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A Highest state of SU(2) algebra

  1. Oct 6, 2016 #1
    Let the generators of the SU(2) algebra be ##\tau_{1}##, ##\tau_{2}## and ##\tau_{3}##.

    Consider an ##N## dimensional representation, which means that the ##\tau_{i}## are ##N \times N## matrices which act on some ##N##-dimensional vector space.

    Consider the ladder operators ##\tau_{\pm}=\tau_{1}\pm i\tau_{2}##. Let's work in a basis where ##\tau_{3}## is diagonal, and let ##|m\rangle## denote a unit normalized eigenstate of ##\tau_{3}## with eigenvalue ##m##.

    Now, I can show that ##\tau_{\pm}## can raise/lower the eigenvalue of ##|m\rangle## by 1, since ##\tau_{3}(\tau_{\pm}|m\rangle)=(m\pm 1)(\tau_{\pm}|m\rangle)##.

    I can also show that ##\tau_{-}## can annihilate the state ##|1\rangle##, since ##\tau_{3}(\tau_{-}|1\rangle)=(1-1)(\tau_{-}|1\rangle)=0##

    How do I show that ##\tau_{+}## also annihilates some state?
     
  2. jcsd
  3. Oct 6, 2016 #2
    This sounds a bit handwavy, but you have a finite-dimensional vector space - so the number of eigenstates of ##\tau_{3}## is finite. So it is not possible to keep applying ##\tau_{+}## to raise the state indefinitely; it must terminate at some maximum ##|m\rangle##.
     
  4. Oct 6, 2016 #3
    How about the following:

    Consider the highest state ##|j\rangle## labelled by its eigenvalue ##j##. Then, ##\tau_{3}(\tau_{+}|j\rangle)=(j+1)(\tau_{+}|j\rangle)##.

    But ##|j\rangle## is the highest state, and ##(j+1) \neq 0##. Therefore, ##\tau_{+}|j\rangle = 0##.

    Is this sound?
     
  5. Oct 6, 2016 #4
    Sounds good to me (that's how I would do it too if I had to write it down mathematically)
     
  6. Oct 6, 2016 #5
    How do I argue that there must be exactly one state annihilated by ##\tau_{+}## and that this is a ##\tau_{3}## eigenstate?

    I have already shown that ##\tau_{+}## annihilates some highest state ##|j\rangle##. All I can do now is to argue that ##\tau_{+}## does not annihilate the lowest state - ##\tau_{+}## cannot annihilate the lowest state, because it can only raise a state, not lower it.

    What are your thoughts?
     
  7. Oct 6, 2016 #6
    Well, if the set of ##\tau_3## eigenstates forms a complete basis of your vector space (as it should), then any state can be represented in terms of these ##\tau_3## eigenstates, and we know that ##\tau_+## only annihilates ##|j\rangle## (among the set of ##\{|m\rangle\}##), so if ##\tau_{+} |\psi\rangle = \sum c_{n} \tau_{+} |n\rangle## vanishes, it must be that all the ##c_n##s vanish for ##n \neq j##.
     
  8. Oct 7, 2016 #7

    stevendaryl

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    I think that a way to get both an upper and lower bound on [itex]\tau_3[/itex] is by realizing that [itex]\tau_1, \tau_3, \tau_3[/itex] commute with [itex]S^2 \equiv (\tau_1)^2 + (\tau_2)^2 + (\tau_3)^2[/itex]. So we can find states that are simultaneously eigenstates of [itex]S^2[/itex] and [itex]\tau_3[/itex]. So let's restrict attention to eigenstates of [itex]S^2[/itex] with eigenvalue [itex]\lambda[/itex]. Then we can write:

    [itex]\tau_+ \tau_- = (\tau_1 + i \tau_2)(\tau_1 - i \tau_2) = [/itex] (using commutation relations and definitions) [itex]S^2 - \tau_3(\tau_3 - 1)[/itex]

    If we let [itex]|m\rangle[/itex] be a state where [itex]\tau_3 |m\rangle = m |m \rangle[/itex], then we know that [itex]\tau_- |m\rangle[/itex] returns a state that is proportional to [itex]|m-1\rangle[/itex]. Let [itex]\alpha_m[/itex] be the constant of proportionality:

    [itex]\tau_- |m\rangle = \alpha_m |m-1\rangle[/itex]

    We can show that [itex]\tau_+ |m-1\rangle = \alpha_m |m\rangle[/itex]. So putting the two together, we have:

    [itex]\tau_+ \tau_- |m \rangle = (\alpha_m)^2 |m\rangle[/itex]

    But we also know that [itex]\tau_+ \tau_- = S^2 - \tau_3(\tau_3 - 1)[/itex]. So we have:

    [itex](S^2 - \tau_3(\tau_3 -1)) |m\rangle = (\alpha_m)^2 |m\rangle[/itex]

    Since we assumed we're dealing with eigenstates of [itex]S^2[/itex] with eigenvalue [itex]\lambda[/itex], and by definition, [itex]\tau_3 |m\rangle = m |m \rangle[/itex], we have:

    [itex](\lambda - m(m-1)) |m\rangle = (\alpha_m)^2 |m\rangle[/itex]

    So [itex]\alpha_m = \sqrt{\lambda - m(m-1)}[/itex]

    So we have:
    [itex]\tau_- |m\rangle = \sqrt{\lambda - m(m-1)} |m-1\rangle[/itex]

    [itex]\tau_+ |m-1\rangle = \sqrt{\lambda - m(m-1)}|m\rangle \Rightarrow \tau_+ |m\rangle = \sqrt{\lambda - m(m+1)}|m+1\rangle[/itex]

    If we require that [itex]\alpha_m[/itex] is real (and why is that?), then it must be that [itex]\lambda \geq m(m-1)[/itex]. If [itex]m[/itex] keeps getting larger, than eventually that constraint will be violated. If [itex]m[/itex] keeps getting smaller, than eventually that constraint will be violated. So if you keep acting with [itex]\tau_-[/itex], eventually you get to zero: For the smallest possible value of [itex]m[/itex], it must be that:

    [itex]\tau_- |m_{min}\rangle = 0 = \sqrt{\lambda - m_{min} (m_{min}-1)} |m_{min}-1\rangle[/itex]

    So [itex]\lambda = m_{min}(m_{min}-1)[/itex]

    Similarly, if you keep acting with [itex]\tau_+[/itex], you will eventually violate the constraint. So there must also be an [itex]m_{max}[/itex] such that

    [itex]\tau_+ |m_{max}\rangle = 0 = \sqrt{\lambda - m_{max}(m_{max} + 1)} |m_{max} + 1\rangle[/itex]

    So [itex]\lambda = m_{max}(m_{max} + 1)[/itex]

    So if we define [itex]m_{max} = j[/itex], and [itex]m_{min} = -j[/itex], then we have:

    [itex]\lambda = j(j+1)[/itex]
    [itex]\tau_+ |j\rangle = 0[/itex]
    [itex]\tau_- |-j\rangle = 0[/itex]
     
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