# Cross product of the quantum mechanical operators $\textbf{p}$ and $\textbf{A}$

While reading Advanced Quantum Mechanics by J.J. Sakurai, chapter: Relativistic Quantum Mechanics of Spin-1/2 Particles, section 3.2 the Dirac Equation, the author states the following identity:

$$\textbf{p}\times\textbf{A}=-i\hbar\left(\nabla\times\textbf{A}\right)-\textbf{A}\times\textbf{p}$$

where $$\textbf{p}$$ is the momentum operator and $$\textbf{A}$$ is the vector potential.

Problem: We know that $$\textbf{p}\equiv-i\hbar\nabla$$. So $$\textbf{p}\times\textbf{A}=-i\hbar\left(\nabla\times\textbf{A}\right)$$. How then the second term appeared in the first equation?

Could anyone please explain how this relation is derived?

Basically whenever you look at quantum mechanical operators, you have to imagine them acting on some object, e.g. a wave function.

In the above example, if you think of $$\vec{p} \times \vec{A}$$ as acting on a wavefunction $$\psi$$, you get the above equation just from the product rule after inserting the spatial representation of the momentum operator $$\vec{p} = \frac{\hbar}{i} \vec{\nabla}$$.

To elaborate, you can use the cross product representation via the epsilon tensor: $$(\vec{p}\times\vec{A})_i = \varepsilon_{ijk}p_jA_k$$, so you have

$$(-i\hbar\vec{\nabla}\times\vec{A})_i\psi = -i\hbar\varepsilon_{ijk}\partial_j(A_k \cdot \psi) = -i\hbar \varepsilon_{ijk}\big[ (\partial_j A_k)\cdot \psi + (\partial_j \psi) \cdot A_k \big]$$

• What if you have $\vec{A}\times\vec{p}$? In this case, in tensor notation, the $A$ comes before $p$ so that the product rule is never applied. Is this still correct? Normally, for non operators, $\vec{A}\times\vec{B}=-\vec{B}\times\vec{A}$ so perhaps this rule does not apply for non-commutative operators? Aug 3, 2018 at 21:21

Momentum $\vec{p}$ is an operator which acts on everything standing to the right of it.

So on starts with the following anti-symmetrization $\vec{p} \times \vec{A} \rightarrow \frac{1}{2} \left[ \vec{p}, \vec{A} \right]$ in order to make the operator Hermitian, and then interprets $\vec{p}$ as the derivative which acts, in particular, on $\vec{A}(\vec{r})$.