Argument for Fermions and Bosons as eigenfunctions of particle exchange operator $P$

Question

Some time earlier, my prof took me through an argument leading to emergence of fermions and bosons by the application of particle exchange operator on the multiparticle wavefunction. I will try to present the argument below:
We have a multiparticle wavefunction written as $\psi(x_1..x_i..x_j..x_n)$ representing a system of $N$ identical particles. And there is a parity operator $P$ which exchanges two particles in our system on $N$ particles. $P$ acts as follows, $$P_{\boldsymbol{i,j}}\psi(x_1...\boldsymbol{x_i}...\boldsymbol{x_j}..x_n) = \psi(x_1...\boldsymbol{x_j}..\boldsymbol{x_i}..x_n)$$ Now, if we operate $P$ on $\psi$ twice, we are supposed to get back the same system we started with before exchanging particles $i$ and $j$, leading to $P^2 = \mathbb{I}$. This tells us that $P$ can have only two possible eigenvalues $\pm1$ and thus we call the eigenfunctions with eigenvalue +1 as bosons and -1 as fermions.

My question is why can't we extend the above argument further to higher even powers of $P$. For example, if we apply the $P$ operator on our system 4 times, we will get an equation of the form $P^4 = \mathbb{I}$. Won't this lead us to two more eigenvalues for $P$, namely $\pm i$. what will we call the eigenfunctions of $P$ in these cases.

I'm not sure if I make much sense here. This has been in my mind since that class and I wish to get it clarified as to where am I going wrong with my argument.
Thanks in advance.

Answer 1

There are two problems here - one with your argument about eigenvalues, and another with the very common but deeply flawed argument for fermions and bosons.

The problem with your eigenvalue argument can be understood in the context of basic algebra. The mistake is in assuming that if I start with some equation and then raise both sides to some power, the solution sets of the two equations remain the same, but that's not true. For example, $z=1$ has one (fairly obvious) solution, but when I square both sides I obtain $z^2=1$, which now has two. A solution to the former is guaranteed to be a solution to the latter, but the converse is not true.

In the context of operator theory, this idea takes the following form. If an operator $\hat A$ has an eigenvalue $\lambda$, then $\hat A^2$ is guaranteed to have $\lambda^2$ as an eigenvalue. However, if $\hat A^2$ has some $\mu$ as an eigenvalue, then we cannot conclude that $\pm \sqrt{\mu}$ are both eigenvalues of $\hat A$. A simple example of this is the identity operator $\mathbb I$, whose only eigenvalue is $1$. $\mathbb I^2$ also has 1 as an eigenvalue (obviously), but we cannot from this conclude that the eigenvalues of $\mathbb I$ are $\pm \sqrt{1} = \pm 1$.

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The flaw in the argument for fermions and bosons is more subtle. The problem is intimately related to the rarely-discussed distinction between states and wavefunctions.

Let's restrict our attention to 2 particles for simplicity. A wavefunction for this system is loosely a function $\psi : \mathbb R^n \times \mathbb R^n \rightarrow \mathbb C$ such that $\int \mathrm d^3 x \mathrm d^3 y \ |\psi(\mathbf x,\mathbf y)|^2 < \infty$. If an observable $\hat A$ has an eigenvalue $\lambda$ corresponding to an eigenvector $\phi_\lambda$, then the probability of measuring $\hat A$ to have that value is given by $$\mathrm{Prob}_\psi(\hat A,\lambda):= \frac{|\langle \psi,\phi_\lambda\rangle|^2}{\langle \psi,\psi\rangle}$$

In that sense, $\psi$ allows us to compute the probability of any measurement outcome we can imagine. This collection of probabilities is what we call a state.

The crucial point is that a given state can be represented by an infinity of different wavefunctions. Its an easy exercise to see that if you look at the probability expression above and then scale $\psi$ by any non-zero complex number, the probability you obtain would be exactly the same. Therefore, a state should not be identified with an individual wavefunction but rather with an entire family of wavefunctions which are non-negative multiples of one another. In this spirit, we define the projective Hilbert space $\mathscr P(\mathcal H)$ to be the set of equivalence classes of elements of $\mathcal H$, where two elements of $\mathcal H$ belong to the same class if each is a non-zero multiple of the other.

Now, it turns out that working directly with some $\Psi\in\mathscr P(\mathcal H)$ is enormously inconvenient. Therefore, in practice we always simply choose a concrete representative $\psi\in \mathcal H$ and work with that instead. Ordinarily this causes no particular issues, and so one can be forgiven for forgetting that $\psi$ is merely a representative of a state, but sometimes the distinction is crucial, and this is one of those times.

Note that any linear or antilinear map $M:\mathcal H \rightarrow \mathcal H$ induces a well-defined map $\widetilde M:\mathscr P(\mathcal H) \rightarrow \mathscr P(\mathcal H)$ on the space of states (note that $\widetilde M$ is not linear, because $\mathscr P(\mathcal H)$ is not a vector space!). Because working at the level of $\mathcal H$ is vastly more convenient than working at the level of $\mathscr P(\mathcal H)$, this is normally what we would do to define a transformation on a state. However, there is great ambiguity here! For instance, if two operators $\hat A$ and $\hat B$ are related by $\hat A = c \hat B$ where $c$ is a non-zero complex number, then these two distinct operators on $\mathcal H$ induce precisely the same transformation of $\mathscr P(\mathcal H)$.

This leads us to the problem with your instructor's argument. We want to define the action of the permutation group $S_2$ on the space of states, and so it is natural to define its action on $\mathcal H$ first. But the family of pseudo-permutation operators $$\hat P_\theta \psi(\mathbf x,\mathbf y) = e^{i\theta} \psi(\mathbf y,\mathbf x)$$ all induce precisely the same permutation map on $\mathscr P(\mathcal H)$. You have arbitrarily chosen $\hat P_0$ as the correct permutation operator but there is no physical justification for doing this. Since $\hat P_\theta^2 = e^{2i\theta}\mathbb I \neq \mathbb I$ in general, the rest of the argument falls apart.

In summary, if we want to define the particle permutation map on the space of physical states $\mathscr P(\mathcal H)$, we can choose any of the family of pseudo-permutation operators $P_\theta$ at the level of the Hilbert space $\mathcal H$. In the absence of any further constraints, all of these operators do the same job. Since $\hat P_\theta^2 \neq \mathbb I$ in general, the requirement that the state be invariant under particle interchange implies only that the wavefunction be an eigenfunction of $\hat P_\theta$, without constraining those eigenvalues to be $\pm 1$.

Answer 2

All eigenfunctions of $P$ must satisfy $$ P^n \psi = \lambda^n \psi $$ where $\lambda$ is the relevant eigenvalue, for all values of $n$. In particular they must satisfy the equation for the equation for $n = 2$, giving the eigenvalues $\pm1$. If you plug these values into the higher order equations you will find they are also satisfied, for example $$ P^4 \psi = \hat{1}^2 \psi = (\pm1)^4 \psi $$ On the other hand you cannot go in the other direction. Your "extra" solutions $\pm i$ satisfy the equation for $n = 4k$, $k\in\mathbb{Z}$, but they do not satisfy the equation for $n = 2$

Answer 3

Mathematically, doing a transposition twice in a row gives the identity, i.e. $P_{ij}^2=\hat 1$. You cannot escape this mathematical fact, and it's baked in the permutation group. If you don't want to use the permutation group, then you're just doing something different than permutation symmetry.

Note that not all permutations satisfy $P^2=\hat 1$, and in that way the permutations of two objects is quite boring and not in any way representative of the structure of the permutation group for $n$ objects.

For instance, in 3-particle systems, the 3-cycle $P_{123}$ which takes $(a,b,c)\to (c,a,b)$ so that $P_{123}P_{123}(a,b,c)=(b,a,c)$ clearly does not satisfy $P_{123}^2\ne \hat 1$. However transpositions, permutations which act on only two entries in a list, you will certainly have $P^2=\hat 1$ irrespective of the number of particles.

Of course part of the strategy in using the permutation group is that you have all the known machinery for that group available from the math literature, and it makes sense physically to "permute" coordinates using this group, at least in the simplest cases (you could do braiding instead of permutations for fancier systems like anyons).