# Proof that eigenvalues of (^L+)n(ˆL)nâ‰¥0(\hat{L^+})^n (\hat{L})^n â‰¥ 0

Hilbert space \mathbb{L^2}\mathbb{L^2}. So my thoughts:

n=0n=0 – true.
Let \hat\alpha = (\hat{L^+})^n (\hat{L})^n \hat\alpha = (\hat{L^+})^n (\hat{L})^n

For n+1n+1: \ \hat{L^+} \hat\alpha \hat{L} \phi = \lambda \phi = \hat\alpha \phi\ \hat{L^+} \hat\alpha \hat{L} \phi = \lambda \phi = \hat\alpha \phi since its just unitary. But for n=0n=0 eigenvalues are just 11, no? So following my proof for (\hat{L^+})^n (\hat{L})^n(\hat{L^+})^n (\hat{L})^n eigenvalues are 11 too.

upd: which are false as I can see from now.

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Which Hilbert space? What’s \hat L\hat L? Is \ ^+\ ^+ taking the adjoint?
– Roland
Oct 20 at 21:56

@Roland There were nothing much about space in the task, but lets assume its \mathbb{L}^2\mathbb{L}^2. \hat{L}\hat{L} is some linear operator on that space. Yes, ^+^+ means adjoint. I guess I need to use ^\dagger^\dagger
– EzWin
Oct 20 at 21:59

I don’t see where you get unitarity from. What if \hat L\hat L has a kernel?
– Roland
Oct 20 at 22:03

@Roland, ok I guess I forgot that I need \hat{L}\hat{L} to be unitary, damn. Then I have no idea of proof yet. There is nothing said about \hat{L}\hat{L}, its just any operator.
– EzWin
Oct 20 at 22:05

The claim is false for L:x \mapsto 0L:x \mapsto 0, or any operator with a kernel, for instance f \mapsto -f”f \mapsto -f” on the Hilbert space L^2(-\pi,\pi)L^2(-\pi,\pi) (domain is given by sufficiently often differentiable functions)
– Roland
Oct 20 at 22:11

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By induction. Base case n=0n=0 is trival. Set M = (L^*)^{n-1} L^{n-1}M = (L^*)^{n-1} L^{n-1}. M \ge 0M \ge 0 by the induction hypothesis. We have (L^*)^{n} L^{n} = L^* M L(L^*)^{n} L^{n} = L^* M L, and

\langle L^* MLx, x\rangle = \langle M(Lx), Lx\rangle \ge 0,

\langle L^* MLx, x\rangle = \langle M(Lx), Lx\rangle \ge 0,

which finishes the inductive step.

Sorry, but is \lambda = \langle \hat L x , x \rangle\lambda = \langle \hat L x , x \rangle?
– EzWin
Oct 21 at 0:06

I am not sure what your notation is. If xx is an unit eigenvector with eigenvalues \lambda\lambda, then, yes, you are right. The definition of a positive semidefinite operator that I am familiar with is that M \ge 0M \ge 0 if and only if MM is self-adjoint and for every xx, \langle Mx, x \rangle \ge 0\langle Mx, x \rangle \ge 0.
– Sasho Nikolov
Oct 21 at 0:14

The question was to proof that all the eigenvalues of that operator are non-negative. So my notation is like \lambda\lambda for any eigenvalue and \phi\phi for eigenfunction (eigenvector).
– EzWin
Oct 21 at 0:27

Well, in that case, take \phi\phi to be an eigenfunction with eigenvalue \lambda\lambda and norm 1, from the proof above you have that 0 \le \langle M\phi, \phi \rangle = \langle \lambda \phi, \phi \rangle = \lambda0 \le \langle M\phi, \phi \rangle = \langle \lambda \phi, \phi \rangle = \lambda
– Sasho Nikolov
Oct 21 at 0:52

yep I got it already. Thank you.
– EzWin
Oct 21 at 0:55