Do creation operators for different spins commute?

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In a Hamiltonian involving fermion creation and annihilation operators, the question arises whether operators for different spins commute. The discussion confirms that these operators do not commute; instead, they anticommute. Specifically, the relation c_{k_1,\uparrow}^{\dagger} c_{k_2,\downarrow}^{\dagger} equals -c_{k_2,\downarrow}^{\dagger} c_{k_1,\uparrow}^{\dagger}. This behavior is consistent with the properties of fermions, where all commutation relations from bosonic theory are replaced by anticommutators. The conclusion reassures that there are no additional subtleties regarding spin in this context.
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Say I have a hamiltonian with fermion creation / annihilation operators like this:

\sum_{k_1,k_2,k_3,k_4} c_{k_1,\uparrow}^{\dagger} c_{k_2,\downarrow}^{\dagger} c_{k_3,\downarrow} c_{k_4,\uparrow}

where the k's are momenta and the arrows indicate spin up / spin down. Can I commute operators for different spins? That is, does

c_{k_1,\uparrow}^{\dagger} c_{k_2,\downarrow}^{\dagger} = c_{k_2,\downarrow}^{\dagger} c_{k_1,\uparrow}^{\dagger}

Or do I pick up a minus sign as usual? Thanks!
 
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The operators anticommute. Specifically all of the commutation relations of the bosonic theory carry over to fermions provided that we replace commutators by anticommutators.
 
Yay! That's what I'd hoped for, but was afraid there might be some subtlety with spin that I wasn't picking up on. Thanks!
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA

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