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I am a past physics student and wanted to revise the rudiments of many body theory, in particular as related to materials physics.

I have a doubt about the definition of creation and annihilation operators. Let's call them $C^{+}_{\lambda}$ and $C_{\lambda}$.

We consider the fermionic (eletronic) case, where the creation operator creates an electron with wavefunction $\psi_{\lambda}(x)$. We have in mind an Hamiltonian $H$ with interactions.

Now, calling $S(*,..,*)$ the Slater determinant operation, I think that apart from normalization we know that $C^{+}_{\lambda}$ acts something like:

$$C_{\lambda}^{+} S(\psi_{\lambda_1},...,\psi_{\lambda_N}) \sim S(\psi_{\lambda},\psi_{\lambda_1},...,\psi_{\lambda_N}) \tag{1}$$

(depending on where we insert $\psi_{\lambda}$ we would have a different sign).

Now my question is:

  • Suppose that we want to understand how $C^{+}_{\lambda}$ acts on a generic antisymmetric funtion $f(x_1,..,x_n)$ that is not provided as a slater determinant (e.g. the ground state wavefunction of $H$ should generally fall in this case). I guess the way to go would be to expand:

$$f(x_1,..,x_n)=\sum_{\lambda_1,..,\lambda_n} \alpha_{\lambda_1,..,\lambda_n} S(\psi_{\lambda_1},...,\psi_{\lambda_n})$$

and proceed by linearity:

$$C_{\lambda}^{+} f(x_1,..,x_n)=\sum_{\lambda_1,..,\lambda_n} \alpha_{\lambda_1,..,\lambda_n} S(\psi_{\lambda,}\psi_{\lambda_1},...,\psi_{\lambda_n}) \tag{2}$$

But the wavefunction $f$ does not know about the wavefunctions $\psi$, so that we could also have expanded in an other single particle bases:

$$f(x_1,..,x_n)=\sum_{\mu_1,..,\mu_n} \beta_{\mu_1,..,\mu_n} S(\phi_{\mu_1},...,\phi_{\mu_n})$$

and definining:

$$C_{\lambda}^{+} f(x_1,..,x_n)=\sum_{\mu_1,..,\mu_n} \beta_{\mu_1,..,\mu_n} S(\psi_{\lambda,}\phi_{\mu_1},...,\phi_{\mu_n}) \tag{3}$$

and this result looks different so that we would choose the first definition (Eq. (2)). Expanding in another bases seems to lead to a different result. Any relevant mistake up to now?

How do I have to interpret the fact that we need to expand $f$ with the same single particle bases of $C_{\lambda}$ in order to evaluate the operator? Maybe I have to consider as if in the same wavefunction there are "hidden" many single particle states, according to the representation that I use, and that $C^+_{\lambda}$ is "probing" somehow the "hidden" single particle states of a certain type? (the onces associated with the $\lambda$ quantum numbers). Does this interpretation makes sense?

Maybe once upon the time I knew the answer to these doubts, sorry if the question is too trivial but wanted to know if I am missing something important...

EDIT: actually now looking at the formulas maybe it is not impossible that the results of (2) and (3) are equal. Maybe the Slater determinant changes so that the coefficients balance... I will try to check and will update the question if I find something more convincing in one direction or the other... I guess that according to the result than the physical interpretation of the formulas may differ...

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  • $\begingroup$ So I haven't worked this through in detail, but I would expect the equality of these expressions to follow from the fact that determinants are linear in the columns of their argument $\endgroup$ Jan 30, 2023 at 12:40
  • $\begingroup$ @BySymmetry I think you are right. I started doing the calculation but I need some time to fix all the indexes. I will try to finalize and write an answer as an exercise. Right now I think that the result should hold. For me this property is very interesting even if maybe some years ago I would have found it trivial :D. $\endgroup$
    – Thomas
    Jan 30, 2023 at 13:30
  • $\begingroup$ @BySymmetry I tried to sketch a calculation in an answer. What do you think? $\endgroup$
    – Thomas
    Jan 30, 2023 at 14:39

2 Answers 2

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If the $\psi_n(x)$ are an orthonormal c-number basis set then we can expand the Fermi field $\hat \Psi(x)$ as $$ \hat \Psi(x)= \sum_n \hat a_n \psi_n(x) $$ and find that the field anti-commutation relation $ \{ \hat \Psi(x),\hat \Psi^\dagger(y)\}= \delta(x-y)$ leads to $\{\hat a_n, \hat a^\dagger_m\}= \delta_{nm}$. If you choose a different basis set, the field $\hat \Psi(x)$ is unchanged, but the $\hat a_n$ change to new set $\hat a'_n$ so that $$ \hat \Psi(x)= \sum_a \hat a'_n \psi'_n(x) $$ Then
$$ \hat a'_n= O_{nm} \hat a_m $$ where $O_{nm}$ is an (infinite) orthogonal matrix. The orthogonal property $O^TO={\mathbb I}$ shows that $$ \{\hat a'_n, \hat a'^\dagger_m\}= \delta_{nm} $$ still holds.

The connection to the Slater determinant picture is that the Slater determiant function is given $$ S[x_1,x_2,x_3,\ldots x_N]= \langle 0|\hat \Psi(x_1)\hat \Psi(x_2)\hat \Psi(x_3)\ldots\hat \Psi(x_N) |S\rangle $$ where $|S\rangle$ is the corresponding abstract Fock-space state.

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  • $\begingroup$ Thanks. Give me some time to think more about the link with my question and answer, I am reviewing now these concepts and need some time to get again into it... $\endgroup$
    – Thomas
    Jan 30, 2023 at 14:33
  • $\begingroup$ I have at the moment issues in linking your answer to my question since in QFT the Fermi field is postulated and the creation/construction operators follow without reference to an explicit form of the underlying Hillbert space whereas in many body theory the hilbert space is the functional one a la Schrodinger and the Fermi field and creation operators follow by explicit definition $\endgroup$
    – Thomas
    Jan 30, 2023 at 17:02
  • $\begingroup$ I am following the second construction and I think you are using the first one. Since they should be equivalent I should be able to translate your answer in my terms but I am not able to do it right now. Could you please link more explicitely your answer to my question? $\endgroup$
    – Thomas
    Jan 30, 2023 at 17:04
  • $\begingroup$ I added some text to relat the Slater determinsnt with the abstract Fock-space picture. $\endgroup$
    – mike stone
    Jan 30, 2023 at 17:55
  • $\begingroup$ Thans for the inputs! I will work ok them. Ah if you find some mistake in the answer I posted please let me know... $\endgroup$
    – Thomas
    Jan 30, 2023 at 19:43
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I think the formulas (2) and (3) of the original post are equivalent as suggested in the comments.

I change slightly the notation for the indexes and use Einstein notation. If we define the matrix $O$ so that:

$$\phi_i=O_{i,j} \psi_j \tag{1}$$

than we have for the Slater determinants:

$$S(\phi_{i_1},..,\phi_{i_n})=O_{i_1j_1}...O_{i_nj_n}S(\psi_{j_1},..,\psi_{j_n}) \tag{2}$$

Using this we can show that if $f(x_1,..,x_n)=\alpha_{i_1,..i_k}S(\phi_{i_1},..,\phi_{i_n})=\beta_{j_1,..j_n}S(\psi_{j_1},..,\psi_{j_n})$, than:

$$\beta_{j_1,..j_n}=\alpha_{i_1,..i_k}O_{i_1j_1}...O_{i_nj_n}\tag{3}$$

Now we start from the definition of the construction operator associated to a wafunction $\rho$:

$$C_\rho^+ f \equiv\beta_{j_1,..j_n} S(\rho,\psi_{j_1},..,\psi_{j_n}) \tag{4}$$

We have for the Slater determinants: $$S(\rho,\psi_{j_1},..,\psi_{j_n})=O^{-1}_{j_1i_1}...O^{-1}_{j_ni_n}S(\rho,\phi_{i_1},..,\phi_{i_n}) \tag{5}$$

Inserting in [4] the relations [5] and [3] we obtain:

$$C_\rho^+ f = \alpha_{i_1,..i_n} S(\rho,\phi_{i_1},..,\phi_{i_n})$$

so that the expression/form of the construction operator is independent of the bases chosen to expand the function $f$.

NB1: some prefactors are missing probably but I hope the arguments are not affected.

NB2: in theory this works only for fermions. But probably the multilinearity used here is valid also for bosons?

UPDATE RELATING TO ANSWER OF @mike stone:

We try to use the relation between construction operators and see if we get a similar picture. Here I am not using Einstein notation. Let's call $a_i^+$ the construction operators for the $\phi$ functions. Than for the function $\rho$ we have:

$$C^+_{\rho}=\sum_i \langle \phi_i|\rho \rangle a^+_i$$

Further we have the Slater determinant expansion for $f$ (NB: we have to restrict the indices otherwise the determinants are not indepdendent. I will not be careful with this issue in the following, but I guess it can be fixed/formalized with some effort):

$$f=\sum_{i_1,..,i_n} \alpha_{i_1,..,i_n}a_{i_1}^+...a_{i_n}^+|0\rangle$$

This means that:

$$C^+_{\rho} f=\sum_{i,i_1,..,i_n} \langle \phi_i|\rho \rangle \alpha_{i_1,..,i_n} a^+_i a_{i_1}^+...a_{i_n}^+|0\rangle$$

Going back to Slater:

$$C^+_{\rho} f=\sum_{i,i_1,..,i_n} \langle \phi_i|\rho \rangle \alpha_{i_1,..,i_n} S(\phi_i,\phi_{i_1},..\phi_{i_n})$$

and by linearity:

$$C^+_{\rho} f=\sum_{i_1,..,i_n} \alpha_{i_1,..,i_n} S(\sum_i \langle \phi_i|\rho \rangle \phi_i,\phi_{i_1},..\phi_{i_n})=\sum_{i_1,..,i_n} \alpha_{i_1,..,i_n} S(\rho,\phi_{i_1},..\phi_{i_n}) \tag{6}$$

Formula [6] generalizes the defininition of creation operator often found in books:

$$a_i^+ S(\phi_{i_1},..\phi_{i_n})=S(\phi_i,\phi_{i_1},..\phi_{i_n})$$

[6] is valid for every bases set $\phi$, and independently of whether the wavefunction $\rho$ belongs to this set or not. In particular we have that when $f$ is a Slater determinant:

$$C_\rho^+ S(\phi_{i_1},..\phi_{i_n})=S(\rho,\phi_{i_1},..\phi_{i_n}) \tag{7}$$

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