- #1
MathematicalPhysicist
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the question:
Let {|u>,|v>} be a basis for a linear space, suppose that <u|v>=0, then prove that:
[tex]A|v>=<A>I|v>+\delta A|u>[/tex]
where, A is hermitian operator, and [tex]<A>=<v|A|v>,\delta A= A-<A>I[/tex]
where I is the identity operator.
my attempt at solution:
basically, from the definitions i need to prove that [tex]\delta A|u>=\delta A|v>[/tex]
now, obviously <A>=<v|A|v>=[tex]\int du <v|u><u|A|v>[/tex]=0
cause <u|v> equals zero, so we actually need to prove that A|u>=A|v>
but A|u>=u|u> where u is the eigenvalue of |u>, then if we multiply it by <u| we get:
<u|A|u>=u (i guess that |u> and |v> are normalised), but from the same assertion as above we get that <u|A|u>=0 and so u=0, and so is v=0, so we get that A|u>=A|v>=0.
is this correct or totally mamabo jambo as i think?
Let {|u>,|v>} be a basis for a linear space, suppose that <u|v>=0, then prove that:
[tex]A|v>=<A>I|v>+\delta A|u>[/tex]
where, A is hermitian operator, and [tex]<A>=<v|A|v>,\delta A= A-<A>I[/tex]
where I is the identity operator.
my attempt at solution:
basically, from the definitions i need to prove that [tex]\delta A|u>=\delta A|v>[/tex]
now, obviously <A>=<v|A|v>=[tex]\int du <v|u><u|A|v>[/tex]=0
cause <u|v> equals zero, so we actually need to prove that A|u>=A|v>
but A|u>=u|u> where u is the eigenvalue of |u>, then if we multiply it by <u| we get:
<u|A|u>=u (i guess that |u> and |v> are normalised), but from the same assertion as above we get that <u|A|u>=0 and so u=0, and so is v=0, so we get that A|u>=A|v>=0.
is this correct or totally mamabo jambo as i think?