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State Vectors as elements of Hilbert Space

  1. Aug 20, 2015 #1
    The state, ##| S\rangle##, say, of a system is represented as a vector in a Hilbert space.

    ##\psi (x, t)## is the representation of the state vector in the position eigenbasis; ##\psi (p, t)## in the momentum eigenbasis et cetera. That is, ##\psi (x, t) = \langle x|S\rangle##; ##\psi (p, t) = \langle p|S\rangle##.

    Now, suppose i expand ##\psi(x,t)## as ##\psi(x,t) = \sum_{n=0}^\infty a_n \cos(\frac{n\pi x} {L})##.
    Similarly for ##\psi (p, t)##.
    [Assuming Dirichlet conditions hold, ##a\leq x\leq b## etc]

    I can now interpret this as: ##\psi(x, t)## is an element of a Hilbert space, with the ##\cos(\frac{n\pi x} {L})##'s as basis. [The ##\cos(\frac{n\pi p} {L'})##'s as basis for ##\psi (p, t)##].

    How do i reconcile these two interpretations? ##\psi (x, t)## as the coefficient in the expansion of ##| S\rangle## in the ##|x\rangle## eigenbasis as opposed to an element of a Hilbert space expanded in the ##\cos(\frac{n\pi x} {L})## basis with coefficients ##a_n##.

    Should i look upon it as an expansion of an expansion (whatever that means)? Please help!
     
  2. jcsd
  3. Aug 20, 2015 #2
    It is an expansion of an expansion.
    The actual vector in the Hilbert space is $$|S\rangle = \int dx |x\rangle \langle x|S\rangle = \int dx \psi(x) |x\rangle.$$ (I leave out the time for simplicity.)
    Now if I have another basis ##|n\rangle##, I can insert the identity again in that expression.
    $$ \int dx \sum_n \langle x|n\rangle \langle n|S\rangle |x\rangle$$
    In your case, ##\langle x|n\rangle = \cos (\frac{n \pi x}{L})## and
    $$\langle n|S \rangle = \int dx \langle n|x\rangle \langle x|S \rangle = \int dx \cos (\frac{n \pi x}{L}) \psi (x) = a_n.$$

    From Ballentine section 4.1
    This is like in relativity where people sometimes call a set of coefficients themselves a vector (or a tensor). But pedantically speaking, it is ##|S\rangle## that is an element of the Hilbert space.
     
  4. Aug 20, 2015 #3
    Thanks for the reply. That helped me get a handle on the problem.
    One question, what 'set' of functions are we talking about? ##\{\psi(x)\}## is one function, defined by the coefficients at each value of the variable x, right?
     
  5. Aug 20, 2015 #4
    You're welcome. I think by ##\{\psi(x)\}## (as opposed to just ##\psi(x)##) Ballentine means a set of wavefunctions, each coming from a different ##|\psi\rangle##. Does that make sense?

    EDIT: I looked at Ballentine again and realize he doesn't use the notation for set, but he does say that ##\psi(x)## is a function and "the vector space consists of the functions ##\psi(x)##." So I guess I put in the set notation automatically without thinking because that's how I interpret it.
     
    Last edited: Aug 20, 2015
  6. Aug 20, 2015 #5
    Yes, i think that is what he means. But, what is that set of functions exactly? What are its elements?

    ##\psi (x)## is simple the set of ordered pairs {##(x_1, \psi (x_1)), (x_2, \psi (x_2)), ... ##} [x is a continuous variable, so i cant really write it like this. But, just to get the point across.]

    So, the thing i'm not being able to comprehend is the nature of the set {##\psi (x)##} as there's just 1 function, ##\psi (x)##!
     
  7. Aug 20, 2015 #6
    To make the notation clear, there should be an additional label like this: ##\{\psi_n(x)\}##.
     
  8. Aug 20, 2015 #7
    But, what would that indicate? What is ##\{\psi_1(x)\}## for example? The coefficient of the ##|x_1\rangle## in the expansion of ##|S\rangle##? But, that is just a number!
     
  9. Aug 20, 2015 #8
    By ##\{\psi_n(x)\}##, I mean ##\{\psi_n(x) = \langle x|\psi_n\rangle|n=1,2,\dots,N\}## or something like that, with ##|\psi_1\rangle, |\psi_2\rangle,\dots## (Edit: or ##|S_1\rangle, |S_2\rangle,\dots##) being vectors in the Hilbert space (and each ##\psi_n(x)## is a function with different values at different ##x##, as you understand).
     
    Last edited: Aug 20, 2015
  10. Aug 20, 2015 #9
    I think i'll have to mull it over a bit more. Will post when i have a clearer picture of this. Thanks for patiently answering my questions! :)
     
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