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Angular momentum eigenstates

  1. Sep 24, 2013 #1

    A_B

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    Hi,

    Both Ballentine in "Quantum Mechanics - a modern development" pages 160-162 and Sakurai in "Modern Quantum Mechanics" pages 193-196 use essentialy the same argument to show the existence of a set of eigenvectors of J² and J_z with integer spaced values of the J_z eigenvalues for fixed J² eigenvalue.

    I get everything up to the result -b_max <= a <= b_max (following Sakurai's notation). But then (in Ballentine) it is claimed that the difference between b_max and -b_max must be an integer. In Sakurai the equivalent (actually stronger) claim is "Clearly, we must be able to reach |a, b_max> by applying J_+ succesively to |a, b_min>".

    It is not obvious to me that this should be so. Can anyone help me see this?


    Thanks,

    A_B
     
  2. jcsd
  3. Sep 24, 2013 #2
    if you have a number and you reduce it by 1 everytime and you have to reach the negative of number in this way.it is possible only when that number is an integer or half of an integer.it simply follows from the relation that
    N-(-N) is an integer.
     
  4. Sep 25, 2013 #3

    A_B

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    Yes, but there is no obvious reason why repeated application of J_+ on |a, b_min> should eventually reach |a, b_max>. For example, if b_max = 3.2[itex]\hbar[/itex] say, then a = 13.44[itex]\hbar[/itex]². The states |13.44[itex]\hbar[/itex]², 3.2[itex]\hbar[/itex]> and |13.44[itex]\hbar[/itex]², -3.2[itex]\hbar[/itex]> are eigenstates by hypothesis, and they give rise to two separate chains of eigenstates, for example

    0 = J_+ |13.44[itex]\hbar[/itex]², 2.8[itex]\hbar[/itex]> = (J_+)² |13.44[itex]\hbar[/itex]², 1.8[itex]\hbar[/itex]> = ... = (J_+)^7 |13.44[itex]\hbar[/itex]², -3.2[itex]\hbar[/itex]>

    and

    0 = J_- |13.44[itex]\hbar[/itex]², -0.8[itex]\hbar[/itex]> = (J_-)² |13.44[itex]\hbar[/itex]², 0.2[itex]\hbar[/itex]> = ... = (J_-)^5 |13.44[itex]\hbar[/itex]², 3.2[itex]\hbar[/itex]>


    I don't see how this situation would lead to a contradiction of anything proved before.
     
  5. Sep 25, 2013 #4
    you are not understanding here.you should be able to go to bmax from bmin by applying the ladder operator.bmax and bmin are simply j and -j.so j should be half integer or full integer.it must go +1 in every step by using ladder operator.
     
  6. Sep 25, 2013 #5

    A_B

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    I can't find proof that indeed |a, b_max> = (J_+)^n |a, -b_max> for some n. That is my problem
     
  7. Sep 25, 2013 #6

    stevendaryl

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    If you started, for example, with a state with [itex]j=1, m=\frac{1}{3}[/itex] and applied the [itex]J_{+}[/itex] operator, you would get to the state [itex]j=1, m=\frac{4}{3}[/itex]. But that's impossible, since [itex]m[/itex] must be less than or equal to [itex]j[/itex]. So there has to be a maximal value for [itex]m[/itex] such that it is impossible to get a new state by applying [itex]J_{+}[/itex] to [itex]|j,m\rangle[/itex].

    So we reason as follows:

    If [itex]|j,m\rangle[/itex] is any state, then either [itex]J_+ |j,m\rangle = 0[/itex] or [itex]J_+ |j,m\rangle \propto |j,m+1\rangle[/itex]. The latter is impossible if [itex]m[/itex] is maximal. So when [itex]m[/itex] is maximal, it must be that [itex]J_+ |j,m\rangle = 0[/itex]. This is only possible if [itex]m=j[/itex].

    So we conclude: For any state [itex]|j,m\rangle[/itex], there is an [itex]n \geq 0[/itex] such that [itex]m+n = j[/itex].
     
  8. Sep 25, 2013 #7

    A_B

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    Thanks for your responses.
    Turns out I was confused by a silly mistake. I'll use Ballentine's notation like in stevendaryl's post, because it's cleaner.

    The state with maximum m value is |β, j>, the state with minimal m value is |β, k> then β = j(j+1) and β = k(k-1) so j(j+1)=k(k-1). Then it is concluded that k=-j, but k = j+1 is also a solution. The second solution can be disregarded because k <= j by hypothesis. I mistakenly wrote down k = 1-j as the second solution and therefore couldn't eliminate the second solution.

    Thanks
     
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