# Tag Info

6

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, ...

5

The term canonical gives it away. The canonical ensemble density matrix $\rho$ is defined as follows in terms of the Hamiltonian $H$ and inverse temperature $\beta = 1/kT$: \begin{align} \rho(\beta) = \frac{1}{Z(\beta)}e^{-\beta H}, \qquad Z(\beta) = \mathrm{tr}(e^{-\beta H}) \end{align} Then the canonical ensemble average of any observable $O$ is given ...

4

It seems you're coming at entropy from a thermodynamics standpoint. This is completely consistent with (and, at the macro scale, equivalent to) the statistical derivation of entropy, but you might find the statistical version more intuitive, if the thermodynamic version is causing you issues. I warn you, statistical physics is both math-heavy and takes some ...

3

Let's recall basics of classical and quantum mechanics for non-statistical systems. In classical Hamiltonian mechanics, one models the non-statistical state of a system as a point in phase space $\mathcal P$. If the configuration space (space of spatial positions) of the system is $N$-dimensional, then the phase space is $2N$ dimensional because the state ...

3

The reason for using microstates is that it is the only way to come up with quantum statistics, but the grand-canonical potential is defined also for a classical system. And you are right, one should take into acount the state with zero particles, let me show why on a simple example. Let us take for instance a system in which $N$ indistiguishable ...

2

Here's a close analogy which (besides having has its own practical use) can shed some light, if you're looking at the problem from the probabilistic-mathematical point of view. Imagine we play the following game ($A$) : we throw $N$ balls into $K$ boxes, with uniform probability. Let call the result (or configuration) $X=(x_1,x_2 ...x_k)$, where $x_i$ is ...

2

[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.] The analogy between Euclidean quantum field theories and equilibrium statistical mechanics is exact, once ...

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

The uncertainty principle is a fundamental property of quantum systems, and is not a statement about observational success. No particle either free or in crystal can have zero momentum otherwise a nonsensical infinity is required for the standard deviation of position $\Delta x$, in the uncertainty principle $\Delta x \Delta p \geq \hbar / 2$. $0 \cdot ... 2 If in your system the number of of photons is non conserved, the Gibbs free energy cannot depend on the number of photons. So you will have $$\mu_{\gamma}=\left(\frac{\partial G}{\partial N_{\gamma}}\right)_{T,P}=0$$ This also implies that$\mu_{matter}+\mu_{\gamma}=\mu_{matter}$. However, in systems that conserve the number of photons, you return to the ... 2 In general, both IQHE and FQHE are rigid quantum states, whose rigidness is protected by the finite energy gap ($h\omega$for IQHE) between the ground state(s) and the exited states. Finite temperature can support excitations to overcome the gap, which destroys the rigidness of the state. Under finite temperature, the quantization of the Hall conductivity is ... 1 Check the derivation of the Boltzmann distribution from the microcanonical ensemble on the Wikipedia page "Maxwell-Boltzmann Statistics". We suppose that the "wealth classes" of individuals are discretised, so that, for example, we find the number of individuals with$m_1 = \$500$, the number with $m_2 = \$10000$and so forth: as an approximation we ... 1 As I had written in the comments, it is the second term that you have gotten incorrect. Focusing exclusively on this term (leaving aside the$1/8$factor), we have $$|\psi_2\rangle \langle \psi_2| = \frac{1}{3}(|00\rangle \langle 00|+ |10\rangle \langle 10|+ |11\rangle \langle 11| - |00\rangle \langle 10|-|10\rangle \langle 00| + \ ...),$$ where I have ... 1$J$can be determined experimentally by knowing, for example,$T_c$. Through a theoretical model including the structure and the species involved you can estimate the value for$J$through the overlap integrals of the electrons involved. Here you can read more about exchange interaction, which is the basics behind magnetism (which is a quantum phenomena). 1 Setting$\Delta=3/2$is useful only to ensure the correct behavior of the second limit. The first limit is given by the condition$f(t,0)\propto t^2$. Because$f(t,h)=t^2 g_f(h/t^\Delta)$, you directly get that$g_f(x)\to {\rm const}$for$x\to 0$. For the sake of completeness, let's do the other case and show that we must have$\Delta=3/2$. You know that ... 1 If you call$ \chi $the exergy (availability) then$ \chi = U + p_o V - T_o S $where$p_o, T_o$are the pressure and temperature of the environment (and are assumed to be constant). To find the maximum amount of useful work that can be extracted form the system it is sufficient to analyze reversible processes only so that$ dU=TdS-pdV $and then the exergy ... 1 Note that in the sum $$=\sum_{S_1=\pm 1}...\sum_{S_N=\pm 1}\langle S_2| T_{_{NN}}^{\dagger}|S_1\rangle\langle S_1| T_{_{NNN}}|S_3\rangle\langle S_3| T_{_{NN}}^{\dagger}| S_2\rangle\langle S_2| T_{_{NNN}}|S_4\rangle...\langle S_1| T_{_{NN}}^{\dagger}|S_N\rangle\langle S_N| T_{_{NNN}}|S_2\rangle$$ every pair$|S_i\rangle\langle S_i|\$ occurs twice with some ...

1

How familiar are you with exact solutions to the 2D Ising model? If you don't know that forwards and backwards, I would probably start with some textbooks that cover that material in depth. This would teach you some of the basics of transfer-matrix techniques and some other tricks about estimating eigenvalues, the thermodynamic limit, and so on. The book ...

1

The problem is essentially the same problem as trying to define information entropy for a continuous probability distribution. You end up with an entropy value that has an offset depending on what units you chose for your random quantity. It is unfortunate, but the problem really stems from the fact that the number of possible physical states is uncountably ...

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