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Question about parity

  1. Jan 5, 2009 #1
    I'm reading a book: B.R. Martin's Nuclear and Particle Physics - An Introduction.

    In section 1.3.1 on Parity, it states the following:

    When dealing with p·x in the exponent, which should p and x be treated as - vectors or operators?

    Suppose I work it out.

    If exp[tex][i([/tex]p·x[tex]-Et)][/tex] is an eigenfunction of momentum, then I assume that means that:

    [tex]\hat{p}\psi([/tex]x,t)=[tex]-i\hbar\frac{\partial}{\partial x}[/tex]exp[tex][i([/tex]p·x[tex]-Et)][/tex]

    Now if p and x are treated as vectors, then I will have to rewrite p·x as pxcosθ, which would lead to [tex]\hbar p \cos{\theta}[/tex]exp[tex][i([/tex]p·x[tex]-Et)][/tex], right?

    Then the author goes on to say
    I don't really understand that. Because if I did the above math correctly, then if p = 0, then [tex]\hbar p \cos{\theta}[/tex]exp[tex][i([/tex]p·x[tex]-Et)][/tex] would also be 0.

    Can someone clear this up for me? Thanks.
    Last edited: Jan 5, 2009
  2. jcsd
  3. Jan 5, 2009 #2


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    Sorry, don't know that one.

    They should be vectors, because you are taking the inner product.

    Actually that doesn't say much. Basically you just wrote out what [itex]\hat p[/itex] looks like in "real"-space. Being an eigenfunction means that
    [tex]\hat p \psi(x, t) = p \psi(x, t)[/tex]
    where the p on the right hand side is just a number. If you want to write it in real space, then you get
    [tex]-i \hbar \frac{\partial}{\partial x} e^{i \vec p \cdot \vec x - E t} = P e^{i \vec p \cdot \vec x - E t} [/tex]
    I have called the eigenvalue P here, because from a mathematical point of view, it is just a number which needn't have anything to do with the momentum vector [itex]\vec p[/itex]. In linear algebra, you would conventionally use [itex]\lambda[/itex] instead of P.

    That is a possibility, where you go to polar coordinates in which you express a vector [itex]\vec p[/itex] in terms of a length [itex]p[/itex] and two angles [itex]\theta, \phi[/itex] and you choose your axes such that one of the variables [itex]\theta[/itex] is precisely the angle between [itex]\vec p[/itex] and [itex]\vec x[/itex] (that is, you make [itex]\vec p[/itex] lie along the z-axis).

    You did the math correctly. Note that if [itex]\vec p = \vec 0[/itex], then in your polar coordinates in which p is the radial coordinate, also [itex]p = 0[/itex]. So even after multiplying by [itex]\hbar \cos\theta e^\cdots[/itex] it will be zero. That entire exercise was just to show that indeed the exponential is an eigenfunction of the momentum operator, and you can read of the eigenvalue (it's the lambda in [itex]\hat p \psi = \lambda(p) \psi[/itex]).

    I think the point the author is trying to make is, that if the momentum vector is zero, there is no dependence on [itex]\vec x[/itex] anymore. So if you flip all the spatial coordinates [itex]\vec x \to -\vec x[/itex] (meaning x to -x, y to -y, z to -z, t to t) then the eigenfunction won't change.

    Hope that lifts some confusion?
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