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Homework Help: Operators and eigenstates

  1. Oct 25, 2003 #1
    Operators and eigenstates (updated with new question)

    Hi, I encountered the following HW problem which really confuses me. Could anyone please explain it to me? Thank you so much!

    The result of applying a Hermitian operator B to a normalized vector |1> is generally of the form:

    B|1> = b|1> + c|2>

    where b and c are numerical coefficients and |2> is a normalized vector orthogonal to |1>.

    My question is: Why B|1> must have the above form? Does it mean if |1> is an eigenstate of B, then b=!0 and c=0? But what if |1> is not an eigenstate of B?

    I also need to find the expectation value of B (<1|B|1> ), but I think I got this part:

    <1|B|1> = <1|b|1> + <1|c|2> = b<1|1> + c<1|2> = b

    since |1> and |2> are orthogonal and they're both normalized. Does that look right?
     
    Last edited: Oct 25, 2003
  2. jcsd
  3. Oct 25, 2003 #2
    an operator takes a vector in a space and turns it into another vector in (not necesarily) another space. if that space is spanned by the vectors |1> and |2>m then b|1> + c|2> is the general form a vector in that space. if you don't know how B looks like, then you have to assume this general form. If you find something oput about B, like |1> being an eigenstate, then c must be 0 as the two base vectors should be orthonormal.

    you're absolutely right. trust yourself.
     
  4. Oct 25, 2003 #3

    Hurkyl

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    Because we can "compute" b, c, and |2>.

    You know how to compute b already; b = <1|B|1>. Can you hazard a guess as to how to compute c and |2>?
     
  5. Oct 25, 2003 #4
    Thank you very much, Sonty and Hurkyl. Now I understand the question. However, I have another one -

    I'm trying to find <1|(B-<B>)2|1>. I broke it up like this:

    <1|(B-<B>)2|1> = <1|B2|1> - <1|2bB|1> + <1|b2|1>, since we already found that <B> = b.

    <1|2bB|1> = 2b <1|B|1> = 2b2
    <1|b2|1> = b2

    But I'm not too sure about <1|B2|1>....should I do something like this ? (Note: b, c are real constants)

    <1|B2|1> = <1|B* B|1>
    = (<1|b + <2|c)(b|1> + c|2>)
    =<1|b2|1> + <1|bc|2> + <2|cb|1> + <2|c2|2>
    = b2 + c2 (since the orthogonal terms cancel)

    So <1|(B-<B>)2|1>
    = b2 + c2 - 2b2 + b2
    = c2
     
  6. Oct 25, 2003 #5

    Hurkyl

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    That looks right.

    Incidentally, you don't need to use hermiticity for this one...

    <1|(B^2)|1> = <1|BB|1> = <1|B(b|1>) = b<1|B|1> = b^2

    so if you ever work with a nonhermitian transformation, you can still compute things like this.
     
  7. Oct 25, 2003 #6
    Wait....I think my answer doesn't look right. You got <1|(B^2)|1> = b^2, but I got

    <1|B^2|1> = <1|B* B|1>
    = (<1|b + <2|c)(b|1> + c|2> )
    =<1|b^2|1> + <1|bc|2> + <2|cb|1> + <2|c^2|2>
    = b^2 + c^2

    Hm......what's wrong with my way of doing it then? Thanks again :smile:
     
  8. Oct 26, 2003 #7
    nothing. it's just that Hurkyl is sometimes in a hurry and skips some terms :smile:
    the thing I wonder about is when did we say that <B>=b=<1|B|1>? I have this strange feeling that <B>=b+c...
     
  9. Oct 26, 2003 #8

    Hurkyl

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    Somehow, when writing my post, I substituted into my head another problem where |1> was an eigenvector of B with eigenvalue b. :frown:

    For your problem, hermiticity is needed because you don't know what B|2> is. Silly me!
     
  10. Oct 26, 2003 #9
    I got <B> like the following, and <B> is defined to be <1|B|1> in the problem :).

    B|1> = b|1> + c|2>
    <1|B|1> = <1|b|1> + <1|c|2> = b<1|1> + c<1|2> = b

    I hope it's right......

    Thanks :smile:!
     
  11. Oct 26, 2003 #10
    Yeah, I did silly things all the time. But thanks though, now I understand this problem a lot better.
     
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