My professor defined these operators, $P_{T}=\rho_1+\rho_2$, and, $x_{cm}=\frac{x_1+x_2}{2}$, in class and said something along the lines of, “These are genuine momentum-position operators.” Now, I recognize that these are classically, the total momentum and center of mass, but what makes them “genuine momentum-position operators”?

For that matter, how would one define an operator which genuinely belongs to a particular class of operators? For example, in the addition of angular momentum, we define an operator, $$\vec{J}=\vec{J_1}+\vec{J_2}.$$ How can we say that this is truly an angular momentum operator? How do we know?

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    $\begingroup$ Have you considered the possibility that it's a flair of language which doesn't mean all that much? $\endgroup$ – Emilio Pisanty Feb 5 '17 at 17:37
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    $\begingroup$ You can show that they satisfy the canonical commutation relations, so they're as 'genuine' a pair of momentum and position operators as the usual ones are. I don't think he meant anything deeper than that. $\endgroup$ – knzhou Feb 5 '17 at 17:39
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    $\begingroup$ This could mean any number of things. My best guess is that he means that they obey the canonical commutation relations $[x_\mathrm{cm}, P_T] = \imath \hbar$. This is, at any rate, something you would want to be true of a set of quantities called position and momentum in QM. $\endgroup$ – By Symmetry Feb 5 '17 at 17:40

If I understand your question, an operator is said to belong to a particular class of operators if it satisfies a particular commutation relation characteristic to all operators of a class. For example, position-momentum operators satisfy, $$[\hat{x},\hat{p}]=i\hbar.$$

Consider the position and momentum operators, your professor defined. We have $$[x_{cm},P_T]=[\frac{x_1}{2}+\frac{x_2}{2},p_1+p_2]=[\frac{x_1}{2},p_1]+[\frac{x_2}{2},p_2]=\frac{1}{2}i\hbar+\frac{1}{2}i\hbar=i\hbar.$$

Similarly, the commutation relation obeyed by angular momentum operators is, $$[L_i,L_j]=i\hbar\epsilon_{ijk}L_k,$$ where $\epsilon_{ijk}$ is the Levi-Civita symbol.

Using this, we can verify that $J=J_1+J_2=J_{1x}+J_{2x}+J_{1y}+J_{2y}+J_{1z}+J_{2z}$ is truly an angular momentum operator. Consider the commutation, $$[J_x,J_y]=[J_{1x}+J_{2x},J_{1y}+J_{2y}].$$ This evaluates to, $$[J_{1x},J_{1y}]+[J_{2x},J_{2y}]=i\hbar J_{1z}+i\hbar J_{2z}=i\hbar(J_{1z}+J_{2z})=i\hbar J_{z},$$ which we expect if $J$ is a true angular momentum operator.

Similar arguments can be made for other operators (i.e. the raising/lowering operators, etc.). Hope that helps.

  • $\begingroup$ Wait. Wouldn't $[\frac{x_1}{2}+\frac{x_2}{2},p_1+p_2]$ also have terms $[\frac{x_1}{2},p_2]+[\frac{x_2}{2},p_1]$? $\endgroup$ – W. Ryan Feb 5 '17 at 18:10
  • $\begingroup$ Yes. But those would just evaluate to $0$, so I ignored them. The momentum of the second particle has no bearing on the position of the first, and vice versa. $\endgroup$ – quanticbolt Feb 5 '17 at 18:13
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    $\begingroup$ Of course assuming the two particles are not correlated at all, i.e., are not parts of a combined system, i.e., are not interacting and never were. $\endgroup$ – Bob Bee Feb 6 '17 at 4:17
  • $\begingroup$ Yes. I should have stated that assumption. $\endgroup$ – quanticbolt Feb 6 '17 at 11:11

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