We know the supersymmetry (SUSY) charge $Q$ satisfies the following relation respect to fermion parity operator $(-1)^F$:
$$
(-1)^F Q + Q (-1)^F :=\{Q, (-1)^F \} =0
$$
which defines the anti-commutator to be zero.


The requirement of SUSY charge $Q$ includes that


1. $Q$ is a Hermitian operator.
2. $[Q,H]=0$, $Q$ commutes with the Hamiltonian $H$ operator. $H$ is also Hermitian.
3. $Q^2$ is bounded from below. (Usually proportional to the Hamiltonian $H$ operator.)


Usually, in the literature, $Q$ is a linear and unitary operator. But can we have $Q$ to be instead antilinear and antiunitary?



> 
> My question is about the following, can we introduce a (new) SUSY charge called $Q'$ satisfy the additional less-common properties (other than satisfying the previous common properties mentioned above):
> 
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> 


4. $Q'$ is an [antilinear](https://en.wikipedia.org/wiki/Antilinear_map) operator.
5. $Q'$ is an [antiunitary](https://en.wikipedia.org/wiki/Antiunitary_operator) operator.


Note that the (Hermitian) adjoint of the $Q'$ is also an antilinear and antiunitary operator. In fact, the (Hermitian) adjoint of the $Q'$ can be made to be the same $Q'$; thus $Q'$ can be regarded as Hermitian, or $Q'=Q'^\dagger$. See for example: <https://physics.stackexchange.com/q/45227/12813>.


Also, the product of two antilinear and antiunitary operators $Q'^2$ become a linear and unitary operator. Such as the complex conjugation (antilinear and antiunitary) $K$, whose square $K^2=+1$ is an identity (linear and unitary). Thus obeying conditions 4. and 5., do not seem to conflict with conditions 1.2.3. earlier.



> 
> Also, are there existing or previous literature introducing SUSY charge $Q'$ to be also antilinear and antiunitary?
> 
> 
> 



Supersymmetry charge $Q$ as anti-linear and anti-unitary operator
Suppose you are given a super Hilbert space $\mathcal{H} = \mathcal{H}\_0 \oplus \mathcal{H}\_1$, with bosonic and fermionic subspaces $\mathcal{H}\_0$ and $\mathcal{H}\_1$ respectively. Define a new super Hilbert space $\mathcal{H}' = \mathcal{H}\_0 \oplus \overline{\mathcal{H}\_1}$, where you have complex-conjugated the fermionic subspace but left the bosonic subspace intact. The space of even unitary operators on $\mathcal{H}$ is $U\_0(\mathcal{H}) = U(\mathcal{H}\_0) \times U(\mathcal{H}\_1) \cong U(\mathcal{H}\_0) \times U(\overline{\mathcal{H}\_1}) = U\_0(\mathcal{H}')$. But an odd operator on $\mathcal{H}$ takes $\mathcal{H}\_0$ to $\mathcal{H}\_1$. As a result, the odd *linear* operators on $\mathcal{H}$ are the same as odd *antilinear* operators on $\mathcal{H}'$, and vice versa.


As a result, your "antilinear susy algebra" has *the same* representation theory as the usual linear susy algebra. There is no benefit to making the change that you suggest, no new examples, and only an emotional cost that you have to handle nonlinear operators.




---


I remark that $Q^2 \propto \hat H$ is never unitary in examples: it is an unbounded self-adjoint operator, with spectrum bounded below (as you say). Similarly, $Q$ is never (anti)unitary. Rather, the supersymmetry operator $Q$ should be required to be an (unbounded) *odd self-adjoint* operator. There are various equally-valid conventions for the meaning of "odd self-adjoint", and they affect the proportionality constant in the expectation that $Q^2 \propto \hat H$. The issue is the following conflict. We expect that the Jordan product, and in particular the square, of self adjoint operators is self-adjoint. But we also expect that the Lie product of self-adjoint operators is *skew*-adjoint. Well, is $Q^2$ the Jordan product of $Q$ with itself (as it would be in the even case) or the Lie product (since $[Q,Q] = QQ - (-1)^{|Q||Q|} QQ$)? For further details of the different possible conventions, I recommend Section 23 of [Greg Moore's notes on Linear Algebra](http://www.physics.rutgers.edu/%7Egmoore/618Spring2019/GTLect2-LinearAlgebra-2019.pdf).


