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State vectors, projection matrices

  1. Oct 8, 2007 #1
    1. The problem statement, all variables and given/known data

    How do I prove that if,

    [tex]|\vec{u_1}><\vec{u_1}| + |\vec{u_2}><\vec{u_2}| = I[/tex],

    where 'I' is the indentity matrix, that [itex]u_1[/itex] and [itex]u_2[/itex] are orthogonal and normalized?

    Can anybody get me started?
     
    Last edited: Oct 8, 2007
  2. jcsd
  3. Oct 8, 2007 #2

    Hurkyl

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    You can't, with the information you've stated.
     
  4. Oct 8, 2007 #3
    Here's the exact question from the problem set, just in case:

    Show that given two arbitrary vectors u1 , u2 ∈ C^2 such that the associated projectors satisfy Pu1 + Pu2 = 1 then it holds that the vectors u1, u2 are orthogonal and normalized.

    Is it really impossible?
     
  5. Oct 8, 2007 #4

    Hurkyl

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    Oh bah. I read the problem backwards. :frown:

    Well, since yo'ure given an equation with I in it, it will probably be useful to throw I into every relevant equation you can think of.
     
  6. Oct 8, 2007 #5
    That's the thing, I can't think of any relevant equations besides the two inner product equations for a normalized vector and two orthogonal vectors.
     
  7. Oct 8, 2007 #6
    What if I multiplied by |u2> on the right-hand side of each term. Would I be correct in saying that in order for this new equation to hold true u1 and u2 have to be orthogonal and u2 would have to be normalized? Then I can do it again with u1.
     
  8. Oct 8, 2007 #7

    Hurkyl

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    Well, then throw an I in them! What do you get?
     
  9. Oct 8, 2007 #8
    I think I would get:

    <u1|u1>I = I
    <u2|u2>I = I
    <u1|u2>I = 0
    <u2|u1>I = 0

    Does this make sense? Is it incorrect to do it like I said in post #6?

    If I substituted the expression in my OP for 'I' in these equations, and they still hold true does it prove what I need?
     
  10. Oct 8, 2007 #9

    Hurkyl

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    You put the I in the wrong place. It's an operator; it should operate on things!

    I suppose that you can put an I there, intending there to be scalar multiplication, but that doesn't do anything useful here.
     
  11. Oct 8, 2007 #10
    Ooh

    <u1|I|u1> = I
    <u2|I|u2> = I
    <u1|I|u2> = 0
    <u2|I|u1> = 0

    Like that then?
     
  12. Oct 8, 2007 #11

    Hurkyl

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    Yes -- those are the statements you want to prove!

    Incidentally, this is a trick you want to take to heart; it's a very useful thing. (And not just in linear algebra; in analysis, you might use a partition of unity in similar ways)
     
  13. Oct 8, 2007 #12
    I'll try to work out the algebra.

    ...I think I'm slllllowly getting my head around this quantum stuff.

    Thanks for your help.
     
  14. Oct 8, 2007 #13

    Hurkyl

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    Hrm...

    This is surely what you want to do, but finishing off the proof is less straightforward than I thought it was.
     
  15. Oct 8, 2007 #14

    Hurkyl

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    Shame on me for not fully thinking this through! :redface: You can do this... if you can first prove that |u1> and |u2> are linearly independent. (do you see why?)
     
  16. Oct 8, 2007 #15
    I see why |u1> and |u2> would have to be linearly independent. I'm not sure how I'd prove that, though. Wouldn't I have to do it based solely on the fact that adding the projection matrices gives the identity matrix?
     
  17. Oct 8, 2007 #16
    When I substitute [tex]|\vec{u_1}><\vec{u_1}| + |\vec{u_2}><\vec{u_2}| = I[/tex]

    into the LHS of
    <u1|I|u1> = I
    <u1|I|u2> = 0
    I get
    1 = I
    <u1|u2> + <u1|u2> = 0
    respectively

    This doesn't seem right... ? It does imply that<u1|u2> = 0, though.
     
  18. Oct 9, 2007 #17

    Hurkyl

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    Assume they are dependent, see what happens. Rank might be involved.


    I don't see how you got that. And remember that <u1|I|u2> = 0 is what you're trying to prove.
     
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