- #1
Petar Mali
- 290
- 0
Vick's theorem help us to find average value of product of even number of operators. For example look the case of four Bose operators
[tex]\langle \hat{b}_1\hat{b}_2\hat{b}_3\hat{b}_4 \rangle =\langle \hat{b}_1\hat{b}_2 \rangle \langle\hat{b}_3\hat{b}_4 \rangle +\langle \hat{b}_1\hat{b}_3 \rangle \langle\hat{b}_2\hat{b}_4 \rangle +\langle \hat{b}_1\hat{b}_4 \rangle \langle\hat{b}_2\hat{b}_3 \rangle [/tex]
In some cases when we have just product of four Bose operators we use decoupling
[tex]\hat{b}^+_i\hat{b}_i\hat{b}^+_j\hat{b}_j=\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_j\hat{b}_j+\langle \hat{b}^+_j\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_j\hat{b}_i+\langle \hat{b}^+_j\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j[/tex]
Why not
[tex]\hat{b}^+_i\hat{b}_i\hat{b}^+_j\hat{b}_j=\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_j\hat{b}_j+\langle \hat{b}^+_j\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i+\langle \hat{b}^+_i\hat{b}^+_j \rangle \hat{b}^+_i\hat{b}_j+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}^+_j +\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_j\hat{b}_i+\langle \hat{b}^+_j\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j[/tex]?
similarly
[tex]\hat{b}^+_i\hat{b}^+_i\hat{b}_i\hat{b}_j=\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j+\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i[/tex]
and no
[tex]\hat{b}^+_i\hat{b}^+_i\hat{b}_i\hat{b}_j=\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j+\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i+\langle \hat{b}^+_i\hat{b}^+_i \rangle \hat{b}_i\hat{b}_j+\langle \hat{b}_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}^+_i [/tex]
Thanks for your answer!
[tex]\langle \hat{b}_1\hat{b}_2\hat{b}_3\hat{b}_4 \rangle =\langle \hat{b}_1\hat{b}_2 \rangle \langle\hat{b}_3\hat{b}_4 \rangle +\langle \hat{b}_1\hat{b}_3 \rangle \langle\hat{b}_2\hat{b}_4 \rangle +\langle \hat{b}_1\hat{b}_4 \rangle \langle\hat{b}_2\hat{b}_3 \rangle [/tex]
In some cases when we have just product of four Bose operators we use decoupling
[tex]\hat{b}^+_i\hat{b}_i\hat{b}^+_j\hat{b}_j=\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_j\hat{b}_j+\langle \hat{b}^+_j\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_j\hat{b}_i+\langle \hat{b}^+_j\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j[/tex]
Why not
[tex]\hat{b}^+_i\hat{b}_i\hat{b}^+_j\hat{b}_j=\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_j\hat{b}_j+\langle \hat{b}^+_j\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i+\langle \hat{b}^+_i\hat{b}^+_j \rangle \hat{b}^+_i\hat{b}_j+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}^+_j +\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_j\hat{b}_i+\langle \hat{b}^+_j\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j[/tex]?
similarly
[tex]\hat{b}^+_i\hat{b}^+_i\hat{b}_i\hat{b}_j=\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j+\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i[/tex]
and no
[tex]\hat{b}^+_i\hat{b}^+_i\hat{b}_i\hat{b}_j=\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j+\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i+\langle \hat{b}^+_i\hat{b}^+_i \rangle \hat{b}_i\hat{b}_j+\langle \hat{b}_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}^+_i [/tex]
Thanks for your answer!