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I Singlet and triplet spin states - the normalisation constant

  1. Mar 20, 2016 #1
    The triplet spin state with a normalisation constant of 1/√2 and the singlet spin state with the same normalisation constant... Where on earth is this normalisation constant derived from? I've been scouring the Griffiths intro to quantum mechanics text book and cant find info on it.
     
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  3. Mar 20, 2016 #2

    blue_leaf77

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    Start from the triplet state with parallel up spins ##|1,1\rangle = |1/2,1/2\rangle##. To obtain the triplet state with ##m=0##, ##|1,0\rangle##, apply the lowering operator ##S_-=S_{1-}+S_{2-}## on both sides of the triplet parallel spins equation before.
     
  4. Mar 20, 2016 #3
    |1,1⟩=|1/2,1/2⟩
    Im confused with what you mean with that statement. How can a state with total s=1 and m=1 equal a state with s=1/2 and m=1/2, these are orthogonal states? Parallel up spins states I believed to be |1,1> and |1.-1>

    On page 174 I see this. Is this along the lines of what you mean?
    S±| s m > = ħ √ ((s(s+1) - m(m±1) ) |s (m±1)>
     
  5. Mar 20, 2016 #4

    blue_leaf77

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    In
    $$
    |1,1\rangle = |1/2,1/2\rangle
    $$
    the left hand side belongs to ##|s,m\rangle## (eigenstate of the operators ##S^2## and ##S_z##) notation, while the right hand side belongs to ##|m_1,m_2\rangle## (eigenstate of ##S_{1z}## and ##S_{2z}##) notation.
     
  6. Mar 20, 2016 #5
    Im confused, I don't think thats the notation I've seen in this book
     
  7. Mar 20, 2016 #6

    blue_leaf77

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    Ok then write down here all four composite states of two spin one-half particles (triplet and singlet states) that your book uses.
     
  8. Mar 20, 2016 #7

    blue_leaf77

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    Yes, this is for the left hand side, for the right hand side you need to apply ##S_{1-}+S_{2-}##.
     
  9. Mar 20, 2016 #8
    Triplet s=1
    |1 1> = up up
    |1 0> = 1/√2 ( up down + down up)
    |1 -1> = down down

    Singlet s =0
    |0,0> = 1 /√2 ( up down - down up)

    Im being noob here... |1/2 1/2> as being up up? So |1 1> = |1/2 1/2>

    I see what you are telling me to do on page 187 it writes
    S_ (up up) = (S1_ up) up + up (S2_up)= (ħ down) up + up (ħ down) = ħ(down up + up down)
    So the lowering operator has been applied But this is not |1 0 > ... how come?
     
  10. Mar 20, 2016 #9

    blue_leaf77

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    Yes, exactly.
    Be careful with the notation. To be consistent, the state in the left side should have been written as |1 1> instead of (up up).
    So, you have the equation
    $$S_-|1,1\rangle = \hbar( |\downarrow \uparrow\rangle + |\uparrow \downarrow\rangle)$$
    But you still have to evaluate the left side ##S_-|1,1\rangle##. What does it give you?
     
  11. Mar 20, 2016 #10

    Charles Link

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    Perhaps I can help with the notation here: You have a system of two electrons and the S=1 Ms=1 triplet state ## |1 ,1>= |1/2,1/2>_1 |1/2,1/2>_2 ## (where the notation in each case is ## |s,m_s> ## and the subscript indicates the specific electron. A shortened notation was introduced in post #2 because the s=1/2 in all cases). The lowering operators mentioned in post #2 are for the individual electrons and each one only operates on its own electron. For the other electron it treats the state as a constant. (As I was writing this, I see a follow-up post. Looks like you have it figured out.)
     
  12. Mar 21, 2016 #11
    I fell asleep trying to work the problem out last night and still no wiser to where 1/√2 comes from.
     
  13. Mar 21, 2016 #12

    vanhees71

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    The problem is to decompose the the two-spin states ##|s_1,m_2;s_2,m_2 \rangle=|s_1,m_1 \rangle \otimes |s_2,m_2 \rangle## into irreducible representations of the total-spin states ##|S,M \rangle##. For ##s_1=s_2=1/2## it's easy to guess them, but it's also a good example to derive them systematically. You can obviously start with the state with ##M=1##. This you can obviously only get when ##m_1=m_2=1/2##, i.e., you have
    $$|S=1,M=1 \rangle=|1/2,1/2;1/2,1/2 \rangle.$$
    Now you can apply the lowering operator ##\hat{S}_-=\hat{s}_{1-}+\hat{s}_{2-}##, which together with the normalization to 1 leads to
    $$|S=1,M=0 \rangle = \sqrt{\frac{1}{2}} (|1/2,-1/2;1/2,1/2 \rangle + |1/2,1/2;1/2,-1/2 \rangle)$$
    Applying the lowering operator once more you get
    $$|S=1,M=-1 \rangle = |1/2,-1/2;1/2,-1/2 \rangle.$$
    Now the spin space is 4-dimensional, and we have found three orthonormal basis vectors. Now we have just to find one more, which is
    $$|S=0,M=0 \rangle = \sqrt{\frac{1}{2}} (|1/2,1/2;1/2,-1/2 \rangle-|1/2,-1/2;1/2,1/2 \rangle).$$
     
  14. Mar 21, 2016 #13
    Whats the requirement of normalisation to 1 in your first step?

    I saw this from another thread.....
    normalising ψ=|1,−1> is easy as ψ∗=<1,−1|
    and then ψ∗ψ=<1,−1|1,−1>=2
    which gives ψ= 1√2 |1,−1> for the normalised ket.

    I think i'm lacking understanding the requirement for normalisation in ket states and how they are worked out. I just cant find any good text on it to help me understand it
     
    Last edited: Mar 21, 2016
  15. Mar 21, 2016 #14

    vanhees71

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    It's just convenience. If you have a orthonormal basis of eigenvectors of an operator representing an observable and normalized state kets then the modulus squared of their product is the probability to find the observable to have the corresponding eigenvalue when measuring it. That's why one usually normalizes the vectors in quantum mechanics.
     
  16. Mar 21, 2016 #15

    PeroK

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    I think there is a weakness in this section of Griffiths' book. He ought to say more about the formal background of spin: what Hilbert space is being used, including how the inner product is defined. For a composite spin system, you need the direct sum of two Hilbert spaces. Griffiths is very woolly on this point. He says "this notation may not be elegant but it does the job". It may do the job, but it gives the reader no idea of the mathematical justification for the rules governing operators on composite spin.

    It may be worth looking for a source that covers the linear algebra of composite spin in more depth. Otherwise, you may just have to accept some things without being able to work out why. In this case, you are supposed to accept that ##\uparrow \uparrow## etc. are normalised - hence the factor of ##1/\sqrt{2}## - but Griffiths doesn't define the inner product on the composite space at any stage (as far as I recall).

    This may help:

    https://en.wikipedia.org/wiki/Hilbert_space#Direct_sums

    Moreover, the whole concept of Glebsch-Gordon coefficients may be much easier to grasp if you realise they represent the relationship between the two natural bases for the composite system, in terms of the direct sum of Hilbert Spaces.

    Finally, when you come to combining spin with the position wavefunction, again the lack of formal background in Griffiths may lead to some confusion about what is going on.
     
  17. Mar 21, 2016 #16

    blue_leaf77

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    Have you calculated ##S_-|S=1,M=1\rangle## (now I'm using vanhees's notation)?
    Despite being mathematically acceptable, that's not how the relation between the spin states in composite vector space is defined in quantum mechanics.
     
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