# Tag Info

23

The partition function is strongly related to a very useful tool in probability theory called the moment generating function(al) of the probability distribution. For any probability distribution $p$ of some random variable $X$, the generating function $\mathcal{M}(z)$ is defined as being: \mathcal{M}(z) \equiv \langle e^{zX}\rangle ...

11

I think it will depend the kind of statistical mechanics. For classical statistical mechanics, there is no time, so it is really hard to imagine a nice physical picture of the propagation of something. But nevertheless we still talk of loops as propagating "particles" (we give the "momenta", for instance, which is conserved, etc.). Interestingly, ...

9

I prefer to see it in the following way: $$Z_{quantum} \equiv \sum_{m} e^{-\beta E_m}$$ Where $m$ is a quantum microstate eigenstate of the Hamiltonian. Now, you can split the sum into two parts; a sum over quantum microstates that yields the same energy eigenvalue $E_n$ and a sum over all possible values of $E_n$: ...

8

For the partition sum, you have so sum $e^{-E}$ ($T=1$) over all possible eigenstates of the system where $E$ is the energy of the corresponding state. Two bosons can be in the 10 states $|kl\rangle$, with $1\leq k \leq l \leq 4$ where we accounted for the degeneracy by introducing an additional state with $E_4 =2E$. The corresponding partition sum reads ...

8

$h$ 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 ...

8

There is quite a big controversy these days about the correct definition of the entropy in the microcanonical ensemble (the debate between the Gibbs and Boltzmann entropy), which is closely related to the question. Everyone agrees that the correct definition of the density matrix is given by $$\rho(E)=\frac{\delta(E-H)}{\omega(E)},$$ where $H$ is the ...

8

The partition function contains so much information because it is directly related to the free energy, $$F = - k_B T \ln(Z) \, .$$ The physical assumption behind considering $F$ as a thermodynamic potential is that the statistics of the system as described by the canonical ensemble. In turn, the applicability of the canonical ensemble is a direct ...

7

Consider a quantum system with state (Hilbert) space $\mathcal H$. For simplicity, let the Hamiltonian $H$ of the system have discrete spectrum so that there exists a basis $|n\rangle$ with $n=0,1,2,\dots$ for the state space consisting of eigenvectors of the Hamiltonian. Let $\epsilon_n$ denote the energy corresponding to each eigenvector $|n\rangle$, ...

6

In my naive view, this is merely a mathematical trick that should not be taken too seriously in term of physical interpretation. After all, a "Wick rotation" applied to the Schrodinger equation yields a diffusion equation. This is helpful for some mathematical problems but the physics it describes is very very different from quantum mechanics; not even ...

6

I do not think Mainwood makes any argument against what he calls the "theoreticians case", much less a compelling one. The "theoretician's case" is that phase transitions do not exist in finite size systems but only as features which become infinitely sharp in the infinite size limit (also user10001's comment). In fact Mainwood briefly dismisses the case and ...

6

Assume that the generating functional is given by a sum of all possible diagrams, i.e. $$Z(J)=\Sigma_{n_i} D_{n_i}.$$ Furthermore, assume that each diagram D is given by a product of connected diagrams $C_i$, i.e. a diagram D can be disconnected. We will write this as $$D_{n_i}=\Pi_i\frac{1}{n_i!}C_i^{n_i},$$ where dividing by $n_i!$ amounts for a ...

4

Since $n=\left(n_1,\,n_2,\ldots\right)$ with each $n_i\in\{0,\,1,\,2,\,\ldots\}$, then you can write the sum as $$\sum_{n_\alpha}=\sum_{n_1\in\{0,\,1,\ldots\}}\sum_{n_2\in\{0,\,1,\ldots\}}\cdots\sum_{n_m\in\{0,\,1,\ldots\}}$$ while your product is, $$\prod_\alpha e^{n_\alpha \beta}=e^{n_1\beta}e^{n_2\beta}\cdots e^{n_m\beta}$$ Thus, we have $$... 4 I assume the system to be N particles with potential energy U(\vec{R}) and kinetic energy \frac{1}{2}(\dot{\vec{R}},M\dot{\vec{R}}) where R is a 3N-dimensional vector in the configuration space and M is the mass matrix. In particular, I assume that the particle numbers are fixed -- there is no interactions that create new particles etc. To avoid ... 4 There are several different relations between Chern-Simons/WZW models, and there are several way to show these. A nice paper doing this in a concrete way is Elitzur et al Nucl.Phys. B326 (1989) 108. The Chern-Simons theory on a compact spatial manifold give rise to a finite dimensional Hilbert space (only global degrees of freedom) which turns out to be ... 4 The definition of the partition function is$$ Z = \sum_\mathbf{q} e^{-\beta E_\Sigma(\mathbf{q})} \qquad (1) $$where \mathbf{q} is the set of quantum numbers describing the microscopical state of the system, E_\Sigma(\mathbf{q}) is the energy of the system when it is in that microscopical state, \beta = 1/(k_B T) In your case \mathbf{q} is the ... 4 Here is a sketched proof of the inequality. The problem is to show that$$ \sum_n\langle \phi_n|e^{-\beta \hat{H}}|\phi_n\rangle ~\stackrel{?}{\geq}~ \sum_n e^{-\beta\langle \phi_n|\hat{H}|\phi_n\rangle} ,\qquad\qquad (1)$$where the Hamiltonian \hat{H} is a selfadjoint operator, and |\phi_n\rangle  denote orthonormal basis vectors in the Hilbert ... 4 There are many problems here. First, one typically takes \beta = 1/T and so you want a partition function like$$ Z(\beta) = \sum_n \exp(-\beta E_n)\,. $$The next technical problem is that \exp(-\beta E_1) + \exp(-\beta E_2) \neq \exp[-\beta(E_1+E_2)] as you claim it does. These two issues aside, the resolution to your problem comes from noting that ... 3 Now, what does it mean for A to be negative? It isn't negative in general. The mistake you're doing is to assume that the logarithm is positive. But \ln Z may be both positive and negative depending on whether Z is greater than one or smaller than one. Both options are possible because Z=\sum \exp(-E_i/kT) and if E_i is smaller than kT (and ... 3 I think LHS of eqn 2.7 is normalized, meaning \frac{1}{Z}\int\mathcal{D}\phi \mathcal{O} exp\left(-S_{E}\left[\phi\right]\right) evaluated on \mathcal{M}_{n} If you put \mathcal{O} =1, you get 1. But Z itself is proportional to the correlation function of the two primary fields ref: http://arxiv.org/abs/hep-th/0405152 sect IIIA Hope this is ... 3 In the US almost all data from federal funded research is already available freely and has been for a long time. The bigger issue is the usefullness of the data. How much use is a raw memory dump of the output of an LHC detector? On the other hand howmuch time/money/effort is the experimenter expected to put into putting the data into a useful format, ... 3 The character \chi_R : G \to \mathbb{C} of a representation R is defined by \chi_R(U) = Tr_R(U), namely by taking the trace of U in the representation R. See for example Appendix A of Aharony et al.. Then the equation you write seems reasonable: presumably there are N_f hypermultiplets, each with fundamental and anti-fundamental fields that ... 3 The differences will show up in the correlation functions because the correlation functions "know" about the group under which the states transform. For example, the first excited level of both CFTs, one with E_8\times E_8 (HE) and one with SO(32) (HO), contains 248+248=32\times 31/(2\times 1)=496 states (and therefore the corresponding operators ... 3 I think one way to understand why this works is that the spectrum of energy levels E_i has undergone a sort of transform (analogous to Laplace transform) which results in the partition function Z(T). In principle if you know the function Z(T) you can reverse the process, and reconstruct the original spectrum of energy levels. As such, all information ... 3 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 ... 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 ...

3

The only trick here is getting used to how discrete sums are turned into integrals. Suppose you let energy be a function of momentum $p$ and position $q$. Then you can rewrite the discrete quantum partition function as $Z_{quantum}=\sum_{p,q}e^{- \beta E(p,q)},$ where the sum is over each of the $N$ positions and $N$ momenta, and the only challenge is how ...

3

I guess, one could start by considering the microcanonical and canonical ensembles as entirely different concepts. For they represent different statistical ensembles: in the former the energy of the system is fixed while in the second the energy can span all possible values allowed by the energy spectrum of the system with some penalty attributed to high ...

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