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Spin angular momentum operators

  1. Aug 21, 2007 #1
    The context of this question is chemistry but I think that it contains enough quantum mechanics to warrent posting it here instead of in the chemistry forum.

    Go to section 2.1.1 at the following site:

    http://tesla.ccrc.uga.edu/publications/papers/qrevbiophys_v33p371.pdf [Broken]

    I am confused about how you go from equation (2) in 2.1.1 which contains the Hamiltonian operator on the left side and the spin angular momentum operators on the right side to equation (3) in 2.1.2 without any operators in the equation at all.

    Is this just the result of applying both sides to a wavefunction or something?
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Aug 22, 2007 #2


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    Apparently if (2) was mediated, the spin part produces 1 by mediation, that is, of course of there's any connection between (2) and (3) at all. Actually the spin part is time independent and, since it doesn't appear in (3), i guess there's no connection between the 2 formulae.
  4. Aug 22, 2007 #3
    What do you mean there is no connection between (2) and (3)?

    They are nearly identical.

    I just want to know how you get rid of the operators when you go from (2) to (3)?
  5. Aug 22, 2007 #4


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    They mean a time average, the way they've written it. But the spin part, from (2) appears to be time-independent, so if the 2 formulae are connected one to another, it follows that from (2) to (3) they've passed through time-averaging and setting the spin operators product to 1 (unit operator, probably).

    Perrhaps i'm not seeing something; i hope someone else can join us in this thread.
  6. Aug 22, 2007 #5
    OK. I could see that if the spin operators are unitary. However, I am also confused about why there is an [tex]H^D[/tex] operator on the LHS of (2) and only D on the LHS of (3).

    D is the coupling value so I am confused about how you go from an operator to a value.
  7. Aug 22, 2007 #6
    I reread it and I think equation (3) is the expectation value for the Hamiltonian operator in equation (2). So, now my question is how did they arrive at this expectation value? What happened to the spin angular momentum operators?? Do unitary operators always disappear in an expectation value calculation? Are they even unitary operators?
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