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Kets, bras, hermiticity etc.

  1. Oct 27, 2005 #1
    I'm reading in Sakurai's 1st chapter that this follows from the "associative axiom":

    [tex]
    \langle\beta|\cdot\left(X|\alpha\rangle\right) = \left(\langle\beta|X\right)\cdot|\alpha\rangle
    [/tex]

    so we might as well write [tex]\langle\beta|X|\alpha\rangle[/tex]. I know this is basic stuff, but I thought this notation only made sense when X is hermitian since when you let X act on the bra instead of the ket you must take the hermitian conjugate. Like this:

    [tex]
    \langle\beta|\cdot\left(X|\alpha\rangle\right) = \left(\langle\beta|X^\dagger\right)\cdot|\alpha\rangle
    [/tex]

    It's pretty bad that I'm in trouble already in the first chapter (exam next thursday).
     
  2. jcsd
  3. Oct 27, 2005 #2
    I believe the Hermitian conjugate arises only when one is finding the dual bra of a ket, and vice versa. So, the associativity axiom applies as Sakurai describes since we are not taking the conjugate of anything.
     
  4. Oct 27, 2005 #3

    Hurkyl

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    [tex]\langle\Psi|X = \langle X^* \Psi|[/tex]
    [tex](X|\Psi\rangle)^* = \langle\Psi|X^*[/tex]
     
  5. Oct 27, 2005 #4

    Hans de Vries

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    [tex]
    \langle\beta|\cdot\left(X|\alpha\rangle\right) = \left(\langle\beta|X\right)\cdot|\alpha\rangle
    [/tex]


    It works even if α and β are different but only because everything is added
    together to a single scalar at the end.

    You can see it more easily if you work it out for simplified vectors α and β
    and matrix X, where α and β both have only a single non-zero element.
    X now has only a single relevant element while all other positions may be zero.

    Work it out for this elementary case. Then have a look how the general case
    is a just a linear combination of this simple case. Remember that an operator
    acting "to the left" works as being transposed.

    You'll see that in the first case you sum over the rows and then the columns
    while in the second case it is columns first and then rows. The end result is
    the same.


    Regards, Hans
     
  6. Oct 28, 2005 #5
    Yeah, that's pretty much the definition of what the Hermitian conjugate is, is that [tex] \langle \Phi |(\Omega |\Psi \rangle) = (\langle \Phi | \Omega^{\dagger}) |\Psi \rangle [/tex]. Most mathematicians consider this to be the DEFINITION for a Hermitian conjugate, is the change in the operator that makes that statement true.
     
  7. Oct 31, 2005 #6

    dextercioby

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    Here's the trick:

    [tex] \langle \alpha ,\hat{X} \beta\rangle [/tex]
    is to be understood as follows:

    it is the scalar product between the vector [itex] \alpha [/itex] and the vector which results when applying the linear operator [itex] \hat{X} [/itex] on an arbitrary vector from its domain [itex] \beta [/itex].
    But at the same time
    it is the value of the linear functional [itex] F_{\alpha} [/itex] (which is continuous on the image of the linear operator [itex] \hat{X}[/itex]) when acting on the vector [itex] \beta [/itex].

    What Sakurai is saying is that this complex number is equal to this number
    [tex] \left (\tilde{\hat{X}} F_{\alpha}\right) \beta [/tex]
    which is nothing but the value of the linear functional obtained when applying the dual operator [itex] \tilde{\hat{X}} [/itex] on the linear functional [itex] F_{\alpha}[/itex] when acting on the vector [itex] \beta [/itex] .

    That's all.

    Daniel.
     
    Last edited: Oct 31, 2005
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