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In the question I have L = 1 and s = 1/2. First I had to find the quantum state for the highest m = m$_l$ + m$_s$ value which I did. To find the quantum states for the next highest m value I used the lowering operator of j, as can be seen in the picture below. I understand that there has to be another state for that m value, but I don't understand how they get the values for $\alpha$ and $\beta$.

They say they find it by via orthogonality but whatever I try I can't get their values. They also state that the values can be found by using a ladder operation $j_+$ on the state which is zero. I think this is equal to zero because the quantum state corresponding to j = 1/2 and m = 3/2 doesn't exist, is this correct? But also using this method I can't find the values for $\alpha$ and $\beta$.

So I have a few questions :)

1) Why is the other quantum state orthogonal to the first, and how do I use this to find the $\alpha$ and $\beta$ values?

2) Is my assumption about the state j = 1/2 and m = 3/2 correct

3) How do they find the $\alpha$ and $\beta$ values for the ladder operation method?

Thanks so much!

enter image description here

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  • $\begingroup$ could you provide more info on the reference you are using and the previous details in the calculation...? for example where and how are $\alpha, \beta$ defined? you are right in that $-j \leq m \leq j$ and therefore never greater than $j$ but more that that you need to give more detail. $\endgroup$ – Nelson Vanegas A. May 20 at 2:16
  • $\begingroup$ You should note that the notation is $|j, m\rangle$ on the LHS and $|m_l;m_s\rangle$ on the RHS. $\endgroup$ – JEB May 20 at 3:01
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(1) The $|j,m\rangle$ form an orthonormal basis, with:

$$ \langle j,m|j',m'\rangle =\delta_{jj'}\delta_{mm'}$$

(2) Yes:

$$ \hat J^+|j,j\rangle = 0$$

(3) The only two states contributing to $|j,\frac 1 2\rangle$ are $A=|1,-\frac 1 2\rangle$ and $B=|0,+\frac 1 2\rangle$, and we know:

$$|\frac 3 2, \frac 1 2\rangle = \sqrt{\frac 1 3}A + \sqrt{\frac 2 3}B$$

so if: $$|\frac 1 2, \frac 1 2\rangle = \alpha A + \beta B$$

then (1) tells (using real coefficients):

$$ \langle\frac 3 2, \frac 1 2|\frac 1 2, \frac 1 2\rangle = \sqrt{\frac 1 3}\alpha + \sqrt{\frac 2 3}\beta = 0 $$

so $\alpha = -\sqrt{2/3}$ and $\beta=\sqrt{1/3}$.

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