Linear algebra question (using braket notation).

  1. May 31, 2008 #1
    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?
     
  2. jcsd
  3. May 31, 2008 #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.
     
  4. May 31, 2008 #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.
     
  5. May 31, 2008 #4

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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: May 31, 2008
  6. May 31, 2008 #5

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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: May 31, 2008
  7. Jun 2, 2008 #6
    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Linear algebra question (using braket notation).
  1. QM - linear algebra (Replies: 14)

  2. A but linear algebra (Replies: 0)

Loading...