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Linear algebra question (using braket notation).

  • #1
MathematicalPhysicist
Gold Member
4,189
168
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?
 

Answers and Replies

  • #2
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
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]A|v> = \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
George Jones
Staff Emeritus
Science Advisor
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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].
 
Last edited:
  • #5
George Jones
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Science Advisor
Gold Member
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so your assertion is true iff [itex]\delta A \left| v\right> = \delta A \left| u \right>[/itex].
Oops, I see that you already know this.

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]
 
Last edited:
  • #6
MathematicalPhysicist
Gold Member
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well i think that delta A is zero, becasue delta A is: sqrt(<v|A^2|v>-<A>^2) (and not an operator as i first thought)
and:
<v|A^2|v>=beta^2
cause A is hermitian (becasue we know that hermitian matrices can always be diagonalize, by lagrange method of squaring) then A|v>=v|v> where v is its eigenvalue, and A|u>=u|u>
from here its obvious that: <v|A|v>=v and <v|A|u>=<u|A|v>=0, and thus we get what we need.

is my reasoning correct?
 

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