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In quantum mechanics we learn that an operator in a basis can be represented as $$\hat{A}=\sum\limits_{\alpha,\beta}A_{\alpha\beta}|\alpha\rangle\langle\beta|.$$ But in many-body physics we suddenly write $$\hat{A}=\sum\limits_{\alpha,\beta}A_{\alpha\beta}a_\alpha^\dagger a_\beta$$

Any idea how to reach to the second from the first. I got something this $$\hat{A}=\sum\limits_{\alpha,\beta}A_{\alpha\beta}|\alpha\rangle\langle\beta|=\sum\limits_{\alpha,\beta}A_{\alpha\beta}a_\alpha^\dagger|0\rangle\langle0| a_\beta$$ Any idea how to show this? Please help

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    $\begingroup$ These two are not related in such a simple way. Why should they, anyways? $\endgroup$ – Norbert Schuch Apr 12 at 15:04
  • $\begingroup$ I thought I can arrive at the second from the first because that is how we represent operators in QM $\endgroup$ – mithusengupta123 Apr 12 at 15:06
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    $\begingroup$ For fermions and bosons, it is not clear a priori what $|\alpha\rangle$ is supposed to be. $\endgroup$ – Norbert Schuch Apr 12 at 15:10
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    $\begingroup$ "Quantum Theory of Many-Particle Systems" by Fetter and Walecka covers the transition from first quantization to second quantization in full details, in the first chapter. As for your example, it is a matter of how you label many-body states. $\endgroup$ – wcc Apr 12 at 15:14
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You first equation contains an operator $\hat A$ acting in the single-particle Hilbert space ${\mathcal H}_1$. The second equation contains an operator $\hat A$ that acts on the many particle Fock space. The two operators are not the same therefore, although they are related in that the second is induced in a natural way from the first..

A Fock space is built built by taking sums of tensor-products of copies of the single-particle space: $$ {\mathcal H}_{\rm Fock}= \sum_{N=0}^\infty \{{\mathcal H}_1\}^{\otimes N}, $$ where $$ \{{\mathcal H}_1\}^{\otimes N}={\mathcal H}_1\otimes {\mathcal H}_1\otimes \ldots {\mathcal H}_1,\quad \hbox{N factors}. $$ The tensor product should be symmetrised or antsymmetrised for Bosons or Fermions, respectively. If this sounds mysterious, the suggestion by IamAStudent is a good one.

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