Factor of 1/2 in Hubbard Hamiltonian?

AI Thread Summary
The 1/2 factor in the Hubbard Hamiltonian addresses the issue of double counting states in quantum mechanics. This arises from the anticommutation relations of fermionic operators, which lead to redundancy in the summation of terms. Specifically, the relevant terms include combinations of creation and annihilation operators that yield the same physical state. The 1/2 factor effectively corrects for this redundancy, ensuring accurate calculations. Overall, this adjustment is crucial for maintaining the integrity of the Hamiltonian's representation.
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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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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.
 
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SisypheanZealot said:
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.
 
No problem
 
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