# How do we know that only symmetrical and antisymmetrical states occur in nature?

I have a question concerning exactly how we get to the following two conclusions in quantum mechanics (which are both experimentally obtained as I understand it):

• Identical particles are indistinguishable
• All wave functions of identical particles are either symmetric or antisymmetric

I believe those two things are not the same.

Here is my understanding at the moment: The indistinguishability of identical particles can experimentally be obtained by showing that entropy does not increase when mixing identical gases.

So far, so good. This leads us to the conclusion that, in a mathematical formulation $$[\hat{A}, \hat{U}_{\pi}] = \hat{0} \ \forall \ \pi,$$ where $$\hat{U}_{\pi}$$ is the operator that applies the permutation $$\pi$$ to a state and $$\hat{A}$$ is any operator. In words, this relation tells us that any expectation value (aka anything that is measurable) stays the same when the role of two identical particles is switched.

Specifically, with $$\hat{A} = \hat{H}$$, we know that any eigenstate of $$\hat{U}_{\pi}$$ stays an eigenstate under time evolution and that we can simultaneously diagonalize any operator together with $$\hat{U}_{\pi}$$.

My first question is: This relation alone does not tell us that linear combinations of eigenstates of $$\hat{U}_{\pi}$$ are forbidden, right? So from this alone, we cannot derive the second conclusion I listed in the beginning.

However, I have read that from the absence of mixing entropy, the second conclusion can be gained. How is that possible? From what I have read, it has something to do with the fact that the increase in entropy would stem from exchange degeneracy if the irreducible representation of the permutation group within an eigenspace of an operator would be more than one dimensional. I do not exactly understand what that means.

So in general, my second question is if we need other experimental verification to gain the second conclusion, whether the missing mixing entropy suffices or if my thoughts are wrong altogether at some point.

Keep in mind, I am not talking about the spin-statistics theorem. I know that without it, we don't know how to choose which subspace (symmetric or antisymmetric) to choose for the description of certain particles.

• They are not the only cases in two dimensions. Look up anyons. Commented Aug 7 at 0:00
• The entire symmetric/antisymmetric wavefunction picture is an unfortunate and confusing attempt to undo a fundamental problem with thinking of particles as distinguishable objects in the first place. There is a much better way, as explained in this other question and its accepted answer. Commented Aug 7 at 4:22
• @DanielSank fair enough but there are many instances where the symmetry of the state (and using symmetrization to "restore" indistinguishability) carries with it important qualitative information. There's a reason why chemists are (more or less) attached to using Slater determinant of states: their problems are not so easy using field theory and 2nd quantization methods. Commented Aug 7 at 16:56
• This is especially true as we are now talking about "partially distinguishable" photons: see this recent paper where bosonic commutation relations must be modified to account for the polarization overlap. Of course, you can do this with fields but the explanation looses some of its clarity. Commented Aug 7 at 16:59
• @Takitoli It is because we want to make statistical entropy function obey that constraint, and have similar value to thermodynamic entropy which obeys it based on a convention, that we define statistical entropy as $k_B \ln \frac{W}{N^N}$ or $k_B \ln \frac{W}{N!}$. This "fix" makes the function homogenous of first degree in $U,V,N$. But it is purely formal alteration to be consistent with convention in thermodynamics. Commented Aug 8 at 11:26

This is phenomenological. Pauli introduced the exclusion principle (for electrons) to explain some data of atomic spectra, and this was eventually upgraded to an anti-symmetry property for indistinguishable fermions, and experimentally verified in other systems (nuclei for instance).

For bosons the reason is also phenomenological, possibly most spectacularly demonstrated in the Hong-Ou-Mandel effect, where the sign of the interference (destructive or constructive) depends on the statistics of the particles (in HOM photons thus bosons), or in Bose-Einstein condensates.

There are all kinds of data - scattering data of identical particles for instance - which are only explained assuming constituents satisfy either fermionic or bosonic statistics, pinning down the linear combinations as symmetric or antisymmetric only.

Mixed symmetry generally messes up average values, and one can show that states must transform by a one-dimensional representation of the permutation group, and only the fully symmetric or fully antisymmetric representations are 1-dimensional; if not complete indistinghuishability is lost.

So: we “know” this is holds because this classification helps understand and predict physical phenomena, and there is no evidence to show it is false or something else is needed…

… possible except, as alluded to in a comment, for anyons although they are possible in 2D only and are actually “quasi-particles”.

• Should mention spin stats theorem too Commented Aug 7 at 1:09

To get a deeper answer to your question, observe that exchange symmetry is a stop-gap method to try to understand identical particles using non-relativistic quantum theory. But identical particles are just a way of making reference to excitations of quantum fields, and those excitations are described by quantum field theory.

The use of states which are either symmetric or antisymmetric with respect to exchange of labels is a mathematical method within non-relativistic quantum mechanics which succeeds in capturing some relevant aspects of what is going on. The more general theory here is quantum field theory, and in that subject the exchange symmetry is handled differently.

In field theory if you want to write a state with a fixed number of particle you can write the state vector in the notation $$| \psi \rangle = | n_1, n_2, n_3, \cdots \rangle$$ where $$n_i$$ gives the degree of excitation of the $$i$$'th mode of the field. In particle language we would say there are $$n_i$$ particles in the $$i$$'th state. Note, in this notation we never attach labels to the particles (any more than one would attach labels to all the millijoules that exist in a joule of energy). Rather, we attach labels to the field modes, specifying things like direction of propagation, polarization, etc.. So you see there is never any need to say 'the state is symmetric with respect to exchange of particle labels' or 'the state is antisymmetric with respect to exchange of particle labels' because the particles were never labelled in the first place. In field theory instead we have an expression for the commutator or anticommutator between raising and lowering operators, and this results in a restriction on the numbers $$n_i$$. For Bosons each $$n_i$$ can take any non-negative integer value. For Fermions each $$n_i$$ can only take the values $$0$$ or $$1$$.

This answers the question in that it shows that both bullet points in the question are an artefact of using a somewhat ad-hoc method to cobble together a theory which captures the low-energy limit of a deeper theory.

• Thank you! Im am always wondering if teaching us all about "particles" when introducing us to quantum mechanics is poisoning our understanding of the underlying theory in some way Commented Aug 7 at 15:18
• Actually, the formalism of 2nd quantization simply displaces the difficulty and hides it in algebraic format. Plus, it is often highly inconvenient. For instance, the ground state of two fermions in an attractive potential is $f^\dagger_{0+}f^\dagger_{0-}\vert 0\rangle$, where the 2nd index is spin. Cute, but you have no way of knowing that the resulting spin of the two fermions must be $0$ unless you do quite a bit of gymnastics. If we add a $\delta(x)$-type perturbation, good luck expressing this in 2nd-quantized form. Commented Aug 7 at 16:51
• I'm not against 2nd quantization: it is often extremely useful, but it does have its limitations. Understanding the permutation symmetry of states can be essential to understanding the physics of this state, and the permutation symmetry can (and sometimes must) be broken explicitly into a a space and spin part, and the 2nd-quantized formalism is poorly adapted to that sort of analysis. Commented Aug 7 at 16:51