# Is $X\otimes X$ not the simultaneous position operator?

I had thought that $$X\otimes X$$ would be the operator on $$H_1\otimes H_2$$ to simultaneously measure the x-positions of two particles.

But there seems to be something wrong with this -- for a given eigenvalue $$z$$, there is an entire subspace $$\mathrm{Span}\left(|x\rangle\otimes|z/x\rangle\right)$$ associated with it, so we don't get precise positions from measuring it, just the product of positions $$x_1x_2$$.

If so, what's the right operator representing "simultaneous measurement" of the x-positions? Is that even possible -- to have "vector eigenvalues"? Or do we just need a spacefilling curve or something?

• A single operator always has single numbers as eigenvalues. I don't understand how you think there should be a single operator that "simultaneously" measures both positions. – ACuriousMind Jul 13 '19 at 11:36

It's easier to think about one particle first. In a one-dimensional case, $$\hat{x}$$ is the position operator. When we move to three dimensions, the position operator must have vector eigenvalues, because position is vector-valued. This is achieved by using a vector operator, $$\hat{\mathbf{x}} \equiv \begin{pmatrix} \hat{x} \\ \hat{y} \\ \hat{z} \end{pmatrix}$$ which is a vector of operators. There is of course no difference when you have two particles. In this case their positions are described by six parameters, so you need a six-element vector, $$\hat{\mathbf{X}} \equiv \begin{pmatrix} \hat{x}_1 \\ \hat{y}_1 \\ \hat{z}_1 \\ \hat{x}_2 \\ \hat{y}_2 \\ \hat{z}_2 \end{pmatrix}.$$ The operator $$\hat{x}_1 \otimes \hat{x}_2$$ doesn't represent a simultaneous measurement of $$\hat{x}_1$$ and $$\hat{x}_2$$, it represents measuring the single quantity $$x_1 x_2$$.

Let $$\Omega$$ denote any set of one or more mutually commuting self-adjoint operators, such as the observables corresponding to the $$x$$-coordinates of two particles. Let $$\Omega'$$ denote the commutant, which is the set of all oeprators that commute with everything in $$\Omega$$. Then let $$\Omega''$$ denote the double commutant, which is the set of all operators that commute with everything that commutes with everything in $$\Omega$$. (That's not a typo.) We can also describe $$\Omega''$$ as the (commutative) von Neumann algebra generated by $$\Omega$$.

(Technically, a von Neumann algebra contains only bounded operators, and operators like $$X$$ are unbounded, but that technicality doesn't affect the spirit of this answer.)

If the operators in $$\Omega$$ qualify as observables in the given model, then the algebra $$\Omega''$$ contains all of the projection operators that we need to characterize the possible outcomes of a simultaneous measurement of the observables in $$\Omega$$.

Here's the key: Every commutative von Neumann algebra is generated by a single self-adjoint operator . That proves the existence of a single self-adjoint operator representing the simultaneous measurement of all of the observables in $$\Omega$$. Actually constructing such an operator is a different problem, and it probably wouldn't be useful. (As explained in knzhou's answer, $$X\otimes X$$ doesn't work.) Using separate operators, one for each coordinate of each particle, is more convenient.

References:

 EP10 on page 23 in Jones (2009), "Von Neumann Algebras," https://math.berkeley.edu/~vfr/VonNeumann2009.pdf

 Lemma 1 in Suzuki and Saitô (1963), "On the operators which generate continuous von Neumann algebras," https://projecteuclid.org/euclid.tmj/1178243811

Let me first point out that this is not a conceptual issue but a "quirk" in the old formalism of quantum theory. Ask yourself this: if real numbers $$x$$ are the eigenvalues of position operator of a single particle then how the eigenvalues of a position operator of two particles should look like? If your answer is somethin like that $$(x_1,x_2)$$ then how do you write an operator with eigenvalues that are an ordered pair of 2 real numbers?

In the modern formalism we stop thinking about measurements in terms of eigenvalues and start thinking about them in terms of outcomes. Distinct outcomes come from distinct eigenspaces of the operator. The eigenvalues themselves are only labels that you may or may not define. In your case if $$P_x^1\otimes I$$ and $$I\otimes P_x^2$$ are the projections on eigenspaces of position for each particle then $$P_{xy}^{12}:=P_x^1\otimes P_y^2$$ are the projections on eigenspaces of position for both particles. The observable can be defined as $$X^{12}=\sum_{xy}\lambda_{xy} P_x^1P_y^2$$ where you can choose the eigenvalues $$\lambda_{xy}$$ however you want as long as they are distinct for all $$xy$$. The point is that the observable is not important, its the projections that are important.

A different approach is to think about it operationally. These are two observables that commute with each other. In the lab you would measure one particle, write down the outcome $$x_1$$ and immediately measure the second and write down the outcome $$x_2$$. Then you have the pair $$(x_1,x_2)$$ and you would say I have measured the simultaneous position of two particles. This suggests that if you have multiple commuting observables you don't have to come up with a whole new observable that measures both of them, you can just say I measure the set of commuting observables $$\{X_1, X_2,X_3,...\}$$