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I'm working through Shankar's Principles of Quantum Mechanics, and I think I have hit a confusion over identical particles. The book refers to 'measuring the position' of two bosons to be $x_1$ and $x_2$, thereby concluding that the state of the system is $$|{\Psi}\rangle = \frac{1}{\sqrt{2}}\left(|x_1 x_2\rangle + |x_2 x_1\rangle\right).$$

But how can we measure the position of both particles? What operator are we using? I see that $$X_1\otimes X_2|{\Psi}\rangle = x_1x_2|{\Psi}\rangle,$$ but this just seems to be a measurement of the product of the two positions. I have an even bigger confusion if we measure the position of a particles using the operator $X_1 \otimes I$. For then, must the state not collapse into one of the $|x_1x_2\rangle$ or $|x_2x_1\rangle$ eigenstates, violating the symmetry of the two bosons? Shouldn't the operator $X_1 \otimes I$ give rise to some sensible observable, as it is Hermitian?

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  • $\begingroup$ Related: Do quantum measurements destroy the symmetrization requirement? $\endgroup$ Commented Aug 2, 2021 at 14:00
  • $\begingroup$ There are no particles in quantum mechanics. There are only quanta of energy. A quantum system that contains two indistinguishable quanta can therefor be measured twice before it has lost all of its energy, but unlike in the case of entanglement we are not looking at the remaining single quantum wave function of the already measured system (no Alice and Bob games for you, helium!). $\endgroup$ Commented Apr 23, 2023 at 2:19

3 Answers 3

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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 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 explained 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).

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  • $\begingroup$ Does this imply that no matter how many times a person measures the many-particle system, it is still possible for the identical particles to “teleport”, i.e. switch places (not that it violates SR, as no one could possibly tell a “teleportation” happened)? Since any of the Fock State we collapse the system into still have equal probability for every combination of particles to occupy the given single-particle state of the Fock State. $\endgroup$ Commented Sep 23, 2021 at 17:09
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    $\begingroup$ @JY_Decipherer_ Nothing teleports. The question "which particle is in which location" is meaningless in this model, because this model does not have any observables attached to individual particles. If it did, then they wouldn't be what physicists call "identical bosons." (I don't like that language, because the colloquial meaming of the word "identical" does not capture the right concept at all.) $\endgroup$ Commented Sep 23, 2021 at 17:55
  • $\begingroup$ Does that mean when a particle is measured to be at position x, what is really measured is just the superposition of all particles in the universe identical to the one measured being at the position x? $\endgroup$ Commented Sep 24, 2021 at 8:24
  • $\begingroup$ @JY_Decipherer_ That question seems to be using the word "superposition" in a non-standard way, so I don't know exactly what you're asking. Suppose we start with a two-particle state $\psi(x_1,x_2)=\psi(x_2,x_1)$ and then do a measurement to count the number of particles in a given region $R$. The outcome is $n\in\{0,1,2\}$, and based on the outcome, we project the state onto the part of $\psi(x_1,x_2)$ that is only nonzero when exactly $n$ of the arguments $\{x_1,x_2\}$ are inside $R$. The projected state is still symmetric. $\endgroup$ Commented Sep 24, 2021 at 13:18
  • $\begingroup$ By "superposition", I meant that in an n-particle system, for every position measured to have a particle in the region 𝑅, there is a 1/n probability for each particle (of the n particles in the system) to be at that position. So the situation isn't "which particle is in which location", but rather "all particles are equally possible to be in which position". And to clarify, "there is 1/n probability for each particle to be at that position" are mutually exclusive events. $\endgroup$ Commented Sep 24, 2021 at 17:19
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But how can we measure the position of both particles?

We just assume an information input that one particle is at $x_1$, 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 two particles could be inferred to be there at those points of space from inspecting a photograph, if they left a trace of droplets/bubbles which can be assigned time and spatial coordinates. Or the particles 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, then the two particles could be distinguishable by their history of positions at previous times. If first particle (first meaning first in our tracking records) was in state $|a\rangle$, and the second particle at state $|b\rangle$, then the appropriate state for these distinguishable particles would be written as $|a,b\rangle$, or $|a\rangle \otimes |b\rangle$. This state captures the knowledge "the particle 1 is in state $a$ and the particle 2 is in state $b$".

But if the particles 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 of the combined system must imply same things for both particles. One way to write down such state is either by symmetrization or anti-symmetrization of $|a,b\rangle$. For bosons, symmetrization is used, for reasons that are hard to explain here. So the appropriate state for two particles and two different states but where the particles are indistinguishable is

$$ \psi = \frac{1}{\sqrt{2}}\bigg(|a\rangle\otimes |b\rangle + |b\rangle\otimes |a\rangle \bigg). $$

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 $x_a,x_b$ where the first particle is at position $x_a$ and the other at position $x_b$? For a single-dimensional coordinate $x$, such operator acts on functions of two variables $\psi(x_1,x_2)$ and its action has to result in a two-component vector multiplying the psi function:

$$ \hat{O} \psi(x_1,x_2) = \left(\array{ x_1 \\ x_2} \right) \psi(x_1,x_2) $$

These are really two simple equations, but we can write them as single "vector" 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 vector operator, which can be also written using the tensor product notation: $$ \left(\array{ \hat{x}_1 \\ \hat{x}_2}\right) = \hat{x}_1\mathbf{e}^{(1)}_{1} \otimes \mathbf{1}^{(2)} + \mathbf{1}^{(1)}\otimes\hat{x}_2\mathbf{e}^{(2)}_{1}. $$ Here $\mathbf{e}^{(1)}_{k}$ is the basis vector along the $k$-th axis in the coordinate space of the first particle, and $\mathbf{e}^{(2)}_{k}$ is the same for the second particle. Together these 6 basis vectors span 6-dimensional coordinate space of the combined system.

In the eigenvalue equation above, each operator component extracts only eigenvalues for "its" particle subspace of the whole coordinate space.

Let't try this for a more complicated example: if the position measurement results are three-dimensional, the psi function for two particles depends on 6 coordinates $x_1, y_1, z_1, x_2 , y_2, z_2$ and the sought operator acts on all those coordinates:

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

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

$$ \left(\array{\hat{x}_1 \\ \hat{y}_1 \\ \hat{z}_1 \\ \hat{x}_2 \\ \hat{y}_2 \\ \hat{z}_2}\right) = \left(\array{\hat{x}_1 \\\hat{y}_1\\\hat{z}_1}\right) \otimes \mathbf{1}^{(2)} + \mathbf{1}^{(1)}\otimes \left(\array{\hat{x}_2 \\\hat{y}_2\\\hat{z}_2}\right) = $$

$$ = \hat{\mathbf r}_1 \otimes \mathbf{1}^{(2)} + \mathbf{1}^{(1)}\otimes\hat{\mathbf r}_2. $$

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Chris
    Commented Sep 1, 2022 at 20:32
  • $\begingroup$ Ok, I saw the replies in the chat and the edit, and I think it's good. I would still emphasize this applies to a system of two non-identical particles. In your own words, "In my answer above, I was considering two distinguishable particles, 1 and 2. What I really meant should be better written as follows. The vector position operator for the combined system of distinguishable particles is $\sum_k \hat{x}_{1,k} \mathbf{e}_{1,k}\otimes \mathbf{1}_2 + \mathbf{1}_1\otimes \sum_k \hat{x}_{2,k} \mathbf{e}_{2,k}$. [...]" $\endgroup$ Commented Sep 2, 2022 at 6:43
  • $\begingroup$ "[...] However, applying the operator to symmetrized state $|a\rangle \otimes |b \rangle + |b\rangle \otimes |a\rangle$ reveals that this is not an eigenstate of that operator. This is as expected, symmetrized or antisymmetrized states do not describe state of definite position of either particle." $\endgroup$ Commented Sep 2, 2022 at 6:44
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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 $x_2$.

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    $\begingroup$ Why must I take the inner product with a symmetric bra? What is wrong with saying $|x_1x_2\rangle$ is an eigenstate of $X_A\otimes I_B$, so we can take the inner product with this bra, and get an answer that is not symmetric? $\endgroup$
    – awsomeguy
    Commented Apr 26, 2020 at 8:03
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    $\begingroup$ @awsomeguy Because $X_A \otimes I_B$ is not actually a linear operator on a bosonic Hilbert space: it takes symmetric states to nonsymmetric states that lie outside of the Hilbert space. For bosonic systems, you can really only talk about the projection $P_B^\dagger (X_A \otimes I_B) P_B = \frac{1}{2} X_A \otimes I_B + \frac{1}{2} I_A \otimes X_B$ onto the bosonic Hilbert space, where $P_B$ is the projection operator from the non-symmetrized Hilbert space to the bosonic subspace. Calculating the action of this projection on operators gets pretty cumbersome. Since only matrix elements are ... $\endgroup$
    – tparker
    Commented Apr 26, 2020 at 15:16
  • $\begingroup$ ... physically measurable at the end of the day, you don't actually need to calculate the projection on the operator; as long as you only calculate matrix elements between symmetric states in the bosonic Hilbert space, you'll get the same answer. $\endgroup$
    – tparker
    Commented Apr 26, 2020 at 15:18
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    $\begingroup$ By the way, the reason that my answer and Chiral Anomaly's look very different is that we are using different formalisms for many-body QM. Chiral Anomaly is using the formalism of second quantization/Fock space. I am working in the "particle basis" (rather the Fock basis, which is kind of like a position basis). The Fock space is more natural to use when you have indefinite particle number (basically always the case in relativistic QFT), but the particle basis is sometimes more convenient when you have fixed particle number (sometimes the case in nonrelativistic QM). $\endgroup$
    – tparker
    Commented Apr 26, 2020 at 15:31

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