Measuring the position of identical particles and wavefunction collapse

Preview of the answer

To "measure the positions" of two bosons, we can fill space with a bunch of particle-counting observables localized in pointlike regions, and we can measure all of these observables simultaneously. (They commute with each other.) Even if we start with a generic two-particle state, the result will be a state of the form shown in the question — that is, an eigenstate of all of the localized particle-counting operators.

Intuition

When we deal with bosons (or fermions), we're really using one of the key ideas from quantum field theory (QFT): observables are tied to regions of space, not to particles. This is true in both relativistic and non-relativistic QFT, and it is the key to answering the question.

QFT has observables representing detectors that count the number of particles of a given species in a given region of space. This makes sense no matter how many particles are in the state, and it is compatible with the fact that the particles of a given bosonic species are indistinguishable.

We're implicitly using such observables in single-particle quantum mechanics, too, when we use the familiar "position observable". When we measure a sequence of observables separated from each other in time, we account for each measurement's outcome by projecting the state onto one of that observable's eigenspaces. The observable's eigenspaces represent the possible outcomes of the measurement. The associated eigenvalues are just convenient labels used to define things like expectation values and standard deviations. So, as far as the general principles of quantum theory are concerned, an observable might as well just be a collection of mutually orthogonal subspaces of the Hilbert space — or the operators that project onto those subspaces.

Applying that perspective to the position observable in single-particle quantum mechanics shows that the position observable is really just a collection of detectors, one per point in space, with eigenvalues (labels) conveniently chosen to be equal to the coordinate of the point where the detector sits. These detector-observables generalize nicely to states with $N$ indistinguishable particles. The position observable does not.

When we talk about measuring the positions of identical particles, we're really talking about placing detectors in specific regions of space to count the number of particles in those regions. The position information comes from knowing where we placed the detectors, just like it does in the real world.

The math

This is a customized review of the formalism of non-relativistic QFT. Work in one-dimensional space for simplicity. A system of any number of "identical" bosons is described by a single field operator $\varphi(x)$ associated with each spatial point $x$, together with its adjoint $\varphi^\dagger(x)$. These operators satisfy \begin{gather} \big[\varphi(x),\,\varphi^\dagger(y)\big] = \delta(x-y) \\ \big[\varphi(x),\,\varphi(y)\big] = 0. \tag{1} \end{gather} The symmetry that makes them bosons is implicit in these commutation relations. All observables are constructed from the field operators $\varphi(x)$ and $\varphi^\dagger(x)$.

Let $|0\rangle$ denote the vacuum state, with no particles. This state satisfies $\varphi(x)|0\rangle=0$. Each application of $\varphi^\dagger(x)$ adds a particle at $x$. The two-particle state shown in the question is $$ |\Psi\rangle = \varphi^\dagger(x_1)\varphi^\dagger(x_2)|0\rangle. \tag{2} $$ The commutation relations imply that this is the same as $$ |\Psi\rangle = \varphi^\dagger(x_2)\varphi^\dagger(x_1)|0\rangle, \tag{3} $$ so the symmetry is automatically enforced: in this way of formulating the model, we can't even write down a non-symmetric state. Now let $R$ denote some finite region of space, and consider the observable $$ D(R) = \int_R dx\ \varphi^\dagger(x)\varphi(x) \tag{4} $$ where the integral is over the region $R$. This observable represents a detector that it counts the number of particles in the region $R$. For example, when applied to the state (2), it gives $$ D(R)|\Psi\rangle = n|\Psi\rangle \tag{5} $$ where $n\in\{0,1,2\}$ is the number of particles in the region $R$. To derive (5), use the commutation relations (1) together with $\varphi(x)|0\rangle=0$.

If the state has just one particle, then we can use a "position observable" as explain earlier, like this: $$ X = \int dx\ x\, \varphi^\dagger(x)\varphi(x). \tag{6} $$ This is essentially a bunch of detection operators (4), each associated with an infinitesimal region $R$ (a single point), and weighted by the coordinate $x$ of that region. When acting on the single-particle state $$ |x\rangle = \varphi^\dagger(x)|0\rangle, \tag{7} $$ this gives $$ X|x\rangle = x|x\rangle. \tag{8} $$ But when acting on a multi-particle state like (2)-(3), the observable (6) is not as useful: it measures the average $x$-coordinate of all of the particles in the system, which is not what we want. We want the observables (4), which count the number of particles in a given region of space. That's the best we can do, because the particles are indistinguishable.

The answer

To "measure the positions" of two bosons, we can fill space with a bunch of particle-counters (4) with pointlike regions $R$ and measure all of these observables simultaneously. (We can do this because the commutation relations (1) imply that all of these observables commute with each other.) Even if we start with a generic two-particle state $$ \int dx_1\,dx_2\ f(x_1,x_2)\varphi^\dagger(x_1)\varphi^\dagger(x_2)|0\rangle, \tag{9} $$ the result will be a state of the form (2)-(3) — that is, an eigenstate of all of the detection operators (4).

But how can we measure the position of both particles?

We just assume an information input that one boson is at $x_1$, and another at $x_2$ and then propose the appropriate state vector for this knowledge. Unfortunately, how this kind of knowledge can be obtained is usually not very clear from QT textbooks, the goal there is to learn the formalism and established methods of its application, not experimental physics or why the theory works in that particular way.

The bosons could be inferred to be there at those points of space from inspecting a photograph, if the bosons left a trace of droplets/bubbles which can be assigned time and spatial coordinates. Or the bosons could have been prepared to be in those places, for example, by shooting them there through a tube from a particle accelerator. If the history of the particles until the time of position determination is kept, the bosons could be distinguishable by their history, so the appropriate state would be $|x_1x_2\rangle$.

But if the bosons are too close to each other beyond the ability of measurements to track their identity, or their history isn't known, they can't be distinguished and the appropriate state must imply same things for both particles. This can be done either by symmetrization or anti-symmetrization of $|x_1x_2\rangle$. For bosons, symmetrization is used.

What operator are we using?

For what - for measuring? None. Measuring is not done by operators! Operators are mathematical concepts that are associated with extracting expected average value of physical quantity for given $\Psi$. Or with eigenvalue equations that define valid values of those quantities.

You can ask, which operator has eigenstates where one particle is at some position $x_1$ and the other at $x_2$, for all possible values of $x_1,x_2$. For a single-dimensional $x$, such operator acts on psi functions that depend on two variables \psi{x,x')$ and the eigenvalue is a two-component vector:

$$ \hat{O} \psi(x,x') = \left(\array{ x_1 \\ x_2} \right) \psi(x,x') $$ These are really two equations, but we can write them as single equation using the column/matrix notation.

So the sought operator $\hat{O}$ is not the product $\hat{x}_1 \otimes \hat{x}_2$, but a two-component operator, which can be also written using the tensor product notation: $$ \left(\array{ \hat{x}_1 \\ \hat{x}_2}\right) = \hat{x}_1 \otimes \mathbf{1} + \mathbf{1}\otimes\hat{x}_2. $$ In the eigenvalue equation above, each component operator extracts only eigenvalues for "its" particle subspace of the whole coordinate space.

Let't try this for more complicated example: if the position measurement results are three-dimensional, the psi function for two particles depends on 6 coordinates $x, y, z, x' , y', z'$ and the sought operator acts on all those coordinates:

$$ \left(\array{\hat{x}_1 \\ \hat{y}_1 \\ \hat{z}_1 \\ \hat{x}_2 \\ \hat{y}_2 \\ \hat{z}_2}\right) \Psi(x, y, z, x' , y', z') = \left(\array{x_1 \\ y_1 \\ z_1 \\ x_2 \\ y_2 \\ z_2} \right) \Psi(x, y, z, x' , y', z') $$ These are really 6 equations, but we can write them as one 6-dimensional equation.

The operator on the left hand side is not tensor product of the particle operators, but sum of two operators that act on independent particle 3D spaces:

$$ \left(\array{\hat{x}_1 \\ \hat{y}_1 \\ \hat{z}_1 \\ \hat{x}_2 \\ \hat{y}_2 \\ \hat{z}_2}\right) = \left(\array{x_1 \\y_1\\z_1}\right) \otimes \mathbf{1} + \mathbf{1}\otimes \left(\array{x_2 \\y_2\\z_2}\right) = \hat{\mathbf r}_1 \otimes \mathbf{1} + \mathbf{1}\otimes\hat{\mathbf r}_2. $$

Good question - this notation gave me no end of confusion when I was learning about identical particles. For me, the thing that cleared things up was carefully distinguishing between labels that indicated particles and labels that indicate position. In Shankar's notation, subscript numbers indicate positions (e.g. "five meters to the right of the origin), with no reference to which particle might be at that position. He's implicitly working on the basis of the tensor product of single-particle wavefunctions, so I think it's helpful to explicitly label the different particles with a different label, say capital letters.

So when he says $$|{\Psi}\rangle = \frac{1}{\sqrt{2}}\left(|x_1 x_2\rangle + |x_2 x_1\rangle\right),$$ he's really referring to two particles $A$ and $B$ and means $$|{\Psi}\rangle = \frac{1}{\sqrt{2}}\left(|x_1\rangle_A \otimes |x_2\rangle_B + |x_2 \rangle_A \otimes |x_1\rangle_B\right),$$ or even more explicitly, $$|{\Psi}\rangle = \frac{1}{\sqrt{2}}\left(|\text{particle $A$ is at position $x_1$ and particle $B$ is at position $x_2$}\rangle + |\text{particle $A$ is at position $x_2$ and particle $B$ is at position $x_1$}\rangle\right).$$

The point of the "symmetrization" is that the state is left invariant if the labels "A" and "B" are switched, which is what we mean by "the particles are exchanged".

So the relevant operators are not in fact $X_1$ and $X_2$ but instead $X_A$ and $X_B$ - the operator indices identify a particle, not a position, and mathematically identify the "slot" in the tensor product (the first or the second one) which is being acted on.

The action of $X_A \otimes I_B$ on this entangled state $|\psi\rangle$ yields $$(X_A \otimes I_B) \frac{1}{\sqrt{2}}\left(|x_1\rangle_A \otimes |x_2\rangle_B + |x_2 \rangle_A \otimes |x_1\rangle_B\right) \\= \frac{1}{\sqrt{2}}\left(x_1 |x_1\rangle_A \otimes |x_2\rangle_B + x_2 |x_2 \rangle_A \otimes |x_1\rangle_B\right),$$ which indeed formally no longer lies in the bosonic Hilbert space. Formally, you need to project this operator back into the symmetric Hilbert space. But in practice, only inner products are actually measurable, so when you take the inner product of this state with a bosonic (i.e. symmetric) bra, you'll get that that takes care of the symmetrization for you and your answer will indeed be symmetric in $x_1$ and $x_2$.