Tag Info

New answers tagged

1

Okay, this is actually pretty straightforward, but I don't know where to start. Review: What's a partition function? Let's step back and derive what we're talking about: what is a partition function? So we have a system which takes on a set of energy levels with degeneracies $\left \{(E_i, g_i) \right \}.$ We know that your system $s$ is in contact with ...


2

For some function $f$ of $x$, the logarithmic derivative is simply $$ \frac{\mathrm{d}\log f}{\mathrm{d}\log x} = \frac{x}{f} \frac{\mathrm{d}f}{\mathrm{d}x}. $$ You can check that this follows from the chain rule applied to $\mathrm{d}g/\mathrm{d}y$, where $g = \log f$ and $y = \log x$. This is a common thing to see in astrophysics, since if we have a power ...


1

You're actually dealing with the Potts model, which is a slight generalization of Ising. Not that it really matters, as you won't need any results from Potts. The point of mean field theory is typically to make each site independent of their neighbors, which allows you to evaluate the partition function by only iterating through the possible states of one ...


1

The sign is governed by the convention - whether the volume is of your system in consideration or not. If you decrease the volume of your system - you increase the energy of your system, so you require for total energy change to be positive. If the volume is describing your system then $ dV <0 $ and so $dE=-PdV>0$ is the correct expression If the ...


1

The entropy of a gas does not simply depend on the number of ways to arrange the particles that make up the gas but also on the number of ways of distributing the available energy between those particles. In an adiabatic expansion there is no heat transfer but he gas does do work on its surroundings. This reduces the internal, and so reduces then number of ...


2

The addition of a constant energy to a hamiltonian is not a problem for the computation of the partition function. For example in the canonical ensemble, taking $H=H'+E_0$ and $Z'$ to be the partition function of the system with Hamiltonian H': $Z = \mu(e^{-\beta H}) = \mu(e^{- \beta H' - \beta E_0}) = \mu(e^{- \beta H'} e^{ - \beta E_0}) = e^{ - \beta ...


1

For the $h$ factor: (the $N!$ is the Gibbs correction as already explained by others ) Step 1: Semi-classical setting The $h$ factor can actually be understood in a semiclassical way. We are talking about classical ideal gas using classical phase space formulation of classical statistical mechanics, but still from a semi-classical point of view we are not ...


2

Now, in my book there is not this factor $1/h^{3N} N!$. Where this factor comes from? Why do we need to include it there? $\mathbf{1/h^{3N}}~$: Some people like the value of the partition integral to be independent of units, so they add the denominator $h^{3N}$ to the formula where $h$ has the same units as $pq$. This makes the expression ...


0

It is true that there is no correct a priori way of deriving the "correct" entropy as the notion of "correct" will depend upon what we mean by this word. However, if statistical mechanics introduces a quantity that has the same name as the most important quantity of thermodynamics, I suspect it has, at the very least, to carry a meaning very close to the ...


5

$h$ factor The factor of $1/h^{3N}$ is a total hack. The integral over phase space has dimensions, whereas $Z$ only makes sense if it's dimensionless. The $h$ factors are there to make $Z$ dimensionless. Suppose you have a system with only one particle in one dimension. Then the integral in phase space goes over one position variable, $dq$ and one momentum ...


1

I'm not completely sure why you think that the bosonic mode fails, but it seems to me that the answer is definitely yes. The system is solvable in both the finite-dimensional and the bosonic case; the problem with the bosonic case is that the solution is ugly, because the hamiltonian is ugly. Take a hamiltonian of the form $$ H=E_0S+S\sum_k (g_k ...


2

When calculating expectation values, you need to know a few things: What is my random variable? What is my distribution function? What is my desired quantity in terms of the random variable? The general form in one dimension would look like this $$ \langle G(x) \rangle = \frac{\int_{x_{min}}^{x_{max}} f(x) G(x) dx}{\int_{x_{min}}^{x_{max}} f(x) dx} \, ...


0

Entropy of a system can increase either by entropy transfer or due to entropy generation. If we live in a world where every process is reversible then entropy will never be generated only transfered from one system to other. Entropy generation is due to irreversibilities in a process.Lets consider water taken inside an adiabatic container we do some work on ...


4

First let me repeat what Yvan Velenik says in the comment: The terminology is somewhat unfortunate, because you don't need that much statistics, rather you'll need some probability theory. To elaborate, quoting Wikipedia, Statistics is the study of the collection, analysis, interpretation, presentation, and organization of data. [...] Statistics deals ...


1

Some of the mathematical aspects of the Liouville operator can be found in the second book by Reed and Simon, in section X.14 (it is not a comprehensive account, but it gives the basic ideas and proofs). In the notes at the end of chapter X, in the part dedicated to section X.14, there is also a quite extensive bibliography that may be useful.


3

First of all you can directly write $p$ as a function of $V$ without solving a quadratic equation: $$ p = \frac{N k_B T}{V- N b } - a \frac{N}{V} $$ Then you take a look at the interesting points $$\left(\frac{\partial p}{\partial V}\right)_{T,N} = 0 $$ The solutions to this equation tell you where the bulk modulus (or its inverse, the compressibility) ...


0

I would say that the question is still not very well defined, as it is important if in the high-temperature (classical) limit spin is conserved or not. If spin is conserved (think strong magnetic fields $B\gg T$), then in the classical limit spinful quantum gases become a mixture of classical gases, as coherent superpositions between spin states are not ...


0

my question is why is $<\hat{n}(\vec{r})>=n$ Presumably you are defining $n=N/V$. In that case, from the definition: $$ <\hat n(\vec r)>=\int d^3x_1d^3x_2\ldots d^3x_N\Psi^*(\vec x_1,... \vec x_N)\sum_{i}\delta(\vec x_i-\vec r)\Psi(\vec x_1,... \vec x_N) $$ $$ =N\int d^3x_2\ldots d^3x_N |\Psi|^2(\vec r,x_2,\ldots,x_N) $$ On the other hand, ...


1

Imagine you have a string of irregularly placed (that is, not evenly spaced) christmas lights, and you want to calculate the total power emitted by all of them. There are two ways you can do this. One way is to sum over all lights the power emitted by a light. Another way would be divide the string of lights into segments of length $\Delta x$, and then sum ...


1

Assuming there is a typo in the equation, this is the Van der Waals equation of state. You need to plot the isotherms on a $PV$ diagram for different values of $T$, and you will find that the equation predicts a region of liquid phase where the isotherms develop both a maximum and minimum. I won't say more since this is obviously a homework question. If you ...


0

I would just like to add this: in ordinary substances, we have the system in thermal equilibrium with the environment, and with a fixed number of particles. However we go to the grand canonical system for efficiency, where we do not have to restrict ourselves to a fixed particle number $N_{0}$, and that simplifies calculations enormously. In this, the system ...


2

You will probably find the Stirling approximation useful here. For your purposes it will be sufficient to use the form $\ln N! \approx N\ln N - N$, which is valid for $N\gg 1$.


1

You have to find out what kind of statistical ensemble you are dealing with. As soon as you know that, you can get the corresponding thermodynamical potential from the knowledge of the partition function. When you know the potential, you know everything! EDIT: Since you don't know to which ensemble this partition function $\mathcal{Z}$ belongs to you ...


2

There are multiple ways to justify this identification. The first one is to recognize that in thermodynamics the Helmholtz free energy is the right thermodynamic potential to figure out spontaneous evolution of your thermodynamic system (in virtue of the second principle of thermodynamics) at fixed (N,V,T). In statistical mechanics, this is expressed by ...


1

I think what they could mean is that $\vec{\alpha}\vec{\alpha}$ is a second rank tensor that is contracted with $\overline{g}$. I saw this notation being used in the context of electrodynamics before. It is used to get a simple notation for multi-dimensional Taylor series. So we get $$\Delta S=\overline{g}:\vec{\alpha}\vec{\alpha} \equiv \sum_{ij} ...


1

Normally we do NOT calculate the phase space density of a system. In the phase space formulation of classical statistical mechanics, the phase space density $\rho(p,q;t)$ has its specified form for different ensembles. Normally for systems at equilibrium the density $\rho$ has no explicit time dependence and thus we work with $\rho(p,q)$. (1) For ...


1

For a classical Maxwell-Boltzmann gas, the partition function is given by $$Z=\sum_i g_i e^{-\beta \epsilon_i}$$ where $g_i$ is the degeneracy. And the probability for each level to be occupied by one particle is given by $$P_i=\frac{g_i e^{-\beta \epsilon_i}}{Z}$$ The partition function function in the above expression is essentially a normalization factor ...


3

The partition function is not in general equal to one. It is a normalization constant, i.e. the probability of being in a configuration $\{\sigma\}$ is $$ P(\{\sigma\}) = \frac{e^{-H(\{\sigma\})/k_B T}}{Z} $$ where $$ Z(T) = \sum_{\{\sigma\}} e^{-H(\{\sigma\})/k_B T} $$ Since multiple configuration can have the same energy, you can define (with a little ...


4

In statistical mechanics, at least when you can ignore the spin of the particles you're dealing with, the occupation number of a quantum state (that is, the number of particles in the state, or probability of a particle being found in the state) is proportional to $e^{-E/kT}$, where $E$ is the energy of the state. If you have a harmonic oscillator at ...


1

The first thing we can do is to split up $\Gamma$ according to the number of particles in the given states. Let $\gamma_N$ be a state with $N$ particles. The grand canonical partition function is then \begin{align} \mathcal{Z} = & \sum_\Gamma \exp\left(-\beta(\mathcal{H} - \mu N)\right)\\ =& \sum_{N=0}^\infty\exp\left(\beta \mu N ...


1

Start from the general definition, how this thing is set up in practice: You start from the mean energy of a system in contact with a thermal reservoir. The systems of the representative statistical ensemble are distributed over the entire number of possibilities, in accordance with the canonical ensemble $$P_i = C \exp(-\beta E_i) = \frac{\exp(-\beta ...


3

It seems to me that you just cannot tell the difference between a Bose condensate and nothing in this case. What will change if you add some photons or phonons with zero energy to the system? No characteristics of the system will change. So it seems to me we have no criterion to decide if there is a Bose condensate in this case, and what's more important, it ...


1

But what is the physical interpretation of the partition function and it's significance to Thermodynamics? I'm seeking a simple yet understandable intuition. The partition function has one simple physical interpretation in terms of Thermodynamic functions: Its natural log is proportional to the Free Energy (the proportionality constant is the ...


0

Duality transformation does not preserve topological order, and hence not preserving the topological entanglement entropy. The quantum Ising model has no topological entanglement entropy. See this related question for more discussions.


-1

The significance is actually quite easy to explain at an elementary level. Namely, the partition function contains roughly all the information the thermodynamic system carries. For example, the expected energy of a system: $$\langle E\rangle =\sum E_i p_i=\sum_i\frac{E_ie^{-E_i/k_BT}}{e^{-E_i/k_BT}}=\frac{-1}{Z}\frac{\partial Z}{\partial (1/k_BT)}$$. The ...


0

Let me try to formulate the question more precisely and then give my answer. The ``phase'' $\theta$ of a BEC is introduced as $\langle \psi_N|\hat{a}|\psi_{N+1}\rangle = |\psi|e^{i\theta}$, where $|\psi_N\rangle$ is the ground state being occupied by $N$ bosons. Because the occupation number of the condensed state $|\psi_N\rangle$ is of the order $N$, the ...


1

Partition functions are a measure of the allowed volume in (microscopic-)configuration space for the system, and as such they are the normalizing function for probabilities expressed as volumes in configuration space (and assuming the applicability of the ergodic hypothesis). I know that this is very abstract, but it is also very general.


17

$\hbar$ does not need to appear in classical statistical mechanics. You are free to replace it with any quantity with units of angular momentum, say $\hbar_{\mathrm{C}}$. As long as this is choosen smaller than the size you can experimentally probe (i.e., as long as you don't ask questions of the theory that contain structure on this length scale or below) ...


1

A proper derivation of the Boltzmann equation from non-equilibrium quantum field theory (which will give the factors $1\pm f$ in the weak coupling, quasi-particle dominated, limit) is a difficult problem. The standard reference is Kadanoff and Baym, Quantum Statistical Mechanics. The standard approach in introductory text books (and indeed, historically, ...



Top 50 recent answers are included