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Homework Help: How measureable parameters associate with operators

  1. Feb 25, 2010 #1
    1. The problem statement, all variables and given/known data
    in quantum mechanics, if we have a wave function, and an operator, we
    can know the eigenvalue from the eigen equation:[tex]\hat{F}[/tex][tex]\phi[/tex]=f[tex]\phi[/tex]. but how we obtain the mathematical form of operator [tex]\hat{F}[/tex]?
    2. Relevant equations

    [tex]\hat{x}[/tex] [tex]\rightarrow[/tex] x ???
    [tex]\hat{p}[/tex] [tex]\rightarrow[/tex] -ih [tex]\partial[/tex]/[tex]\partial[/tex]x ???
  2. jcsd
  3. Feb 25, 2010 #2
    It depends on what you are doing. I mean, mathematically what you are doing is projecting a vector (your eigenvector) onto the space you want to. You can do just about any projection you want, the question is whether it will mean anything. I can go into more depth if you'd like, but I have a feeling that doing some research first will help you take your general question to something more specific.
  4. Feb 25, 2010 #3
    thanks Mindscrape!
    to be specific, my question is how the following associations are constructed:
    [tex]\hat{x}[/tex] [tex]\rightarrow[/tex] x ???
    [tex]\hat{p}[/tex] [tex]\rightarrow[/tex] -ih [tex]\partial[/tex]/[tex]\partial[/tex]x ???
  5. Feb 26, 2010 #4
    I'm not quite sure of your math level, so let me just tell you how it works in the way I know. If I don't answer your questions completely I suggest you see if you can find one of these books:

    Introduction to Quantum Mechanics by Griffiths
    Quantum Mechanics: Concepts and Applications by Zettili (sp.?)
    Principles of Quantum Mechanics by Shankar

    So, as you may or may not know, there are some conditions that have to apply to valid eigenfunctions
    Square Integrable [tex]\int |f(x)|^2 dx < \inf[/tex]
    and observables
    Hermitian [tex]\langle f|\hat{A} g \rangle = \langle \hat{A} f| g \rangle[/tex]

    So basically where we figure out the position operator is by probability. We want to use that operator to figure out where the most likely spot the wavefunction will be appear. We also know that from probability that all the outcomes have to add up to 1, there just can't be greater than 100% chance of things happening (it's our way of 'normalizing' outcomes). So there must be some probability density that describes all the probabilities and that has to add to 1.
    [tex]1=\int \rho(x) dx[/tex]
    In discrete mathematics if we wanted an average we would do
    [tex]\langle x \rangle = \frac{\sum{j N(j)}}{N} = \sum(j P(j))[/tex]
    It translates in such a way that
    [tex] \langle j \rangle = \int j \rho (j) dj [/tex]
    [tex] \langle x \rangle = \int x \rho(x) dx [/tex]
    and so really the probability density [itex]\rho[/itex] is related to the wavefunction, or also called the eigenfucntion, in the way I mentioned earlier
    [tex] 1 = \int |\psi(x)|^2 dx [/tex]
    put it all together
    [tex]\langle x \rangle = \int x |\psi(x)|^2 dx[/tex]
    which is to say that this gives us the most probable location the wavefunction will collapse to.
    In full notation it would be (sometimes people leave the hat off)
    \langle \psi | hat{x} \psi \rangle = \int x |\psi(x)|^2 dx

    For momentum, let's just relate it back to classical mechanics
    just the same in quantum mechanics
    [tex]\hat{p} = \frac{d \langle x \rangle}{dt}[/tex]
    using some integration by parts and schrodinger's equation we eventually see
    [tex]\hat{p} = -i \hbar \frac{\partial}{\partial x}[/tex]

  6. Mar 1, 2010 #5
    sorry to be late. it is really cool!
    thanks again!
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