Linear algebra question (using braket notation).

[MathematicalPhysicist]

the question:

Let {|u>,|v>} be a basis for a linear space, suppose that <u|v>=0, then prove that:
$$A|v>=<A>I|v>+\delta A|u>$$
where, A is hermitian operator, and $$<A>=<v|A|v>,\delta A= A-<A>I$$
where I is the identity operator.

my attempt at solution:
basically, from the definitions i need to prove that $$\delta A|u>=\delta A|v>$$
now, obviously <A>=<v|A|v>=$$\int du <v|u><u|A|v>$$=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?

 

[DavidWhitbeck]

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.

 

[DavidWhitbeck]

The completeness relation in this problem is

$$|u><u| + |v><v| = 1$$ since u and v form an orthonormal basis.

And you can say that also since they are a basis that there exists complex numbers $\alpha, \beta$ such that

$$A|v> = \alpha |u> + \beta |v>$$

It is very easy to show that

$$\beta = <A>$$

And then you're left with the other term, and maybe there the completeness relation is helpful.

 

[George Jones]

It's easy to find an example that contradicts the assertion that you're trying to prove. If fact, since

$$A \left| v \right> = \left< A \right> \left| v \right> + \delta A \left| u \right>$$

is equivalent to

$$A \left| v \right> - \left< A \right> \left| v \right> = \delta A \left| u \right>$$

and, from the given,

$$A \left| v \right> - \left< A \right> \left| v \right> = \left( A - \left< A \right> I \right) \left| v \right> = \delta A \left| v \right>,$$

so your assertion is true iff $\delta A \left| v\right> = \delta A \left| u \right>$.

Maybe you're actually trying to prove the trivial

$$A \left| v \right> = \left< A \right> \left| v \right> + \delta A \left| v \right>$$.

 

[George Jones]

so your assertion is true iff $\delta A \left| v\right> = \delta A \left| u \right>$.

Oops, I see that you already know this.

Cleary,the equations $A \left| u \right> = \left| u \right>$ and $A \left| v \right> = 0$ define a unique Hermitian operator $A$. Now,

$$\left< A \right> = 0$$

$$\delta A = A$$.

Therefore,

$$\delta A \left| u \right> = A \left| u \right> = \left| u \right>$$

$$\delta A \left| v \right> = A \left| v \right> = 0.$$

 

[MathematicalPhysicist]

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?