Calculation of Clebsch-Gordan coefficients

[PineApple2]

Homework Statement

I have two spin-1/2 particles and I need to calculate their Clebsch-Gordan coefficients.

Homework Equations

The Attempt at a Solution

I followed the procedure of applying [itex]J_- [/itex] to [itex]|{j,m}\rangle [/itex]
and [itex]J_{1-}[/itex] and [itex]J_{2-}[/itex] to [itex]|{m_1,m_2}\rangle [/itex] and comparing them. I got correctly
[itex]\langle{1,1}|{1/2,1/2}\rangle =1 [/itex],
[itex]\langle{1,0}|{1/2,-1/2}\rangle=1/\sqrt{2} [/itex],
[itex]\langle{1,0}|{-1/2,1/2}\rangle =1/\sqrt{2} [/itex],
[itex]\langle{1,-1}|{-1/2,-1/2}\rangle=1 [/itex].
Now I want to find [itex]\langle{0,0}|{1/2,-1/2}\rangle [/itex]
and [itex]\langle{0,0}|{-1/2,1/2}\rangle [/itex].
Therefore I denoted [itex]|{0,0}\rangle = \alpha|{1/2,-1/2}\rangle + \beta|{-1/2,1/2}\rangle [/itex]
and used the normalization condition [itex]|\alpha|^2 + |\beta|^2 = 1[/itex] and orthogonality to the [itex]|{1,0}\rangle[/itex] state. I got the equation
[itex]|\alpha|^2 = 1/2[/itex] from which there are 2 options:
[itex]\alpha = 1/\sqrt{2}[/itex] and [itex]\alpha = -1/\sqrt{2}[/itex] (only real coefficients by convention). How do I know which is the right option out of the two?

Thanks!

 

[vela]

The overall sign of ##\alpha## doesn't really matter. What's important is the relative phase between ##\alpha## and ##\beta##. What combination of ##\alpha## and ##\beta## will yield a state orthogonal to ##|1, 0\rangle##?

 

[PineApple2]

Ok. I agree. But the standard table has a certain convention. How do I know how to pick the sign to fit this convention?

 

[vela]

According to this page, this is the convention:

The Condon-Shortley convention is that the highest m-state of the larger component angular momentum is assigned a positive coefficient.

Still seems a bit ambiguous in your case, though.

 

[paranormal]

I would like to first commend you on your correct application of the Clebsch-Gordan coefficients for spin-1/2 particles. Your approach is correct and your results are also correct.

Now, to answer your question about which option is the right one for \alpha, we need to consider the physical interpretation of the Clebsch-Gordan coefficients. These coefficients represent the probability amplitudes for the combined state of two particles with individual spins j_1 and j_2 to have a total spin j. In this case, we are looking at the combined state of two spin-1/2 particles to have a total spin of 0.

Since the total spin of the combined state must be zero, we can rule out the option of \alpha = 1/\sqrt{2} as it would result in a non-zero total spin. This leaves us with the option of \alpha = -1/\sqrt{2}, which would result in a total spin of 0 when combined with \beta = 1/\sqrt{2}.

Therefore, the correct option for \alpha is \alpha = -1/\sqrt{2}.

I hope this helps clarify your doubt. Keep up the good work!