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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

$$\frac{AB+BA}{2}.$$

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

$$\frac{\vec{p}\cdot\frac{\vec{r}}{r}+\frac{\vec{r}}{r}\cdot\vec{p}}{2},$$

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

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Maybe I'm exposing my ignorance here, but I don't think I understand why the symmetrized combination $(\hat A\hat B+\hat B\hat A)/2$ should be the physically correct quantum operator corresponding to the classical observable $AB$. I don't even think I'm convinced that one could derive a "correct" choice of operator from first principles. For any $c$, the linear combination $c\hat A\hat B+(1-c)\hat B\hat A$ would reduce to $AB$ in the classical limit. What does it even mean to say that the particular choice $c=1/2$ is "correct"? The correspondence principle is a useful guide, not an algorithm. –  Ted Bunn May 1 '11 at 18:20
    
@Ted Ive elaborated a bit on this in a new answer. please feel free to correct if im wrong. –  yayu May 1 '11 at 18:46
1  
@Ted In my opinion what constitutes a "correct" combination is that the quantum mechanical operator should be an observable as well. In your notation, if both $A$ and $B$ are hermitian, then a hermitian choice is possible only with $c=1/2$. If they are not, then the combination would be $1/2(AB +(AB)^+)$ as this is the simplest hermitian that can be constructed. –  yayu May 1 '11 at 19:25
    
Oh, of course! I should have figured that out. You're right. –  Ted Bunn May 1 '11 at 21:39

2 Answers 2

up vote 4 down vote accepted

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.

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this is incorrect, the actual form is $$p_r = \frac{\hbar}{i}(\frac{\partial}{\partial r}+\frac{1}{r})$$ –  yayu May 1 '11 at 17:56
    
@yayu: hm, then I'll have to investigate this further. –  David Z May 1 '11 at 17:59
    
on second thought, your expression is correct for 1 dimensional space. For D dimensional euclidean space the expression is $$p_r = \frac{\hbar}{i}(\frac{\partial}{\partial r}+\frac{D-1}{2r})$$ Your expression is correct in 1D. Please make a trivial edit so I may atleast remove my downvote. –  yayu May 1 '11 at 18:04
    
@yayu: I did a bit of research and made an edit to reflect that. –  David Z May 1 '11 at 18:19

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

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

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

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

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

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

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Dear @yayu: So your initial guess in the question is correct after all! –  Qmechanic May 2 '11 at 20:55
    
@Qmechanic yes, but I'd thought that this was an ad hoc choice and missed the notion of hermiticity which actually guides the construction. –  yayu May 7 '11 at 6:02
    
@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) –  yayu May 7 '11 at 6:05

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