I'm getting imaginary eigenvalues of $X^2-P^2$ This is hermitian but I'm getting imaginary eigenvalues. Start with the commutator:
$$[X^2-P^2,X+P]$$
$$=[X^2,P]-[P^2,X]$$
$$=X[X,P]+[X,P]X-(P[P,X]+[P,X]P)$$
$$=2i(X+P)$$
Now consider the ket $|E\rangle$ such that $(X^2-P^2)|E\rangle=E|E\rangle$
$$(X^2-P^2)(X+P)|E\rangle$$
$$=(X+P)(X^2-P^2)|E\rangle+[X^2-P^2, X+P] |E\rangle$$
$$=E(X+P) |E\rangle + 2i (X+P)|E\rangle$$
The above proves that the ket $(X+P)|E\rangle$ has eigenvalue $E+2i$, which is imaginary. How can it be?
 A: The question is formulated in a such vague manner that it is not possible to properly answer.
This is a mathematical issue and to answer it is necessary to fix all mathematical hypotheses.
First of all: What is the domain of $P^2-X^2$?
A standard domain is the space of Schwartz functions. On that dense domain the operator is symmetric and commutes with the standard complex conjugation. As a consequence, it admits selfadjoint extensions, but this fact it is quite irrelevant here. I do not expect that there is a unique selfadjoint extension (I think that the problem has been completely analyzed in the literature, but I cannot check it now).
However, all manipulations with commutators which appear in the post are valid in the Schwartz space, since it is invariant under $P$ and $X$  and I assume that $|E\rangle$ is an eigenvector in that domain. The existence of $|E\rangle$ can be checked directly by solving the corresponding differential equation (actually I do not expect that such a vector exists).
Since symmetric operators have real point spectrum, the conclusion in the post  is untenable and the only possibility is that $(X+P)|E\rangle =0$.
I repeat, with the given domain, everything can be directly checked from scratch and I think that simply the initial $|E\rangle$ does not exist in the Schwartz space.
More exotic choices of the domain or intepretations of $|E\rangle$ (e.g. a Schwartz distribution) should be analyzed case by case with precise definitions and without them it is impossible to answer.
