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Existence of adjoint of an antilinear operaor, time reversal

  1. Nov 27, 2012 #1
    The time reversal operator [itex]T[/itex] is an antiunitary operator, and I saw [itex]T^\dagger[/itex] in many places
    (for example when some guy is doing a "time reversal" [itex]THT^\dagger[/itex]),
    but I wonder if there is a well-defined adjoint for an antilinear operator.

    Suppose we have an antilinear operator [itex]A[/itex] such that
    $$
    A(c_1|\psi_1\rangle+c_2|\psi_2\rangle)=c_1^*A|\psi_1\rangle+c_2^*A|\psi_2\rangle
    $$
    for any two kets [itex]|\psi_1\rangle,|\psi_2\rangle[/itex] and any two complex numbers [itex]c_1^*, c_2^*[/itex].

    And below is my reason for questioning the existence of [itex]A^\dagger[/itex]:

    Let's calculate [itex]\langle \phi|cA^\dagger|\psi\rangle[/itex].
    On the one hand, obviously
    $$
    \langle \phi|cA^\dagger|\psi\rangle=c\langle \phi|A^\dagger|\psi\rangle
    $$
    But on the other hand,
    $$
    \langle \phi|cA^\dagger|\psi\rangle =\langle \psi|Ac^*|\phi\rangle^*=\langle \psi|cA|\phi\rangle^*=c^*\langle \psi|A|\phi\rangle^*=c^*\langle \phi|A^\dagger|\psi\rangle
    $$,
    from which we deduce that [itex]c\langle \phi|A^\dagger|\psi\rangle=c^*\langle \phi|A^\dagger|\psi\rangle[/itex], almost always false, and thus a contradiction!

    So where did I go wrong if indeed [itex]A^\dagger[/itex] exists?
     
  2. jcsd
  3. Nov 28, 2012 #2

    dextercioby

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    Science Advisor
    Homework Helper

    It's not reccomandable to use bra-ket notation in the absence of self-adjoint linear operators.

    It no more difficult to define the adjoint of an antilinear operator than it is to define it for a linear one. The scalar product involved in the definition is the same, there's no scalar in it which would make the difference between antilinear and linear fpr the operator involved.

    Try to rewrite your chain of 5 equalities without bra's and ket's.
     
  4. Nov 29, 2012 #3

    DrDu

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    Science Advisor

    I would be careful already here. You should make clear whether the operator acts on the bra or on the ket. It is not evident whether the two possibilities coincide.
     
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