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I'm trying to reproduce the computations of this paper and I'm running into some troubles because I'm rather new to Mathematica.

In the paper's page 4, in figure 2 the authors show a plot of a function. The function is as follows.

First define

$$\mathcal{D}(t,\theta)=1+\frac{1-t^2}{2}\sum_{n=0}^\infty t^{2n}(1+(n+1)(1-t^2))\log\left[\frac{1-t^2}{2}t^{2n}(1+(n+1)(1-t^2))\right]-\frac{1}{2}\sum_{i=\pm}\operatorname{Tr}(\rho_{R|i}\log \rho_{R|i})$$

The plot I'm trying to construct is $\mathcal{D}(\tanh r,\pi/2)$. In the above we have:

$$\rho_{R|\pm}=\dfrac{1-t^2}{2}\left(1\pm\cos\theta)M_{00}+(1\mp\cos\theta)M_{11}\pm\sin\theta M_{10}\pm\sin\theta M_{01}\right)$$

The matrices $M_{ij}$ above as well as the matrices $\rho_{R|i}$ were discussed in this question and the Mathematica code that generates them is given by @Roman's answer:

M00[nmax_Integer,t_] := SparseArray[Band[{1,1}] -> Table[t^(2n), {n,0,nmax}]]
M11[nmax_Integer,t_] := (1-t^2)*SparseArray[Band[{1,1}] -> Table[n*t^(2(n-1)), {n,0,nmax}]]
M01[nmax_Integer,t_] := Sqrt[1-t^2]*SparseArray[Band[{1,2}] -> Table[Sqrt[n+1]*t^(2n), {n,0,nmax-1}], {nmax+1,nmax+1}]
M10[nmax_Integer,t_] := Transpose[M01[nmax,t]]

rhoplus[nmax_Integer,t_,th_] := (1-t^2)/2*((1+Cos[th])*M00[nmax,t]+(1-Cos[th])*M11[nmax,t]+Sin[th]*(M10[nmax,t]+M01[nmax,t]))
rhominus[nmax_Integer,t_,th_]:= (1-t^2)/2*((1-Cos[th])*M00[nmax,t]+(1+Cos[th])*M11[nmax,t]-Sin[th]*(M10[nmax,t]+M01[nmax,t]))

With this we can compute the traces from the eigenvalues of these matrices. As also discussed in the question, we do:

s[x_] = Piecewise[{{x*Log[2,x], 0<x<1}}]
EntropyPlus[nmax_Integer, t_, th_] := Total[s /@ Eigenvalues[rhoplus[nmax,t,th]]]
EntropyMinus[nmax_Integer, t_, th_] := Total[s /@ Eigenvalues[rhominus[nmax,t,th]]]

I then computed the other sum with a cutoff of $N = 100$:

SumAuxElement[t_][n_Integer] := ((1 - t^2)/2)*t^(2 n)*(1 + (n + 1) (1 - t^2))
SumAuxList[nmax_Integer, t_] := Array[SumAuxElement[t], nmax + 1, {0, nmax}]
SumAux[t_,nmax_Integer]:=Total[s /@ SumAuxList[nmax, t]]

First I defined one auxilairy $n$-th element of the sum, assembled those in a list and called total on it using the $s$ function as with the eigenvalues. The minus sign is because on the function $s$ we have one minus that doesn't appear on the second term of $\mathcal{D}$.

I finally defined the function to be plotted:

Discord[nmax_Integer, t_, th_] := 1 + SumAux[t,nmax] - 1/2 (EntropyPlus[nmax, t, th] + EntropyMinus[nmax, t, th])

Plot[Discord[100, Tanh[r], Pi/2], {r, 0, 2.5}]

And the graph I get is:

enter image description here

Which is not the correct one (see below the green line):

enter image description here

So what am I doing wrong? I'm confident on the answer to the other question so that the last term in $\mathcal{D}$ is probably right. I believe my issue is with that infinite sum.

Why am I not getting the right plot here? What should I modify in this code?

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  • $\begingroup$ Why do you hardcode the cutoff at 100 in SumAux instead of using the nmax parameter? $\endgroup$ – Roman Jan 26 at 7:58
  • $\begingroup$ @Roman, I did it when testing, later I've removed it and used the variable instead. I've updated the code in the post. $\endgroup$ – user1620696 Jan 26 at 13:06
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It seems there is a problem with the signs in your expressions. Try using this (although the Piecewise is not needed)

s[x_] = Piecewise[{{x*Log[2,x], 0<x<1}}];

and then

SumAux[t_, nmax_Integer] := Total[s /@ SumAuxList[100, t]]

These should give you the paper's graph. Note you should type "SumAux[t, nmax]" instead of just "SumAux[t]", but since you got the first graph I suppose you already have! :)

Discord[nmax_Integer, t_, th_] := 1 + SumAux[t,nmax] - 1/2 (EntropyPlus[nmax, t, th] + EntropyMinus[nmax, t, th])

Trying for $nmax = 300$, you get the following output:

enter image description here

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  • $\begingroup$ The Piecewise is needed because sometimes numerical inaccuracies give slightly negative eigenvalues. Take, for example, a vector v = RandomReal[{-1, 1}, 10] and its density matrix R = KroneckerProduct[v, v]. The eigenvalues of R should be all zero except one of them. Numerically, though, they fluctuate around zero: Eigenvalues[R]. This creates trouble for the s function if defined without Piecewise. $\endgroup$ – Roman Jan 26 at 7:55
  • $\begingroup$ I think the paper uses the natural logarithm, not the two's logarithm, in the definition of the entropy: use s[x_] = Piecewise[{{x*Log[x], 0 < x < 1}}] instead of what you've used. $\endgroup$ – Roman Jan 26 at 8:19
  • $\begingroup$ There indeed was one sign issue. The last term in the equation has a minus in front. Only when the minus is included in the sum we get the sum of two entropies. I left the minus sign together with the entropies in the code. I've corrected these issues. The function s now doesn't include the sign and it is placed where it is required. The graph got better, but it is still wrong. I've tried setting a higher cutoff (300) but it took forever to build the graph and didn't change almost. $\endgroup$ – user1620696 Jan 26 at 13:29
  • $\begingroup$ I see @Roman! Indeed these expressions numerically are rather problematic. Mathematica complains about a precision loss even for nmax as low as 40. @user1620696, the notch you notice is due to convergence issues. You will notice that as you move the nmax higher, the notch moves towards the left. Trying nmax=300 gives the output I have uploaded, which I think is pretty similar to the one in the paper. $\endgroup$ – Sotiris Jan 26 at 15:23
  • $\begingroup$ @Sotiris thanks, I've found what was wrong all along (apart from the sign mistake): the plot I got had the $y$ axis starting at $0.7$ and I thought it was starting at $0.0$ giving me the impression the graph was totally wrong. I've made the axis start at $0.0$ with PlotRange and got the same result as you, which agrees with the paper. Thanks a lot ! $\endgroup$ – user1620696 Jan 28 at 0:02

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