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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

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, ...


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 ...


5

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 ...


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

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

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

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 $|n\rangle$, ...


3

[By statistical mechanics I mean classical statistical mechanics throughout this answer. If you are curious to think about the complications added with making the statistical side of the story quantum mechanical, that sounds like a very good exercise. For clarification look at Chap. 3 of "Conformal Field Theory" by Di Francesco et al.] The analogy between ...


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

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 ...


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

It's a 3N dimensional integral, but it reduces to the N-th power of a 3-dimensional integral (and ultimately to the 3N-th power of a 1 dimensional integral), so you probably have a sloppy source. $$ Z = \int (\prod_i d^3p_i) e^{-\beta \sum_i {p_i^2\over 2m}} = \prod_i (\int e^{-\beta {p^2\over 2m}} d^3p) = I^N $$ Where $$ I = \int e^{-\beta {p^2\over 2m}} ...


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

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 ...


2

The idea is correct (it's called the Hubbard-Stratonovich transformation), but I can't say more without the details of the action. It is discussed in any good textbook on quantum field theory for condensed matter. Concerning you questions (if you're okay with the physicist's approach to Grassman numbers) : the product of two Grassmann numbers commutes with ...


2

The second formula is not a generalization of the first, it is a simple consequence of the definition of the twist fields, like the first. These formulas are just defining the formal path integral for the Riemann surface, and then the "twist field" correlation functions are essentially defined by to reproduce the correlation functions. (I will be explicit ...


2

In addition to the polynomial moments, people often consider the Fourier transform of a probability distribution. This is the expected value $$ \int \rho(V) e^{ikV} dV $$ These exponential moments are clearly enough to reconstruct the distribution completely. The polynomial moments are not always enough, but the countexamples are badly behaved. The moment ...


2

Energy has to be bounded below. The relationship between energy & momentum in this case is $E_i = c |p_i|$, not $E_i =c p_i$. So the integral you should be trying is $\int_{-\infty}^\infty e^{-\beta c |p_i|} dp_i$, which converges just fine.


2

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 ...


2

It seems that OP is pondering about the notion of supernumbers, and the generalization of Fubini and Tonnelli's theorems for integration over superdomains and supermanifolds. See e.g. this Phys.SE post and the references listed therein for details. Example: Consider the integral $$\tag{1} ...


2

The classical calculation goes like this. Consider the Hamiltonian of the classical harmonic oscillator $$ H(q,p) = \frac{p^2}{2m} + \frac{1}{2}m\omega^2 x^2 $$ The expression $\Omega(E)$ refers to the volume of the region in phase space corresponding to classical states $(q,p)$ with energies less than or equal to $E$, namely it is the volume of the ...


2

The partition function is a special case of the matrix element $\langle \psi_f| e^{i\sum_\alpha c_\alpha \mathcal{O}_\alpha} | \psi_i\rangle$, where $\psi_f$ and $\psi_i$ are final and initial states, the $\mathcal{O}_\alpha$ are operators that appear in your theory. It's a function of the parameters $c_\alpha$; it's meant to be a generating function for ...


2

First of all, you can simplify this by rewriting it as an indefinite integral of $x = Z/F$ like so: \begin{equation} \int_\epsilon^\infty \frac{ds}{s^3} \left ( \frac{s/2}{\sinh s/2} \right )^2 e^{-s x} = - \frac{1}{4} \int_\epsilon^\infty ds \int_0^x dx \frac{1}{\sinh^2 s/2} e^{-s x} \end{equation} From here, you have roughly \begin{equation} ...


2

The average energy is $$\overline{U}=-\frac{\partial}{\partial\beta}\log(Z)=-\frac{1}{Z}\frac{\partial Z}{\partial\beta}=\frac{\sum_pg_p\epsilon_p\exp(-\epsilon_p\beta)}{\sum_pg_p\exp(-\epsilon_p\beta)}=\sum_p\epsilon_pP_p$$ where $P_p$ is the probability of being in the $p^\text{th}$ state. Multiplying this by $N$ (the total number of particles) and noting ...


1

You ask how can you formally go from the summation over $s$ to the double sum over $s_1$ and $s_2$? As we'll see in a moment, passing to the double sum relies on the mathematical fact (about tensor products of Hilbert spaces) that if $s_1$ labels a basis of states for system $1$, and if $s_2$ labels a basis of states for system $2$, then the set of all ...


1

Why in the heck are you worrying about the difference between 3/2 and 5/2? Your formulae differ not just by 3/2 and 5/2 but also by $\ln(N)$. For the thermodynamic limit, $\ln(N)$ is infinity, and so a lot bigger than 1. The number 3/2 has in this context no meaning, except that it is 5/2-1. The number 5/2 gives the correct entropy correct for the ...


1

The logarithmic relationship is equivalent to $$Z[J]=\exp[iW[J]]$$ where $W$ is the sum of connected diagrams. This formula is trivial to prove via Taylor expansion of the exponential $$\exp(X) = \sum_{n=0}^\infty \frac{X^n}{n!} $$ If we substitute $i$ times the sum of all connected diagrams $iW$ for $X$ in this formula, the term $X^n/n!$ will simply ...


1

If you think for a second about permutations: All permutations are a product of disjoint cycles. So you can write a permutation by multiplying cycles together, then dividing through by the number of ways you can stitch them together because that gives the same permutation. So to get all permutations, you multiply cycles C by each other, which gives ...



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