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Pentru operatorul pozitie $\widehat{x}=\int\limits_{\Bbb{R}}x\cdot \widehat{E }_{(x)}(dx)$ se pot obtine rezultate similare celor de mai sus$.$ Pentru operatorul canonic conjugat: $\widehat{k}=\int\limits_{\Bbb{R}}k\cdot \widehat{E}_{(k)}(dk)$ obtinem:

$$M_1^{(O)}(\widehat{E}_{(k)})=Tr(\widehat{\rho }_{post}^{(c)}\int\limits_{ \Bbb{R}}k\cdot \widehat{E}_{(k)}(dk))=$$

$$= Tr(\int\limits_{\Bbb{R}}dx\widehat{\mathcal{A}}_x\widehat{\rho }\widehat{ \mathcal{A}}_x^{+}\int\limits_{\Bbb{R}}k\cdot \widehat{E}_{(k)}(dk))=$$ (4.4.51)

$$= i\int\limits_{\Bbb{R}}dy[(\frac{\partial \rho }{\partial y^{\prim... ...mits_{\Bbb{R}}dx(\frac{ \partial f}{\partial x^{\prime }})_{(x^{\prime }=x-y)}]$$

$$M_1^{(E)}(\widehat{E}_{(k)}) = Tr(\widehat{\rho }_{post}^{(E,c)}\int \limits_{\Bbb{R}}k\cdot \widehat{E}_{(k)}(dk))=$$

$$= Tr(\int\limits_{\Bbb{R}}dx\widehat{\mathcal{A}}_x^{(E)}\widehat{\... ...idehat{\mathcal{A}}_x^{(E)}\int\limits_{\Bbb{R}}k\cdot \widehat{E} _{(k)}(dk))=$$ (4.4.52)

$$= -\frac i2\int\limits_{\Bbb{R}}dy\cdot \rho (y,y)\int\limits_{\Bbb{R}}dx( \frac{\partial f}{\partial x^{\prime }})_{(x^{\prime }=x-y)}$$

Diferenta dintre (4.4.51) si (4.4.52) este data de termenul $i\int\limits_{\Bbb{R}}(\frac{\partial \rho }{\partial y^{\prime }} )_{(y,y^{\prime }=y)}dy$, care este un termen de memorie pentru coerenta cuantica a starii initiale (inainte de masuratoarea fuzzy a pozitiei), care, totusi, nu depinde de parametrii de fuzzificare.