Simplest non-additive measures of quantum resources

Abstract

Given an arbitrary state $\rho$ and some figure of merit $\mathcal{E}(\rho )$, it is usually a hard problem to determine the value of $\mathcal{E}(\rho ^{\otimes N})$. One noticeable exception is the case of additive measures, for which we simply have $\mathcal{E}(\rho ^{\otimes N}) = Ne$, with $e\equiv \mathcal{E}(\rho )$. In this work, we study measures that can be described by $\mathcal{E}(\rho ^{\otimes N}) =E(e;N) \ne Ne$, that is, measures for which the amount of resources of N copies is still determined by the single real variable e, but in a nonlinear way. If, in addition, the measures are analytic around $e=0$, recurrence relations can be found for the Maclaurin coefficients of E for larger N. As an example, we show that the $\ell _1$-norm of coherence is a non-trivial case of such a behavior.

Appendices

Third-order coefficient solution

To solve the recurrence relation (8), we choose $K=a$ and apply the known solutions for $d_1(a^n)$ and $d_2(a^n)$ [8]:

$$\begin{aligned} d_3(a^n)=a^{\nu (n-1)}d_3(a) + 2[d_2(a)]^2a^{\nu (n-1)}\left( \frac{a^{\nu (n-1)}-1}{a^\nu -1}\right) + d_3(a^{n-1})a^{3\nu }. \end{aligned}$$

Remember that $a^n=N$. The recursive substitution of $d_3(a^{n})$ into itself $l=n-1$ times gives:

$$\begin{aligned} d_3(N)= & {} d_3(a)\left[ a^{\nu (n-1)}+a^{\nu (n-2)}a^{3\nu }+\cdots \right] + 2[d_2(a)]^{2} \sum _{l=1}^{n-1} \left\{ a^{\nu (n-l)} \left( \frac{a^{\nu (n-l)}-1}{a^\nu -1}\right) a^{3\nu (l-1)} \right\} \\ \cdots= & {} d_3(a)\sum _{l=1}^{n}a^{\nu (n-l)} a^{3\nu (l-1)} + \frac{2[d_2(a)]^2}{a^\nu -1}\left( \frac{N}{a^3}\right) ^\nu \sum _ {l=1}^{n-1}a^{2\nu l}\left( N^{\nu }a^{-\nu l}-1\right) . \end{aligned}$$

Redefining all sums to start at $l=0$ and using the geometric series $\sum _{l=0}^{n-1}x^l=\frac{1-x^{n}}{1-x}$, we get the third order in (9):

$$\begin{aligned} d_3(N)=d_3(a)\left( \frac{N}{a}\right) ^{\nu }\left( \frac{1-N^{2\nu }}{1-a^{2\nu }}\right) + 2[d_2(a)]^2 \left( \frac{N}{a^{2}}\right) ^\nu \frac{(N^\nu -1)(N^\nu -a^\nu )}{(a^\nu -1)(a^{2\nu }-1)} \end{aligned}$$

Binomial coefficients

We start with (4) for $K=2$ (here the index l is rewritten as $j-k$):

$$\begin{aligned} d_j(N)= & {} \sum _{k=0}^{j-1} d_{j-k}(N/2) \sum _{i=1}^{{{j-1} \atopwithdelims (){j-k-1}}}\pi _i(j,j-k;2) \end{aligned}$$

(20)

Remember that $N \in \mathbb {P}_{2}$, so $\frac{N}{2}$ is an integer. The hypothesis that only $d_1(2)$ and $d_2(2)$ are nonzero means that $\pi _i(j,l;2)$ is a product of combinations of these two quantities only, so we need to consider the composition of j using only the numbers 1 and 2, which has $\tiny {\frac{(j-k)!}{k!(j-2k)!}}$ elements, for k repetitions of the number 2 (the maximum value of k is $\lfloor \frac{j}{2} \rfloor$), so:

$$\begin{aligned} j = \underbrace{1+\cdots +1}_{j-2k\text { times}}+\underbrace{2+\cdots +2}_{k\text { times}} \rightarrow \sum _{i=1}^{{{j-1} \atopwithdelims (){j-k-1}}} \pi _i(j,j-k;2) = {{j-k} \atopwithdelims (){k}} [d_1(2)]^{j-2k} [d_2(2)]^{k}.\nonumber \\ \end{aligned}$$

(21)

Putting (21) into (20) and considering the case $E^{(a=2)}(e)=2e+d_2(2)e^2$, we find the following recurrence relation:

$$\begin{aligned} d_j(N)= & {} \sum _{k=0}^{\lfloor \frac{j}{2} \rfloor } d_{j-k}(N/2){{j-k} \atopwithdelims (){k}} 2^{j-2k} [d_2(2)]^{k}. \end{aligned}$$

Now we test expression $d_j(N)=[d_2(2)]^{j-1}\tiny {{{N} \atopwithdelims (){j}}}$, induced in (11). Using the subset-of-a-subset property [22], we get:

$$\begin{aligned} d_j(N) = [d_2(2)]^{j-1}\sum _{k=0}^{\lfloor \frac{j}{2} \rfloor } {{N/2} \atopwithdelims (){k}}{{N/2-k} \atopwithdelims (){j-2k}} 2^{j-2k}. \end{aligned}$$

(22)

For the evaluation of the sum in (22), we take the integral representation $\tiny {\left( \begin{array}{c} n \\ m \end{array} \right) } = \frac{1}{2\pi i} \oint _{\Gamma } \frac{(1+z)^n}{z^{m+1}} dz$ to solve the problem using the Egorychev method [23, 24], where z is a complex variable and $\Gamma$ is a small closed contour around $z=0$:

$$\begin{aligned} \sum _{k=0}^{\lfloor \frac{j}{2} \rfloor } {N/2 \atopwithdelims ()k}{{N/2-k} \atopwithdelims (){j-2k}} 2^{j-2k} = \sum _{k=0}^{\lfloor \frac{j}{2} \rfloor }\frac{1}{2 \pi i} \oint _{\Gamma } dz \frac{(1+z)^{N/2-k}}{z^{j-2k+1}}{N/2 \atopwithdelims ()k}2^{j-2k}. \end{aligned}$$

Is easy to see that for $\lfloor \frac{j}{2} \rfloor<k<\frac{N}{2}$ the integral vanishes. By making $k=\lfloor \frac{j}{2} \rfloor +l$ (so $l \le \frac{N}{2}- \lfloor \frac{j}{2} \rfloor$), the integrand does not have residue at $z=0$ for any value of $l \in (1,\frac{N}{2}- \lfloor \frac{j}{2} \rfloor )$ and, therefore, we can rewrite the sum with upper limit equal to $\frac{N}{2}$. Then, we can use the binomial theorem:

$$\begin{aligned} \frac{1}{2 \pi i} \oint _{\Gamma } dz \frac{(1+z)^{N/2}}{z^{j+1}}2^j \sum _{k=0}^{\frac{N}{2}}{N/2 \atopwithdelims ()k} \left( \frac{z^2}{4(1+z)} \right) ^k= & {} \oint _{\Gamma } dz \frac{{(1+z)^{N/2}}}{z^{j+1}}2^j \left[ \frac{(z+2)^2}{4{(1+z)}}\right] ^{\frac{N}{2}}. \end{aligned}$$

After some cancellations, we make a simple change of variables $z\,\rightarrow \,2z'$ ($\Gamma \rightarrow \Gamma '$) and the integral becomes exactly the Cauchy integral representation of the binomial coefficient, so the proposition is proven.

$$\begin{aligned} d_j(N) = [d_2(2)]^{j-1} {N \atopwithdelims ()j} \end{aligned}$$

(23)