1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?