# Is the product of spin and momentum $\vec S\cdot\vec p$ a scalar operator?

Add the below perturbation: $$V(x)=\lambda\left[\delta^{(3)}(\vec x)\vec S\cdot\vec p+\vec S\cdot\vec p \delta^{(3)}(\vec x)\right]$$ into the Hamiltonian of electron in hydrogen, where $$\vec S,\vec p$$ are spin and momentum operator. Good quantum numbers are $$n,l,s,j,m_j$$ due to spin-orbital coupling.

Is $$\vec S\cdot\vec p$$ a scalar operator? A scalar operator is an operator that doesn't change under rotation. Although $$\vec S\cdot\vec p$$ is inner product of two vector operators, rotation in 3D-space may cause the spin to take a minus sign (projective representation of $$SO(3)$$).

• As currently written, it seems than an answer of "no" should be sufficient for you question, for precisely the reason you mentioned (spin is a spinor, not ordinary vector). Maybe you should edit your question to get at what youreally want to learn here Dec 23, 2022 at 4:53

You appear to be confusing two different sources of minus signs that arise in the context of spin-$$\frac{1}{2}$$ particles.
The operator $$\vec{S}\cdot\vec{p}$$ is a pseudoscalar. That means precisely that it is invariant under rotations, which are elements of the group $$SO(3)$$, but not the other elements of $$O(3)$$. The spin (or any angular momentum variable, including the classical orbital angular momentum $$\vec{L}=\vec{r}\times\vec{p}\,$$) is not a true (polar) vector but an axial vector (or pseudovector). It changes sign under an improper element of the orthogonal group $$O(3)$$ (that is, an element with determinant $$-1$$ instead of $$+1$$). Every improper element can be written as a combination of a rotation $$R\in SO(3)$$ and the spatial inversion $$i$$ that takes $$\vec{r}\rightarrow-\vec{r}$$.
This behavior under $$i$$ (taking $$\vec{v}\rightarrow-\vec{v}\,$$) is the defining characteristic of a true, polar vector. Both position $$\vec{r}$$ and momentum $$\vec{p}$$ transform this way. However, since $$\vec{L}=\vec{r}\times\vec{p}$$ is linear in both $$\vec{r}$$ and $$\vec{p}$$, it transforms under $$i$$ without a change of sign, $$\vec{L}\rightarrow\vec{L}$$, making it an axial vector. Generally, the cross product of two vectors is an axial vector, which the cross product of a axial vector and a true vector is another true vector. (The most important example of this is the Lorentz force $$\vec{F}_{m}=q\vec{v}\times\vec{B}$$. The velocity and force are true, polar vectors, but the magnetic field $$\vec{B}$$ is an axial vector.)
Similarly, the dot product of two vectors is a scalar, but the dot product of a vector and a pseudovector, like $$\vec{S}\cdot\vec{p}$$ is pseudoscalar. A scalar like $$p^{2}=\vec{p}\cdot\vec{p}$$ is invariant under the spatial inversion $$i$$ [or any other improper element of $$O(3)$$]. However, $$\vec{S}\cdot\vec{p}$$ changes sign, $$\vec{S}\cdot\vec{p}\rightarrow-\vec{S}\cdot\vec{p}$$ under $$i$$; this happens because $$\vec{p}$$ changes sign, but $$\vec{S}$$ does not. There is nothing special about the involvement of spin-$$\frac{1}{2}$$ fermions involved in this. The quantity $$\vec{S}\cdot\vec{p}$$ is a still a pseudovector for a spin-1 boson, or any other particle with an intrinsic spin. (A Russel-Saunders spin-orbit coupling $$\vec{L}\cdot\vec{S}$$, the dot product of two axial vectors, is also a true scalar, since in that case, neither $$\vec{L}$$ nor $$\vec{S}$$ changes sign under spatial inversion.)
You seem to have been confused in your question by appearance of another minus sign that appears in the quantum mechanics of spin-$$\frac{1}{2}$$ particles and which is specific to fermions with half-integer spins. Under a $$2\pi$$ rotation, the a spinor wave function $$|\psi\rangle=\left[ \begin{array}{c} \psi_{\uparrow} \\ \psi_{\downarrow} \end{array}\right]$$ acquires an overal minus sign: $$R_{2\pi}|\psi\rangle=-|\psi\rangle$$. However, this negative sign, while it is indirectly observable in certain ways, cancels out in the evaluation of a quantity like the spin $$\vec{S}$$. The expectation value of the spin is $$\langle\vec{S}\rangle=\langle\psi|\vec{S}|\psi\rangle$$, and under the rotation which gives $$|\psi\rangle\rightarrow-|\psi\rangle$$ and $$\langle\psi|\rightarrow-\langle\psi|$$, there are two negative signs which cancel out.