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Vector without basis

  1. Dec 2, 2009 #1
    Inspired by the Dirac bra-ket notation I came to think that an ordinary Euclidean vector must be expressible without reference to a basis. But if I specify the length and angle of a vector, I have to refer this angle to some particular direction. Isn't this the same as choosing a basis?

    Edit: Well I guess length+angle is just polar coordinates ...
     
  2. jcsd
  3. Dec 3, 2009 #2
    Even with bra-ket notation you don't really express a vector without a basis. You just attach some letter to it [tex] | \Psi > [/tex]. Think about it, does it really has any meaning to you? At some point you will have to express it some how (via some basis) to really work with it.

    This is the same way with euclidean vectors. You can write something like [tex] \vec{v} [/tex] forever, and talk about it "interactions" with other anonymous [tex]\vec{u}[/tex]'s but what does that give you until you write it down in some basis? (And defining it through interaction with other vectors is just like selecting a basis).
     
    Last edited: Dec 3, 2009
  4. Dec 3, 2009 #3
    If we have the commutator between operators [itex]\hat x[/itex] and [itex]\hat p[/itex] (position and momentum), we can derive the eigenvalues of the operator [itex]\hat x[/itex] and label the corresponding eigenvectors using the eigenvalues. But what are these eigenvectors REALLY? What do they look like? We know nothing about them, because they are sort of the first vectors in the universe. How will you express them in a basis, if you have no other vectors? But we can express other vectors as linear combinations of them, and this leads to the wave function. Does this make sense?
     
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