There exists a classical definition of the orbital angular momentum, $$\mathbf{L}=\mathbf{r}\times\mathbf{p}$$ Due to this definition, the following quantity is identically zero, $$\boldsymbol{\nabla}\cdot\mathbf{L}=0$$ because the divergence of a curl is zero. I presume this result carries over to quantum mechanics as well, that is, $$\langle\phi|\boldsymbol{\nabla}\cdot\mathbf{\hat{L}}|\psi\rangle=0$$ and is true for any two arbitrary states $|\phi\rangle$ and $|\psi\rangle$. Now I am not sure whether there exists any relationship analogous to $\mathbf{L}=\mathbf{r}\times\mathbf{p}$ for spin angular momentum. So my questions are these,

  • What is $\boldsymbol{\nabla}\cdot\mathbf{S}$? Does it identically vanish?
  • What is the quantum version of it, i.e., what is $\langle\phi|\boldsymbol{\nabla}\cdot\mathbf{\hat{S}}|\psi\rangle$?

I'm most interested in knowing whether they vanish identically or not even if any analytical expression for them is not available.


2 Answers 2


In terms of quantum mechanical operators, the gradient is essentially the momentum operator, $$\mathbf{p} = - i \nabla.$$ So your first statement is equivalent to the assertion that $\mathbf{p} \cdot \mathbf{L} = 0$. This is true, but it's not as simple to prove as you think, because the vectors here contain noncommuting operators. We have $$\mathbf{p} \cdot \mathbf{L} = \epsilon_{ijk} p_k r_i p_j = \epsilon_{ijk} p_k (p_j r_i + [r_i, p_j]).$$ The first term vanishes by the antisymmetry of $\epsilon_{ijk}$, while the second contributes $i \delta_{ij} \epsilon_{ijk} p_k$ by the canonical comutation relation, and hence also vanishes.

At the level of nonrelativistic quantum mechanics, $\mathbf{S}$ has no dependence on space at all, it's purely an operator on spinor space. The operators $\mathbf{p}$ and $\mathbf{S}$ commute, so $\mathbf{p} \cdot \mathbf{S}$ can be interpreted straightforwardly: just as in classical mechanics, it's the dot product of momentum and spin. The operator certainly isn't zero.

At the level of quantum field theory, you can go further by forming a spin density operator $\tilde{\mathbf{S}}(\mathbf{x})$, which is conceptually distinct from the single-particle spin operator $\mathbf{S}$. In this case the spin density has a position argument, so it would be meaningful to consider $\nabla \cdot \tilde{\mathbf{S}}$, but this quantity doesn't have to vanish. (If it's nonzero, it just means that spin is net pointing outward from a point.) Similarly, you can define an orbital angular momentum density $\tilde{\mathbf{L}}(\mathbf{x})$, and its divergence doesn't have to vanish either.


When considering the mechanics of a point particle, the quantity $\nabla \cdot (\mathbf r \times \mathbf p)$ is not well-defined.

The familiar gradient, curl, and divergence operators are objects which act on fields (vector fields in the case of curl and divergence, and scalar fields in the case of the gradient). For example, in Cartesian coordinates,

$$div(A) = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z} $$ $$=\lim_{\epsilon\rightarrow 0} \left[\frac{A_x(x+\epsilon,y,z)-A_x(x,y,z))}{\epsilon}\right] + \ldots $$

Fundamentally, such an operation only makes sense if you can evaluate $A_x(x+\epsilon,y,z)$ and subtract $A_x(x,y,z)$ - in other words, you require $A_x$ to be a quantity which takes some value at every position.

But now consider $\mathbf r \times \mathbf p$, where $\mathbf r = \mathbf r(t)$ and $\mathbf p=\mathbf p(t)$ are the position and momentum of a point-like object at time $t$. How could we apply the divergence operator to this quantity? How does one "evaluate" this vector quantity at neighboring positions and then subtract?

The answer is that you cannot. There is no meaningful sense in which you can calculate a spatial derivative of a vector quantity which is not a field, and so even though $\nabla \cdot (\mathbf r \times \mathbf p)$ may look like a reasonble thing to talk about at first glance, it ultimately is not.

As a side note, if you are talking about the flow of some kind of fluid which has mass density $\rho(\mathbf r)$ and flow velocity $\mathbf u(\mathbf r)$, then the quantity $\boldsymbol \ell = \mathbf r \times (\rho \mathbf u)$ is the angular momentum density of the fluid (calculated with respect to the coordinate origin), and this is a vector field which can be differentiated using the divergence and curl operators. It is rarely spoken of - it is usually much less interesting than the vorticity $\boldsymbol \omega = \nabla \times \mathbf u$ - but it is at least a well-defined quantity.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.