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Generators and QM mechanics

  1. Apr 1, 2014 #1
    I have a basic question that still is not clear to me. In QM, various operators are generators of something (ie. momentum is a generator of linear translations, Ang. Mom the generator of Rotations and such)

    Question 1: Let us look at L, or a component, say Lz. When we operate on a wavefunction with Lz it seems terminology implies we rotate our system and get a value, But that means we changed it and then get some result. However, we can't really know how we rotated the x-y axis since probability says we don't know how those axis are rotated, we know it is rotated, but not how? Is that correct logic? Seems kind of weird for lack of a better word.

    Then apply that to an experiment, what are you doing to the system to measure this generator of rotations? Are you applying a magnetic field to rotate the system? I think I am also missing the connection between math and application.
  2. jcsd
  3. Apr 1, 2014 #2


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    Staff: Mentor

    If you haven't gotten hold of a copy of Ballentine... Do so.
  4. Apr 1, 2014 #3
    No, this is not correct thinking. First, rotations about z are generated by Lz via exponentiation, [itex]e^{-i\theta L_z}[/itex], so that is what you would use on the state to rotate it. Second, there is no uncertainty in the rotation. It's determined by theta.
  5. Apr 1, 2014 #4


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    No. This has nothing to do with measurements. We have a coordinate system ##x^i## and we apply a rotation ##R^{i}{}{}_{j}x^j##. The rotation matrix ##R^{i}{}{}_j## is a three-dimensional coordinate representation of some element of the rotation group. Associated with each such ##R^{i}{}{}_j## is an infinite-dimensional unitary representation ##\hat{U}## which acts on states according to ##\hat{U} \psi(x) = \psi(R^{-1}x)##; ##\hat{U}## will turn out to be given by the exponential of the rotation generators ##\hat{U} = e^{-i \theta\hat{n}\cdot \hat{J}}## where ##\hat{n}## is the axis of rotation and ##\theta## is the rotation angle. If the rotation matrices ##R^{i}{}{}_{j}## correspond to infinitesimal rotations then the associated transformation of states will just be given by the rotation generators ##\hat{J}## as per ##\hat{J_i}\psi = -i\hbar \epsilon_{ijk}x_j \partial_k \psi## which can be easily verified by expanding ##\hat{U} \psi(x) = \psi(R^{-1}x)## to first order in the infinitesimal rotation angle. Again there are no measurements involved here-we are just transforming states of our system.

    ##\hat{J}## is an observable-it corresponds to the angular momentum operator. What we measure are the eigenvalues of its components, just like with any observable. These correspond to the orbital angular momentum eigenvalues of a system (for a one-component state) and additionally the spin eigenvalues (for a multi-component state). We are not forcing anything to rotate-the system simply has rotational degrees of freedom (relative to a coordinate system for the orbital angular momentum and independent of any for the spin) that are manifest in the eigenvalues of the components of ##\hat{J}## yielded through measurement; it is only here that probabilities come into play, in the usual way.

    An extremely simple system you should review is the rigid rotor: http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/rotqm.html
  6. Apr 1, 2014 #5
    Ok, so I am working on understanding this. I didn't follow the above part of your answer. I get a rotation matrix and its 3D in R3 space. However how do you go from rotation matrix to infinite dimensional unitary representation U. and why do you have it before the wavefunction on LHS but the R^{-1} after it on RHS. That is more math related, but may help clear things up.

    Sorry for my ignorance, I should/need to understand this better. So for the second part, what you are saying is that by being able to measure an eigenvalue of J, you are learning both that it has rotational degrees of freedom, and that the system as such then contains J in its Hamiltonian? (Assuming of course that you did not know H before measurement)

    And degree of freedom in QM is more general than just spacial degrees? So spin 1/2 has 2 degrees of freedom which add to the freedom the system has to be in different states? I will keep studying the rotor example.
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