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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. $H = \sum c_i^\dagger + c_i$ ? 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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    $\begingroup$ I don't have a general mathematical answer, but I think a Hamiltonian like (c_i^\dagger + c_i) -with the second term added to make it Hermitian- violates it being a closed system as they would inject particles into, or remove them from, the system $\endgroup$
    – RGWinston
    Mar 16 '17 at 21:34
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    $\begingroup$ @RGWinston With that same reasoning you would also call the effective Hamiltonian for a superconductor ($H = \cdots + c^\dagger c^\dagger + \cdots$) an open system. So my question is, why do even effective fermionic Hamiltonians for closed systems have $P$ symmetry? $\endgroup$ Mar 16 '17 at 21:37
  • $\begingroup$ Hmm... I think we do have to think of superconductors in the effective picture as an open systems, because they don't have a conserved particle number. The BCS wavefunction is a superposition of all possible different even particle numbers. $\endgroup$
    – RGWinston
    Mar 16 '17 at 21:49
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    $\begingroup$ Great question, which I've wondered myself and never gotten a good answer two. Someone once argued to me that condensed-matter effective Hamiltonian degrees of freedom are collective excitations of actual Standard Model electrons, and the Standard Model Hamiltonian doesn't contain any terms that violate fermion parity so the effective Hamiltonians can't either. I'm not sure I buy that argument though. $\endgroup$
    – tparker
    May 25 '17 at 3:07
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    $\begingroup$ @tparker This would still allow for spontaneous symmetry breaking. I mean, there are many effective models we know and love which have less symmetries than our microscopic standard model. $\endgroup$ May 25 '17 at 23:52
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I am not sure if this is a complete answer, but it might help others. It suggests that the fundamental reason is locality.

The question we want to address is: why am I not ''allowed'' to write down a term in the Hamiltonian that is odd in the number of fermionic operators, e.g. $H = \sum_n \left( c_n + c_n^\dagger\right)$?

This might look like a local Hamiltonian $H = \sum_n H_n$ with $H_n = c_n + c_n^\dagger$, but suppose one has a state $|\psi\rangle$ such that $\langle \psi | H_n | \psi \rangle = \alpha \neq 0$. (For convenience we can take the state to be translation-invariant.) Then if we are to regard $H_n$ as a physical local observable, then by locality we would require that $$ \lim_{|n-m| \to \infty} \langle H_n H_m \rangle = \langle H_m \rangle \langle H_m \rangle $$ However, for the above state the right-hand side is a non-zero number, whereas the left-hand side of course anti-commuting in $n$ and $m$. Contradiction.

Conclusion: locality forbids a Hamiltonian that does not commute with fermionic parity symmetry.

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  • $\begingroup$ I can't reproduce your main equation. I get that for a two-fermion system, under the ordering convention $$| \psi \rangle = \sum_{i_1, i_2} c_{i_1, i_2} (a_1^\dagger)^{i_1} (a_2^\dagger)^{i_2} | 0 \rangle,$$ $\langle \psi | H_2 | \psi \rangle = 2\ \text{Re}(c_{01}^* c_{00} - c_{11}^* c_{10})$, but $\langle \psi | a_1 H_2 a_1^\dagger | \psi \rangle = 2\ \text{Re}(c_{10}^* c_{00})$. $\endgroup$
    – tparker
    Jun 1 '17 at 3:02
  • $\begingroup$ Also, there are states of indefinite fermion parity that have $\langle \psi | H_n | \psi \rangle$ for all $n$. For example, the state $$| \psi \rangle = \frac{1}{2} e^{i \phi} \left[ (\cos \theta + \sin \theta) (-| 00 \rangle + | 11 \rangle) + (-\cos \theta + \sin \theta) (| 0 1 \rangle + | 1 0 \rangle) \right]$$ for any angle $\theta$ for a two-fermion system. $\endgroup$
    – tparker
    Jun 1 '17 at 3:59
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    $\begingroup$ @tparker Thanks for pointing out that algebra mistake! I have reshaped the argument as to not rely on that equation. I do think the general message I was trying to make still holds. $\endgroup$ Jun 1 '17 at 8:59
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You are correct that the reason comes down to locality, but wrong in the specifics.

An operator is defined to be "local" if it only acts on a region whose diameter is independent of the total system size. But there is an unfortunate ambiguity in the phrase "acts on". If a Hilbert space factorizes into a tensor product $\mathcal{H}_A \otimes \mathcal{H}_B$ (or a subspace of that tensor product, like the bosonic and fermionic Fock spaces), then it seems like the phrase "an operator $O$ only acts on region A" would mean that it never changes the state of subsystem $B$ - that is, for every direct product vector $| a \rangle \otimes | b \rangle \in \mathcal{H}_A \otimes \mathcal{H}_B$, $O(| a \rangle \otimes | b \rangle) = | a' \rangle \otimes | b \rangle$ for some vector $| a' \rangle \in \mathcal{H}_A$. Fermionic ladder operators indeed satisy this property. But this is not what we usually mean by "$O$ only acts on region A". Instead, we mean that $O = O_A \otimes I_B$ for some operator $O_A$ acting on $\mathcal{H}_A$. This is a strictly stronger definition, because it means that the vector $| a' \rangle$ cannot depend on $| b \rangle$, but only on $| a \rangle$. The operator $O$ can't even "know about" the state of system $B$, even if it doesn't change it.

Fermion parity-violating operators don't satisfy this stronger requirement that they must factorize into the tensor product of a local operator and the identity. No matter how far apart sites $i$ and $j$ are, you can't write $a_i = A_i \otimes I_j$ or $a_j = I_i \otimes A_j$ for any single-site operators $A_i$ or $A_j$ - because if you could, then $a_i a_j$ and $a_j a_i$ would both equal $A_i \otimes A_j$, and $a_i$ and $a_j$ would commute rather than anticommute, in violation of the canonical anticommutation relations. (I'm being a bit schematic in my notation by neglecting the rest of the system besides sites $i$ and $j$.) In fact, the support of any fermion parity-violating term must be the entire system, because in order to get the signs right, the operator must "know about" states arbitrary far away even if it doesn't actually change the state of the system on those faraway sites. (This is, for example, why there's no way to write the Jordan-Wigner transformation so as to transform $a_i$ in a finite product of Pauli matrices.)

Whether or not the fact that these formal mathematical ladder operators don't factorize into tensor products with finite support reflects a true lack of locality in the physical particles, is a question that I'll leave to philosophers. But the fact is that practically speaking, physically realistic systems don't seem to be described by parity-violating Hamiltonians.

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One way to see this is to remember that fermions have half-integer spin. The Hamiltonian should be invariant under 360-degree rotation. Any term with an odd number of fermionic terms will pick up a minus sign that it shouldn't have.

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  • $\begingroup$ I appreciate the reply, but it doesn't get to the core of things. In condensed matter systems we are happy to think in terms of effective fermions, which can often be spinless. $\endgroup$ Mar 16 '17 at 21:59
  • $\begingroup$ @RubenVerresen I've just learned something here. What fermionic properties remain then? $\endgroup$ Mar 16 '17 at 22:02
  • $\begingroup$ @ZeroTheHero In a condensed matter context, we define fermions as operators $c_i$ which obey the fermionic commutation relations, i.e. $\{ c_i,c_j^\dagger \} = \delta_{ij}$ and $\{ c_i,c_j \} = 0$, where the subscript labels different fermions (e.g. at different sites of your system). $\endgroup$ Mar 16 '17 at 22:05
  • $\begingroup$ @RubenVerresen obviously with no reference to spin... Thanks. $\endgroup$ Mar 16 '17 at 22:06
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Fermionic Hamiltonians do not always obey fermionic parity symmetry.

The terms mentioned in the question appear in several physical models, for instance in the case of a single potential scatterer on a Tomonaga-Luttinger liquid (see Eq.(140) of https://arxiv.org/abs/cond-mat/9805275). The original Hamiltonian is usually conserving fermion parity, however, a transformation to quasi-particles generates the terms mentioned in the question (notice that these are Majorana fermions). This transformation (in the case I mentioned above) is called refermionization and nonlinear in the original fields. But the resulting fields are still fermions. Hence the answer.

You can then use the Matveev trick (cited in the paper above) to diagonalize the Hamiltonian if the other terms are quadratic (there is also literature on something like a generalized Bogoliubov transformation how to diagonalize the Hamiltonian but I keep forgetting the reference).

Edit: In regard to the other answers given, I would like to mention that the Hamiltonian before the described transformation perfectly respects locality.

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What I cant tell you from the example that you wrote for $H$ is that for a closed system, one usually conserve the number of particles. That $H$ will not conserve and in principle your Fock space have to be infinite. I have seen some models in quantum optics with terms similarly to your $H$ (if you are interested I can send you the paper references).

The simpler way to see second quantization is like an algebraic way to solve many-body problems, and instead of carrying the fermionic symmetry in the form of Slater determinants, you carry it step by step by using appropiate anti-commutation operators. You can read the first chapters of Stefanucci's Nonequilibrium Many-Body Theory of Quantum Systems (the book is very amenable, despite of its name).

If you want the hardcore mathematical approach, look for CAR algebras. But I think it can confuse more than clarify when one is beginning to learn the subject (no stringy argumments as far as I know).

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