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Consider an interaction of the type (on a lattice) $$H=\sum_{\langle i,j\rangle}\left\{\left[\alpha c_{i\uparrow}^\dagger c_{j\uparrow}+h.c.\right]+\left[\beta c_{i\downarrow}^\dagger c_{j\downarrow}+h.c.\right]+\left(\gamma c_{i\uparrow}^\dagger c_{j\downarrow}+h.c.\right)+\left(\delta c_{i\downarrow}^\dagger c_{j\uparrow}+h.c.\right)\right\}$$ Here, $c_{i\sigma}^\dagger$ is the electron creation operator at the lattice site $i$ in spin state $\sigma$. The symbols $\alpha,\beta,\gamma$ and $\delta$ are four complex coeffeciients.

The first term in each of the two first brackets is spin-violating interaction because it destroys an electron at the site in the spin-down state but creates an electron at another site in the spin-up state. However, adding the hermitian conjugate term makes both $(...)$ interactions spin-conserving when considered as a whole. Thus it seems like, spin conservation is automatic by the hermiticity requirement of the Hamiltonian. My question is if we consider two-body interactions, does the requirement of hermeticity automatically ensure spin-conserving interaction? I hope that I have rightly interpreted the terms.

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The first two terms of the Hamiltonian are spin-conserving: one operator delets an electron with spun up/down, and the other creates another electron with the same spin, so that the total spin remains unchanges. This is the same as with charge conservation: we remove one electron, but create another one.

The last two terms are however non-spin conserving. Hermitian Hamiltonians are not encessarily spin-conserving - any Hamiltonian with spin-orbit coupling can serve as an example. Secodn quantization does not change this in any way.

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  • $\begingroup$ Thanks. Is charge conservation ensured by requiring hermiticity? It would be very unphysical if an electron at site $i$ is destryted with no compensating electron being created somewhere else. $\endgroup$
    – Solidification
    Jun 25, 2021 at 17:20
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    $\begingroup$ No, operator $c^\dagger + c$ is hermitian, but does not conserve charge (and the number of particles). Hermicity in QM is required so that the eigenvalues are real quantities. $\endgroup$
    – Roger V.
    Jun 25, 2021 at 17:24
  • $\begingroup$ Yes, I know that. But $c, c^\dagger$ do not appear in isolation in a many-body Hamiltonian. We usually consider one-body or two-body operators to begin with. $\endgroup$
    – Solidification
    Jun 25, 2021 at 17:26
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    $\begingroup$ They do appear, in the chapters on bose-einstein condensation and superconductivity. In some formalisms they are also widely used as source fields. Also, beyond many-body texts, e.g., when treating atom-photon interactions. Also when using slave-boson and other similar approaches. $\endgroup$
    – Roger V.
    Jun 25, 2021 at 17:33
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    $\begingroup$ @mithusengupta123 yes, indeed. But there are important things to remember: 1) it should correspond to correct physical situation, 2) some standard results do not hold or need to be modified - e.g., Lehmann representation heavily uses particle conservation. $\endgroup$
    – Roger V.
    Jun 25, 2021 at 18:30
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The answer is certainly no. Indeed, your argument can be made extremely generally about any quantum number, and (if true) would imply that every quantum number is conserved. Just because your Hamiltonian contains one term that flips spins up and another term that flips spins down, doesn't mean that spin is conserved.

As a much simpler example, consider the Hamiltonian $$ H = \sigma^x = | {\uparrow} \rangle \langle {\downarrow} | + | {\downarrow} \rangle \langle {\uparrow} | . $$ This Hamiltonian contains one term which flips spins from up to down, and another which flips spins from down to up. But $\sigma^z$ is certainly not a conserved quantity, since $[\sigma^z, H] \neq 0$. Indeed, it is easy to see that if you initialize your state as $| \psi(0) \rangle = | {\uparrow} \rangle$, then at later times you will have $$ | \psi(t) \rangle = \cos t | {\uparrow} \rangle - i \sin t | {\downarrow} \rangle $$ so that the $z$-spin oscillates between up and down.

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  • $\begingroup$ Thanks. Can you say that if we consider one-body and two-body operators, then hermitian Hamiltonian automatically leads to charge conserving interactions? $\endgroup$
    – Solidification
    Jun 25, 2021 at 17:29
  • $\begingroup$ Of course not. The Hamiltonian I wrote above can easily be written as one-body operator in second quantization, and it does not conserve charge. $\endgroup$
    – Zack
    Jun 25, 2021 at 17:34
  • $\begingroup$ But the charge must be conserved. How do you explain that? Does it mean the Hamiltonian you wrote is unphysical? $\endgroup$
    – Solidification
    Jun 25, 2021 at 17:40
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    $\begingroup$ Not at all, it's perfectly easy to realize the Hamiltonian I wrote. Charge is conserved because particle number is conserved; both one-body and two-body operators are particle number conserving. Perhaps it pays to be clear about what we're talking about: when we say that spin is conserved, what we mean is that $\langle S^z \rangle$ does not evolve in time, where $S^z$ is the total $z$ spin. The Hamiltonian $H = c^{\dagger}_{\uparrow} c_{\downarrow} + h.c.$ (which is the 2nd quantized version of my above Hamiltonian) conserves particle number, but not $S^z$. $\endgroup$
    – Zack
    Jun 25, 2021 at 18:13

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