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Dirac notation confusion

  1. Oct 21, 2013 #1
    1.) an inner product of a state vector represent by <[itex]\psi[/itex]|[itex]\psi[/itex]>. sometimes the notation is like <[itex]\phi[/itex]|[itex]\psi[/itex]> is mean transfer from state |[itex]\psi[/itex]> to <[itex]\phi[/itex]|.it mean the former 1 do not transfer the state? what is the difference between both?
    2.) what is mean by <x|[itex]\psi[/itex]>? is it mean x(position) mean the final state we reach where we extract form |[itex]\psi[/itex]> ? from eigen value equation [itex]\hat{x}[/itex] |[itex]\psi[/itex]>=[itex]\lambda[/itex]|[itex]\psi[/itex]> where [itex]\lambda[/itex] is the eigen value of position. so why do we need <x|[itex]\psi[/itex]>?
  2. jcsd
  3. Oct 21, 2013 #2


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    Staff: Mentor

    It is similar to the dot product of ordinary vectors. It is not a "transfer", but rather a projection. ##\langle \psi | \psi \rangle## is the "length" of the state ##\psi##. It is equal to 1 if the state is normalized.

    I will not go into the fact that ##\langle x |## is not a proper bra, but basically writing
    ##\langle x | \psi \rangle## allows to given a representation of the state ##| \psi \rangle## as a wave function of the cartesian coordinate ##x##,
    \psi(x) = \langle x | \psi \rangle

    ##| \psi \rangle## is an arbitrary quantum state, not an eigenvector of the ##\hat{x}## operator.
  4. Nov 1, 2013 #3
    is projection kinda same as the ψ "collapse" to 1 of the eigenstate [itex]\phi[/itex]?
  5. Nov 1, 2013 #4


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    Science Advisor

    No. Do you know anything about dot-products from linear/vector algebra? The <|> inner products are just generalization of dot-product. Phi and psi are vectors, and [itex]\langle \phi | \psi \rangle[/itex] is just a dot-product between the two. Of course, in the continuous case, the sum over components becomes an integral over position. Hence if you are given [itex]\langle x | \psi \rangle = \psi(x)[/itex] and [itex]\langle x | \phi \rangle = \phi(x)[/itex], then [itex]\langle \phi | \psi \rangle = \int \phi^*(x)\psi(x) dx[/itex].
  6. Nov 3, 2013 #5
    But if $\phi$ is an eigenfunction of $\psi$, then:
    \langle \phi | \psi \rangle
    is the transition amplitude, right? (I think that is maybe what the OP is asking.)
  7. Nov 3, 2013 #6


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    [itex]|\phi\rangle[/itex] is a vector. So is [itex]|\psi\rangle[/itex]. So one can't be an eigen vector of the other. Because they are both vectors. To have an eigen vector, you must be dealing with a map from a Hilbert space to Hilbert space. Such a map is called an operator. So if A is an operator, then for every [itex]|\psi\rangle[/itex] there exists a corresponding vector [itex]|A\psi\rangle[/itex] such that [itex]|A\psi\rangle = A|\psi\rangle[/itex]. Furthermore, if [itex]|\psi\rangle[/itex] is an eigen vector of A, then [itex]A|\psi\rangle = \lambda |\psi\rangle[/itex] for some [itex]\lambda[/itex].
  8. Nov 4, 2013 #7
    Yeah, you are completely right. Sorry. I meant if [itex]|\phi\rangle[/itex] is a basisvector of the Hilbert space, then my statement is correct?
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