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&gt;=&lt;A&gt;I|v&gt;+\delta A|u&gt;
where, A is hermitian operator, and &lt;A&gt;=&lt;v|A|v&gt;,\delta A= A-&lt;A&gt;I
where I is the identity operator.

my attempt at solution:
basically, from the definitions i need to prove that \delta A|u&gt;=\delta A|v&gt;
now, obviously <A>=<v|A|v>=\int du &lt;v|u&gt;&lt;u|A|v&gt;=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&gt;&lt;u| + |v&gt;&lt;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&gt; = \alpha |u&gt; + \beta |v&gt;

It is very easy to show that

\beta = &lt;A&gt;

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

<br /> A \left| v \right&gt; = \left&lt; A \right&gt; \left| v \right&gt; + \delta A \left| u \right&gt;<br />

is equivalent to

<br /> A \left| v \right&gt; - \left&lt; A \right&gt; \left| v \right&gt; = \delta A \left| u \right&gt;<br />

and, from the given,

<br /> A \left| v \right&gt; - \left&lt; A \right&gt; \left| v \right&gt; = \left( A - \left&lt; A \right&gt; I \right) \left| v \right&gt; = \delta A \left| v \right&gt;,<br />

so your assertion is true iff \delta A \left| v\right&gt; = \delta A \left| u \right&gt;.

Maybe you're actually trying to prove the trivial

<br /> A \left| v \right&gt; = \left&lt; A \right&gt; \left| v \right&gt; + \delta A \left| v \right&gt;<br />.

 

[George Jones]

so your assertion is true iff \delta A \left| v\right&gt; = \delta A \left| u \right&gt;.

Oops, I see that you already know this.

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

<br /> \left&lt; A \right&gt; = 0<br />

<br /> \delta A = A<br />.

Therefore,

<br /> \delta A \left| u \right&gt; = A \left| u \right&gt; = \left| u \right&gt;<br />

<br /> \delta A \left| v \right&gt; = A \left| v \right&gt; = 0.<br />

 

[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?