
#1
May3108, 02:43 AM

P: 3,177

the question:
Let {u>,v>} be a basis for a linear space, suppose that <uv>=0, then prove that: [tex]Av>=<A>Iv>+\delta Au>[/tex] where, A is hermitian operator, and [tex]<A>=<vAv>,\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 Au>=\delta Av>[/tex] now, obviously <A>=<vAv>=[tex]\int du <vu><uAv>[/tex]=0 cause <uv> equals zero, so we actually need to prove that Au>=Av> but Au>=uu> where u is the eigenvalue of u>, then if we multiply it by <u we get: <uAu>=u (i guess that u> and v> are normalised), but from the same assertion as above we get that <uAu>=0 and so u=0, and so is v=0, so we get that Au>=Av>=0. is this correct or totally mamabo jambo as i think? 



#2
May3108, 09:43 AM

P: 352

Oh no there is something wrong in your argument. The vector u is orthogonal to v, it is a specific vector. When you invoke completeness you are summing over all vectors, and not just the ones orthogonal to v. It is not true that <A> = 0.




#3
May3108, 09:48 AM

P: 352

The completeness relation in this problem is
[tex]u><u + v><v = 1[/tex] since u and v form an orthonormal basis. And you can say that also since they are a basis that there exists complex numbers [itex]\alpha, \beta[/itex] such that [tex]Av> = \alpha u> + \beta v>[/tex] It is very easy to show that [tex]\beta = <A>[/tex] And then you're left with the other term, and maybe there the completeness relation is helpful. 



#4
May3108, 12:38 PM

Mentor
P: 6,044

Linear algebra question (using braket notation).
It's easy to find an example that contradicts the assertion that you're trying to prove. If fact, since
[tex] A \left v \right> = \left< A \right> \left v \right> + \delta A \left u \right> [/tex] is equivalent to [tex] A \left v \right>  \left< A \right> \left v \right> = \delta A \left u \right> [/tex] and, from the given, [tex] A \left v \right>  \left< A \right> \left v \right> = \left( A  \left< A \right> I \right) \left v \right> = \delta A \left v \right>, [/tex] so your assertion is true iff [itex]\delta A \left v\right> = \delta A \left u \right>[/itex]. Maybe you're actually trying to prove the trivial [tex] A \left v \right> = \left< A \right> \left v \right> + \delta A \left v \right> [/tex]. 



#5
May3108, 12:53 PM

Mentor
P: 6,044

Cleary,the equations [itex]A \left u \right> = \left u \right>[/itex] and [itex]A \left v \right> = 0[/itex] define a unique Hermitian operator [itex]A[/itex]. Now, [tex] \left< A \right> = 0 [/tex] [tex] \delta A = A [/tex]. Therefore, [tex] \delta A \left u \right> = A \left u \right> = \left u \right> [/tex] [tex] \delta A \left v \right> = A \left v \right> = 0. [/tex] 



#6
Jun208, 10:23 AM

P: 3,177

well i think that delta A is zero, becasue delta A is: sqrt(<vA^2v><A>^2) (and not an operator as i first thought)
and: <vA^2v>=beta^2 cause A is hermitian (becasue we know that hermitian matrices can always be diagonalize, by lagrange method of squaring) then Av>=vv> where v is its eigenvalue, and Au>=uu> from here its obvious that: <vAv>=v and <vAu>=<uAv>=0, and thus we get what we need. is my reasoning correct? 


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