# Tag Info

25

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 \left\langle e^{zX}\right\...

13

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 $E_n$: \begin{...

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, ...

10

$h$ 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 ...

9

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 (...

9

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 ...

8

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 $|n\rangle$, ...

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 ...

7

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 ...

7

The path integral involving the Nambu-Goto square root in the exponent is a very complex animal. Especially in the Minkowski signature, there is no totally universal method to define or calculate the path integrals with such general exponents. So if you want to make sense out of such path integrals at all, you need to manipulate it in ways that are ...

6

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 ...

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 ...

5

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 ...

5

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 ...

5

The partition function $Z[J]$, both in QM and in CM, is underdetermined: any multiple of $Z[J]$ gives rise to the same dynamics. This means that $Z[0]$ is arbitrary, and is usually set to one: $$Z[0]\equiv 1 \tag{1}$$ effectively getting rid of vacuum diagrams, that is, we set $H|\Omega\rangle=0$. In other words: the energy of the vacuum is not measurable ...

5

While the trace is invariant under a transform to another basis, you need to take into account here that the coherent state basis is not an orthogonal basis and it is overcomplete. We can evaluate the trace of an operator $A$ by inserting identity operators in front of and after the operator and then using resolution of identity in terms of the coherent ...

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 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 ...

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 ...

4

You derivation is correct. It proves that no substance at T=0K can be an ideal gas, nor any of the other systems you may have in mind whose partition function doesn't have the right limiting behavior. Indeed, all known materials are solid or fluid at sufficiently low temperature. But it hasn't have to be a perfect crystal either: Really perfect crystals ...

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

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  \sum_{n_\...

4

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

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

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

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, ...

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