I Are certain combinations of quantum numbers (basis vectors) forbidden?

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Certain combinations of quantum numbers, such as ##\ket{n l s m_l m_j}##, are not strictly forbidden but are less desirable due to the relationship ##m_j = m_l + m_s##, which can invalidate some states for spin-1/2 particles. The transformation from the standard basis ##\ket{n l s m_l m_s}## to ##\ket{n l s m_l m_j}## is possible, but it requires careful consideration of the allowed values of the magnetic quantum numbers. Knowing all three angular momentum quantum numbers (l, s, j) allows for the determination of at most one magnetic quantum number, while knowing all three magnetic quantum numbers permits the determination of at most two angular momentum quantum numbers. The intrinsic nature of the spin quantum number further constrains the combinations of quantum numbers available. Overall, while these combinations are not forbidden, they come with specific limitations based on quantum mechanics principles.
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The electron's wavefunction is usually expressed in the standard basis {n, l, m_l, s, m_s}, but how to express it in the basis {n, l, m_l, s, m_j} ? (Note that m_s is replaced with m_j.) Or is it that certain combinations of quantum numbers are forbidden?
I've seen the hydrogen electron's wavefunction expressed in the basis ##\ket{n l s m_l m_s}## or ##\ket{n l s j m_j}##, but so far, never in ##\ket{n l s m_l m_j}##. My question is, are certain combinations of quantum numbers, eg, ##\ket{n l s m_l m_j}##, forbidden?

If ##\ket{n l s m_l m_j}## is not forbidden, how do we get it from the standard basis ##\ket{n l s m_l m_s}##?

I know how to get ##\ket{n l s j m_j}## from ##\ket{n l s m_l m_s}## using Clebsch-Gordan coefficients:
Screenshot 2024-07-12 at 5.29.38 AM.png


where ##J=L+S##.
##J## is the total angular momentum.

But other than that, I do not know how to express the wavefunction in other bases.
 
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##|nlsm_lm_s\rangle## is equivalent to ##|nlsm_lm_j\rangle## due to the relation ##m_j=m_l+m_s##. This relation does render some ##|nlsm_lm_j\rangle## states invalid, though, eg., for a spin-1/2 particle we have the requirement ##m_j = m_l \pm 1/2##. I think this just makes ##|nlsm_lm_j\rangle## a less-desirable way to write out the state, though it is equivalent.

In general, with the addition of angular momentum, if you know all three angular momentum quantum numbers (i.e. l, s, and j), you may know at most one magnetic quantum number. If you know all three magnetic quantum numbers, you may know at most two angular momentum quantum numbers. You can know less information, but not more (with the exception of the "top" or "bottom" states, where ##j=l+s## and ##m_j = \pm j##).

Also, in the case of adding orbital angular momentum and spin, we always know the s quantum number, because it is an intrinsic property of the particle, so that also effectively limits what combination of quantum numbers you can have.
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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