How to obtain the "anti-symmetry" state explicitly?

Question

For a $l=1$ and $s=1/2$ system ($l$ is orbital angular momentum and $s$ is spin angular momentum), $|j=3/2, m_j=3/2rangle$ state can be written as:
$$|j=3/2, m_j=3/2rangle=(Y_1^1)_{uparrow}=frac{1}{sqrt{2}}(p_x^uparrow+ip_y^uparrow)sim (frac{x+i y}{r})_uparrow$$
where $Y_1^1$ is complex sphere harmonic function with $l=1, m_l=1$. Assuming that the system has inversion symmetry, and we can check that this state is invariant under the inversion operation $I$:
$$I:rrightarrow-r  I |j=3/2, m_j=3/2rangle= (frac{-x-i y}{-r})_uparrow=|j=3/2, m_j=3/2rangle $$
which means  $|j=3/2, m_j=3/2rangle$ is actually "gerade". Now, I want to construct the "ungerade" partner  $|j=3/2, m_j=3/2rangle^u$, which means this partner will gives $-1$ under the inversion operation $I$:
$$I|j=3/2, m_j=3/2rangle^u=-|j=3/2, m_j=3/2rangle^u$$
Loosely speaking, these two states can be understood as the relation of "bond" and "anti-bond". But I do not know how to obtain the explicit expression of $|j=3/2, m_j=3/2rangle^u$.

user200143 · Answer

Gerade and Ungerade tell whether or not molecular orbitals are symmetric or antisymmetric under reflection about a center of symmetry. An Ungerade orbital typically does not have the same quantum numbers as a gerade orbital.
Similarly, the angular momentum functions you have written have good parity. This is given by the $ell$ value. If it is even, the spherical harmonics are unchanged under reflection. If it is odd, they change sign. You have a sign error where you have written $r=sqrt{x^2+y^2+z^2}$ changing sign under parity, which obviously it does not. Your $ell=1$ state will therefore have odd parity. If you want to construct an even parity j=3/2 state, you will need to add l=2 to s=1/2. This can give a j=3/2, m=3/2, l=2 state with appropriate Clebsch-Gordon coefficients.

How to obtain the "anti-symmetry" state explicitly?

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