Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Angular Momentum eigenstates, and tensor products

  1. Aug 5, 2010 #1
    This is taken from a text problem, but I am putting it in this section because I think my question goes beyond the problem itself:

    If a particle has a wave function psi = A*R(r)*cos2 (theta), for example, then if I want to find the probability that its angular momentum is l I would find the absolute value squared of

    P(l) = <L=l|psi> .

    But the eigenstates of the angular momentum operator are products of exp(i*l*theta/h) and an unspecified function of r. If I do the above integral, don't I have to know what the function of r in the eigenfunction for L is? (I mean Lz, to be precise).

    This got me thinking--in this case, |psi> is a tensor product of kets from two Hilbert Spaces, a space with elements depending on r only and a space with elements depending on theta only, correct?

    When such a product is put into the r, theta basis, the resulting wave function is a product of a function of r and a function of theta. So does this mean that all wave functions are like this, i.e. products of functions of one variable? In other words, it would be impossible to have a wave function like exp(i*r*theta), for instance? (Aside from the normalization problem)

    Thanks for your time.
     
  2. jcsd
  3. Aug 6, 2010 #2
    Yes, you can regard the (spinless, for simplicity) particle's state space as a tensor product of functions of r, theta and phi. This doesn't mean that every state function is a product of 3 functions that depend separately on the 3 variables. Rather, you can expand each state function in something like

    [tex]\psi(r,\theta,\phi)=\sum_{abc}c^{abc}f_a(r)g_b(\theta)h_c(\phi)[/tex]

    this is precisely the definition of tensor product.
     
  4. Aug 6, 2010 #3
    Ah, and in the case of a function like psi = cos(r*theta*phi) or something, the expansion would be the Taylor Series. thanks dude
     
  5. Aug 7, 2010 #4
    Or would it? I'm not sure I get it, actually.
     
  6. Aug 7, 2010 #5
    It's not quite a Taylor series, its a different type of expansion. For the theta and phi you take the spherical harmonics, which are a base of the functions defined on the unit spherical surface. For the r part you can take a whole set of functions that are a base of the (0, infinity) interval (Legendere functions, Bessel functions, etc), and you get something like

    [tex]\psi(r,\theta,\phi)=\sum_{nlm}c^{nlm}J_n(r)Y_{lm}(\theta,\phi)[/tex]

    The way to calculate the coefficients is to performs certain integrals. The meaning is essentially the same of Taylor expansion, only the base is different. In the case of taylor expansion the base are polinomials

    [tex]1[/tex]
    [tex]r,\theta,\phi[/tex]
    [tex]r^2,\theta^2,\phi^2,r\theta,r\phi,\theta\phi[/tex]

    [tex]\dots[/tex]

    in other cases you have more complicated yet more useful functions.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Angular Momentum eigenstates, and tensor products
  1. Momentum eigenstate (Replies: 3)

Loading...