Hermiticity depends on the space that the operator is acting on. You may have already seen this when (if) you dealt with angular momentum. Neither x nor p need be Hermitian, in general. What is the allowed range of x?... I don't see why -x^4 is not hermitian. Since we know that p^2 is hermitian (from free particle problem), all we need to do is to investigate -x^4, which has to be hermitian from the integration you have shown. Maybe I miss anything?
Well, I would say that the functions that compose the Hilbert space on which the Hamiltonian is defined are subject to boundary conditions.All Hamiltonian are subjected to boundary condition.
If x→∞, then of course x^{4}→∞. There is nothing wrong with this, and, in particular, this does not preclude the Hermiticity of x^{4} nor H. On the contrary, there would be something severely wrong with a finite x^{4} as x→∞.... when x goes to infinity (... we will have problem to show the Hermiticity of -x^4 in momentum representation, there partial integration will just yield infinity.)
Extending x into the complex plane is not a PT operation. Usually, PxP = -x, and TxT = x*. I only put the "*" on the time-reversed x because you are suggesting an analytical continuation of the eigenfunctions of x into the complex plane. Normally, TxT = x. In particular, if x is already real-valued, then PTxPT = -x, which is also real-valued.So if we extend x into complex plane (PT symmetry), it is clear that -x^4 is not Hermitian, because x is complex.
I started reading that article, but it may be too far over my head. I don't think I agree with it, though. For example, PTpPT = p, but p has negative eigenvalues, so I must be missing part of their argument.See arXiv:physics/9712001, Bender.
The thing is we are still treating x as real, which if we use WKB to approximate the asymtotic eigenstate, you will notice that it just blows to infinity when x goes to infinity.That is what I am trying to say, if the Hamiltonian produces non L2 integrable function, then it is a not a well defined operator, hence the problem of hermiticity does not even have a place to start with....So, if I simply require j_{if}(a) = j_{if}(b) for all i,f in my Hilbert space, then doesn't that make H Hermitian?
Your statement makes me ponder on another question: does one replaces the operator as real valued variable when going from quantum mechanics to classical mechanics? Or is complex variable also allowed? The problem is if we treat x and p as complex, then H=p^2-x^4 is not PT symmetry, as [H,PT][tex]\neq[/tex]0Extending x into the complex....particular, if x is already real-valued, then PTxPT = -x, which is also real-valued.
By "it" you apparently mean the "wavenumber", that is, how quickly the function oscillates in x. I thought that you were talking about x^{4} . Anyway, the WKB method assumes the existence of the eigenfunction of H. So, if the WKB method does not give you a valid eigenfunction, then all that means is that you cannot use the WKB method to determine a valid eigenfunction. I don't think that you can use this argument to demonstrate the nonexistence of valid eigenfunctions.... if we use WKB to approximate the asymtotic eigenstate, you will notice that it just blows to infinity when x goes to infinity.
I don't know what you mean by "the Hamiltonian produces [a] ... function". The QM Hamiltonian acts on a wavefunction in order to generate time translation of that wavefunction, but QM posits the existence of the wavefunction independently of the Hamiltonian (i.e. you are free to specify an initial state). I will assume that you meant that the wavefunction on which the Hamiltonian acts must be L_{2}(ℝ^{n}) (where I've been assuming that n = 1 for this problem). In that case, either you have been VERY frustrated with QM, or you have just recently decided to impose this condition on the "Hilbert" space of QM. See if you can figure out why I think this.... if the Hamiltonian produces non L2 integrable function, then it is a not a well defined operator, hence the problem of hermiticity does not even have a place to start with.
CM variables may be complexified (e.g. ict in early formulations of relativity). You just have to decide how to interpret results in terms of observables. The correspondence between CM and QM does involve a complex phase, one way or another.... does one replaces the operator as real valued variable when going from quantum mechanics to classical mechanics? Or is complex variable also allowed?
Basically, I don't think we should be talking about "complex-valued" x and p.The problem is if we treat x and p as complex, then H=p^2-x^4 is not PT symmetry, as [H,PT][tex]\neq[/tex]0