# How to find an orthogonal quantum state? [closed]

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! • 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. – Nelson Vanegas A. May 20 at 2:16
• You should note that the notation is $|j, m\rangle$ on the LHS and $|m_l;m_s\rangle$ on the RHS. – JEB May 20 at 3:01

(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 $$\beta=\sqrt{1/3}$$.