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

Triplets/Singlets and applying lowering S operator

  1. May 2, 2015 #1

    Maylis

    User Avatar
    Gold Member

    upload_2015-5-2_16-1-36.png
    Hello, I am going through this and I am totally confused. Where do they come up with ##\mid 1 \hspace {0.02 in} 0 \rangle = \frac {1}{\sqrt{2}}(\uparrow \downarrow + \downarrow \uparrow)##? They just use the lowering operator, but I'm wondering if the switch in order from equation 4.177 and the one right above it (not numbered) is significant?

    Also, why is this thing called a "triplet" (I see three rows in equation 4.177, is that why?)

    I guess my problem is to go from ## \mid s \hspace{0.02 in} m \rangle## to some sort of up down representation. How do I get the singlet, ##\mid 0 \hspace{0.02 in} 0 \rangle##?
     
    Last edited: May 2, 2015
  2. jcsd
  3. May 2, 2015 #2
    The notation |SM> refers to total spin numbers.
    The arrow notation refers to individual m values, individual s=1/2) being understood.
    The |00> state is the combination orthogonal to |10>.
     
  4. May 2, 2015 #3

    Maylis

    User Avatar
    Gold Member

    Okay, is there some way to go from an arbitrary ##\mid s \hspace {0.02 in} m \rangle## to the up/down arrow combination? I read everything you wrote previously, but I still don't know how to go from the spin eigenstate to the up/down arrow
     
  5. May 2, 2015 #4

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    The way the author derives the ##|10>## state is he makes use of the property of lowering operator ##S_-##. Lowering (and raising) operators have the property that when it acts on an angular momentum eigenstate, it will lower (raise) the m value by one unit while retaining the angular momentum quantum number (in that notation s). So knowing that ##|11> = \uparrow \uparrow##, you can apply ##S_-## to that state to get ##|10>##. If you want to check if ##|10>## is really how it was written in that equation you can try to make ##S^2 = (S_1+S_2)^2## and ##S_z = S_{1z}+S_{2z}## act on the arrow up and down expression for ##|10>##, and see what expression they ended up respectively, although the algebra might get quite tedious.
     
  6. May 2, 2015 #5
    Yes there is. If you have N spins of s=1/2, first write down the |N/2 N/2> state, that is all N arrows point up. Then apply the lowering operator to get the |N/2 N/2-1> state, etc. Now you have the |N/2 m> states. To get the |N/2-1 m> multiplets first construct the |N/2-1 N/2-1> states. These are the N-1 combinations orthogonal to |N/2 N/2-1>. Apply S- repeatedly to get the multiplets. And so on.
    Check at the end if you have 2^N states.
     
  7. May 2, 2015 #6
    It is the other way around. The reason that you cannot go from |Sm> to arrows is that there are many ways to create an |Sm> state. Instead, go from the arrow notation to the spin eigenstate. An alternative to the construction algorithm above is to take all states with n+ up and n- down arrows, construct the matrices of S^2 = Sz^2 + (S+S- + S-S+)/2. Diagonalize the matrix and you will have found the m = n+ - n- members of all multiplets.
     
    Last edited: May 2, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Triplets/Singlets and applying lowering S operator
Loading...