Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Why is the spin of this state equal to one?

  1. Apr 22, 2016 #1
    I've just come across the spin states of a two electron system. There are 4 states possible and I am a little confused as to why the state below has values of s=1 m_s=0?


    where α(i) and β(i) tell us if the particle has +ve or -ve z component of spin respectively.

    I don't quite understand why s=1 as the sum of the spins for this would be (+1/2)(-1/2)+(+1/2)(-1/2)

    http://www.tcm.phy.cam.ac.uk/~bds10/aqp/lec11_compressed.pdf a similar thing can be seen here on slide 10.
  2. jcsd
  3. Apr 22, 2016 #2


    User Avatar

    Staff: Mentor

    That's not correct. First, you are mixing up the total spin and the projection of the spin along z (the spin of an electron is always 1/2, never -1/2).

    Second, what is the result of ##\hat{S} \alpha## or ##\hat{S} \beta##?
  4. Apr 22, 2016 #3


    User Avatar
    Science Advisor

    What you calculated, is not the sum of the spins but the sum of the z-component of the spins. The z-component can take on the values -1, 0, 1 for a particle with spin 1.
  5. Apr 22, 2016 #4
    Do hunds rules not say that the S= sum of all m_s values?
  6. Apr 22, 2016 #5
    Why is the spin of this state s=1 then? What am I missing?
  7. Apr 22, 2016 #6


    User Avatar
    Staff Emeritus
    Science Advisor

    If we let [itex]\vec{S} = \vec{S_1} + \vec{S_2}[/itex], where [itex]\vec{S}[/itex] is the total spin, and [itex]\vec{S_1}[/itex] and [itex]\vec{S_2}[/itex] are the spins of the two electrons, then we have:

    [itex]S^2 = (S_1)^2 + (S_2)^2 + 2 \vec{S_1} \cdot \vec{S_2}[/itex]

    If you let this operator act on the state, you will find that yields value [itex]2 \hbar^2[/itex], which is consistent with an angular momentum of [itex]1 \hbar[/itex]. (Remember, the total angular momentum is quantized to have value [itex]S^2 = s(s+1) \hbar^2[/itex], so [itex]S^2 = 2 \hbar^2 \Rightarrow s = 1[/itex])
  8. Apr 22, 2016 #7


    User Avatar
    Staff Emeritus
    Science Advisor

    No, you have the total spin, [itex]\vec{S}[/itex], which has components [itex]S_x, S_y, S_z[/itex]. You have the individual particle spins [itex]\vec{S_1}[/itex] and [itex]\vec{S_2}[/itex], which have components [itex](S_1)_x, (S_1)_y, (S_1)_z, (S_2)_x, (S_2)_y, (S_2)_z[/itex].

    It is true that [itex]S_z = (S_1)_z + (S_2)_z = 0[/itex], but that doesn't imply that [itex]S_x[/itex] and [itex]S_y[/itex] are zero.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted