Cazul discret

Calculam momentele observabilei $\widehat{E}$ pentru cele doua stari (4.3.44) si (4.3.36):

$$M_1^{(O)} = Tr(\widehat{\rho }_{post}^{(c)}\sum\limits_{m\in \mathbb{Z} }m\ve... ...angle \langle m\vert)=\sum\limits_{m,k\in \mathbb{Z}}m{\omega _{km}}\rho _{mm}=$$

$$= \sum\limits_{k\in \mathbb{Z}}{\omega _{km}}\sum\limits_{m\in \mathbb{Z} }m\rho _{mm}=\sum\limits_{m\in \mathbb{Z}}m\rho _{mm}=$$

$$= Tr(\widehat{\rho }\sum\limits_{m\in \mathbb{Z}}m\vert m\rangle \langle m\vert)=M_1$$

Observam ca $M_1^{(O)}$, si de asemenea toate momentele superioare, nu depind de parametrii de fuzzificare. In celalalt caz, dimpotriva obtinem:

$$M_1^{(E)} = Tr(\widehat{\rho }_{post}^{(E,c)}\sum\limits_{m\in \mathbb{Z} }m\... ...angle \langle m\vert)=\sum\limits_{m,k\in \mathbb{Z}}m{\omega _{mk}}\rho _{kk}=$$

$$= \sum\limits_{k\in \mathbb{Z}}k\rho _{kk}+\sum\limits_{m,u\in \mathbb{Z}}u{ \omega _{m+u,m}}\rho _{mm}$$

Fie $\sum\limits_{u\in \mathbb{Z}}u{\omega _{m+u,m}=}M_1({\omega ,m)}$. Pentru distributii omogene, adica ${\omega _{m+u,m}=}\omega _u$, notam $M_1({ \omega ,m)=}M_1^{({\omega )}}$. Atunci:

$$M_1^{(E)}=M_1+M_1^{({\omega )}}$$

si, in general:

$$M_n^{(E)}=\sum\limits_{k=0}^nC_n^kM_kM_{n-k}^{({\omega )}}$$

care este o formula bine-cunoscuta a fizicii statistice.

Se pot efectua si calcule directe pentru entropia liniara:

$$S=Tr(\widehat{\rho }-\widehat{\rho }^2)=1-Tr(\widehat{\rho }^2)$$ (4.4.47)

Pentru (4.3.44) obtinem:

$$S^{(O)}=1-\sum\limits_{n,k\in \mathbb{Z}}(\sum\limits_{m\in \mathbb{Z}}\sqrt{ {\omega _{mk}\omega _{mn}}})^2\rho _{kn}\rho _{nk}$$ (4.4.48)

iar pentru (4.3.36):

$$S^{(E)}=1-\sum\limits_{n,k\in \mathbb{Z}}(\sum\limits_{m\in \mathbb{Z}}\sqrt{ {\omega _{mk}\omega _{mn}}})^2\rho _{nn}\rho _{kk}$$ (4.4.49)