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:


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?


2 Answers 2


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

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

  • $\begingroup$ 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? $\endgroup$
    – user279043
    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 $\vec{A}(\vec{r})$.


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