look at this system, or a classical way to apply the grand canonical ensemble ?

Many thanks in advance.

 

[binbagsss]

On other thoughts, perhaps in order to make my application of [2] valid, I should take a limit of the continuous canonical ensemble once computed, one that will reduce it to it's quantum canonical ensemble, to a certain order relative to that limit. Could this possibly be a low temperature limit?

 

[binbagsss]

or perhaps change the summation over $N$ to an integral in [2]?

 

[mastrofoffi]

The grand canonical ensemble is not just a 'quantum thing', and the link you posted doesn't seem to say such a thing. If you don't have a reference textbook this might help clear things up for you: https://itp.uni-frankfurt.de/~valenti/WS13-14/all_1314_chap10.pdf
At the end of this chapter you'll also find a worked example for the classical ideal gas: you might want to work that out by yourself, check your results, and then come back to this problem.
Also, post #2 doesn't make any sense to me(besides what you actually meant, note that quantum mechanics is not a limit case of classical mechanics, it's the other way around, so there's no way you can ever turn a classical quantity into a quantum one through a limit process) and post #3 is clearly not a good idea(you understand why?)

To address your question about [1], I didn't check the calculations so I'm assuming it's all correct, notice that you have a bunch of stuff to the power n, divided by n!, so
$$ \sum_{N=0}^\infty \frac{x^N}{N!} $$
Does this look familiar?

 

[TSny]

I think all of @mastrofoffi 's comments are excellent.

@binbagsss , the expression $z_1 = \frac{\sqrt{2\pi}}{h\beta c}V\frac{\Gamma(d)}{\Gamma(\frac{d}{2})}$ doesn't appear to be quite correct. Shouldn't $z_1$ be dimensionless? Maybe you just have a typographical error.

 

[binbagsss]

and post #3 is clearly not a good idea(you understand why?)

$N$ is the number of particles in an energy state of $E$, and so changing the summation to an integral would be doing the previous point? trying to take a limit to get a quantum system to a classical?

$$ \sum_{N=0}^\infty \frac{x^N}{N!} $$
Does this look familiar?

indeed

 

[binbagsss]

I think all of @mastrofoffi 's comments are excellent.

@binbagsss , the expression $z_1 = \frac{\sqrt{2\pi}}{h\beta c}V\frac{\Gamma(d)}{\Gamma(\frac{d}{2})}$ doesn't appear to be quite correct. Shouldn't $z_1$ be dimensionless? Maybe you just have a typographical error.

yes apologies typographical , the denominator of the first fraction should be raised to the power of d

 

[binbagsss]

At the end of this chapter you'll also find a worked example for the classical ideal gas: you might want to work that out by yourself, check your results,

okay, many thanks for this link
so relation [2] is general, so with this is my method outlined in the OP still wrong, I assume from your suggestion? To compute the classical canonical ensemble and then apply this?

 

[mastrofoffi]

$N$ is the number of particles in an energy state of $E$, and so changing the summation to an integral would be doing the previous point? trying to take a limit to get a quantum system to a classical?

Leave quantum mechanics aside, there's no quantum stuff going on here. Forget this particular exercise for a second: in a most general setting, when do you use a summation and when do you use an integral?

okay, many thanks for this link
so relation [2] is general, so with this is my method outlined in the OP still wrong, I assume from your suggestion? To compute the classical canonical ensemble and then apply this?

Not sure what you mean; so first of all i think you should get your namings right to avoid misunderstandings: you are not 'computing the ensemble', you are computing the partition function, a function which completely describes the statistics of your system. Every system, depending on which physical quantities it is allowed to exchange with the environment, can be assigned a partition function, and each partition function represents an ensemble.
In your case, [2] is the definition of the grand-canonical partition function and it is defined in terms of the canonical partition function, so yes you can calculate the canonical partition function of the system and then apply [2] to get the grand-canonical. And this always works, that's the cool thing: for every problem of this kind you'll just have to calculate the partition function of your choice and then extract the desired thermodynamical quantities from that.

I think all of @mastrofoffi 's comments are excellent.

Thank you, I appreciate that.
I'm actually just a student and I'm taking Statistical Mechanics right now, so I guess I can relate to OP's difficulties on the matter; trying to help also gives me a chance to check my understanding of the subject.

 

[binbagsss]

Leave quantum mechanics aside, there's no quantum stuff going on here. Forget this particular exercise for a second: in a most general setting, when do you use a summation and when do you use an integral?

Discrete - summation
Cts- integral.

So I was thinking quantumly the energy spectrum is discrete, and classically cts. but rather is this the energy levels, and nothing to do with the number of particles that may occupy an energy level- this remains the possibly of being infinite, for bosons? But doens't a summation imply this is always countable- is this always countable?

 

[mastrofoffi]

Don't care about spectrums, bosons, and unicorns: none of this is mentioned in your text. Particle number $N$ sounds like a discrete quantity to me, so the summation will be fine I guess.
The summation in [2] has nothing to do with energy levels, check again how the grand-canonical ensemble is defined if you have doubts. When you use an ensemble you are theoretically creating infinite copies of your system in different states and assigning, through the ensemble distribution, a probability value to each of these states; in the grand-canonical ensemble one of the quantities which is allowed to fluctuate(or better, one of the quantities that the system is allowed to exchange with the environment) is the number of particles $N$, so when you compute the grand-canonical partition function you have to consider all the possible values of $N$ that your system, in all its different copies, can take. How do you do that? You just sum over $N$ from $0$ to $\infty$. If you have troubles understanding [2] I think you should give a look at the derivation of the formula, that will make things crystal clear.

 

[TSny]

yes apologies typographical , the denominator of the first fraction should be raised to the power of d

Yes. But I think the numerator of the first fraction also needs some modification.

 

[binbagsss]

Yes. But I think the numerator of the first fraction also needs some modification.

it sure does. :)

 

[binbagsss]

is the number of particles $N$, .

Apologies to re-bump.
So the derivation to get (2) that relates the canonical ensemble to the grand canonical ensemble uses that $\sum_{n=0}e^{-\beta E_n} = \sum_E \Omega(E,N) e^{-\beta E}$ where $\Omega(E,N)$ is enabling to account for degenaracy of an energy level.

So I think I may be getting a bit mixed up because we use $n$ and $N$. So does $n$ label an energy state, but $N$ counts particles

 

[binbagsss]

Don't care about spectrums, bosons, and unicorns: none of this is mentioned in your text. Particle number $N$ sounds like a discrete quantity to me, so the summation will be fine I guess.
.

Still a tad confused with this.So under the classical limit, there are some quantities such as energy that go from discrete to cts, and others, such as the number that remain discrete?

 

[TSny]

Apologies to re-bump.
So the derivation to get (2) that relates the canonical ensemble to the grand canonical ensemble uses that $\sum_{n=0}e^{-\beta E_n} = \sum_E \Omega(E,N) e^{-\beta E}$ where $\Omega(E,N)$ is enabling to account for degenaracy of an energy level.

So I think I may be getting a bit mixed up because we use $n$ and $N$. So does $n$ label an energy state, but $N$ counts particles

Yes. The first summation is over all the individual states. Some states will have the same energy. The second summation is over the energy levels. N is the number of particles.

 

[TSny]

Still a tad confused with this.So under the classical limit, there are some quantities such as energy that go from discrete to cts, and others, such as the number that remain discrete?

Yes. You can treat the energy as varying continuously. N is discrete.

 

[binbagsss]

The grand canonical ensemble is not just a 'quantum thing', and the link you posted doesn't seem to say such a thing. If you don't have a reference textbook this might help clear things up for you: https://itp.uni-frankfurt.de/~valenti/WS13-14/all_1314_chap10.pdf
At the end of this chapter you'll also find a worked example for the classical ideal gas: you might want to work that out by yourself, check your results, and then come back to this problem.

Hi , apologies this link now seems to be down, do you happen to any others given the classical integration expression for the grand canonical ensemble.

Given that the quantum is $\sum\limits_{n=0}^{n=\infty} e^-{\beta(E_n-\mu n_n)}$
I suspect the integral expression (in d-dimensions) is:

$\int \Pi_{i=1}^{i=d} (\frac{dp dq}{h}) e^{\beta(H(p,q)-\mu n_n)}$

(Apologies this is probbaly a stupid question, but, since $n_n$ is the number of particles in the energy state $E_n$, please correct me if I am wrong here, I believe that $n_n = n_n(p,q)$ (well really $p$ since it must be a function of $E$ and $E=p^2/2m$) , can I just check this is right? Basically I can't use $e^{A+B}=e^Ae^B$ to take this outside the integral because it has $p$ depdence? ).

Also, I am wondering what it's expression looks like in terms of a density of energy states integral expression , let me call the density of energy states $g(E)$. My lecture notes give an expresion for the canonical ensemble partition function in terms of $g(E)$ but not for the grand canonical ensemble partition function.

The canonical ensemble partition function expression is given as follows:

$g(E) = \int\frac{dp dq }{h} \delta(H(p,q)-E)$ (1)
and then
$Z(\beta, V,N)= \int dE g(E) e^{-\beta E}$

For the grand canonical ensemble I guess the expression for $g(E)$ (1) above would remain the same and we would have (where $\tilda{Z}$ denotes the grand canonical partition function:

$\tilda{Z}(\beta,V,\mu)= \int dE g(E) e^{E-\mu n}$

can someone let me know if I'm on the right track or not? In particular any links to the grand canonical ensemble as an integral, and it's expression in terms of some sort of density of states function such as $g(E)$, I say some sort of cause I'm just wondering whether this should almost be some function of $z=e^{\beta \mu}$ compared to the canonical ensemble partition function, thanks.

 

[binbagsss]

apologies i'd also like to ask how you find $\Tilda{Z}$ for $N$ particles and 1 particle, how the expression would differ, and what my integral expression would be above- is it for N particles or just one particle.

For instance in the canonical ensemble function $Z_{N} = Z^N/N!$ if the particles are all indistinguishable. Is there something similar that holds for the grand canonical ensemble?

Now, if I am to use the relation between the grand canonical partition function and the canonical partition function, I can simply plug in $Z_1$ or $Z_N$ for the canonical partition function (using (2) in the OP), and I believe this will give the corresponding N particle or 1 particle expression for the grand partition function- would this be correct? , however I don't know how you would relate a one particle grand canonical partition function and an N particle grand partition function without using this relation (2) and just talking to the grand canonical expression alone- quantum or classically.many thanks

 

[binbagsss]

apologies i'd also like to ask how you find $\Tilda{Z}$ for $N$ particles and 1 particle, how the expression would differ, and what my integral expression would be above- is it for N particles or just one particle.

For instance in the canonical ensemble function $Z_{N} = Z^N/N!$ if the particles are all indistinguishable. Is there something similar that holds for the grand canonical ensemble?

Now, if I am to use the relation between the grand canonical partition function and the canonical partition function, I can simply plug in $Z_1$ or $Z_N$ for the canonical partition function (using (2) in the OP), and I believe this will give the corresponding N particle or 1 particle expression for the grand partition function- would this be correct? , however I don't know how you would relate a one particle grand canonical partition function and an N particle grand partition function without using this relation (2) and just talking to the grand canonical expression alone- quantum or classically.many thanks

Anyone ?
Many thanks

 

[binbagsss]

Yes. The first summation is over all the individual states. Some states will have the same energy. The second summation is over the energy levels. N is the number of particles.

apologies confused again. Is N the number of particles of the system or is N the number of particles of a given energy E . I see the degeneracy factor is written as $\Omega(E,N)$ but perhaps this would be clearer if the notation was $\Omega(E,N_{E})$ , or is it the number of particles of the entire system?

 

[TSny]

apologies confused again. Is N the number of particles of the system or is N the number of particles of a given energy E . I see the degeneracy factor is written as $\Omega(E,N)$ but perhaps this would be clearer if the notation was $\Omega(E,N_{E})$ , or is it the number of particles of the entire system?

It's the number of particles of the system. In post #14 you wrote

$\sum_{n=0}e^{-\beta E_n} = \sum_E \Omega(E,N) e^{-\beta E}$ where $\Omega(E,N)$ is enabling to account for degenaracy of an energy level.##

$\displaystyle \sum_{n=0}e^{-\beta E_n}$ represents the canonical partition function, $z_N$. The canonical partition function is for a system with a fixed number of particles $N$. The summation index $n$ labels individual microstates of the N-particle system. In order to avoid confusion between $n$ and $N$, use a different label for the states, say $l$. So,

$z_N = \displaystyle \sum_le^{-\beta E_l}$

Each term in the sum represents a contribution from an individual microstate. $E_l$ is the energy of the N-particle system when the system is in microstate $l$. Microstates with the same energy contribute the same amount to the sum.

You can rewrite $z_N$ as a sum over all the allowed energy values $E_1, E_2, E_3, ...E_k, ...$ of the system. If $g(E_k)$ denotes the number of microstates that have the energy $E_k$, then

$z_N = \displaystyle \sum_k g(E_k)e^{-\beta E_k}$

Here, the index $k$ labels energy levels, not microstates.

The possible microstates of the N-particle system depend on the number of particles N. (Microstates of a two-particle system are different than microstates of a one-particle system.) Likewise, the possible energy values, $E_k$, depend on $N$ and the degeneracy factors $g(k)$ also depend on the value of $N$. Thus, we could write $E_{N,k}$ instead of $E_k$ and $g({N, k})$ instead of $g(k)$. Then,

$z_N = \displaystyle \sum_k g(E_{N,k})e^{-\beta E_{N,k}}$

The degeneracy factor $g(E_{N,k})$ is the same as your $\Omega(E,N)$. With all of these subscripts, things look messy. So, most people don't bother with the $N$ index when writing the canonical partition function. It is understood that you are dealing with a system of fixed number of particles. Thus, people will generally write

$z_N = \displaystyle \sum_k g(E_k)e^{-\beta E_k}$

with the understanding that $g(k)$ and $E_k$ are for a system with a fixed number of particles.

However, for the grand canonical partition function, $Z$, the number of particles N of the system is not fixed.

$Z = \displaystyle \sum_{microstates \, l}e^{\beta N_l \mu-\beta E_l}$

Here, you are summing over all microstates for all possible values of $N$. $N_l$ is the number of particles of the system for microstate $l$. The sum will include contributions from all of the microstates of the system when N = 1, plus contributions from all of the microstates of the system when N = 2, plus contributions from all of the microstates of the system for N = 3, etc. It even includes a contribution for N = 0 since N = 0 is possible.

Thus, an equivalent way to write $Z$ is $$Z = \displaystyle \sum_{N=0}^\infty \displaystyle \sum_{\;states \, l_N}e^{\beta N \mu-\beta E_{l_N}} $$ The index $l_N$ is used to label the different microstates that have $N$ particles.

This can then be written as $$Z = \displaystyle\sum_{N=0}^\infty e^{\beta N \mu} \displaystyle \sum_{\; states \, l_N}e^{-\beta E_{l_N}} = \displaystyle\sum_{N=0}^\infty e^{\beta N \mu} z_N$$