Do creation operators for different spins commute?

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SUMMARY

The discussion confirms that fermion creation and annihilation operators for different spins do not commute; instead, they anticommute. Specifically, the relation c_{k_1,\uparrow}^{\dagger} c_{k_2,\downarrow}^{\dagger} = -c_{k_2,\downarrow}^{\dagger} c_{k_1,\uparrow}^{\dagger} holds true due to the properties of fermionic operators. The key takeaway is that all commutation relations from bosonic theories are applicable to fermions when substituting commutators with anticommutators.

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  • Familiarity with quantum mechanics and Hamiltonians
  • Knowledge of commutation and anticommutation relations
  • Basic grasp of spin statistics in quantum theory
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  • Study the implications of anticommutation relations in quantum field theory
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  • Learn about the mathematical framework of Hamiltonians in quantum mechanics
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Physicists, quantum mechanics students, and researchers in quantum field theory who are working with fermionic systems and need to understand operator behavior in relation to spin.

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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!
 

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