# reaching from $\hat{A}=A_{\alpha\beta}|\alpha\rangle\langle\beta|$ to $\hat{A}=A_{\alpha\beta}a_\alpha^\dagger a_\beta$

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

• These two are not related in such a simple way. Why should they, anyways? – Norbert Schuch Apr 12 '19 at 15:04
• I thought I can arrive at the second from the first because that is how we represent operators in QM – mithusengupta123 Apr 12 '19 at 15:06
• For fermions and bosons, it is not clear a priori what $|\alpha\rangle$ is supposed to be. – Norbert Schuch Apr 12 '19 at 15:10
• "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. – wcc Apr 12 '19 at 15:14

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.