I Calculating the number of energy states using momentum space

[JohnnyGui]

Which lecturer? The slides? Or the video? From what I saw in the video, he never got to any formula involving $E_F$ at all. He only gave formulas with $\alpha$ and $\beta$ in them, and never gave an equation for $E_F$ in terms of $\alpha$.

The lecturer from the video. It is correct that he didn't give an equation involving $E_F$, but I substitued his derived $\alpha$ with $\frac{E_F}{kT}$ since that's what's sheet $20$ states from the slides source.

Perhaps another source that I found would help me, this source (a Powerpoint) derived the formula just as the video, $n_j = e^{-\alpha - \beta E_j}$ and says on sheet $21$ that $-\alpha = \frac{E_F}{kT}$ instead of $\alpha = \frac{E_F}{kT}$ such that it ends up with the same formula as the previous mentioned slides source.

I noticed that whether you derive the formula as $(1)$ $n_j = e^{-\alpha - \beta E_j}$ or $(2)$ $n_j = e^{\alpha - \beta E_j}$ depends on how you apply Lagrange's Constants during the derivation. And that if you have derived it as $(1)$ then you should somehow assume that $-\alpha = \frac{E_F}{kT}$, whereas when you derived it as $(2)$, then you should assume that $\alpha = \frac{E_F}{kT}$. I just don't get how these assumptions about $\alpha$ are made based on how Lagrange Constants are applied.

 

[PeterDonis]

I substitued his derived $\alpha$ with $\frac{E_F}{kT}$ since that's what's sheet 2020 states from the slides source.

And that makes no sense, because the slides started with the opposite sign for $\alpha$. So if you flip the sign of $\alpha$ in the formulas in the video, you have to also flip the sign of $\alpha$ in the formula for $E_F$ in terms of $\alpha$. You can't take a piece of one derivation and a piece of another and put them together if they started with opposite sign choices.

Note that the slides don't derive that formula: they say it's in another section. If you look at the derivation of that formula, you should find that it requires the same choice of sign for $\alpha$ that the slides made, which is the opposite of the choice that the video made.

Perhaps another source that I found would help me

Yes as you can see, it requires the opposite sign for $\alpha$ in the formula for $E_F$ in terms of $\alpha$.

I just don't get how these assumptions about $\alpha$ are made based on how Lagrange Constants are applied.

Huh? $\alpha$ is a Lagrange multiplier, and signs of Lagrange multipliers can always be chosen arbitrarily. That choice of sign isn't an "assumption". It's an arbitrary choice. You just have to make sure the choice is consistent throughout your entire derivation.

 

[PeterDonis]

if you have derived it as $(1)$ then you should somehow assume that $-\alpha = \frac{E_F}{kT}$, whereas when you derived it as $(2)$, then you should assume that $\alpha = \frac{E_F}{kT}$.

There aren't two "assumptions". There is only one. The formula for $\alpha$ in terms of $E_F$ and $kT$ isn't "assumed", it's derived, and its derivation has to start with the same sign choice for $\alpha$ as a Lagrange multiplier as the other derivations that are being done, like the derivation of the formula for $n_j$ in terms of $\alpha$ and $\beta$.

 

[JohnnyGui]

Huh? αα\alpha is a Lagrange multiplier.

Yes, I know that $\alpha$ is a Lagrange multiplier.

and signs of Lagrange multipliers can always be chosen arbitrarily

This is what I didn't know, so it cleared this out for me. Thanks.

I've got one last question. When degeneracy is taken into account in the case of discrete energylevels, the formula is written as:
$$n(\epsilon_j) = g_j \cdot \frac{N}{Z} \cdot e^{-\frac{\epsilon_j}{kT}}$$
Earlier, I have derived that the number of quantum states $Q_j$ for a certain energy $\epsilon_j$ is equal to:
$$Q_j = \frac{1}{2} \pi \cdot \frac{\epsilon_j \cdot 8mL^2}{h^2}$$
This is based on calculating the surface of an 8th a sphere surface in n-space.
Since this formula gives the number of quantum states for a specific energy $\epsilon_j$, can $g_j$ be substituted by this formula such that:
$$n(\epsilon_j) = \frac{1}{2} \pi \cdot \frac{\epsilon_j \cdot 8mL^2}{h^2} \cdot \frac{N}{Z} \cdot e^{-\frac{\epsilon_j}{kT}}$$
?

 

[PeterDonis]

can $g_j$ be substituted by this formula

I haven't checked your formula for $g_j$, but in general, yes, if you have a formula for $g_j$ you can substitute it into the formula for $n$.

 

[JohnnyGui]

I haven't checked your formula for $g_j$, but in general, yes, if you have a formula for $g_j$ you can substitute it into the formula for $n$.

Thanks for verifying.

I do find it a bit peculiar that, when energy is considered discrete, the formula takes the form of $n(\epsilon_j) = g_j \cdot e^{-\alpha - \epsilon_j\beta}$ with $g_j$ being the formula in my previous post, but when energy is considered continuous, then $g_j$ has to be substituted by the density of states formula times $d\epsilon_j$ (previously shown), while the part $e^{-\alpha - \epsilon_j\beta}$ can still be kept for the continuous approach.

Why shouldn't the part $e^{-\alpha - \epsilon_j\beta}$ be substituted by a formula as a function of energy that gives the "density of particle numbers" times $d\epsilon_j$, just like how it is done for the degeneracy?

 

[PeterDonis]

Why shouldn't the part $^{-\alpha - \epsilon_j\beta}$ be substituted by a formula as a function of energy

Because it already is a formula as a function of energy.

that gives the "density of particle numbers" times $d\epsilon_j$,

The exponential factor is a function of energy that gives the density of particle numbers. In the case where there is no degeneracy, that's the only factor there is. In the case where there is degeneracy, there is an additional factor of $g_j$ because of the degeneracy. The $d \epsilon_j$ doesn't come from $g_j$ or the exponential; it comes from the fact that you're doing an integral instead of a discrete sum.

 

[JohnnyGui]

The exponential factor is a function of energy that gives the density of particle numbers.

Hold on, this makes me reason the other way round; if the exponential factor $e^{-\alpha - \epsilon_j\beta}$ does give the density of particle numbers, doesn't the summation of that same exponential factor $e^{-\alpha - \epsilon_j\beta}$ (when energy is considered discrete) mean that you're adding particle number densities instead of the real particle numbers in each energy level?

 

[PeterDonis]

if the exponential factor $e^{-\alpha - \epsilon_j\beta}$ does give the density of particle numbers, doesn't the summation of that same exponential factor $e^{-\alpha - \epsilon_j\beta}$ (when energy is considered discrete) mean that you're adding particle number densities instead of the real particle numbers in each energy level?

Yes, of course. The exponential of any negative number is less than 1, so it has to be a density, not a particle number.

 

[JohnnyGui]

Yes, of course. The exponential of any negative number is less than 1, so it has to be a density, not a particle number

And yet, for discrete energy levels $e^{-\alpha - \epsilon_j\beta} = n_j$ in which $n_j$ is the particle number at a certain energylevel, not a density? Sorry but I'm lost...

 

[PeterDonis]

for discrete energy levels $e^{-\alpha - \epsilon_j\beta} = n_j$ in which $n_j$ is the particle number at a certain energy level, not a density?

No, it's a density. More precisely, it's a fraction of particles in that energy level, or a statistical probability of a particle being in that energy level.

 

[JohnnyGui]

No, it's a density. More precisely, it's a fraction of particles in that energy level, or a statistical probability of a particle being in that energy level.

Huh? But in your post #99 you said that the fraction of particles at a paticular energy is $n_j$ divided by $N$, not just $n_j$. Here's a snippet of your quote regarding this:

In which $e^{-\alpha} = \frac{N}{Z}$

 

[PeterDonis]

in your post #99 you said that the fraction of particles at a paticular energy is $n_j$ divided by $N$, not just $n_j$.

You're confusing yourself by using the symbol $n$ to refer to different things.

In what you quoted from me from post #99, $n_j$ obviously refers to the exponential of a negative number times $N$. So that would be an actual number of particles. Dividing that by $N$ just leaves the exponential of a negative number, which, as I said, has to be less than $1$ so it can't be a particle number, it has to be a fraction or a probability.

But in post #126, you used $n$ to refer to just the exponential of a negative number, all by itself. So obviously that can't be a particle number, as above.

 

[JohnnyGui]

You're confusing yourself by using the symbol $n$ to refer to different things.

In what you quoted from me from post #99, $n_j$ obviously refers to the exponential of a negative number times $N$. So that would be an actual number of particles. Dividing that by $N$ just leaves the exponential of a negative number, which, as I said, has to be less than $1$ so it can't be a particle number, it has to be a fraction or a probability.

But in post #126, you used $n$ to refer to just the exponential of a negative number, all by itself. So obviously that can't be a particle number, as above.

My intention in post #126 with writing it as $n(\epsilon_j)$ instead of $n_j$ is because energy is being considered continuous so that one can not dedicate a certain particle number $n_j$ to a certain discrete energy level $\epsilon_j$.

My mentioned formula in post #126 for $n(\epsilon_j) = g_j \cdot e^{-\alpha - \epsilon_j\beta}$ is stating that this should give the particle number, because $e^{-\alpha} = \frac{N}{Z}$. So $n(\epsilon_j)$ must be the particle number and not the density.
If $e^{-\alpha}$ is not $\frac{N}{Z}$ in the continuous approach, then that's a whole different story which I don't understand how it changed formula-wise.

Is $e^{-\alpha}$ in the continuous approach then just a value that has to be further derived and is assumed to give the density when multiplied with $e^{-\beta \epsilon}$?

 

[PeterDonis]

My intention in post #126 with writing it as $n(\epsilon_j)$ instead of $n_j$ is because energy is being considered continuous so that one can not dedicate a certain particle number $n_j$ to a certain discrete energy level ϵj\epsilon_j.

You're confusing yourself with this too. The difference between continuous and discrete is irrelevant to the issue that you're struggling with.

$e^{-\alpha} = \frac{N}{Z}$.

Ah, so now you're confusing $e^{- \alpha}$ with $e^{- \alpha - \beta E_j}$. And I see that you've been mixing those together for the last few posts. And you have managed to confuse me as well in the process.

I'm not going to try an disentangle all of that. Instead, let's start with the final distribution function, as given in the slides (from the summary, the next to last slide):

$$
f(E_j) = \frac{N_j}{g_j} = e^{\frac{- \left( E_j - E_F \right)}{k_B T}}
$$

What is this? Well, since it's a fraction $N_j / g_j$, it's clearly a number of particles--the number in a single state of energy $E_j$ where there is a number $g_j$ of degenerate states with that energy. (This easily simplifies to the non-degenerate case $g_j = 1$.)

But what about that exponential of a negative number on the right? Well, the numerator of that number (the argument of the exponential) is $- \left( E_j - E_F \right)$. In other words, it's minus the difference between the energy of the state, $E_j$, and the Fermi energy. But if there are no chemical reactions taking place--i.e., there is only a single type of particle present, or the temperature is too low for any reactions between multiple particle types to take place--then $E_j < E_F$! Which means that the argument of the exponential is minus a negative number, i.e., a positive number.

In other words, this is telling us that the average number of particles in a single state of energy $E_j$ is more than one if $E_j$ is below the Fermi energy. If $E_j$ exactly equals the Fermi energy, then there is exactly one particle in that state (on average); and if $E_j$ is greater than the Fermi energy, there is on average less than one particle in that state.

Now, "average number of particles" is easier to interpret if the states are discrete. But the formula itself works just fine if the states are continuous. If you struggle with how to interpret "average number of particles" for continuous states, then you can, as I said in what you quoted from post #99, just divide $f(E_j)$ by $N$, the total number of particles, to get a formula that looks like this

$$
\frac{f(E_j)}{N} = \frac{N_j}{N g_j} = \frac{1}{N} e^{\frac{- \left( E_j - E_F \right)}{k_B T}}
$$

which has a simple interpretation as the statistical probability that a randomly selected particle will have energy $E_j$.

Now, you keep on separating that exponential factor into two pieces. You write it in terms of $\alpha$ and $\beta$, but we know what those actually turn out to equal so there's no need to obfuscate. We can just write:

$$
f(E_j) = \frac{N_j}{g_j} = e^{\frac{E_F}{k_B T}} e^{\frac{- E_j}{k_B T}}
$$

And if you want, you can go back and dig up that $e^{\frac{E_F}{k_B T}} = N / Z$ and rewrite this as

$$
f(E_j) = \frac{N_j}{g_j} = \frac{N}{Z} e^{\frac{- E_j}{k_B T}}
$$

which, again, clearly has an interpretation as "number of particles" (because $N$ appears explicitly on the right and we take the exponential of a negative number to get a fraction, with the partition function as an additional divisor). And the "divide by $N$" form of this would be

$$
\frac{f(E_j)}{N} = \frac{N_j}{N g_j} = \frac{1}{Z} e^{\frac{- E_j}{k_B T}}
$$

which, again, clearly has an interpretation as "fraction of particles".

I really don't care which one of these you want to look at; they are all valid formulas and they all have well-defined meanings. But I think you need to take care to be clear about which of them you are talking about. Otherwise you will just confuse yourself, and, as I noted above, you will end up confusing others (e.g., me) as well.

 

[JohnnyGui]

@PeterDonis Thank you for the extensive explanation. Everything that you posted makes perfect sense to me. But only to the extent of when energy is considered discrete.

Please let me elaborate. Are all of those formulas also valid when energy is considered continuous? If according to you, yes, then why can't I find the $Z$, the $g_j$ being substituted by the states density and $e^{\frac{- E_j}{k_B T}}$ being the only parameter that is kept from your formulas for the continuous approach?

 

[PeterDonis]

Are all of those formulas also valid when energy is considered continuous?

Yes, except that there is no well-defined concept of "degeneracy" for a continuous spectrum of energy states, so there is no $g_j$ (basically you have to set $g_j = 1$ so it drops out of the formulas), and that you have to calculate the partition function $Z$ correctly for the continuous spectrum.

why can't I find the $Z$, the $g_j$ being substituted by the states density and $e^{\frac{- E_j}{k_B T}}$ being the only parameter that is kept from your formulas for the continuous approach?

I have no idea what you are trying to ask here.

 

[JohnnyGui]

Yes, except that there is no well-defined concept of "degeneracy" for a continuous spectrum of energy states, so there is no $g_j$

This is a good start to ask part of my question you didn't understand. I have difficulty grasping this quote. If the degeneracy for a particular discrete energy level is equal to $g_j = \frac{4 \pi mL^2 \cdot E_j}{h^2}$, why can't the number of quantum states in the continuous approach be $g(E) = \frac{4 \pi mL^2 \cdot E}{h^2} \cdot dE$? Why should it be the density of states integrated instead?

 

[PeterDonis]

why can't the number of quantum states in the continuous approach be

There is no such thing as "number of quantum states" if the spectrum is continuous. See below.

Why should it be the density of states integrated instead?

The thing you wrote down as $g(E)$ is a "density of states", not a number of states. (I'm not saying it's necessarily the right density of states formula; I haven't checked it. But anything of the form "some function of $E$ times $dE$" is a density of states: it's a thing you integrate over a continuous range of $E$. Whereas a "number of states" is something you add up over a set of discrete values of $E$.)

 

[JohnnyGui]

The thing you wrote down as g(E)g(E)g(E) is a "density of states", not a number of states. (I'm not saying it's necessarily the right density of states formula; I haven't checked it. But anything of the form "some function of EEE times dEdEdE" is a density of states: it's a thing you integrate over a continuous range of EEE. Whereas a "number of states" is something you add up over a set of discrete values of EEE.)

Ah, the function $g_j$ is formulated as the surface of 1/8th of a sphere in n-dimensions, and thus gives the number of quantum states for a particular energy level. So I assume integrating that function would not give the correct outcome for the continuous approach.

What I'd need for the continuous approach is a derivative of a function for the cumulative number of quantum states (which $g_j$ is not). This would be the derivative of the volume of 1/8th of an n-sphere, which is the density as you said.
$$D(E) = \frac{V \cdot 2^{2.5}\pi \cdot m^{1.5}}{h^3} \cdot \sqrt{E}$$
This is the formula that I should multiply with $dE$ in order to get the number of quantum states between $E \geq E + dE$.
Now that we have the number of quantum states for the continuous approach: $D(E) \cdot dE$, I'd need to multiply this with a function $F(E)$ that gives the number of paticles between $E \geq E + dE$ per 1 quantum state, so that I can calculate the number of particles between $E \geq E + dE$

It seems so that this function is not necessarily deduced by reasoning that the function for the discrete appraoch can also be used for the continuous approach, as you said in your post #99. It is rather reasoned that this function $F(E)$ must be proportional to $e^{-\beta E}$ because the probability density function is also proportional to $e^{-\beta E}$. This reasoning is given on this wiki and also here. It is apparently a property of statistical mechanics. The constant can then be derived through normalization (integrating the probability density to infinity and normalizing it to 1)

The property of the function being proportional to $e^{-\beta E}$ is my exact problem that I can't seem to understand. I don't know if this is deduced from the discrete approach and how this is done. I have tried reading the explanations in the links but I still have trouble grasping it. Perhaps you could give a better example or explanation for this?

 

[PeterDonis]

It is rather reasoned that this function $F(E)$ must be proportional to $e^{-\beta E}$ because the probability density function is also proportional to $e^{-\beta E}$. This reasoning is given on this wiki and also here.

The stack exchange link answer is basically the same reasoning that was in one of the slides you linked to (the only difference is that that answer considers the possibility of there being other constants of the motion besides energy). Which illustrates that that basic line of reasoning works whether the spectrum of energy states is discrete or continuous.

 

[JohnnyGui]

The stack exchange link answer is basically the same reasoning that was in one of the slides you linked to (the only difference is that that answer considers the possibility of there being other constants of the motion besides energy). Which illustrates that that basic line of reasoning works whether the spectrum of energy states is discrete or continuous.

I'm not sure which slide source you mean, I couldn't find an explanation in the slides about the pobability density being proportional to $e^{-\beta E}$. Furthermore, the fact that $n(E) = \frac{N}{Z} \cdot e^{-\beta E}$ can also be used for the continuous approach while it is initially derived based on the discrete approach still surprises me.

How can a specific value of energy be filled in that formula if energy is continuous in the first place?

 

[PeterDonis]

I'm not sure which slide source you mean

The one you linked to in post #121.

I couldn't find an explanation in the slides about the pobability density being proportional to $e^{-\beta E}$.

Deriving that formula doesn't count as an explanation? The number of particles with energy $E$, divided by the total number of particles, is the probability.

How can a specific value of energy be filled in that formula if energy is continuous in the first place?

Huh? A continuous range of values of $E$ is still a set of possible values of $E$. Just pick one.

 

[JohnnyGui]

Huh? A continuous range of values of EEE is still a set of possible values of EEE. Just pick one.

I thought in case of continuous energy one could only speak of $E \geq E + dE$. Hence the formula for the continuous approach consists at least of the States Density that gets integrated.
It doesn't contain $\frac{N}{Z}$ for the number of particles per 1 quantum state.

 

[PeterDonis]

I thought in case of continuous energy one could only speak of $E \geq E + dE$.

First, this doesn't even make sense if $dE$ is positive, which is the usual assumption.

Second, I don't know why you would think this. You can pick a single value of $E$ out of a continuous set just as you can pick one out of a discrete set. In both cases you're picking a single value out of a set of values.

the formula for the continuous approach consists at least of the States Density that gets integrated.

Yes, because you have to integrate over the entire range of possible values of $E$ in order to normalize it (i.e., in order to find the partition function $Z$ that goes in the denominator).

It doesn't contain $\frac{N}{Z}$ for the number of particles per 1 quantum state.

You're getting mixed up. Go back and look at post #99, where I summarized all of the different formulas we've thrown around in this discussion. The first formula I gave in that post is one for the number of particles with a particular energy. What factor does it have in it?

 

[JohnnyGui]

First, this doesn't even make sense if dEdEdE is positive, which is the usual assumption.

Second, I don't know why you would think this. You can pick a single value of EEE out of a continuous set just as you can pick one out of a discrete set. In both cases you're picking a single value out of a set of values

I'm thinking this because of an analogue with probability density. The probability to have, for example, a particle at exactly location $x$ is practically zero because $x$ is a continuous variable. One could therefore only speak of a probability between $x$ and $x+dx$ which is calculated by multiplying the cumulative probability density with $dx$. Another example is found here which is where I got this way of thinking from.

Just like in this case one could only speak of a number of particles between energy $E$ and $E+dE$ because energy is considered continuous, and therefore one could only calculate the number of particles in this case by multiplying a cumulative particle density function with $dE$ and not use the discrete function.

Please explain why this way of thinking is incorrect.

 

[PeterDonis]

The probability to have, for example, a particle at exactly location $x$ is practically zero because $x$ is a continuous variable.

This is not correct.

Suppose I have a Gaussian probability distribution for position: a particle's probability to be at position $x$ is $(1 / \sqrt{\pi} ) e^{-x^2}$. This is a perfectly normalized probability distribution and I can plug any value of $x$ I want into it and get a valid answer back that is not zero. I don't need to multiply it by $dx$ or anything like that.

The fact that $x$ is a continuous variable does not mean the probability of finding a particle at exactly $x$ is zero. It just means that the integral over all possible values of $x$ of whatver probability distribution function we have must be $1$. The function I gave above meets that requirement (do the integral and see).

I think you have been confused by reading some popular presentations on probability that use sloppy language.

 

[JohnnyGui]

This is not correct.

Suppose I have a Gaussian probability distribution for position: a particle's probability to be at position $x$ is $(1 / \sqrt{\pi} ) e^{-x^2}$. This is a perfectly normalized probability distribution and I can plug any value of $x$ I want into it and get a valid answer back that is not zero. I don't need to multiply it by $dx$ or anything like that.

The fact that $x$ is a continuous variable does not mean the probability of finding a particle at exactly $x$ is zero. It just means that the integral over all possible values of $x$ of whatver probability distribution function we have must be $1$. The function I gave above meets that requirement (do the integral and see).

I think you have been confused by reading some popular presentations on probability that use sloppy language.

Hmm, this makes me wonder though. Why is the formula for the Maxwell-Boltzmann Distribution then always written in terms of a probability density times $dE$ which gives the probability for a particle between energy $E \geq E+dE$, while like you said, one should be able use the previously discussed formula to calculate the probability at a specific $E$, even in the continuous approach?

 

[PeterDonis]

Why is the formula for the Maxwell-Boltzmann Distribution then always written in terms of a probability density times $dE$

It is if it is inside an integral because it's inside an integral. I have never seen it written with a $dE$ if it's not in an integral.

 

[JohnnyGui]

It is if it is inside an integral because it's inside an integral. I have never seen it written with a $dE$ if it's not in an integral.

Someone else answered this by saying that the $dE$ is always needed purely for the Density of States function $D(E)$. In a continuous approach, you'd need a function that gives the number of states over a range, e.g. the volume of quantum states within $dE$. There is no exact quantum states function in the continuous approach without the need to integrate it.
According to him, the formula for the number of particles $n(E)$ in the continuous approach is:
$$n(E) = D(E) \cdot dE \cdot \frac{N}{Z} \cdot e^{-\beta E}$$
I have never seen a number of particles function in the continuous approach without the need to integrate one of its functions.