marcus said:
He shows that for all pure states of the quantum Rindler horizon it is identically true that
∂A/4 = ∂E/T
The argument that this extends by linearity to superpositions---to mixed states of the quantum Rindler horizon, and large assemblies thereof---is not made explicitly. But a relevant observation is made immediately after equation (20) on page 4:
"Notice that the entropy density is independent of the acceleration a, or equivalently from the distance from the horizon."
This opens the way to our concluding that ∂A/4 = ∂E/T applies as well to mixed states and collections thereof.
I went through the statistical computation of the entropy in Bianchi's polymer model using the energy constraint. I was wrong when I said that there would be a rescaling of the Lagrange multiplier by a factor of \gamma. The important point is that the energy constraint is not merely equivalent to the area constraint, they are in fact exactly the same. From equation (9) of the present paper (let's call it B12 for Bianchi-2012),
E = \sum_f \hbar\gamma j_f a = \frac{a}{8\pi G} \sum_f A_f = \frac{aA_H}{8\pi G}.
We can rewrite this constraint as
\sum_f j_f = \frac{A_H}{8\pi \gamma G\hbar}.
This is the same as the area constraint used in equation (16) of the polymer paper (B10 for Bianchi-2010), except, as previously discussed, here we are using a slightly different basis where the eigenvalues of |\vec{L}| are j rather than \sqrt{j(j+1)}. You can check that the distribution of states depends only on the degeneracy and not on the precise eigenvalue, so the rest of equ (16) is unchanged.
I'd already gone through all of the math in that section of the paper before realizing that there weren't any numerical differences between the constraints, so I might as well report on the result. The derivation of the entropy only differs in the numerical value of the constants derived there. This is the effect of the different eigenvalue spectrum. For example, the occupation numbers at equilibrium satisfy
p_j \equiv \frac{N_j^*}{N^*} \approx (2j+1) e^{-\mu^* j}.
Imposing the normalization requirement
\sum_j p_j =1
can be done by approximating the sums by an integral. I find that \mu^* is the solution to
\int_0^\infty dx(x+2) e^{-\mu^* (x+1)/2} =1.
This leads to the equation
4(\mu^*+1) e^{-\mu^*/2} = (\mu^*)^2,
which has a numerical solution at
\mu^* \sim 2.086.
This is a little bit different from the value obtained in B10, but in a reasonable neighborhood given the similarity of the normalization constraints.
Similarly, the constant
\alpha^* = \sum_j j p_j = \frac{1}{4} \int_0^\infty dx(x+1)(x+2) e^{-\mu^* (x+1)/2}
= \frac{e^{-\mu^*/2} }{(\mu^*)^3} ( (\mu^*)^2 + 2 \mu^* + 4) \sim 0.486.
The leading term in the entropy is once again
S = \frac{\kappa}{4G\hbar} \frac{\mu^*}{2\pi\gamma} A_H,
so that we require
\gamma =\frac{\mu^*}{2\pi} \sim 0.322.
Presumably the difference between this value of the Immirzi parameter and earlier results is due to the difference in the spacing between the area eigenvalues \sqrt{j(j+1)} vs j. We are effectively using slightly different scales to quantize the area operator.
. And as you said earlier, it is completely new!
). What I meant and should have said is that already in equation (9), which is not semiclassical, you can see there is no dependence on Immirzi in the leading term, which is the area term. The other two terms can be considered corrections, that appear in the full quantum version of the entropy equation, namely equation (9).