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 $SO(3)$).

  • $\begingroup$ 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 $\endgroup$
    – KF Gauss
    Dec 23, 2022 at 4:53

1 Answer 1


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.


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