If I have a system of $N$ non-interacting fermions, I can write the wave function of the ground-state of the system using a Slater determinant

$$ \Phi_{0}(\textbf{r}_{1}, ..., \textbf{r}_{N}) = \frac{1}{\sqrt{N!}} \begin{vmatrix} \phi_{k_{1}}(\textbf{r}_{1}) & \cdots & \phi_{k_{N}}(\textbf{r}_{1})\\ \vdots & \ddots & \vdots \\ \phi_{k_{1}}(\textbf{r}_{N}) & \cdots & \phi_{k_{N}}(\textbf{r}_{N}) \end{vmatrix}, $$

where $\phi_{k}$ is the single-particle wave function of one fermion. I can define operators $a_{k}$ and $a^{\dagger}_{k}$ that create and annihilate fermions in the state $\phi_{k}$ when they act in the state of the non-interacting system.

But, if now I am interested in the state of N interacting fermions, I can write the total wave function as a linear combination of all possible Slater determinants

$$ \Psi_{0}(\textbf{r}_{1}, ..., \textbf{r}_{N}) = \sum_{n} C_{n} \Phi_{n} (\textbf{r}_{1}, ..., \textbf{r}_{N}).$$

I can interpret $a^{\dagger}\Phi_{0}$ as the state of a system of $N$ fermions with an add fermion in state $\phi_{k}$. My question is: Can I do the same interpretation when I apply one of the creation or annihilation operators to $\Psi$? Can I say that $a^{\dagger}_{k}\Psi_{0}$ is a state of $N$ interacting fermions with an add fermion in state $\phi_{k}$?


1 Answer 1


In a strictly non-relativistic model, one in which the canonical annihilation operators actually do annihilate the true vacuum state, the answer is yes: Each application of a canonical creation operator adds another particle to the state, and this is true with or without interactions.

Since we're talking about creation/annihilation operators, a QFT formulation is convenient. In a typical strictly non-relativistic model, the canonical number operator commutes with all observables, even if the Hamiltonian includes interaction terms, like this one: $$ H\sim \int d^3x\ \psi^\dagger(x)\frac{-\nabla^2}{2m}\psi(x) + \int d^3x\, d^3y\ \psi^\dagger(x)\psi^\dagger(y)V(x-y)\psi(y)\psi(x) $$ with field operators $\psi(x)$ satisfying $\{\psi(x),\psi^\dagger(y)\}\sim\delta^3(x-y)$. The canonical number operator in this case is $\int d^3x\ \psi^\dagger(x)\psi(x)$.

For each $N$, states obtained from the vacuum by applying a product of $N$ creation operators $\psi^\dagger$ are in different superselection sectors: different values of $N$ are not connected to each other by any observables. (This is a stronger statement than merely saying that the number of particles is conserved, which only requires that it commutes with the Hamiltonian.) In this sense, a strictly non-relativistic QFT such as the one illustrated above is like a collection of separate theories, one for each possible number $N$ of particles. The canonical creation/annihilation operators take you back and forth between those separate theories.

Again, this is for strictly non-relativistic models. Relativistic models are a different story.


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