I am currently reading this paper: Entropy Inequalities by Araki and Lieb (project Euclid link). And I am not able to understand one step:

$${\rm Tr}^{123}\left(\rho^{12}\rho^{23}\right)={\rm Tr}^{2}\left(\rho^{2}\right)^2$$

I understand that

$$\rho^{12}={\rm Tr}^{3}\left(\rho^{123}\right)= \sum_i\left\langle x_1,x_2,e_i | \rho^{123}|y_1,y_2,e_i\right\rangle$$


$$\rho^{23}={\rm Tr}^{1}\left(\rho^{123}\right)=\sum_j \left\langle e_j,x_2,x_3 | \rho^{123}|e_j,y_2,y_3\right\rangle$$

So if I plug in these last two expressions in the upper expression I get:

$${\rm Tr}^{123}\left(\sum_{i,j}\left\langle x_1,x_2,e_i | \rho^{123}|y_1,y_2,e_i\right\rangle \left\langle e_j,x_2,x_3 | \rho^{123}|e_j,y_2,y_3\right\rangle\right)$$

But I am not able to simply this expression any further.

  • $\begingroup$ The way you write it now (your 2nd eq), $\rho^{12}$ is a number (which depends on $x_1$, $x_2$, etc.). It would be more clear if you would expand $\rho^{12}$ in a basis, i.e., add $\sum_{x_1,x_2,y_1,y_2}\lvert x_1,x_2\rangle\langle y_1,y_2\rvert$ on the r.h.s., and similarly in the next equation. Then, try to express the last equation using these expansions of $\rho^{12}$ and $\rho^{23}$. $\endgroup$ Feb 4 '15 at 19:35
  • $\begingroup$ Ok so here is the correct derivation. $$\rho^{123}=\sum_{ijlmn} \lambda_{ijklmn} \left|x_j,y_j,z_k\right\rangle\left\langle x_l,y_m,z_n \right|$$ It turns out: $$\rho^{12}=Tr^3(\rho^{123})=\sum_o \left\langle z_o \right|\rho^{123}\left|z_o\right\rangle=\sum_{ijlmo} \lambda_{ijolmo} \left|x_i,y_j\right\rangle\left\langle x_l,y_m\right|$$ Similarly $$\rho^{23}=\sum_{jlmno} \lambda_{ojkomn} \left|y_j, z_k\right\rangle\left\langle y_m,z_n\right| $$ $\endgroup$
    – onephys
    Feb 11 '15 at 10:09
  • $\begingroup$ Now $$Tr^{123}[\rho^{12}\rho^{23}]=\sum_{abc}\left\langle x_a,y_b,y_c\right|\left(\sum_{ijlmno} \lambda_{ijolmo} \left|x_i,y_j\right\rangle\left\langle x_l,y_m\right| \sum_{defgh} \lambda_{hdehfg} \left|y_d,z_e\right\rangle\left\langle y_f,z_g\right|\right) \left|x_a,y_b,z_c\right\rangle$$ $\endgroup$
    – onephys
    Feb 11 '15 at 10:09
  • $\begingroup$ Now with some slight abuse of notation. $$=\sum_{abcijlmodefgh} \left\langle x_a,y_b,z_c|x_i,y_j\right\rangle\left\langle x_l,y_m|y_d,z_e\right\rangle\left\langle y_f,z_g|x_a,y_b,z_c\right\rangle$$ $$=\sum_{abcijlmodefgh}\lambda_{ijolmo}\lambda_{hdehfg}\delta_{ai}\delta_{ce}\delta_{la}\delta_{gc} \left\langle y_b|y_i\right\rangle\left\langle y_m|y_d\right\rangle\left\langle y_f|y_b\right\rangle$$ $\endgroup$
    – onephys
    Feb 11 '15 at 10:09
  • $\begingroup$ $$=\sum_{bjlmodfgh}\lambda_{ljolmo}\lambda_{hdghfg} \left\langle y_b|y_a\right\rangle\left\langle y_m|y_d\right\rangle\left\langle y_f|y_b\right\rangle$$ Taking the brakets with the $y_b$'s out and a simple rearrangement of summations shows that this is equal to: $$=Tr^2(\rho^2\rho^2)$$ $\endgroup$
    – onephys
    Feb 11 '15 at 10:10

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