- #1
fluidistic
Gold Member
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Hi guys,
I don't really know how to explain my question...
I've heard my professor saying something like "for ##\hat H## to be self-adjoint, Psi and its first derivative with respect to x must be absolutely continuous."
However when I consider the problem of the particle in a box, the fact that Psi(x) is worth 0 at the edges seems to be due to the fact that Psi(x) is continuous and since it's worth 0 outside the box, it's 0 on the edges. However in that case, the derivative of Psi(x) is not continuous: in a 1 dimension box the eigenfunctions are either sines or cosines depending upon their energy. The derivative evaluated at the edges of a cosine or a sine that is worth 0 at the edges is clearly not 0. However outside the box Psi(x)=0=constant so that Psi'(x)=0. Hence the non continuity of Psi'(x).
So if I understood well, this means that the Hamiltonian is not self-adjoint in this case. (I'm not even sure it's well defined since ##V(\hat x )## is infinite).
So I don't really understand what my prof. was saying. In which case(s) what he said does hold?
I don't really know how to explain my question...
I've heard my professor saying something like "for ##\hat H## to be self-adjoint, Psi and its first derivative with respect to x must be absolutely continuous."
However when I consider the problem of the particle in a box, the fact that Psi(x) is worth 0 at the edges seems to be due to the fact that Psi(x) is continuous and since it's worth 0 outside the box, it's 0 on the edges. However in that case, the derivative of Psi(x) is not continuous: in a 1 dimension box the eigenfunctions are either sines or cosines depending upon their energy. The derivative evaluated at the edges of a cosine or a sine that is worth 0 at the edges is clearly not 0. However outside the box Psi(x)=0=constant so that Psi'(x)=0. Hence the non continuity of Psi'(x).
So if I understood well, this means that the Hamiltonian is not self-adjoint in this case. (I'm not even sure it's well defined since ##V(\hat x )## is infinite).
So I don't really understand what my prof. was saying. In which case(s) what he said does hold?