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Question: what is the connection between {$UU^*$} and {$\lambda$}? Is it possible to prove that knowledge of {$\lambda$} determines {$UU^*$}? Eg, does {$\lambda=-1$} imply {$UU^*=-1$} ? The converse is obviously not true, eg double the unit cell size (although is that due to an incorrect definition of the local parity operator? is it -1 in that case?)

Comparison with Kramers theorem. Suppose T^2 = -1, and [H,T] = 0, H hermitian. Then H is even-degenerate. Proof: let x be an eigenvector of H. Then Hx = Ex. Let x' = Tx. Then Hx' = HTx = THx = ETx = Ex'. Moreover, x and x' are orthogonal: <x|Tx> = <xT^\dagger|T^2 x>^* (follows from anti-linear nature of T) = -<xT^\dagger|x>^* = -<x|Tx>. What is antilinear here?

Forgetting the local parity for the moment, can we prove that if we have an eigenvalue -1, so that

{$A^s X A^{s*} = -X$}

then A must be even dimension? There is obviously no product state that satisfies this. Basis for X:

{$G A^s G^\dagger X' G^* A^{s*} G^T = -X'$}

let {$X = G^\dagger X' G^*$}

so that X' = G X G^T

If X is real, then we can diagonalize it. If X is complex-symmetric then we can diagonalize it with G unitary, in which case we can in fact make X real and positive. This is the Takagi Factorization. So with the correct normalization we have

A^s A^{s*} = -I

A obviously cannot be symmetric. Could be skew-symmetric, then A^\dagger = -A^*, and AA^* is negative-definite.

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Page last modified on August 29, 2017, at 06:17 AM