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Bra's and Ket's independent of basis?

  1. Jul 23, 2015 #1
    I'm just learning about the whole Dirac notation stuff and I have come across the fact that bra's and ket's are somehow independent of bases. Or rather that they do not need the specification of a basis. I really don't understand this from a vector point of view. Maybe that is the problem, should I not be thinking about these like vectors?
    E.g. I understand that [itex]\mathbf e_i[/itex] needs a basis, but for some reason [itex]\mid e_i \rangle[/itex] doesn't?

    Thanks in advance!
     
  2. jcsd
  3. Jul 23, 2015 #2

    wle

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    It's probably meant in the sense that a ket is a basis-independent concept as opposed to, for instance, its components in some particular basis, i.e., you can expand a ket vector as something like ##\lvert \psi \rangle = \sum_{k} c_{k} \lvert k \rangle## (or e.g. ##\lvert \psi \rangle = \int \mathrm{d}x \, \psi(x) \lvert x \rangle## in the continuous case). The components ##c_{k}## (or ##\psi(x)##) depend on the particular basis, but the ket ##\lvert \psi \rangle## does not.

    It's the same thing as with the relation between (say) a Euclidean vector ##\bar{u}## and its expression as a list of components ##[u_{x}, u_{y}, u_{z}]## in some given Cartesian coordinate system.
     
  4. Jul 23, 2015 #3

    bhobba

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    They are vectors - but its a bit subtler than what you likely have come across in linear algebra due to the fact they have an infinite basis.

    Its full elucidation comes when you delve into the advanced area of Rigged Hilbert Spaces - but that is not recommended for beginners. As an introduction see chapter 1 of Ballentine.

    Thanks
    Bill
     
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