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thisisphysics

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- Homework Statement
- Hey! So I'm reading an intro to QFT book this summer for self-enrichment, it's not for a class., So, the author goes through the Hubbard model a bit, but I'm confused about one of the steps of the derivation, and I'm sure I'm missing something super simple, so any help would be appreciated. This is the derivation: https://imgur.com/8OWuKvX.

So, my question is: what happens to the 1/2? I understand that U=V_iiii, and all the other elements of V are 0, so the sum collapses. The author is simply substituting the number operators in, but where would an extra factor of 2 come in to cancel the 1/2?

- Relevant Equations
- For those that can't open the link, the author essentially writes that the potential energy term of the Hamiltonian is equal to :

$$ \frac{1}{2} \sum_{ijkl\sigma\sigma'} c_{i\sigma}^{\dagger} c_{j\sigma'}^{\dagger} V_{ijkl} c_{k\sigma'}c_{l\sigma} $$ Then, we assume that the potential energy is constant and significant only when electrons are at the same site i.e. $$ U = V_{iiii}$$ So, finally, the potential energy term simplifies to $$U \sum_{i} n_{i, spin up} n_{i, spin down}$$

So, what happens to the factor of 1/2?

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