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Wigner-Eckart theorem

  1. Mar 28, 2008 #1
    Given a rank-k spherically symmetric tensor operator [itex]\hat{T}^{(k)}_q[/itex] (in other words a family of 2k + 1 operators satisfying [itex][J_z,T_q^{(k)}] = q T_{q}^{(k)}[/itex] and [itex]J_{\pm},T_q^{(k)}] = \sqrt{(k\pm q + 11)(k \mp q)}T_{q\pm 1}^{(k)}[/itex] for all k.

    We have the Wigner-Eckart thorem

    [itex]\langle j',m' |T^{(k)}_q|j,m \rangle = \frac{1}{\sqrt{2j+1}}\langle jk; mq | jk; j'm' \rangle\langle j' || T^{(k)} || j \rangle[/itex]

    where the ``double bar'' is independent of m, m' and q.

    I want to calculate the double bar for the electric dipole operator (proportional to the position operator). I'm expecting the answer to be proportional to [itex]\sqrt{2j+1}[/itex].

    The first thing to answer is whether the theorem applies, ie is the position operator an irreducible spherical tensor operator. Secondly, how would I go about computing the double bar in this case?
     
  2. jcsd
  3. Mar 28, 2008 #2

    vanesch

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    I'm a bit rusty on this, so if I tell nonsense, I hope to be corrected.
    But I would say, yes, the position operator is an irreducible spherical tensor operator of spin 1. Only, the 3 components, x, y and z, are not the "m" components. I guess you have to use something like x + iy, x - iy and z.
     
  4. Mar 28, 2008 #3

    blechman

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    As vanesch says: you can use W-E Theorem here as long as your careful about what m value to use ([itex]r^{\pm 1}\propto x\pm iy[/itex], [itex]r^0\propto z[/itex]).

    The WE Theorem does not tell you how to compute the reduced matrix element ("double-bar") - to compute that, you must go ahead and actually do the integral explicitly. The power of WE is that it allows you to RELATE several matrix elements to only one or two reduced matrix elements. So all you have to do is find the Clebch-Gordan coefficients (table) and compute one or two integrals (as opposed to tens of integrals!). In fact, sometimes you don't even have to compute them: if, for example, you are taking the ratio of matrix elements, sometimes the reduced matrix element cancels and you don't have to do a single integral (yay!).
     
  5. Mar 28, 2008 #4
    Thanks for responding guys.

    Can you tell me how you knew that r was an irred spherical tensor and moreover how did you deduce that the components were [itex]x\pm iy,z[/itex].
     
  6. Mar 28, 2008 #5
    How does one evaluate [itex]\langle j'm' | z | jm \rangle [/itex]. This seems a little bit strange because z is a variable which extends to plus or minus infinity whereas the spherical harmonics only have [itex]\theta,\phi[/itex] dependence.
     
  7. Mar 29, 2008 #6

    vanesch

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    That's because you forgot the "non-angular" index. Look at http://en.wikipedia.org/wiki/Wigner-Eckart_theorem
    for instance. The idea is that you have a complete basis, which as a "non-angular" index n (which can consist of several indices if you want), but which are eigenfunctions of L^2 and Lz (the j and m rotation group indices).

    A rotation applied to the state |n,j,m> will then only mix the m-values.

    What is a spherical tensor operator ? (or better, a set of spherically symmetric operators) It is a set of operators T_k that, under rotation, transform within this set, in a linear combinations of themselves, just like a set |j,m> does.
     
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