Projection operator (redundant?) definition

43 Views Asked by At

In his Quantum Mechanics, Ballentine says that “in general, a self-adjoint operator which obeys $\rho^2=\rho$ is a projection operator”. I'm not sure I follow the need for the self-adjoint caveat? I thought that $\rho^2=\rho$ defined a projection operator? Is there any reason he would add this caveat that I'm missing, perhaps in the application to the weird (rigged Hilbert) spaces in which quantum mechanics is done?