Will a vector rotate back to itself under unitary rotation? Let there be an arbitrary complex vector vec(v) in bbbC^D that has unit length defined as sum_i v_i v_i^**=1. Let there be a Hermitian D xx D matrix H. Given arbitrary vec(v) and H, will there always exist a non-zero t in bbbR such that e^(iHt) vec(v) =vec(v)?

autumnunfound91

autumnunfound91

Open question

2022-08-19

Will a vector rotate back to itself under unitary rotation?
Let there be an arbitrary complex vector v C D that has unit length defined as i v i v i = 1. Let there be a Hermitian D × D matrix H. Given arbitrary v and H, will there always exist a non-zero t R such that e i H t v = v ?

Answer & Explanation

klunnalegj6

klunnalegj6

Beginner2022-08-20Added 7 answers

No. Consider H = ( 1 0 0 3 ) . Then
e i H t = ( e i t 0 0 e i 3 t )
And it is now apparent that the vectors v C 2 such that there is some t 0 satisfying e i H t v = v are the ones such that either v 1 = 0 or v 2 = 0
Added: In the same spirit, it isn't hard to show that the hermitian matrices H such that for all unitary vectors v there is some t R { 0 } such that e i t H v = v are exactly the ones such that there is some real number α such that λ α Q for all λ eigenvalues of H.
More precisely, let H = H C n × n and v C n . Let's call V λ = ker ( H λ I ) and let's impose the following equivalence relation on spec H:
λ R μ λ = μ = 0 ( λ 0 μ 0 μ λ Q )
Let's call W λ = μ R λ V μ . Then there is some t R { 0 } such that e i t H v = v if and only if v V 0 + W λ for some λ spec H

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