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I Wave function: vector or scalar?

  1. Apr 5, 2016 #1
    Meant as element of Hilbert space of L^2 functions... etc., the wave function is a vector.

    In the abstract description with kets and operators on these, the wave function is the scalar product between a ket |Psi> and the "eigenkets" |x> of the position operator: psi(x) = <x|Psi>.

    So: psi is a vector or a scalar?

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  3. Apr 5, 2016 #2

    ShayanJ

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    The wave-function(## \psi(x)=\langle x | \psi \rangle ##) is a scalar in the sense that it doesn't change under a coordinate transformation ## x \rightarrow x' ##, i.e. ## \psi'(x')=\psi(x) ##. In this sense, a vector means a collection of components that transform to linear combinations of themselves under a coordinate transformation. The wave-function is not a vector in this sense.
    Here you can read about the notion of a vector space. The wave-function is a member of some vector space, and is said to be a vector in this sense. The vector space that ## \psi(x)=\langle x | \psi \rangle ## is a member of, is the space of square-integrable complex valued functions over ## \mathbb R ## with the appropriate boundary conditions.
     
    Last edited: Apr 5, 2016
  4. Apr 5, 2016 #3
    You are calling two quantities with the same name: the ket |Psi> is the "quantum state", it belongs to the Hilbert space and it is a vector, while the wavefunction Psi(x) is the projection of this state on a basis element (x), and it's a function.
     
  5. Apr 5, 2016 #4

    vanhees71

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    The object ##|\Psi \rangle## is in the abstract Hilbert space ##\mathcal{H}## (up to equivalence there's basically only one separable Hilbert space). The wave function ##\Psi(x)=\langle x|\Psi \rangle## is the component of ##|\Psi \rangle## with respect to the generalized eigenbasis of the position operator.

    In this way there is a one-to-one mapping between the abstract Hilbert space ##\mathcal{H}## and the Hilbert space of square-integrable functions ##L^2(\mathbb{R})##.
     
  6. Apr 5, 2016 #5
    Thanks.
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