Timeline for Why $\sum_{f\in F(n,d)} A_{f}^*A_{f}=\sum_{|\alpha|=n}\frac{n!}{\alpha!}{A^*}^{\alpha}A^{\alpha}?$
Current License: CC BY-SA 3.0
11 events
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| Feb 17, 2018 at 11:07 | comment | added | Pietro Majer | 2.) If you want the formula in the original post, then you don't need $A_i^*A_i=A_iA_i^*$, but just $A_jA_i=A_jA_i$. In this case you apply the expansion formula in another commutative setting, i.e. $X_i\in\mathcal{L}(\mathcal{L}(H))$ is the operator on $\mathcal{L}(H)$ defined by $X_i(L)=A_i^*LA_i$. Then $X_iX_j=X_jX_i$ holds true just because $A_jA_i=A_jA_i$, no need that $A_i$ commute with their adjoints. | |
| Feb 17, 2018 at 11:06 | comment | added | Pietro Majer | There are two different issues: 1.) if you want the last formula you wrote, then you take $X_i:=A^*_iA_i\in\mathcal{L}(H)$ in the expansion of $(X_1+\dots+X_d)^n$, and you need the $2d$ operators, $A_1,\dots,A_d,$ and $A^*_1,\dots,A_d^*,$ all commuting with each other to get the RHS. | |
| Feb 17, 2018 at 8:25 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Feb 17, 2018 at 8:20 | comment | added | Pietro Majer | Are you clear with the expansion of $\big(X_1+X_2+\dots+X_d\big)^n$ in a commuting ring as I wrote? (The multinomial coefficient ${n\choose \alpha}$ is ${n!\over\alpha!}:={n!\over\alpha_1!\alpha_2!\dots\alpha_d!}$). | |
| Feb 6, 2018 at 17:27 | comment | added | Pietro Majer | If $X_j$ is the operator that takes $L$ to $A_j^*L A_j$ then e.g. $X_2X_1$ takes $L$ to $A_2^*A_1^*L A_1 A_2$, and $X_d^{\alpha_d}X_{d-1}^{\alpha_{d-1}}\dots X_2^{\alpha_2}X_1^{\alpha_1}$ takes $L$ to ${A_d^*}^{\alpha_d}\dots {A_2^*}^{\alpha_2}{A_1^*}^{\alpha_1}L A_1^{\alpha_1}A_2^{\alpha_2}\dots A_d^{\alpha_d}$ | |
| Feb 6, 2018 at 15:37 | comment | added | Student | If $X=(A_1^*LA_1,\cdots,A_d^*LA_d)$, why $X^\alpha={A^*}^\alpha LA^\alpha$? Thank you very much for your help | |
| Feb 6, 2018 at 14:35 | comment | added | Pietro Majer | 1) Because it is the operator $(X_1+\dots+X_d)^n$ applied to $L$. 2) Yes, $[d]:=\{1,2,\dots, d\}$, and $[d]^n$ is the same as $F(n,d)$ | |
| Feb 6, 2018 at 12:33 | comment | added | Student | Thank you for your answer. But I don't understand why $$\sum_{f\in [d]^n} A_{f}^*LA_{f}=\displaystyle\sum_{|\alpha|=n}{n\choose \alpha}\ {A^*}^{\alpha}LA^{\alpha}\ .$$ Also what do you mean by $ [d]^n$? Thank you for your help. | |
| Feb 6, 2018 at 11:25 | vote | accept | Student | ||
| Feb 6, 2018 at 7:47 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Feb 6, 2018 at 7:41 | history | answered | Pietro Majer | CC BY-SA 3.0 |