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Homework Help: Squaring a bra

  1. Sep 28, 2007 #1
    Which of the following is true?

    [tex]<\phi|^2 = <\phi|\phi>[/tex]
    [tex]<\phi|^2 = |\phi><\phi|[/tex]

    Or does the order even matter?
  2. jcsd
  3. Sep 28, 2007 #2
    I'm not entirely sure you can "square a bra" like that. But the order certainly matters.
  4. Sep 28, 2007 #3
    Well I guess I should have said how do you multiply a bra by its complex conjugate, can you stick in the front or the back?
  5. Sep 28, 2007 #4
    If you want a scalar, use option one. If you want an operator, use option two. Neither are called "squaring" -- and they're completely different things.
  6. Sep 28, 2007 #5
    <phi|^2 = <phi|<phi| (pretty much meaningless)
    <phi|phi> = 1 (or some other constant)
    |phi><phi| = P_phi (an operator)
  7. Sep 29, 2007 #6


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    I don't think the tensor product is meaningless. One can build multi-particle states from it.
  8. Sep 29, 2007 #7
    Ok well more specifically, my problem is turning this:

    [tex](E_n - E_{n'}) <\phi_{n'}|X|\phi_n> = \frac{i\hbar}{m} <\phi_{n'}|P|\phi_n>[/tex]

    into this:

    [tex]\displaystyle\sum_{n'}^{} (E_n - E_{n'})^2 |<\phi_{n'}|X|\phi_n>|^2 = \frac{\hbar^2}{m^2} <\phi_n|P^2|\phi_n>[/tex]

    using the closure relation. So basically everything is getting squared, but I don't know how to handle the part with the P and it's respective bra and ket. And what is confusing me as well are the two [tex]\phi_n[/tex]'s intead of a [tex]\phi_{n'}[/tex] in there.
  9. Sep 29, 2007 #8


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    is X and P the position and momentum operator?

    What have you tried so far?
  10. Sep 29, 2007 #9


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    Keep the following in mind, and give it a shot:

    [tex] \langle \phi _1 | A | \phi _2 \rangle ^* = \langle \phi _2 | A^{\dagger} | \phi _1 \rangle [/tex]

    If [itex] \{ \phi _i \} [/itex] is a complete basis, then [itex] \sum _i |\phi _i \rangle \langle \phi _i |= \mathbf{1} [/itex]

    You want to multiply every complex number in the upper expression by its complex conjugate and then sum over n'.

    PS: Before doing this, you might want to review the math a little bit. It seems you are unfamiliar with some of the basics of the algebra of a linear vector space.

    For instance:

    A bra is not a complex number (scalar) and hence, can not have a complex conjugate. It does, however, exhibit a dual correspondence to a ket living in a ket space that is dual to the bra space that your bra comes from.
    Last edited: Sep 29, 2007
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