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What is a useful way to talk about eigenstates of the position operator

  1. Mar 17, 2010 #1
    So I've been having a specific major hang-up when it comes to understanding basic quantum mechanics, which is the position operator.

    For the SHO, the time independent Schroedinger's equation looks like

    E\psi = \frac{\hat{p}^2}{2m}\psi + \frac{1}{2}mw^2\hat{x}^2\psi

    Except that usually it isn't written with [tex]\hat{x}[/tex] but with [tex]x[/tex]. What really showed me that I didn't understand this was an extra credit problem on an exam today, where

    E\psi = \frac{\hat{p}^2}{2m}\psi + \frac{1}{2}mw^2z^2\psi + az\sigma_z + b\sigma_z

    and to solve it we needed to show that H and [tex]\sigma_z[/tex] commuted, then show some other business. But to treat z as a variable seemed incorrect, since to do that, we would need to be using a basis of its eigenstates. I also got confused when thinking about the collapse of observing position onto a single position, which would transform the wavefunction in x into a delta function. It seems, then, that delta functions are the eigenstates of our position operator, since an observation collapses our wave function into one. Then writing wave functions in the variable x would be turning that infinite-dimensional basis of delta functions into a different basis. Alternately, if we assume that we are in the state |x>, the quantum number for this state would be x, and so [tex]\hat{x}|x>=x|x>[/tex]

    Is any of this along a good line of though?

    Also, the solutions to eigenstates of H for the SHO write the x operator as a sum of raising and lowering operators. I could probably find the eigenstates for x in terms of the SHO basis. Will the basis of x be the same for any problem, and will it then make [tex]\hat{p}=i\hbar\frac{d}{dx}[/tex] the same when I am using this basis. What should I call this basis, function space?
  2. jcsd
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