I was wondering that if particles are indistinguishable in quantum mechanics, then why do we still express their states $\left| \uparrow \downarrow \right\rangle$, as meaning particle 1 (in the first position) in state "up" and particle 2 (in the second position) in state "down", and so this is a different state than $\left| \downarrow \uparrow \right\rangle$. We then go onto constructing a linear combination which is symmetric and so forth. But we are saying it is symmetric under exchange of particles 1 and 2.

Why don't we just talk about two particles, why do we have to give them labels, even if they don't affect measurement? Why aren't $\left| \uparrow \downarrow \right\rangle$ and $\left| \downarrow \uparrow \right\rangle$ the same state? Surely, if they are truly indistinguishable, labels are just a redundancy. Is there some mathematical reason behind this, or is there a deeper physical reason?

  • $\begingroup$ Let $|a\rangle \otimes |b \rangle$ be a state in $H^{(1)} \otimes H^{(2)}$. The formalism of QM is such that $H^{(1)}$ indeed correspond to the first particle, and $H^{(2)}$ the second, for in general the particles in these Hilbert spaces are not identical. We can then treat the case of identical particles as a special case where the two Hilbert spaces represent indistinguishable particles. And at this point you can either do the canonical thing where we construct symmetric states, or as you suggested, change the entire Hilbert space formalism. The former is clearly easier. $\endgroup$ – zzz May 15 '14 at 14:28
  • $\begingroup$ Some examples of troubles that may come up in what you're suggesting, for example, think about how would you construct operators in such a formalism? $\endgroup$ – zzz May 15 '14 at 14:31

It seems to me like this is basically a question of the form

"Why do we use the model that we do for identical particles in quantum mechanics as opposed to some other one?"

to which the best and ultimate reply is

"Because the model we have has never disagreed with experiment,"

but perhaps I can convince you that there are situations in which the commonly used tensor product formalism + symmetrization (or anti-symmetrization) is natural.

Consider a system of two electrons, one trapped in a box on Earth, and the other trapped in a box on Alpha Centauri.

These two particles are certainly identical; all electrons are, but they are distinguishable in the sense that since they are each trapped in a box, we can call the electron in the Earth box electron 1, and we can call the electron in the Alpha Centauri box electron 2, and we won't confuse ourselves. (See What are the differences between indistinguishable and identical? )

I think you'd agree that in this case, it makes complete sense (and is important) to distinguish between the spin states of each electron by writing tensor product states like $|\uparrow\rangle|\downarrow\rangle$, because when one makes a measurement of the spin state of system, one might ask a question like

"after a spin measurement, was the Earth electron in the spin up state or the spin down state?"

So we see that in certain cases, there is in fact a necessity to have a formalizm that distinguishes between identical particles to reproduce the types of individual subsystem measurements that one would like.

Now, if you had two electrons in a single box, then of course one cannot distinguish between the electrons in the same manner, but if the electrons are, for example, assumed to be non-interacting, then the physics of spin measurements in this case is precisely the same, so it makes sense to use the same model. You could, of course, try to concoct another model that reproduces the predictions of the tensor product model, but why would you when we have one that already has spectacular agreement with experiment?

  • $\begingroup$ Yeah, although I would say that which box they are in is just part of their state, so this no different than saying instead of being in two different boxes, they have two different colors or something. The reason I'd experiment with a different model even if this one agrees with experiment is of course that maybe the new one offers new insights :) $\endgroup$ – guillefix May 15 '14 at 19:15
  • $\begingroup$ @guillefix Then you are primed to study theoretical high energy physics :) $\endgroup$ – joshphysics May 15 '14 at 20:10

I was wondering that if particles are indistinguishable in quantum mechanics ... Why don't we just talk about two particles, why do we have to give them labels, even if they don't affect measurement?

Let's recall what this indistinguishability means in practice: in order to get good agreement of calculations of atomic properties like emission line frequencies with experiments, Hamiltonian eigenfunctions are needed (they are not necessary but they make the calculation tractable). It turns out that ordinary atomic Hamiltonians have such properties (they are real and symmetric...) that their eigenfunctions are all either symmetric or antisymmetric with respect to exchange of two particles' coordinates.

It was found by comparison to measurements that for electrons, antisymmetric wave functions for electrons $\psi(\mathbf r_1, \mathbf r_2)$ should be used.

Now, where is indistinguishability in that? It lies in the fact that solely from an antisymmetric wave function $\psi(\mathbf r_1, \mathbf r_2)$, we cannot tell any difference in properties or behaviour of the particle 1 from the particle 2; the probability density that 1 is at A and 2 is at B is the same as the probability density that 2 is at A and 1 is at B.

It may come as a surprise, but it seems clear that to explain just what indistinguishability means in practice of atomic physics, we need at least two different entities, distinguishable in speech - and only then it can be said that we cannot distinguish them with the description based on the $\psi(\mathbf r_1, \mathbf r_2)$ function alone. It does not necessarily mean two or all electrons in the world are actually one and the same electron or anything metaphysical like that as people sometimes fantasize. Indistinguishability is just practical nuisance.

Why aren't $\left| \uparrow \downarrow \right\rangle$ and $\left| \downarrow \uparrow \right\rangle$ the same state?

These two kets refer to different states by definition; the position of the arrow within the ket matters and is intentionally chosen based on which one of the individual spin-carrying body is meant. We want the formalism to work this way because it allows us to easily describe situations like spin z up at the detector A (first position within the ket), spin down at detector B(second position within the ket). The behaviour you have in mind does happen, but with different ket - the symmetric ket $$ \left|\uparrow\downarrow\right\rangle + \left|\downarrow \uparrow\right\rangle $$ which can be written as $sym\{\uparrow,\downarrow\}$ or $sym\{\downarrow,\uparrow\}$; the order does not matter here.

Surely, if they are truly indistinguishable, labels are just a redundancy. Is there some mathematical reason behind this, or is there a deeper physical reason?

Distinct labels like $\mathbf r_1,\mathbf r_2$ are used for distinct particles partially because they were used already in classical mechanics, in particular in the Hamiltonian mechanics. Schroedinger devised his equation with this Hamiltonian formalism in mind, so it uses distinct coordinates as well. For example, we need them to formulate the Hamiltonian operator for the helium atom

$$ \hat{H} = - \frac{\hbar^2}{2m}\Delta_1 - \frac{\hbar^2}{2m}\Delta_2 + \frac{K}{4\pi}q_1 q_2 \frac{1}{|\mathbf r_1 - \mathbf r_2|} + \frac{K}{4\pi}Qq_1\frac{1}{|\mathbf r_1|} + \frac{K}{4\pi}Qq_2\frac{1}{|\mathbf r_2|}. $$

There is no such thing as Schroedinger's equation with two indistinguishable variables. This should not be regarded as deficiency of his formalism - the variables are not particles after all. There is no difficulty with distinct variables referring to indistinguishable particles.

There is no indistinguishability in the formal objects - the two electrons are labeled by distinct variables. Only when seeking functions that solve the Schroedinger equation $$ \hat{H} \Phi(\mathbf r_1, \mathbf r_2) = E \Phi(\mathbf r_1, \mathbf r_2) $$ it turns out that these are all either symmetric or antisymmetric which means the description of the electron $1$ is the same as the description of the electron 2; one cannot tell anything different about the first second particle than about the second particle from the $\psi$ function alone.

If we want, we may even think that the two electrons are not exactly the same and are actually little bit different (for example, their mass may have variance $10^{-62}$kg, which is beyond our skills to measure). It should be clear that this does not prevent us to use anti-symmetric functions in calculations with advantage, nor to call electrons indistinguishable (when such mass differences are unmeasurable).


The quantum formalism has the useful property that we can use a Hilbert space that is larger than necessary to describe the dynamics of a system, and then fix this by putting constraints on possible states. The treatment of identical particles is an example.

To clarify the issues, consider a quantum computer running a classical computation: a classical lattice gas. In the classical gas model, binary 1's act like identical particles and binary 0's act like empty space. Interchanging two 1's doesn't give a new classical state (e.g., 110 -> 110). But suppose we decide in the quantum implementation to attach a distinct particle label to each particle, and describe particle interactions as if we could keep track of which particle was which. In this case interchanging particles that differ only in their particle label would give a new quantum state. We can fix this inconsistency with the original classical model by defining new basis states, each an equally-weighted sum of equivalent particle states (i.e., symmetrization). Using only these as basis states effectively removes the particle labels, reducing the size of the basis without changing the description of the dynamics.

Now, it's clear that entanglement in a quantum computation doesn't depend on the treatment of identical 1-bits as distinguishable particles! Thus it may well be that there is no deep mathematical reason for this kind of description (symmetrizing to reduce the effective size of the Hilbert space used for describing the dynamics) other than its close connection to classical "follow-the-particles" Lagrangian mechanics. Quantum lattice gas models of quantum field theories don't have this issue.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.