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I am self-studying quantum field theory and have gotten to creation and annihilation operators, respectively denoted $A^\dagger$ and $A$. Conceptually I understand what these objects are, at least on a surface level, which is they increase/decrease the number of particles present by one. In terms of maps: $$A_n: H_s^{\otimes n} \rightarrow H_s^{\otimes (n-1)}\\ A_n^\dagger: H_s^{\otimes n} \rightarrow H_s^{\otimes (n+1)} $$ where $H_s^{\otimes n}$ is the $n$-fold symmetric tensor product of a Hilbert space $H$.

The formula for these operators are: $$A_n(\xi)(\beta)_{i_1,~\ldots,~ i_{n-1}} \equiv \sqrt{n}\sum_{i=1}^\infty \xi_i^* \beta_{i_1,~\ldots,~i_{n-1}, i} \quad \quad \xi \in H, \beta \in H_s^{\otimes n}\\ A_n^\dagger(\eta)(\alpha)_{i_1,~\ldots,~ i_{n+1}} \equiv \frac{1}{\sqrt{n+1}}\sum_{\ell=1}^{n+1} \eta_{i_\ell} \alpha_{i_1,~\ldots,~\hat i_{\ell}, \ldots,~ i_{n+1}} \quad \eta \in H, \alpha \in H_s^{\otimes n}.$$

I have been having trouble breaking down the above formulas. Why is $\eta$ subscripted by $i_\ell$ and why is $\xi$ is just subscripted by $i$?

Also, $\alpha$ and $\beta$ are both $n$-tensors, so how are we taking $\alpha$ to be subscripted by $n+1$ indices in the creation operator, and what is the logic behind attaching an $i$ for $\beta$? Mathematically it seems after the summation the right hand side of both operators will still be $n$-tensors. I must be misunderstanding something with the indices here.

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    $\begingroup$ Which book/ source? $\endgroup$ Commented Sep 25, 2022 at 16:50
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    $\begingroup$ @JasonFunderberker Talagrand's QFT book, it is written for mathematicians which I have a background in as opposed to physics. $\endgroup$
    – CBBAM
    Commented Sep 25, 2022 at 16:51

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  • $\eta$ is subscripted by $i_\ell$ because the indices $i_1,...,i_{n+1}$ are given on the left hand side and we are summing over their sub indices

  • Talagrand gives on p. 79 the example $$\tag{A} A^\dagger_2(\eta)(\alpha)_{i_1i_2i_3}=\frac{1}{\sqrt{3}}(\eta_{i_1}\alpha_{i_2i_3}+ \eta_{i_2}\alpha_{i_1i_3}+\eta_{i_3}\alpha_{i_1i_2}) $$ in which $\ell$ runs through $1,2,3$.

  • To get out of that subindex business perhaps it is helpful to write (A) equivalently as $$\tag{A'} A^\dagger_2(\eta)(\alpha)_{\mu\nu\rho}=\frac{1}{\sqrt{3}}(\eta_{\mu}\alpha_{\nu\rho}+ \eta_{\nu}\alpha_{\mu\rho}+\eta_{\rho}\alpha_{\mu\nu})\,. $$

  • $\alpha,\beta$ are both $n$-tensors. Note hovewer that the $\hat{i_\ell}$ is omitted. We subscript $\alpha$ always only with $n$ indices. See again (A) where $\alpha$ is a $2$-tensor.

  • To $\beta$ we are not attaching an index $i$. We are contracting the $n$-tensor with a vector which results in an $(n-1)$-tensor -as it must because one particle gets destroyed.

  • The $i$ is just a dummy index which for which one could use any symbol that was not used yet. It would be clumsy to use a subscript for $i$.

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    $\begingroup$ Thank you for your reply. So the subindices in $\eta$ is simply to agree with the summation? Because we see $i_1, \ldots, i_{n+1}$ in $\xi$ as well, but in that case we simply have $\xi_i$ (i.e. there is no subindex). Regarding his example on p. 79, on the left hand side he has $i_1, i_2, i_3$, and comparing to the definition of $A^\dagger$, this would imply that $n = 2$. Hence, $\alpha \in H^{\otimes 2}$, so how can we even speak of a third index $i_3$ for $\alpha$? $\endgroup$
    – CBBAM
    Commented Sep 25, 2022 at 17:38
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    $\begingroup$ For $\beta$, what do you mean by contracting the $n$-tensor with a vector? I have not seen this terminology before. $\endgroup$
    – CBBAM
    Commented Sep 25, 2022 at 17:40
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    $\begingroup$ Lots of questions in very short time. See edit of a way avoiding subindices. Where in $\xi$ do you see $i_1,\dots, i_{n+1}$ "as well". $\xi$ is a vector that carries a single index. when you sum over it as we do in the contraction it does not matter how you name that dummy index. Contraction is such a basic business in tensor calculus that I can expect you to google it or learn about it in other books. For example Spacetime and Geometry by Sean Carroll. $\endgroup$
    – Kurt G.
    Commented Sep 25, 2022 at 17:48
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    $\begingroup$ Finally. To $\alpha$ in (A) and (A') we attach always only two indices. I said that before. $\endgroup$
    – Kurt G.
    Commented Sep 25, 2022 at 17:50
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    $\begingroup$ It is correct. In the toy example (A) $\alpha$ is nothing else than a matrix and so carries two indices each of which can loop through at least $n+1$ values. In fact, in QFT these matrices allow loops through $\mathbb N$. I recommend again the Sean Carroll book. Embarking on a self-study of QFT is admirable but also means (with Seven Weinberg) to "go where the mess is". $\endgroup$
    – Kurt G.
    Commented Sep 26, 2022 at 5:41

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