I am a sophomore learning quantum mechanics, and I got a confusion in my study, I tried to search for answers online but failed to find a satisfying answer. Hence, I wanted to post my question here and I would like to apologize in advance for all sorts of possible mistakes of language and typo as I am not a native English speaker.

As far as I have learned, the quantities in quantum mechanics are directly transformed to operators from classical physics (especially from Hamiltonian mechanics as we are always expressing them by $\hat{r}$ and $\hat{p}$). But to express some of the quantities we sometimes need to use the dot product "$\cdot$" and the cross product "$\times$" of vectors. In classical mechanics, we may change their order at will since we do not have to worry about that they might not commute (but a change of sign for cross product though). But in quantum mechanics, we need to be extremely careful for the possible commutators.

Here comes my question. When we are defining physical quantities using dot product, we do not set a specific order for them, then how could we define the order of the the quantities in quantum mechanics if they do not commute? I asked my professor about this and he said he had not yet come across such a case in his life. (I realized this queston as we were discussing the $L$-$S$ coupling today, where there would be $L\cdot S$ which commute for they are in different Hilbert Space)

I want to confirm if there is any example of a quantity with physical meaning that is defined as dot product of two quantities that do not commute? (I could come up with $\hat{\vec{r}}\cdot\hat{\vec{p}}$ but I am not sure if it has any physical meaning.) i.e. :

$$D:=\vec{A}\cdot\vec{B}\qquad [\vec{A},\vec{B}]\neq0$$

If there is such a quantity, how could we figure out the correct order of them? If there is not, is there any deep reason or principle behind it?

If possible, please also make a brief discussion on cross product, which I believe would be more straightforward for its immanent order, but I also find it a little bit tricky as I know the Laplace-Runge-Lentz vector is defined as $$\vec{K}:=\frac{\vec{r}}{r}+\frac{1}{2\mu e^2}(\vec{L}\times\vec{p}-\vec{p}\times\vec{L})$$ where two possible orders both appear in it.


2 Answers 2


You have actually already correctly identified one case of it not commuting.

When you scrutinise how the kinetic energy operator turns into radial and angular kinetic energy operators, that involve the orbital angular momentum $\hat L^2=\left|\vec{\hat r}\times\vec{\hat p}\right|^2=\ell(\ell+1)\hslash^2$, the other part is $\hat p_r=\dfrac12\left(\hat{\vec{\hat r}}\,\cdot\,\vec{\hat p}+\vec{\hat p}\,\cdot\,\hat{\vec{\hat r}}\right)=-i\hslash\left(\hat\partial_r+\dfrac1{\hat r}\right)$ so that $\vec{\hat p}\,^2=\hat p_r^2+\dfrac{\hat L^2}{\hat r^2}$

Note that $\hat{\vec{\hat r}}=\dfrac{\vec{\hat r}}{\left|\vec{\hat r}\right|}$, and that you should separate between operators that return vectors, and unit vectors, and operators that return unit vectors, and the returned eigenvalue, and if they are vectorial quantities or not. My notation makes this very clear.

Also, the form of the radial momentum that I quoted above, is the usual, Hermitian one. $-i\hslash\hat\partial_r$ alone happens to not be Hermitian.

It just so happens that the orbital angular momentum term is devoid of non-commuting terms (you can easily prove this), so that there is no need for explicit care. The Laplace-Runge-Lenz vector happens to be the first non-trivial case that really needed us to scrutinise the commutation relations to fix which version we should be considering.


There is no perfect way to associate a quantum operator to a given classical quantity (such an association is called quantization) such that the poisson bracket is respected. This is known as Groenewold's theorem. See this wikipedia article.

There are many different quantization schemes. The best one (and most used one) is probably the Weyl-Quantization. The Weyl-Quantization is used in your example of the Runge-Lentz vector. The Weyl-Quantization would associate to $\vec{A} \cdot \vec{B}$ the operator $$\frac{1}{2}(A_1 B_1 + B_1 A_1 + A_2B_2 + B_2 A_2 + A_3 B_3 + B_3 A_3)$$ and therefore resolve the ambiguity by considering the average of all possibilities.

Since there is no "right" quantization, we can in principle choose one on a case by case basis to yield a self-adjoint operator and the correct physical results. Allthough the Weyl-Quantization is most often used.

After all the quantum operator is more fundamental than the classical observable and so it might be expected that quantum observables are not in a one to one correspondence to classical ones (after all there exist quantum observables that dont have a classical counterpart like spin).

There is a whole section dedicated to quantization schemes in the book "Quantum Theory for Mathematicians" by Brian Hall (Chapter 13).

  • $\begingroup$ This is really a brand new realm for me, and as the name of the book, very "mathematical". May I understand it simply as "there is no absolute way to define the operator and the Weyl-Quantization would put all possible expressions together and take average"? $\endgroup$ Commented Jun 4, 2023 at 14:27
  • $\begingroup$ @AlexanderZhang yes. Also the Weyl-Quantization is probably the best, because the commutator of two quantum operators obtained from it is equal to $i \hbar$ times the Weyl-Quantization of the Poisson-Bracket of the classical observables, but only if one of the two classical observables is a polynom of order at most 2. So it is "almost" perfect. $\endgroup$
    – jd27
    Commented Jun 4, 2023 at 15:01

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