Can't understand ket notation for spin 1/2

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The discussion clarifies the ket notation for spin-1/2 particles, explaining that each ket contains four elements representing the spin and z-component of spin for two particles. The first two elements indicate the spin of each particle, while the last two specify their z-components, which can be represented as either +1/2 or -1/2. For compactness, the spin value is often omitted, leading to a simpler notation. It is noted that for particles with spins greater than 1/2, the full notation must be used to avoid confusion. The conversation concludes with a mention of a later edition of Griffiths' work that changes the notation.
pepediaz
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Homework Statement
Each particle can have spin up or spin down, so there are four possibilities.
Relevant Equations
It's from Example 4.5, from Griffiths (Quantum Mechanics)
I can't why there are four elements in each ket instead of only two
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pepediaz said:
Homework Statement:: Each particle can have spin up or spin down, so there are four possibilities.
Relevant Equations:: It's from Example 4.5, from Griffiths (Quantum Mechanics)

I can't why there are four elements in each ket instead of only two View attachment 279407
The notation is that the first two 1/2 just indicated the spin of each particle. The last two entries give the z component of the spin, ##S_z## of each particle.
 
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With electrons it is understood that the spin is 1/2 and is often omitted for compactness. In that case instead of ##|\frac{1}{2},\frac{1}{2},s_{1z},s_{2z}\rangle## one can write ##|s_{1z},s_{2z}\rangle## where ##s_{1z},s_{2z}=\pm \frac{1}{2}## or use ##\uparrow\downarrow## as Griffiths does.

If the two spins, or more generally angular momenta, are greater than 1/2, one must be clear about what one means and write the full-blown ##|S_1,S_2,S_{1z},S_{2z}\rangle## to avoid confusion.
 
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Thanks! I see it now!
 
pepediaz said:
Thanks! I see it now!
In the later edition he abandons that notation, in fact. Instead, he has:
$$\uparrow \downarrow \ = \ |\frac 1 2 \frac 1 2 \rangle |\frac 1 2 -\frac 1 2 \rangle $$
 
I want to find the solution to the integral ##\theta = \int_0^{\theta}\frac{du}{\sqrt{(c-u^2 +2u^3)}}## I can see that ##\frac{d^2u}{d\theta^2} = A +Bu+Cu^2## is a Weierstrass elliptic function, which can be generated from ##\Large(\normalsize\frac{du}{d\theta}\Large)\normalsize^2 = c-u^2 +2u^3## (A = 0, B=-1, C=3) So does this make my integral an elliptic integral? I haven't been able to find a table of integrals anywhere which contains an integral of this form so I'm a bit stuck. TerryW

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