# Meaning of the symmetrisation postulate in absence of a proper model

My question is on the use of the concept of indistinguishable particles (in quantum mechanics) in a very general context and in particular in statistical mechanics. I have made clear some of my opinions on this subject in a recent paper but when thinking even more about it, I cannot seem to comprehend how come this indistinguishability feature is so universally used and how it is justified.

I am aware of the questions asked here and there and the answers therein but I think they do not exactly address my question.

I am surprised by the specific and intrinsic role that seems to be played by the permutation operator (say $\hat{\Pi}_{12}$ for swapping particles 1 and 2) when compared to other symmetry operators.

In particular, in most cases that come to my mind (rotation, translation, time inversion, parity, etc...), the symmetries of the system are contained in the Lagrangian or Hamiltonian of the system. In simple quantum mechanics for instance, if a system has a certain symmetry, then its hamiltonian commutes with the corresponding representation of this symmetry group. It is then standard that, when we want to have a complete description of the state, we seek a Complete Set of Commuting Observables that will induce a particular structure to the quantum state.

To take the example of two particles considered in many cases, if the hamiltonian commutes with the permutation operator, it is then natural to seek a solution to the Schrodinger equation which is an eigenvector of $\hat{\Pi}_{12}$ with either eigenvalue $+1$ or $-1$.

Now, I perfectly understand that the reasoning above is not enough to get to the symmetrisation postulate (as it has been well answered in the other questions cited above) but my point is that, while the tendency seems usually to go from the Hamiltonian (which I conceptually equate to the modelling of the problem) to the symmetries and the structure of the quantum state, I am surprised that in the case of "identical" particles, the fermion/boson reasoning always precedes the hamiltonian (at the exception of the Standard Model of Particle Physics). In other words it is essentially overlooked (I think) that the apparent indistinguishability of a bunch of particles could be induced by the (maybe simplified or at least effective) modelling of the problem.

My concern is more about general particles such as composite particles like atoms, molecules etc... where it is sometimes (if not often) assumed that they are indistinguishable in the quantum mechanical sense while I would think that they are almost never actually indistinguishable.

If I take a gas of molecules for instance, I seriously wonder what it means to consider them as indistinguishable in the quantum sense since they can have different conformations or electronic states at any finite temperature (spectroscopists know that very well).

My question is then the following:

For an assembly of conceptually identical composite systems (i.e. identical in their composition), is it enough to provide an effective hamiltonian that commutes with the permutation operator to conclude that the so-described particles are indistinguishable in the quantum mechanical sense? If not, what is the rationale to justify the use of fermionic and bosonic statistics for composite systems?

EDIT: I will try to be a bit more specific

Let us consider for now a 2-particle system with the effective hamiltonian

$\hat{H} \equiv \frac{\hat{P}_1^2}{2m} + \frac{\hat{P}_2^2}{2m} + V(\hat{X}_1) + V(\hat{X}_2)$

where the external field $V$ acts in the same on the two particles at a given location.

As far as I am concerned, this hamiltonian commutes with the operator $\hat{\Pi}_{12}$ that swaps labels $1 \leftrightarrow 2$.

Let us now imagine that the particles under study have another observable feature $\hat{F}$ (color, size, spin projection etc...) so that $[\hat{H}, \hat{F}] = 0$ and that the two particles we are looking at may possibly have a different eigenvalue for the feature $\hat{F}$.

Let us now denote $\{|\phi_n \rangle \}_{n=1,2..}$ the eigenstates of the individual hamiltonian for each particle and $\{|f_{\alpha} \rangle \}_{\alpha=1,2..}$ the eigenvectors of the feature $\hat{F}$.

Are these ingredients enough to recquire the complete state to be symmetric or anti-symmetric? e.g.

$|\Psi \rangle = \frac{1}{\sqrt{2}}\left(|\phi_{n_1},f_{\alpha_1}; \phi_{n_2},f_{\alpha_2} \rangle \pm |\phi_{n_2},f_{\alpha_2}; \phi_{n_1},f_{\alpha_1} \rangle \right)$

I would suppose so and Hector's answer seems to indicate that it is the case as well for two particles and that it seems "natural" to extend the chosen symmetric or anti-symmetric property to the N-particle case (to be eventually tested against experiments).

• What does "indistinguishable in the quantum mechanical sense" mean if not "hamiltonian that commutes with the permutation operator"? – ACuriousMind Apr 9 '15 at 16:07
• To me it means "I have to use the Symmetrization/Anti-Symmetrization Postulate" which, as far as I understand, does not follow necessarily from having the permutation operator commuting with the hamiltonian. Moreover, simply having the permutation operator commuting with the hamiltonian is not enough to make the usual (and controversial?) distinction between distinguishable identical particles and indistinguishable identical particles. – gatsu Apr 9 '15 at 16:19
• Are you aware that the "(anti-)Symmetrization Postulate" (if you mean the assumption that integer spin particles are always bosons etc.) is only a postulate for non-relativistic quantum mechanics? The Spin-statistics theorem of QFT proves that spin and boson/fermion nature are indeed related. – ACuriousMind Apr 9 '15 at 16:23
• That's not what I mean. There is a clear usual distinction between distinguishable identical particles (for which we don't need to symmetrize or anti-symmetrize the quantum states) and indistinguishable identical particles (for which we ought to fully symmetrize or anti-symmetrize the states). In the mind of many people, this distinction clears-up many mysteries surrounding the Gibbs-Paradox in statistical mechanics. My question is how to justify (if it is justifiable) full symmetrization or anti-symmetrization of the states of say Helium atoms in a gas at any finite temperature. – gatsu Apr 9 '15 at 20:06

I think your concern is why to use the fundamental "symmetrization postulate" and not only the hamiltonian symmetry. The thing is that doesn't matter, suppose i'm trying to describe two particles (fermions for example) with Hilbert space $\mathfrak{H}_1$ and $\mathfrak{H}_2$, the Hilbert space of the two particles is
$$\mathfrak{H} = \mathfrak{H}_1\otimes\mathfrak{H}_2$$ Now suppose you have a hamiltonian in this space $$H\,:\,\mathfrak{H} \to \mathfrak{H}$$ And the permutation operator $P$ that acts in $\mathfrak{H}$ by $$P(u\otimes v) = v\otimes u$$ and if the identity operator $Id\,:\,\mathfrak{H} \to \mathfrak{H}$ act as $$Id(u\otimes v) = u\otimes v$$ You can take a quotient hilbert space, using the range of $P+Id$. So now we have $\mathfrak{H}/\sim$ where we have identified $$u\otimes v + v\otimes u \sim 0 \implies u\otimes v \sim - v\otimes u$$ This the usual antisymmetric space you work with the "symmetrization postulate", the class of equivalence of $u\otimes v$ is usually written as $u \wedge v$, to remember the antisymmetrization postulate (and in a matrix setting you set a representant usign slater determinants, but this really is unnecesary if one's knows where is working). In this setting you are working now in another hilbert space to model your system. Now, a valid question is, When my hamiltonian $H\,:\,\mathfrak{H} \to \mathfrak{H}$ can be unambiguosly defined in the cocient space $\mathfrak{H}/\sim$? that is, if $\pi\,:\,\mathfrak{H}\to\mathfrak{H}/\sim$ is the natural projection $\pi(u\otimes v) = u\wedge v$, when $\tilde{H} = \pi \circ H$ is well defined? the answer is only when $H$ conmutes with the permutation operator $P$, that is, only when $P$ is a symmetry of the original hamiltonian. So you get exactly the same results if you use the "symmetrization postulate" (work in a smaller Hilbert space) or if you use a hamiltonian with permutation symmetry. So why do we bother to use this symmetrization postulate? since the only "good" hamiltonians that can be translated to this setting are the ones who conmute with $P$, what i'm actually saying in physical terms is "Every physical hamiltonian for a pair of electrons conmute with the permutation operator". So this symmetry is a fundamental property of particles not a consequence of the dynamics, so i translate this property to the hilbert space itself, not to the hamiltonian. But formally are equivalent, the symmetrization postulate is some kind of "affirmation about the fundamental nature of things", not about some particular dynamics.
• @gatsu Maybe it wasn't so clear in my answer. But what i demostrate is that there is an equivalence of "symmetrization postulate" and "symmetry under permutations of the hamiltonian". In the helium gas, if you model it with a quantum hamiltonian with permutation symmetry (not the only choice because as you say permutation is not a fundamental symmetry in this case) then you can use the symmetrization postulate to get the same results. One usually, in a fundamental way, use "symmetrization postulate" $\implies$ "symmetry under permutations of the hamiltonian", here we use the reverse. – Héctor Apr 9 '15 at 20:58