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In second quantization, let $c_{\alpha}^\dagger$ and $c_\alpha$ be the creation/annihilation operators for a fermionic particle in a quantum state $|\alpha\rangle$. Define $c_\beta^\dagger, c_\beta$ analogously for a state $|\beta\rangle$ with $\beta\ne \alpha$. In Fock space, a state containing one $\alpha$ particle and one $\beta$ particle is $$c_\alpha^\dagger c_\beta^\dagger|\Omega\rangle\sim|1_\alpha,1_\beta,0,...\rangle$$ where $|\Omega\rangle$ is the empty state and ignoring the various constants.

I have a small doubt. Is the product of these operators the usual dot product of matrices, or is it actually a tensor product $c_\alpha^\dagger\otimes c_\beta^\dagger$? After all, these operators belong to different spaces, even if they can be written as the same matrix once a basis is fixed. However, I've never seen this made explicit. Can anyone clear this up for me?

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    $\begingroup$ They are all operators in the same Hilbert space (the Fock space), so it is just ordinary product. $\endgroup$
    – Meng Cheng
    Sep 4, 2022 at 22:29
  • $\begingroup$ Related: physics.stackexchange.com/q/54896/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Sep 5, 2022 at 6:27

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If these were bosonic operators $b_{\alpha}^{\dagger}$, you could rightly regard the product $b_{\alpha}^{\dagger} b_{\beta}^{\dagger}$ as a tensor product. This is because you can think of the bosonic many-body Hilbert space as the tensor product space of many harmonic oscillators, one for each $\alpha$. As a result, you can rightfully think of these operators as being defined by $$ b_{\alpha}^{\dagger} = \mathbb{1} \otimes \ldots \otimes \mathbb{1} \otimes b^{\dagger} \otimes \mathbb{1} \otimes \ldots \otimes \mathbb{1} $$ where the $b^{\dagger}$ appears in the $\alpha$th place; ie, it acts on the $\alpha$th harmonic oscillator mode. As a result, $b_{\alpha}^{\dagger} b_{\beta}^{\dagger}$ for $\alpha \neq \beta$ can be reasonably regarded as either an ordinary product or a tensor product, depending on how you interpret the notation.

For fermions on the other hand, you know that we cannot have the same simple tensor product structure because $c^{\dagger}_{\alpha}$ and $c^{\dagger}_{\beta}$ for $\alpha \neq \beta$ do not commute! Instead, there is a tricky phase factor which makes the operators not quite tensor products. One way to realize the fermion operators explicitly is to start with spin-half operators (ie, Pauli matrices) $\sigma^z$ and $\sigma^{\pm} = \frac{1}{2}(\sigma^x \pm i \sigma^y)$, and to write $c_{\alpha}^{\dagger}$ as $$ c_{\alpha}^{\dagger} = \sigma^z \otimes \ldots \otimes \sigma^z \otimes \sigma^+ \otimes \mathbb{1} \otimes \ldots \otimes \mathbb{1} $$ The string of $\sigma^z$'s on the left is often called the Jordan-Wigner string, and it implements the anticommuting algebra of the fermions. The moral of the story is that when you see an operator $c^{\dagger}_{\alpha} c^{\dagger}_{\beta}$ for $\alpha \neq \beta$, it is not quite correct to regard the product as a tensor product.

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  • $\begingroup$ So, to get this completely straight: I've looked into JW transformations, which allow to write fermionic models such as the Ising model in terms of spin matrices: for example $$H=-J\sum_j \sigma_{j}^x\sigma_{j+1}^x-g\sum_j\sigma_j^z$$ in 1D with a field orthogonal to the coupling of spins. In this case, those products should be interpreted as tensor products, i.e. $\sigma_j^x\sigma_{j+1}^x=\sigma_j^x\otimes\sigma_{j+1}^x$, correct? $\endgroup$
    – Ghost of Ludwig Boltzmann
    Sep 5, 2022 at 11:57
  • $\begingroup$ It's probably better at the end of the day to think of them as ordinary products, and to think of each $\sigma^x_j$ as $\mathbb{1} \otimes \ldots \otimes \mathbb{1} \otimes \sigma^x \otimes \mathbb{1} \otimes \ldots \otimes \mathbb{1}$. But it's not harmful to think of it as a tensor product here, as long as you recognize that $\sigma^x_j \sigma^x_{j+1}$ really means $\mathbb{1} \otimes \ldots \otimes \sigma^x_j \otimes \sigma^x_{j+1} \otimes \ldots \otimes \mathbb{1}$. $\endgroup$
    – Zack
    Sep 5, 2022 at 14:35
  • $\begingroup$ Got it, thank you. Just one last thing: if I were to try and implement this Hamiltonian numerically in C or Python, what would be the most convenient approach? Should I just loop over the sites $j=1,...,N$ and do tensor products of $\mathbb{1} \otimes \ldots \otimes \sigma^x_j \otimes \sigma^x_{j+1} \otimes \ldots \otimes \mathbb{1}$, or are there better approach to deal with these matrices? I've never tried to do explicit calculations with many-body hamiltonians, so I'm wondering about the methods (perhaps this deserves a different question though). $\endgroup$
    – Ghost of Ludwig Boltzmann
    Sep 5, 2022 at 17:03
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    $\begingroup$ If you're trying to do exact diagonalization approaches, then this is exactly what you would do. But keep in mind that the complexity of this approach will explode rapidly: you will not be able to go further than 10-15 sites! There are sometimes better numerical approaches for certain circumstances. For example, for the Ising model you can employ free fermion numerics or matrix product state methods. $\endgroup$
    – Zack
    Sep 5, 2022 at 18:12
  • $\begingroup$ I'll look into that. Thank you, Zack. $\endgroup$
    – Ghost of Ludwig Boltzmann
    Sep 6, 2022 at 9:43

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