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Adjoint of a bra-ket

  1. Sep 13, 2011 #1
    Hello everybody,

    why is the adjoint of a bra-ket like this:

    [itex] < \phi | \psi >^+ = < \psi | \phi >[/itex]

    Is it a definition or can it be derived somehow?

    Thanks :)
  2. jcsd
  3. Sep 13, 2011 #2


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    The scalar product in a (pre-)Hilbert space is a sesquilinear form, i.e., by definition

    [tex]\langle \psi|\phi \rangle = \langle \phi|\psi \rangle^*[/tex]


    [tex]\langle \psi | \alpha \phi_1 + \beta \phi_2 \rangle = \alpha \langle \psi | \phi_1 \rangle + \beta \langle \psi | \phi_2 \rangle.
  4. Sep 14, 2011 #3
    Thank you!
  5. Sep 14, 2011 #4


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    When [itex]\langle\psi|\phi\rangle[/itex] denotes the inner product (or semi-inner product) of [itex]\psi[/itex] and [itex]\phi[/itex], what vanhees71 said is the complete answer. But if it denotes [itex]\langle\psi|[/itex] acting on [itex]|\phi\rangle[/itex], some elaboration is required. [itex]\langle\psi|[/itex] is defined as a function that takes kets to complex numbers. To be more specific, it's defined as the function such that takes [itex]|\phi\rangle[/itex] to [itex]\big(|\psi\rangle,|\phi\rangle\big)[/itex]. (Here I'm using the [itex](\cdot,\cdot)[/itex] notation for the inner product of two kets, to make things more readable). Now we can prove it like this:
    [tex]\langle\psi|\phi\rangle^* =\big(\langle\psi|\big(|\phi\rangle\big)\big)^* =\big(|\psi\rangle,|\phi\rangle\big)^* =\big(|\phi\rangle,|\psi\rangle\big) =\langle\phi|\big(|\psi\rangle\big) =\langle\phi|\psi\rangle[/tex] The equality in the middle is the same identity that vanhees71 mentioned. As he said, it's part of the definition of an inner product.
  6. Sep 14, 2011 #5


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    The bra-ket is not a scalar product, but a short-hand for the action of a linear functional on a vector which yields a complex/real scalar.
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