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...

  • $\begingroup$ 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


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.

  • $\begingroup$ 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

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:


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?


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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