I'm having trouble doing it. I know so far that if we have two Hermitian operators $A$ and $B$ that do not commute, and suppose we wish to find the quantum mechanical Hermitian operator for the product $AB$, then


However, if I have to find an operator equivalent for the radial component of momentum, I am puzzled. It does not come out to be simply


where $\vec{r}$ and $\vec{p}$ are the position and the momentum operator, respectively. Where am I wrong in understanding this?


4 Answers 4


I was able to figure it out, so here goes the clarification for the record.

Classically $$p_r=\hat{D}_r = \frac{\hat{r}}{r} \cdot\hat{p} = \frac{\hbar}{i}\frac{\partial}{\partial r}$$

However $\hat{D}_r$ is not hermitian. Consider the adjoint $$\hat{D}_r^\dagger= \hat{p}\cdot\frac{\hat{r}}{r} =\frac{\hbar}{i} \left ( \frac{\partial}{\partial r}+\frac{2}{r} \right ) $$

Now we know from linear algebra how to construct a hermitian operator from an operator and its adjoint: $$\hat{p}_r = \frac{\hat{D}_r^\dagger+\hat{D}_r}{2}=\frac{\hbar}{i} \left ( \frac{\partial}{\partial r}+\frac{1}{r} \right ) $$

And btw, for those who followed my initial question, don't do the following calculational mistake that I committed:

$$\hat{p}_r = \frac{1}{2}\left(\frac{1}{r}(\vec{r}\cdot\vec{p})+(\vec{p}\cdot\vec{r})\frac{1}{r}\right)\neq \frac{1}{2}\frac{1}{r}(\vec{r}\cdot\vec{p}+\vec{p}\cdot\vec{r})$$

  • $\begingroup$ Dear @yayu: So your initial guess in the question is correct after all! $\endgroup$
    – Qmechanic
    Commented May 2, 2011 at 20:55
  • $\begingroup$ @Qmechanic yes, but I'd thought that this was an ad hoc choice and missed the notion of hermiticity which actually guides the construction. $\endgroup$
    – yayu
    Commented May 7, 2011 at 6:02
  • $\begingroup$ @Qmechanic Also thanks, the same notation for operator and unit vectors can be regarded as one of the culprits for my initial mistake in calculation. (i.e making the final inequality an equality) $\endgroup$
    – yayu
    Commented May 7, 2011 at 6:05

You would have to use the fact that the momentum operator in position space is $\vec{p} = -i\hbar\vec{\nabla}$ and use the definition of the gradient operator in spherical coordinates:

$$\vec{\nabla} = \hat{r}\frac{\partial}{\partial r} + \hat{\theta}\frac{1}{r}\frac{\partial}{\partial\theta} + \hat{\phi}\frac{1}{r\sin\theta}\frac{\partial}{\partial\phi}$$

So the radial component of momentum is

$$p_r = -i\hbar\hat{r}\frac{\partial}{\partial r}$$

However: after a bit of investigation prompted by the comments, I found that in practice this is not used very much. It's more useful to have an operator $p_r'$ that satisfies

$$-\frac{\hbar^2}{2m}\nabla^2 R(r) = \frac{p_r'^2}{2m} R(r)$$

This lets you write the radial component of the time-independent Schrödinger equation as

$$\biggl(\frac{p_r'^2}{2m} + V(r)\biggr)R(r) = E R(r)$$

The action of the radial component of the Laplacian in 3D is

$$\nabla^2 R(r) = \frac{1}{r^2}\frac{\partial}{\partial r}\biggl(r^2\frac{\partial R(r)}{\partial r}\biggr)$$

and if you solve for the operator $p'_r$ that satisfies the definition above, you wind up with

$$p'_r = -i\hbar\biggl(\frac{\partial}{\partial r} + \frac{1}{r}\biggr)$$

This is called the "radial momentum operator." Strictly speaking, it is different from the "radial component of the momentum operator," which is, by definition, $p_r$ as I wrote it above, although I wouldn't be surprised to find people mixing up the terminology relatively often.


Even in one dimension the operator $p_r=-i\partial_r$ on the half line $r>0$ has deficiency indices $(0,1)$. There is thus no way to define it it as a self-adjoint operator. In practical terms this abstract mathematical statement means that there is no set of boundary conditions thta we can impose on the wavefunction $\psi(r)$ that lead to a complete set of eigenfunctions for $p_r$. For example, integration by parts to prove formal hermiticity requires that $\psi(0)=0$ but all potential eigenfunctions are of the form $\psi_k(r)=\exp\{ikr\}$ for some real or complex $k$, and no value of $k$ can make $\psi_k(0)$ be zero.

Since one needs eigenfunctions and eignvalues to assign a probability to the outcome of a measurement of $p_r$ this means that $p_r$ is not an ``observable.''

  • 6
    $\begingroup$ You are just explaining why $\mathrm{i}\partial_r$ isn't Hermitian. The question asks how to construct the correct observable corresponding to radial momentum, not why the naive guess doesn't work. $\endgroup$
    – ACuriousMind
    Commented Nov 5, 2015 at 16:02

1D is indeed not good one for radial operator (and if one uses it then it corresponds to half line with reflection boundary condition at x=0, and then the boundary term vanishes there so again h/i*d/dr is hermitian). But on higher dimensions, e.g. 3, one need to think of inner product with $r^2$ weight function, so it cancels the the boundary term. Of course this gives another term when doing integration by parts, and that is why $d/dr$ alone if not enough for hermiticity.

Of course this corresponds also to the fact that "plane wave" in radial coordinate has some function of $r$ in the denominator, because of energy conservation, e.g. $e^{ikr}/r$ in 3D.

Also see this nice paper, which describes the failure of this opproach in 2D.


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