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Adjoint operator

  1. Dec 11, 2008 #1

    KFC

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    For an opeation as follow

    [tex]\langle \varphi | \hat{A} |\psi\rangle[/tex]

    where [tex]\hat{A}[/tex] is an operator.

    It is no problem to have [tex]\hat{A}[/tex] directly operate on the ket state [tex]|\psi\rangle[/tex], but if I want [tex]\hat{A}[/tex] operates on the bra state [tex]\langle\varphi |[/tex], do I have to take the adjoint of A first? That is

    [tex]\left(\langle \varphi | \hat{A}^\dagger\right) |\psi\rangle = \langle \varphi | \left(\hat{A} |\psi\rangle\right)[/tex]
     
  2. jcsd
  3. Dec 11, 2008 #2

    Fredrik

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    No. Lose the that dagger on the left-hand side. Se #7 in this thread. (In particular the stuff I quoted from another thread).
     
    Last edited: Dec 11, 2008
  4. Dec 12, 2008 #3

    KFC

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    So, as you told in that thread. For

    [tex]{\Large\langle\varphi | A | \psi \rangle}[/tex]

    the opreator A can either operate on [tex]\psi[/tex] or [tex]\varphi[/tex] ?
     
  5. Dec 12, 2008 #4
    Think of it this way, A acts on the thing to the right while [tex]A^{\dagger}[/tex] acts on the thing to the left. If you want to operate on the bra-state [tex]\langle \phi | [/tex] then you do need [tex]A^{\dagger}[/tex] unless A is self adjoint, in which case [tex]A = A^{\dagger}[/tex] and the operator can act either to the left or the right.

    If A is some observable then it must be self-adjoint.
     
  6. Dec 12, 2008 #5

    KFC

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    Got it. In this sense, the following conclusion only true when [tex]A^\dagger = A[/tex], right? That is, if operator A is self-adjoint, it can either operate on the left or the right?

    [tex]\left(\langle \varphi | \hat{A}^\dagger\right) |\psi\rangle = \langle \varphi | \left(\hat{A} |\psi\rangle\right)[/tex]
     
  7. Dec 12, 2008 #6

    Hurkyl

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    This is nothing new, nor special. Think about any functions at all:
    [tex]f(g(x)) = (f \circ g)(x)[/tex]
    i.e. both of these give the same result:
    1. Evaluating g at x, then evaluating f at the result
    2. Composing f and g, and evaluating the result at x

    And similarly, think about matrix arithmetic. I can evaluate the product wAv in any order I choose.
     
  8. Dec 12, 2008 #7

    Fredrik

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    I suppose you can say that, if you mean [itex]|\psi\rangle[/itex] and [itex]\langle\varphi|[/itex].

    The expression [itex]\langle\varphi|A|\psi\rangle[/itex] means [itex]\langle\varphi|(A|\psi\rangle)[/itex], and by definition of the bra, that's the scalar product of [itex]|\varphi\rangle[/itex] and [itex]A|\psi\rangle[/itex]. By definition of the adjoint operator, that scalar product is equal to the scalar product of [itex]A^\dagger|\varphi\rangle[/itex] and [itex]|\psi\rangle[/itex], which in bra-ket notation takes the form [itex](\langle\varphi|A)|\psi\rangle)[/itex]. That's why you can drop the parentheses.


    I don't see a way to interpret this as a correct statement.


    Any operator X satisfies [itex]\langle\varphi|(X|\psi\rangle)[/itex]=[itex](\langle\varphi|X)|\psi\rangle[/itex], so yes, your equation is correct for all [itex]\langle\varphi|[/itex] and [itex]|\psi\rangle[/itex] if and only if A is self-adjoint, but I don't see why you would want to express that as "if operator A is self-adjoint, it can either operate on the left or the right".
     
    Last edited: Dec 12, 2008
  9. Dec 13, 2008 #8

    KFC

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    Oh, there are too many comments here. It becomes more confusing. Fredrik, could you please ask me the following questions again. I read the thread you post and now let's start from

    [tex]\langle f | A | g \rangle = (|f\rangle, A|g\rangle) = (A|g\rangle, |f\rangle)^* = \langle g | A^\dagger | f \rangle ^*[/tex]

    Is this right? In this case (A might not btself-adjoint), [tex]A^\dagger[/tex] should operate to the left or right?


    My second question is : assuming [tex]A^\dagger[/tex] still operates to the right, considering the last term in above equation

    [tex]\langle g | A^\dagger | f \rangle ^* = (|g\rangle, A^\dagger | f \rangle)^* = \left((A^\dagger | f \rangle, |g\rangle)^*\right)^* = (A^\dagger | f \rangle, |g\rangle)[/tex]

    Can we interpret [tex]\langle f | A | g \rangle [/tex] in this way? Firstly, let [tex]A^\dagger[/tex] operates to the ket [tex]|f\rangle[/tex] and then take the inner product with the ket [tex]|g\rangle[/tex], right? In a word, if I want an operator operates to a bra [tex]\langle f|[/tex], I can have its corresponding adjoint operator acting on the corresponding ket [tex]|f\rangle[/tex] to get another ket, after I get the result, I convert that ket back to a corresponding bra, is that right?

    I am asking this question because sometimes I got an expression like [tex]\langle f | A | g \rangle[/tex], I need to keep [tex]|g\rangle[/tex] unchanged while I've already know the eigenvalue problem [tex]A|f\rangle[/tex], that's why I am asking how to make A operate to the left.
     
  10. Dec 13, 2008 #9

    Fredrik

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    Suppose that [itex]A|a\rangle=a|a\rangle[/itex], then what is [itex]A^\dagger|a\rangle[/itex]? Let's find out:

    [tex]A^\dagger|a\rangle=\sum_{a'}|a'\rangle\langle a'|A^\dagger|a\rangle[/tex]

    [tex]\langle a'|A^\dagger|a\rangle=(|a'\rangle,A^\dagger|a\rangle)=(A|a'\rangle,|a\rangle)=(a'|a'\rangle,|a\rangle)=a'^*(|a'\rangle,|a\rangle)=a'^*\delta_{a'a}[/tex]

    [tex]A^\dagger|a\rangle=\sum_{a'}|a'\rangle a'^*\delta_{a'a}=a^*|a\rangle[/tex]

    You can use this to find out what you should do when the eigenstate appears in the form of a bra on the left.

    [tex]\langle a|A|b\rangle=(|a\rangle,A|b\rangle)=(A^\dagger|a\rangle,|b\rangle)=(a^*|a\rangle,|b\rangle)=a(|a\rangle,|b\rangle)=a\langle a|b\rangle[/tex]

    Note that there's no need to ever talk about an operator acting to the left. We don't have to define the "product" of a bra and an operator, but we do it anyway because it's convenient. We can define [itex]\langle f|A[/itex] either by

    [tex](\langle f|A)|g\rangle=\langle f|(A|g\rangle)[/tex]

    or (equivalently) by

    [tex](\langle f|A)|g\rangle=(A^\dagger |f\rangle, |g\rangle)[/tex]

    This definition allows us to interpret the previous result as

    [tex]\langle a|A=\langle a|a[/tex]
     
    Last edited: Dec 13, 2008
  11. Dec 13, 2008 #10

    KFC

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    Got you. Thank you so much!
     
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