# Factor of 1/2 in Hubbard Hamiltonian?

#### thisisphysics

Problem 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?
Above

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#### SisypheanZealot

So, in the majority of cases with Hamiltonians like this the 1/2 term is to deal with the double counting of states. Look at the original summation term $\sum_{ijkl\sigma\sigma'}$. Based on the anticommutation relations $$\lbrace c^{\dagger}_{i},c^{\dagger}_{j}\rbrace=0$$ $$\lbrace c_{i},c_{j}\rbrace=0$$ and $$\lbrace c_{i},c^{\dagger}_{j}\rbrace=\delta_{ij}$$ We see that the relevant terms that remain are $$c^{\dagger}_{i\uparrow}c^{\dagger}_{i\downarrow}c_{i\downarrow}c_{i\uparrow}+c^{\dagger}_{i\downarrow}c^{\dagger}_{i\uparrow}c_{i\uparrow}c_{i\downarrow}$$ Looking back at the anticommutations we see that this becomes $$c^{\dagger}_{i\uparrow}c^{\dagger}_{i\downarrow}c_{i\downarrow}c_{i\uparrow}+c^{\dagger}_{i\uparrow}c^{\dagger}_{i\downarrow}c_{i\downarrow}c_{i\uparrow}$$ So, the states are double counted and as stated the 1/2 takes care of that.

#### thisisphysics

So, in the majority of cases with Hamiltonians like this the 1/2 term is to deal with the double counting of states. Look at the original summation term $\sum_{ijkl\sigma\sigma'}$. Based on the anticommutation relations $$\lbrace c^{\dagger}_{i},c^{\dagger}_{j}\rbrace=0$$ $$\lbrace c_{i},c_{j}\rbrace=0$$ and $$\lbrace c_{i},c^{\dagger}_{j}\rbrace=\delta_{ij}$$ We see that the relevant terms that remain are $$c^{\dagger}_{i\uparrow}c^{\dagger}_{i\downarrow}c_{i\downarrow}c_{i\uparrow}+c^{\dagger}_{i\downarrow}c^{\dagger}_{i\uparrow}c_{i\uparrow}c_{i\downarrow}$$ Looking back at the anticommutations we see that this becomes $$c^{\dagger}_{i\uparrow}c^{\dagger}_{i\downarrow}c_{i\downarrow}c_{i\uparrow}+c^{\dagger}_{i\uparrow}c^{\dagger}_{i\downarrow}c_{i\downarrow}c_{i\uparrow}$$ So, the states are double counted and as stated the 1/2 takes care of that.
Thank you! This makes perfect sense.

#### SisypheanZealot

No problem

"Factor of 1/2 in Hubbard Hamiltonian?"

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