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blue_leaf77

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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)>

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blue_leaf77

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$$

|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.

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Im confused, I don't think thats the notation I've seen in this book

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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.Im confused, I don't think thats the notation I've seen in this book

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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-}##.S±| s m > = ħ √ ((s(s+1) - m(m±1) ) |s (m±1)>

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|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) = (S

So the lowering operator has been applied But this is not |1 0 > ... how come?

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blue_leaf77

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Yes, exactly.|1/2 1/2> as being up up? So |1 1> = |1/2 1/2>

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).S_ (up up) = (S1_ up) up + up (S2_up)

So, you have the equationSo the lowering operator has been applied But this is not |1 0 > ... how come?

$$S_-|1,1\rangle = \hbar( |\downarrow \uparrow\rangle + |\uparrow \downarrow\rangle)$$

But you still have to evaluate the left side ##S_-|1,1\rangle##. What does it give you?

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I fell asleep trying to work the problem out last night and still no wiser to where 1/√2 comes from.

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$$|S=1,M=1 \rangle=|1/2,1/2;1/2,1/2 \rangle.$$

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

$$|S=0,M=0 \rangle = \sqrt{\frac{1}{2}} (|1/2,1/2;1/2,-1/2 \rangle-|1/2,-1/2;1/2,1/2 \rangle).$$

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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

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

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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.I fell asleep trying to work the problem out last night and still no wiser to where 1/√2 comes from.

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.

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blue_leaf77

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Have you calculated ##S_-|S=1,M=1\rangle## (now I'm using vanhees's notation)?where 1/√2 comes from.

Despite being mathematically acceptable, that's not how the relation between the spin states in composite vector space is defined in quantum mechanics.which gives ψ= 1√2 |1,−1> for the normalised ket.

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