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See, I have looked through a bunch of scripts about second quantization on the internet, but everywhere at some point something weird is happening so I get stuck over and over and over again, which is a little bit depressing. So here is the thing: Let‘s assume some single particle operator $\mathcal{O}^{(1)}$. Now taken it is separable acting on a N-particle Hilbert space one can represent the operator by

$$\mathcal{O}^{(1)}=\sum_{i}\mathcal{o}_{i}=\sum_{\alpha,\beta,i}\langle{\alpha}|\mathcal{o}_{i}|{\beta}\rangle|{\alpha}\rangle\langle{\beta}|.$$ Now $|\alpha\rangle={a^{\dagger}}_{\alpha}|0\rangle$ so I could write

$$\mathcal{O}^{(1)}=\sum_{\alpha,\beta,i}\langle{\alpha}|\mathcal{o}_{i}|{\beta}\rangle{a^{\dagger}}_{\alpha}|{0}\rangle\langle{0}|a_{\beta}$$

which in the result should be equal to $$\sum_{\alpha,\beta,i}\langle\alpha|\mathcal{o}_{i}|\beta\rangle a^{\dagger}_{\alpha}a_{\beta}.$$

But I can‘t believe that these two expressions are equal... Can someone please explain to me how I get to the operator expression of second quantization or at least recommend some website or something where it is explained really well ? Thank you in advance!

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  • $\begingroup$ Yeah, I have also looked at this. But here I dont understand the following step: $\langle \Phi‘|\mathcal{O}|\Phi\rangle=\int \mathrm{d^Nx}\,\Phi^*‘(x_1...x_N)\sum_\limits{i=1}^\limits{\infty}\mathcal{o}_{i}(x_i)\Phi(x_1...x_N)=\sum_\limits{\alpha}o_{\alpha}n_{\alpha}\int\mathrm{d^Nx}\Phi^*‘(x_1...x_N)\Phi(x_1...x_N)=\delta_{\Phi‘\Phi}\sum_\limits{\alpha}o_{\alpha}n_{\alpha}$. How does the sum of operators act on $\langle x|\Phi\rangle $? How do i get the $n_{\alpha}$ and the sum over $\alpha$ instead of i? $\endgroup$
    – Lighter
    Commented Mar 7, 2020 at 9:09

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Okay, so I cleaned up the latex a bit (Use \langle and \rangle for the braket notation. Though I wish they'd install the braket package, which would make it even easier).

First, you have a mistake in the first line, and it doesn't quite make sense when going to second quantization anyway: There should be no sum over individual particles any more! In your first line, what sort of basis do the states $|\alpha\rangle$ even refer to? If they are single particle basis states of the operator, we can't use them like that in the many-particle operator.

And that's probably why you run into issues. I'd skip that step entirely. In "first" quantization, we have $\mathcal{O}^{(1)} = \sum_i o_i$ and in second quantization we have

$$\mathcal{O}^{(1)} = \sum_{\alpha, \beta} \langle \alpha | o | \beta \rangle a^\dagger_\alpha a_\beta$$

To prove that they are indeed equivalent, we have to show that they have the same matrix elements in a given basis. We can pick any basis we want, so of course we pick the appropriate many-particle basis of, e.g., Slater determinants.

Some info on this can be found eg here: http://physics.gu.se/~tfkhj/OsloSecondQuant.pdf

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  • $\begingroup$ Okay I pushed the wrong add comment button, look at the comment on the question please:) I dont understand the action of the operator in (78) of the script... $\endgroup$
    – Lighter
    Commented Mar 7, 2020 at 11:32
  • $\begingroup$ Okay so the first step is just inserting the definitions. For the second step, insert the complete set of eigenfunctions of the $\hat o_i(x_i)$, then use the appropriate orthogonality relations. $\endgroup$
    – Lagerbaer
    Commented Mar 9, 2020 at 3:52

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