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Having a hard time thinking about quantum mechanical vector addition.

  1. Nov 29, 2012 #1
    If a neutron in a nucleus is in a 1p state, before splitting this up in to seperate j-state (due to spin-orbit effect) this neutron has 6 possible states.

    l = 1
    m_l = 1, 0 or -1
    m_s = 1/2 or -1/2

    Splitting this in to j-states corresponding to l+s and l-s, as expected there are 6 states.

    j = 3/2 (l+s)
    m_j = 3/2, 1/2, -1/2, -3/2

    j = 1/2 (l-s)
    m_j = 1/2, -1/2.

    I've tried to think of this result classically. How exactly the j-vector is made of the l and s vectors. For a particular j and mj, we have on a vector diagram that the j-vector can revolve in a circle around that particular m_j value. For any particular direction of this j-vector, the l and s vectors add vectorially and can lie anywhere on a circle that precesses around the j-vector.

    But exactly what m_l, m_s state corresponds to which j/m_j state?

    For j = 3/2, m_j = 3/2, the only state that could correspond to this is m_l = 1, m_s = 1/2. Similarly for j = 3/2, m_j = -3/2, we have m_l = -1, m_s = -1/2.

    But what about the other four states? I've tried for a few hours now to figure out how you can deduce this. For j =1/2, m_j = 1/2, there are two possible states which give the correct z-component and, I believe, the correct magnitude (l = 1, m_s = -1/2, or l = 0, m_s = 1/2). So how is it possible to know which of these two it is? Similarly for j = 3/2, m_j = 1/2, there are two possible states which give the correct z-component and, I believe again, the correct magnitude (l = 1, m_s = -1/2, or l = 0, m_s = 1/2). The same two possible states for both j = 3/2, j = 1/2. So I need to know which of l = 1, m_s = -1/2, or l = 0, m_s = 1/2 corresponds to j = 3/2, or j = 1/2. How is it possible to know?
    Last edited: Nov 29, 2012
  2. jcsd
  3. Nov 29, 2012 #2


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    Big mistake. The idea that angular momenta are little vectors that precess about one another went out in the 1930's. It seems like an appealing idea, but it's wrong. The sooner you let go of this crutch and admit to yourself that the problem is quantum mechanical, not classical, the sooner you'll start to understand what's going on. :smile:
    It's neither. It's a quantum mechanical linear combination of the two. To figure it out, you must use stepping operators. Start with the uniquely defined state, j=3/2, m = 3/2. Apply the lowering operator j- = l- + s- to it. The result will be, by definition, the next lowest state, j=3/2, m = 1/2. It will be a superposition of ℓz=1, sz = -1/2 and ℓz=0, sz = +1/2. The coefficients in front of these two states are called Clebsch-Gordan coefficients. The other state, j=1/2, m = 1/2, can now be found as the linear combination which is orthogonal to the one you just found.
  4. Nov 29, 2012 #3


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    Good, this are the highest (lowest) weight states. Now construct operators which reduce m_j by one unit and operate on the highest weight state.
  5. Nov 29, 2012 #4
    Thanks, that makes a lot more sense. I have actually done stepping operators in a different module, I just hadn't thought. I have been thinking classically as I'm supposed to convince myself that the coupling of J-angular momenta of an odd proton and neutron in a nucleus in the highest outer orbitals tends to favour parallel intrinsic spin. It suggests to think of how their j-vectors couple in each case, then break these j-vectors in to their l and s components and examine the alignment of their intrinsic spins. It does state that this is very approximate and quantum mechanically wrong way to think about it, though.
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