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Are there any physical meaning or application of this property? Why nature gives this property to the widely used Fermi-Dirac Function? The property you found (Taylor expansion $\neq$ original function on any interval around the point) has nothing to do with physics or nature, and is not particularly connected to the Fermi-Dirac distribution. It may be ...

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That definition of $\Omega$ only works if you have distinguishable particles and you can put as many as you want in any state. That gives you the Maxwell-Boltzmann distribution. In quantum mechanics, particles of the same type are not distinguishable; e.g. there's no way to tell one electron from another. Moreover, for Fermions (like electrons), you can ...

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Plank's distribution (law) is a specific application of the Bose-Einstein distribution. For example, there is no chemical potential, $\mu$, for photons, so it is missing from Planck's law, although it's in the Bose-Einstein distribution. (The chemical potential only comes into play when you have a fixed number of particles; there is no such restriction for ...

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Planck didn't know Bose-Einstein statistics at the time around 1900. With the existence of minimal unit, or quantization $E=hf$, in mind, he derived the Planck's law which describe the black body radiation. Two decades late, after the establishment of the Bose-Einstein statistics, then it is known that Plank's law is a special case of Bose-Einstein ...

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$E\left(\lambda\right)$ is the energy of one photon of light with wavelength $\lambda$ $f\left(E\left(\lambda\right)\right)$ is the number of photons in a state with wavelength $\lambda$ $D\left(\lambda\right)d\lambda$ is the number of states with wavelengths between $\lambda$ and $\lambda+d\lambda$. ($D\left(\lambda\right)$ is the density of states.) ...

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I'm assuming that this section of the book is talking about the ultraviolet catastrophe, where an ideal black body in thermal equilibrium will emit an infinite amount of power through radiative means. The source goes on to say: The ultraviolet catastrophe results from the equipartition theorem of classical statistical mechanics which states that all ...

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As far as I have seen, there are two views regarding the tension and pressure. If the area of the membrane is fixed and you prevent local dilation of the area then the lagrange multiplier associated with area constraint is just a mathematical term known as surface pressure and it is a spatially varying field, not the same as surface tension. The surface ...

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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 $\Delta x \Delta p \geq \hbar / 2$. $0 \cdot ... 3 The reason for using microstates is that it is the only way to come up with quantum statistics, but the grand-canonical potential is defined also for a classical system. And you are right, one should take into acount the state with zero particles, let me show why on a simple example. Let us take for instance a system in which$N$indistiguishable ... 2 In general, both IQHE and FQHE are rigid quantum states, whose rigidness is protected by the finite energy gap ($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 ... 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 ... 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$\beta = 1/kT: \begin{align} \rho(\beta) = \frac{1}{Z(\beta)}e^{-\beta H}, \qquad Z(\beta) = \mathrm{tr}(e^{-\beta H}) \end{align} Then the canonical ensemble average of any observableO$is given ... 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 ... 0 Since there is no way in which the molecules can be labeled, the particles are indistinguishable. On the other hand, if the assembly is a crystal, the molecules can be labeled in accord with the positions they occupy in the crystal lattice and can be considered distinguishable. 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 ... 0 You can use all the ensemble to both quantum and classical statistics (microcanonical, canonical, grand canonical). Furthermore, quantum statistical physics respects the basic thermodynamics equalities (of the entropy, free energy, etc.). Usually, you get classical statistical physics out of the quantum one in the proper limit (high-temperature/low density, ... 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). 0 If the particles are all the same (we don't need to invoke indistinguishability at that stage I think), then there is no loss of information since the permutation of two particles in a reduced distribution function will ask exactly the same question as before the permutation probability-wise. Note that usually, reduced distribution functions are defined ... 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 ... 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 ... 0 Macrostate- The microstate of a system is the state which can be experimentally observed. It is the state which represents the macroscopic properties of the system, and not the properties of each individual practicle of the system. Microstate- The microstate of a system is the state in which we consider the arrangement of each individual particle of the ... 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 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 $and then the exergy ... 0 I don't think you can calculate the fraction for molecules with exactly 305 m/s speed. Statistical calculations have an inherent assumption about the 'precision'. Exact 305 m/s speed implies your precision for calculation of speeds has become infinite. You can only determine speeds upto some precision and hence there will be a 'dv' factor in your final ... 0 So a "pure" paramagnetic system has a positive magnetic susceptibility and for which the dipoles don't interact. Usually the effect of the "thermal bath" outweighs an applied magnetic field so the net magnetization is 0. I would say that in general the answer is yes seeing as it takes a Squid magnetometer to detect a paramagnetic system, but it is certainly ... 2 In my naive view, this is merely a mathematical trick that should not be taken too seriously in term of physical interpretation. After all, a "Wick rotation" applied to the Schrodinger equation yields a diffusion equation. This is helpful for some mathematical problems but the physics it describes is very very different from quantum mechanics; not even ... 0 I would like to add that it is very important to know what you can find out without knowing the exact solution. Kardar's book "Statistical physics of fields" teaches that in a very engaging way. To get quickly started, I suggest the topics of Scaling theory, and real space renormalization in 1-d ising model. Your motivation in keeping up with Baxter will ... 0 I'm just going to quote Wikipedia here: For the case of two colliding bodies in two dimensions, the overall velocity of each body must be split into two perpendicular velocities: one tangent to the common normal surfaces of the colliding bodies at the point of contact, the other along the line of collision. Since the collision only imparts force along ... 0 If I have a velocity which has some component$v_x$in the$x$-direction, then is there any reason for you to assume you know anything anything about the component of my velocity which might be in a perpendicular direction,$v_y$? No. So you can see that it is reasonable to assume that, if you know my$v_x$, my$v_y$is still unconstrained, i.e. you have ... 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 ... 0 The confusion arises because there are two kinds of classical limits, depending of the system under study. Let's start with fermions, which distribution is$n_F(\epsilon)=\frac{1}{e^{(\epsilon-\mu)/T}+1}$. The first classical limit (corresponding to the case mentioned in the question) is$T\gg \epsilon-\mu\$. This corresponds to the case where the ...

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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 ...

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The introductory paragraph you quote with horror says temperature ''high enough'' to avoid quantum effects. (It did not say anything like ''arbitrarily large''.) If the temperature is too low, things like Bose--Einstein condensation can occur, which invalidate Maxwell--Boltzmann statistics. The temperature should be high enough so that it is unlikely to ...

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See http://arxiv.org/abs/quant-ph/0512105. It gives a derivation of Landauer's Principle from the two postulates of the Second Law of Thermodynamics, and shows how Landauer's Principle follows from the second postulate.

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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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