Hot answers tagged partition-function
5
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 states 10
$|kl\rangle$, 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 ...
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
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 ...
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
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
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 ...
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
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
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
$$ I = \int e^{-\beta {p^2\over 2m}} ...
2
The classical calculation goes like this. Consider the Hamiltonian of the classical harmonic oscillator
$$
H(q,p) = \frac{p^2}{2m} + \frac{1}{2}m\omega^2 x^2
$$
The expression $\Omega(E)$ refers to the volume of the region in phase space corresponding to classical states $(q,p)$ with energies less than or equal to $E$, namely it is the volume of the ...
1
Now, what does it mean for A to be negative?
It isn't negative in general. The mistake you're doing is to assume that the logarithm is positive. But $\ln Z$ may be both positive and negative depending on whether $Z$ is greater than one or smaller than one. Both options are possible because $Z=\sum \exp(-E_i/kT)$ and if $E_i$ is smaller than $kT$ (and ...
1
Your calculation looks fine to me (technically, your partition function should have an extra factor of $e^{-\frac 12 \beta\hbar\omega}$, but this is unimportant, as it cancels in all observables).
Edit: As in the comment by abcXYZ, the probability of finding the system in a state corresponding to any odd value of $n$ is
$$
P(n~\text{is odd}) = (1-x)(x + x^3 ...
1
This is a quantum partition function, not a statistical mechanical partition function. He is just talking about an idealized self-interacting field. If you have a scalar with cubic self interactions, you write the Lagrangian as
$$ \partial_\mu \phi \partial^\mu \phi - \lambda \phi^3 $$
If you fourier transform the field variables, this is
$$ \int_k k^2 ...
1
The answer is in general No, for various reasons.
The WKB approximation is generally not exact for a finite energy-level $N$.
Point 1 is true even if we include the metaplectic correction from the Maslov index.
OP's sought-for formula is not exact for a finite energy-level $N$ even for the simple quantum harmonic oscillator (SHO). (This is despite the ...
1
Well yes, for the ideal gas model, $pV=Nk_BT$, you find
$$U=\frac{3}{2}Nk_BT\propto T,$$
and
$$C_V=\left(\frac{∂U}{∂T}\right)_V=\frac{3}{2}Nk_B=\text{const}.$$
This itself is a violation of the third law.
What does it say to us? The lengthy discussion in the comments of this question might help you understand the problem.
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