Linear algebra question (using braket notation).

1. May 31, 2008

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?

2. May 31, 2008

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.

3. May 31, 2008

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.

4. May 31, 2008

George Jones

Staff Emeritus
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

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

Last edited: May 31, 2008
5. May 31, 2008

George Jones

Staff Emeritus
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,

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

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

Last edited: May 31, 2008
6. Jun 2, 2008

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?