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example hamiltonian was not hermitian
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Ruben Verresen
Ruben Verresen
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Consider a closed system of fermions. Any reasonable fermionic Hamiltonian you might think of will always have an even number of fermionic operators in each term. In other words: $[H,P] = 0$ where $P=(-1)^{N_f}$. My question is: why is this? I am not looking for a flowery description in terms of 'fermions are ends of strings and hence can only be created in pairs'. My question is more mathematical in nature: is there some kind of inconsistency in considering a closed system of fermions with e.g. $H = \sum c_i^\dagger$$H = \sum c_i^\dagger + c_i$ ? Indeed it seems that physically we should see fermionic parity symmetry as a gauge symmetry. But is there a mathematical argument to see that we have to see it as a gauge symmetry to keep things consistent?

It would be enough to argue that any fermionic wavefunction has to have a well-defined fermionic parity. (Indeed: if $[H,P] \neq 0$ then time evolution would take a state with a well-defined parity into one without well-defined parity.) But again, I have no good way of arguing this.

Perhaps the resolution is simply pragmatic: we haven't figured out a way of engineering an interaction term which is not bosonic in nature. Or is there a fundamental reason we can't have such an interaction?

Consider a closed system of fermions. Any reasonable fermionic Hamiltonian you might think of will always have an even number of fermionic operators in each term. In other words: $[H,P] = 0$ where $P=(-1)^{N_f}$. My question is: why is this? I am not looking for a flowery description in terms of 'fermions are ends of strings and hence can only be created in pairs'. My question is more mathematical in nature: is there some kind of inconsistency in considering a closed system of fermions with e.g. $H = \sum c_i^\dagger$ ? Indeed it seems that physically we should see fermionic parity symmetry as a gauge symmetry. But is there a mathematical argument to see that we have to see it as a gauge symmetry to keep things consistent?

It would be enough to argue that any fermionic wavefunction has to have a well-defined fermionic parity. (Indeed: if $[H,P] \neq 0$ then time evolution would take a state with a well-defined parity into one without well-defined parity.) But again, I have no good way of arguing this.

Perhaps the resolution is simply pragmatic: we haven't figured out a way of engineering an interaction term which is not bosonic in nature. Or is there a fundamental reason we can't have such an interaction?

Consider a closed system of fermions. Any reasonable fermionic Hamiltonian you might think of will always have an even number of fermionic operators in each term. In other words: $[H,P] = 0$ where $P=(-1)^{N_f}$. My question is: why is this? I am not looking for a flowery description in terms of 'fermions are ends of strings and hence can only be created in pairs'. My question is more mathematical in nature: is there some kind of inconsistency in considering a closed system of fermions with e.g. $H = \sum c_i^\dagger + c_i$ ? Indeed it seems that physically we should see fermionic parity symmetry as a gauge symmetry. But is there a mathematical argument to see that we have to see it as a gauge symmetry to keep things consistent?

It would be enough to argue that any fermionic wavefunction has to have a well-defined fermionic parity. (Indeed: if $[H,P] \neq 0$ then time evolution would take a state with a well-defined parity into one without well-defined parity.) But again, I have no good way of arguing this.

Perhaps the resolution is simply pragmatic: we haven't figured out a way of engineering an interaction term which is not bosonic in nature. Or is there a fundamental reason we can't have such an interaction?

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Ruben Verresen
Ruben Verresen
  • 9.3k
  • 49
  • 67

Why does a fermionic Hamiltonian always obey fermionic parity symmetry?

Consider a closed system of fermions. Any reasonable fermionic Hamiltonian you might think of will always have an even number of fermionic operators in each term. In other words: $[H,P] = 0$ where $P=(-1)^{N_f}$. My question is: why is this? I am not looking for a flowery description in terms of 'fermions are ends of strings and hence can only be created in pairs'. My question is more mathematical in nature: is there some kind of inconsistency in considering a closed system of fermions with e.g. $H = \sum c_i^\dagger$ ? Indeed it seems that physically we should see fermionic parity symmetry as a gauge symmetry. But is there a mathematical argument to see that we have to see it as a gauge symmetry to keep things consistent?

It would be enough to argue that any fermionic wavefunction has to have a well-defined fermionic parity. (Indeed: if $[H,P] \neq 0$ then time evolution would take a state with a well-defined parity into one without well-defined parity.) But again, I have no good way of arguing this.

Perhaps the resolution is simply pragmatic: we haven't figured out a way of engineering an interaction term which is not bosonic in nature. Or is there a fundamental reason we can't have such an interaction?