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Suppose you already know how to treat oneparticle quantummechanical systems, and now you wish to go big, go statistical, or maybe go to fields. It is therefore necessary to understand systems with multiple (identical) particles, and how to represent them mathematically.
The fundamental hypothesis here is that particles of the same kind are indistinguishable. The question of how reasonable this hypothesis is is “a good question… for another time”, so very much like The Force Awakens’ writers, I will put a shade on that lamp, ignore the issue, and move on ^{1}. In the end, we should not say that “particle $a$ is in state $x$ and particle $b$ is in state $y$”; at most we should say “there are two particles, one in state $x$ and one in state $y$”. How do we go about describing these systems mathematically, given that we already know how to describe a oneparticle system? It seems cumbersome to try and work out how to mathematically write the second statement, whereas the first one seems simpler if we already know how to state that “a particle is in state $x$” and “a particle is in state $y$”.
Then what we will do is write “particle $a$ is in state $x$ and particle $b$ is in state $y$” mathematically, and then find out how to turn this statement into the indistinguishable case “there are two particles, one in state $x$ and one in state $y$”. We will see that the indistinguishable case can be thought of as a “distinguishable case” but with permutation symmetry.
This means that we first have to discuss a little bit what a symmetry is, and then apply this to the permutation symmetry of multiparticle states. We will obtain that there are two types of indistinguishability, one of which naturally takes us to Pauli’s exclusion principle.
Symmetries.zip
Consider an observable $A$ of a system $S$ (this system does not even have to be quantum). Suppose that we have two states $\psi_1,\psi_2$ of this system that are identical if we only compare how they “look” with the $A$ observable (for example, if the observable $A$ is “distance to a point $o$”, then all the points in a circle centered at $o$ are identical for anyone who can only check the $A$ observable). What we mean is that the probability distribution of $A$ in states $\psi_1$ and $\psi_2$ are the same. Under this observable, the states $\psi_1$ and $\psi_2$ are identical. In this case we say that $\psi_1$ and $\psi_2$ are $A$equivalent or $A$symmetric (or symmetric with respect to $A$). There might be some other observable $B$ which can distinguish between $\psi_1$ and $\psi_2$, or otherwise the states $\psi_1$ and $\psi_2$ will, by definition, be exactly the same.
In the context of quantum mechanics, we say that two states ${\left\vert \psi_1 \right\rangle}, {\left\vert \psi_2 \right\rangle}$ are $A$symmetric if the probability distribution of the observable $A$ is the same for both. That is, they are $A$symmetric if and only if for all possible values $a$ of $A$, the cumulative distributions are
\[P\left(A\leq a\middle\vert\psi_1\right)=P\left(A\leq a\middle\vert \psi_2\right).\]Recall then that the probability distribution of an observable $A$ is given by the eigenvalues of an associated Hermitian operator $\hat{A}$. Namely, if ${\left\vert a_1 \right\rangle},{\left\vert a_2 \right\rangle},\dots$ are orthonormal eigenstates with eigenvalues $a_1,a_2,\dots$, then the probability distribution of $A$ given a state $\psi$ is given by^{2}
\[P(A=a_j\vert \psi) = \left\vert{\left\langle a_j \middle\vert \psi\right\rangle}\right\vert^2.\]This implies that two states ${\left\vert \psi_1 \right\rangle},{\left\vert \psi_2 \right\rangle}$ are $A$symmetric if and only if for all eigenvalues $a_j$ of $\hat{A}$,
\[\left\vert{\left\langle a_j \middle\vert \psi_1\right\rangle}\right\vert^2 = \left\vert{\left\langle a_j \middle\vert \psi_2\right\rangle}\right\vert^2.\]We say that a mapping $T:\mathcal{H}\to\mathcal{H}$ is $A$symmetric (or an $A$symmetry) if for all ${\left\vert \psi \right\rangle}\in \mathcal{H}$, the states ${\left\vert \psi \right\rangle}$ and ${\left\vert \psi’ \right\rangle}=T{\left\vert \psi \right\rangle}$ are $A$symmetric. This means that for all eigenvalues $a_j$ of $A$,
\[\left\vert{\left\langle a_j \middle\vert T\middle\psi\right\rangle}\right\vert^2 = \left\vert{\left\langle a_j \middle\vert \psi\right\rangle}\right\vert^2.\]This immediately implies that the expectation value of $A$ is the same for ${\left\vert \psi \right\rangle}$ and ${\left\vert \psi’ \right\rangle}$ (as can be seen by inserting some wellplaced $I$ using the completeness relation for the ${\left\vert a_j \right\rangle}$ eigenstates).
In particular, if $T$ is an $A$symmetry, then for all ${\left\vert \psi \right\rangle}$ and ${\left\vert \psi’ \right\rangle}=T{\left\vert \psi \right\rangle}$ we have that
\[\begin{aligned} \\psi'\^2=\left\vert{\left\langle \psi' \middle\vert \psi'\right\rangle}\right\vert^2 &= \sum_j\left\vert{\left\langle a_j \middle\vert \psi'\right\rangle}\right\vert^2\\ &= \sum_j\left\vert{\left\langle a_j \middle\vert \psi\right\rangle}\right\vert^2\\ &= \left\vert{\left\langle \psi \middle\vert \psi\right\rangle}\right\vert^2 = \\psi\^2. \end{aligned}\]And thus, $T$ is normpreserving. Unfortunately, this fact alone does not say much about $T$. However, if we also require $T$ to be invertible and linear, then it must be a unitary transformation, that is, $T^\dagger T = T T^\dagger = I$, or equivalently ${\left\langle (T\psi) \middle\vert T\phi\right\rangle}={\left\langle \psi \middle\vert \phi\right\rangle}$. This follows from a straightforward application of the polarization identity
\[{\left\langle u \middle\vert v\right\rangle} = \frac{1}{4}\left(\u+v\^2  \uv\^2 +i\uiv\^2 i \u+iv\^2\right).\]Similarly, if we require $T$ to be invertible and additive but antilinear, $T(au + v)= a^*T(u)+T(v)$, then if it preserves norms it must be antiunitary, i.e. ${\left\langle (T\psi) \middle\vert T\phi\right\rangle}={\left\langle \phi \middle\vert \psi\right\rangle}$.
These results are quite similar to Wigner’s theorem, which states that any invertible mapping $T$ that preserves the norms of inner products and that maps rays into rays (i.e. if $v=\alpha u$ for some scalar $\alpha$ then there exists a scalar $\beta$ such that $T(v)=\beta T(u)$), (and whose inverse does too) must be unitary or antiunitary.
What we have shown is that any linear $A$symmetry must be a unitary transformation. In particular, if $\hat{U}$ is unitary $A$symmetry, then the $\hat{U}$conjugate of $\hat{A}$ satisfies
\[\begin{aligned} {\left\langle \psi \middle\vert \hat{U}^{1}\hat{A}\hat{U}\middle\vert\psi\right\rangle} &= {\left\langle \psi \middle\vert \hat{U}^{\dagger}\hat{A}\hat{U}\middle\vert\psi\right\rangle}\\ &={\left\langle (\hat{U}\psi) \middle\vert \hat{A}\middle\vert(\hat{U}\psi)\right\rangle}\\ &= {\left\langle \psi \middle\vert \hat{A}\middle\vert\psi\right\rangle}. \end{aligned}\]This is because $\hat{U}^{\dagger}=\hat{U}^{1}$, and since $\hat{U}$ is an $A$symmetry, it preserves the expectation value of $A$. This holds for any ${\left\vert \psi \right\rangle}\in \mathcal{H}$, therefore $\hat{U}^{1}\hat{A}\hat{U}=\hat{A}$, which means that $\hat{A}\hat{U}=\hat{U}\hat{A}$, or equivalently, $[\hat{A},\hat{U}]=0$.
There are different levels on which we can describe symmetry. The first one we discussed is determined pointwise, that is, only two states being symmetric with respect to an observable. Then, one step further, we described transformations $T$ that are also $A$symmetries. Under these transformations, every pair ${\left\vert \psi \right\rangle},T{\left\vert \psi \right\rangle}$ are $A$symmetric. If this operator is to be linear, then it must also be unitary. Most often, in the standard QM references, the symmetries that are considered are unitary transformations that are symmetric with respect to the Hamiltonian (energy) observable.
We can go one step further and consider symmetries with respect to all observables, which we call absolute symmetries^{3}. Any two states that are related by these symmetries are absolutely indistinguishable. What this means is that if $T$ is an absolute symmetry and ${\left\vert \psi \right\rangle}=T{\left\vert \phi \right\rangle}$, then ${\left\vert \psi \right\rangle}$ and ${\left\vert \phi \right\rangle}$ have the exact same distributions for all observables. This means that they are indistinguishable!^{4}
Of course, if our system allows for these kinds of absolute symmetries, then our state space is defined with redundancies, since we should define the states and the observables in such a way that two states are equal if and only if they have the same distribution for all observables. In truth, absolute symmetries arise from being sloppy in the definition of states.
We will now attempt to characterize absolute symmetries. So let $T$ be an absolute symmetry. Consider any state ${\left\vert \psi \right\rangle}$, and complete it to an orthonormal basis ${\left\vert \psi \right\rangle}={\left\vert \psi_1 \right\rangle},{\left\vert \psi_2 \right\rangle},{\left\vert \psi_3 \right\rangle},\dots$. Since $T$ is an absolute symmetry, for all Hermitian $\hat{A}$, we have that ${\left\langle (T\psi) \middle\vert \hat{A}\middle\vert(T\psi)\right\rangle}={\left\langle \psi \middle\vert \hat{A}\middle\vert\psi\right\rangle}$. In particular, we consider the projection operators with respect to the elements of the basis,
\[P_{\psi_j}={\left\vert \psi_j \right\rangle}{\left\langle \psi_j \right\vert}.\]Then we have that
\[\begin{aligned} {\left\langle (T\psi_i) \middle\vert P_{\psi_j}\middle\vert(T\psi_i)\right\rangle} &= {\left\langle (T\psi_i) \middle\vert \psi_j\right\rangle}{\left\langle \psi_j \middle\vert T\psi_i\right\rangle}\\ &= \left\vert{\left\langle \psi_j \middle\vert T\psi_i\right\rangle} \right\vert^2. \end{aligned}\]On the other hand,
\[\begin{aligned} {\left\langle (T\psi_i) \middle\vert P_{\psi_j}\middle\vert(T\psi_i)\right\rangle} &= {\left\langle \psi_i \middle\vert P_{\psi_j}\middle\vert\psi_i\right\rangle}\\ &= {\left\langle \psi_i \middle\vert \psi_j\right\rangle}{\left\langle \psi_j \middle\vert \psi_i\right\rangle}\\ &= \left\vert {\left\langle \psi_i \middle\vert \psi_j\right\rangle} \right\vert^2\\ &= \delta_{ij}. \end{aligned}\]This means that
\[\left\vert{\left\langle \psi_j \middle\vert T\psi_i\right\rangle} \right\vert^2 = \delta_{ij},\]so that, say, ${\left\langle \psi_j \middle\vert T\psi_j\right\rangle}=e^{i\theta_j}$. From here we obtain that
\[T{\left\vert \psi_j \right\rangle}= \sum_k {\left\langle \psi_k \middle\vert T\psi_j\right\rangle}{\left\vert \psi_k \right\rangle} = e^{i\theta_j}{\left\vert \psi_j \right\rangle}.\]Now we don’t know that $T$ is linear or antilinear, but it certainly does preserve rays. For any given state ${\left\vert \varphi \right\rangle}$ we repeat the previous process and obtain that there exists a unitary scalar $\alpha_\varphi$ such that $T{\left\vert \varphi \right\rangle}=\alpha_\varphi{\left\vert \varphi \right\rangle}$. Then if ${\left\vert \psi \right\rangle}=\beta{\left\vert \phi \right\rangle}$, we have that
\[\begin{aligned} T{\left\vert \psi \right\rangle} = \alpha_{\psi}{\left\vert \psi \right\rangle} = \alpha_{\psi}\beta{\left\vert \phi \right\rangle} = \alpha_{\psi}\beta\alpha_{\phi}^{1}T{\left\vert \phi \right\rangle}. \end{aligned}\]Then $T$ maps rays into rays, and it preserves the norms of inner products,
\[\vert{\left\langle (T\psi) \middle\vert (T\phi)\right\rangle}\vert = \vert\alpha_{\psi}^*\alpha_{\phi}{\left\langle \psi \middle\vert \phi\right\rangle} \vert = \vert{\left\langle \psi \middle\vert \phi\right\rangle}\vert,\]and thus it satisfies the hypotheses of Wigner’s theorem, so $T$ must be unitary or antiunitary. Suppose that it is unitary, so it preserves inner products. Then for any ${\left\vert \psi \right\rangle},{\left\vert \phi \right\rangle}$, it follows that
\[{\left\langle \psi \middle\vert \phi\right\rangle}={\left\langle (T\psi) \middle\vert T\phi\right\rangle}=\alpha_{\psi}^*\alpha_{\phi}{\left\langle \psi \middle\vert \phi\right\rangle}.\]Then $\alpha_{\psi}^*\alpha_{\phi}=1$. However, since $\vert\alpha_{\psi}\vert=1$, then $\alpha_{\psi}^{1}=\alpha_{\psi}^*$, and this implies that $\alpha_{\psi}^{1}\alpha_{\phi}=1$, or rather, $\alpha_{\psi}=\alpha_{\phi}$. In conclusion, if $T$ is a unitary absolute symmetry, then is is multiplication by a unit scalar:
\[T{\left\vert \psi \right\rangle}=\alpha{\left\vert \psi \right\rangle}\quad \forall {\left\vert \psi \right\rangle}.\]And we know that this is a redundancy in our description of the states! In truth, we should work in the projective Hilbert space $\mathbb{P}\mathcal{H}$, in which we assume that vectors which are related by multiplication by any unitary scalar represent one and the same state.
Describing multiparticle states
Suppose that we have two systems $S_1,S_2$, described by Hilbert spaces $\mathcal{H}_1,\mathcal{H}_2$, respectively. The mathematical setting for describing the joint system is the [completion] $\mathcal{H}$ of the tensor product^{5} $\mathcal{H}_1\otimes \mathcal{H}_2$. The inner product in $\mathcal{H}$ is termwise as
\[{\left\langle u_1\otimes u_2 \middle\vert v_1\otimes v_2\right\rangle}:= {\left\langle u_1 \middle\vert v_1\right\rangle}{\left\langle u_2 \middle\vert v_2\right\rangle}.\]There are a few different notations for product elements of $\mathcal{H}$, but we will mostly stick to $\left\vert u_1 u_2 \right\rangle:=\left\vert u_1 \right\rangle\otimes\left\vert u_2 \right\rangle$. Note that not every state (actually very few states) in $\mathcal{H}$ can be written as a tensor product of two states in $\mathcal{H}_1$ and $\mathcal{H}_2$. A general state is a linear combination of states of the form $\left\vert u_1,u_2 \right\rangle$, with $u_i\in \mathcal{H}_i$, and if $\{\left\vert \varphi_i \right\rangle\}$, $\{\left\vert \psi_j \right\rangle\}$ are bases for $\mathcal{H}_1$ and $\mathcal{H}_2$, then $\{\left\vert \varphi_i \psi_j \right\rangle\}$ is a basis for $\mathcal{H}$.
Any observable $\hat{A}$ over the space $ \mathcal{H}_1 $ can be “extended” to an observable over $\mathcal{H}$, by identifying it with $\hat{A}\otimes \text{id}_{\mathcal{H}_2}$. Similarly for observables over $\mathcal{H}_2$.
I will not go into all the nuances and interesting details of these kind of states (e.g. entangled states and such), since what we are interested in is representing indistinguishability.
Indistinguishability defined
For the timebeing, we will only work with $2$particle spaces, and then we will go nuts and go to general $k$particle spaces. Consider the case where $\mathcal{H}_1=\mathcal{H}_2:=\mathcal{H}$, in which case the total Hilbert space is $\mathcal{H}\otimes\mathcal{H}$. By definition, the states ${\left\vert \psi_1\psi_2 \right\rangle}$ and ${\left\vert \psi_2\psi_1 \right\rangle}$ are considered different. However, we want them to represent the same state. There are two ways to go about this:

Scrap everything and rethink a suitable Hilbert space that clearly represents indistinguishable particles without any redundancy, or

force any two states that represent the same configuration (e.g. as above) to be absolutely symmetric.
If you’re all the way down here, you’ll be glad to read that we will not choose, I repeat, not choose option 1. Even more, working with option 2 will actually suggest which Hilbert space to use that would work for option 1. It’s a winwin.
Then we choose option 2. Define a transformation $T$^{6} that acts on product terms as
\[T{\left\vert \psi_1\psi_2 \right\rangle}={\left\vert \psi_2\psi_1 \right\rangle},\]and extend it by linearity. This is a permutation transformation. Since we require $T$ to be a (unitary) absolute symmetry, then by the result of the previous section $T$ must behave as multiplication by some unit scalar $\alpha$, that is, $T=\alpha I$. However, we also have that permuting twice should do nothing (if you switch $1$ and $2$, then switch them again, you get $1$ and $2$), so $T^2=I$. But we also have that $T^2=\alpha^2I$, so this implies that $\alpha^2=1$. Then $\alpha=\pm 1$, that is, $T=\pm I$.
But wait!
This cannot possibly be true for all elements of $\mathcal{H}\otimes\mathcal{H}$! As a counterexample, consider literally any pair of linearly independent elements $\left\vert \psi_1 \right\rangle,\left\vert \psi_2 \right\rangle\in \mathcal{H}$. By definition, the tensor products $\left\vert \psi_1\psi_2 \right\rangle$ and $\left\vert \psi_2\psi_1 \right\rangle$ are not only different, but also linearly independent. This means that
\[T\left\vert \psi_1\psi_2 \right\rangle = \left\vert \psi_2\psi_1 \right\rangle\overset{!!!}{\neq}\pm \left\vert \psi_1\psi_2 \right\rangle.\]So the relation $T=\pm I$ cannot be true for all vectors! Did we arrive at a contradiction? Is it impossible to represent the state space of two identical particles with permutation symmetry? Well… no. We just showed a counterexample where $T{\left\vert \psi_1\psi_2 \right\rangle}\neq \pm{\left\vert \psi_1\psi_2 \right\rangle}$, but there are some vectors in $\mathcal{H}\otimes\mathcal{H}$ where permutation is a symmetry. For example, consider the vector
\[\left\vert \psi \right\rangle = \frac{1}{\sqrt{2}}\left(\left\vert \psi_1\psi_2 \right\rangle+\left\vert \psi_2\psi_1 \right\rangle\right).\]If we apply $T$, then we get
\[\begin{aligned} T\left\vert \psi \right\rangle &= \frac{1}{\sqrt{2}}\left(T\left\vert \psi_1\psi_2 \right\rangle+T\left\vert \psi_2\psi_1 \right\rangle\right)\\ &= \frac{1}{\sqrt{2}}\left(\left\vert \psi_2\psi_1 \right\rangle+\left\vert \psi_1\psi_2 \right\rangle\right)\\ &= \frac{1}{\sqrt{2}}\left(\left\vert \psi_1\psi_2 \right\rangle+\left\vert \psi_2\psi_1 \right\rangle\right)\\ &= \left\vert \psi \right\rangle. \end{aligned}\]Similarly, consider the vector
\[\left\vert \psi' \right\rangle = \frac{1}{\sqrt{2}}\left(\left\vert \psi_1\psi_2 \right\rangle\left\vert \psi_2\psi_1 \right\rangle\right).\]Upon acting with $T$,
\[\begin{aligned} T\left\vert \psi' \right\rangle &= \frac{1}{\sqrt{2}}\left(T\left\vert \psi_1\psi_2 \right\rangleT\left\vert \psi_2\psi_1 \right\rangle\right)\\ &= \frac{1}{\sqrt{2}}\left(\left\vert \psi_2\psi_1 \right\rangle\left\vert \psi_1\psi_2 \right\rangle\right)\\ &= \frac{1}{\sqrt{2}}\left(\left\vert \psi_1\psi_2 \right\rangle\left\vert \psi_2\psi_1 \right\rangle\right)\\ &= \left\vert \psi' \right\rangle. \end{aligned}\]Then for some states, it is true that $T\left\vert \psi \right\rangle=\pm\left\vert \psi \right\rangle$.
What do we do now? We restrict our state space to one where there is a permutation symmetry, and we simply ignore all the other vectors that do not have this symmetry. So we define the symmetric product of $\mathcal{H}$ as the set of all vectors ${\left\vert \psi \right\rangle}\in\mathcal{H}\otimes\mathcal{H}$ for which $T{\left\vert \psi \right\rangle}={\left\vert \psi \right\rangle}$:
\[\mathcal{S}^2\mathcal{H} := \left\{\left\vert \psi \right\rangle\in \mathcal{H}\otimes\mathcal{H}~:~T\left\vert \psi \right\rangle=\left\vert \psi \right\rangle\right\}.\]We call this symmetric product the boson state space^{7} (of two particles). Similarly, we can also define the antisymmetric product of $\mathcal{H}\otimes\mathcal{H}$ as
\[\Lambda^2\mathcal{H} := \left\{\left\vert \psi \right\rangle\in \mathcal{H}\otimes\mathcal{H}~:~T\left\vert \psi \right\rangle=\left\vert \psi \right\rangle\right\},\]and we call this the fermion state space in two particles. These fermionic states have the particularity that two particles cannot be in the same state! That is, the vectors ${\left\vert \psi\psi \right\rangle}$ (which represent two particles in the same state) are not in the fermion state space. This is in stark contrast with the bosonic case, where ${\left\vert \psi\psi \right\rangle}\in \mathcal{S}^2\mathcal{H}$.
The beauty of this is that we are not violating any of the axioms of quantum mechanics by restricting ourselves to this space. The boson and fermion state spaces are Hilbert spaces (one must simply check that they are closed subspaces of $\mathcal{H}\otimes\mathcal{H}$), and in fact, we can even find a nice, juicy basis for them, given that we already have one for $\mathcal{H}$. That comes next week, pinky promise.
There are (spoiler alert) more than two particles in the universe
This astonishing discovery means that we’re going to need a bigger boat. What if we have $k$ particles? The procedure is almost exactly the same as in the previous section, but we have to be a bit more careful. In this case, the ambient space “with redundancies” will be the $k$th tensor product of $\mathcal{H}$, which we write as $\otimes^k\mathcal{H}$.
But now permutations get a lot more complicated since we can exchange any number of particles. In general a $k$permutation is a rearrangement of the numbers $\left\{1,\dots,k\right\}$, which is to say a bijective^{8} function $\sigma:\left\{1,\dots,k\right\}\to\left\{1,\dots,k\right\}$. The set of all $k$permutations is denoted $\mathfrak{S}_k$.
Our requirement of symmetry under permutations can be rewritten as follows: For any permutation $\sigma\in \mathfrak{S}_k$, define a transformation $T_\sigma$ on product vectors as
\[T_\sigma\left\vert \psi_1\cdots\psi_k \right\rangle = \left\vert \psi_{\sigma(1)}\cdots\psi_{\sigma(k)} \right\rangle,\]and extend it to the entire space $\otimes^k\mathcal{H}$ by linearity. As an example, let $k=5$, and consider the permutation \(\sigma:(1,2,3,4,5)\mapsto (2,4,5,1,3).\) Then the action of $T_\sigma$ is
\[T_{\sigma}\left\vert \psi_1\psi_2\psi_3\psi_4\psi_5 \right\rangle = \left\vert \psi_{\sigma(1)}\psi_{\sigma(2)}\psi_{\sigma(3)}\psi_{\sigma(4)}\psi_{\sigma(5)} \right\rangle = \left\vert \psi_2\psi_4\psi_5\psi_1\psi_3 \right\rangle\](and extended by linearity). Note that $\psi_1,\dots,\psi_k$ do not have to be different a priori for this to make sense.
Also note that $\sigma\mapsto T_\sigma$ is a linear action, or what we call a representation of $\mathfrak{S}_k$ on $\otimes^k\mathcal{H}$. This is because $T_\sigma\circ T_{\sigma’}=T_{(\sigma\circ\sigma’)}$ for any pair of permutations $\sigma,\sigma’\in \mathfrak{S}_k$.
Our requirement of permutation invariance is that $T_{\sigma}$ is an absolute (unitary) symmetry for all permutations $\sigma\in\mathfrak{S}_k$. This means that for each $\sigma\in \mathfrak{S}_k$, there is a unit scalar $\alpha_\sigma$ such that $T_\sigma=\alpha_\sigma I$. In particular, let’s choose a transposition, which simply switches two numbers, $\tau:(1,\dots,i,\dots,j,\dots,k)\mapsto(1,\dots,j,\dots,i,\dots,k)$. We write $(i~j)$ for the transposition that switches $i$ and $j$. Then we’re back in known waters, since transpositions are their own inverses! This means that $T_{(i~j)}^2 = I$, which, again, means that $\alpha_{(i~j)}=\pm 1$, as in the twoparticle case. However, this does not mean that it is $1$ for all transpositions or $1$ for all transpositions; it might happen that $\alpha_{\tau}=1$ for some transpositions but $\alpha_{\tau}=1$ for others. Just because this is getting too long, I will leave as an appendix^{9} the proof that, indeed, it should be the same for all transpositions, i.e. $\alpha_{(i~j)}=1$ for all transpositions or $\alpha_{(i~j)}=1$ for all transpositions.
In order to continue, we now need an important result of group theory, which we will not prove, which states that every permutation can be written as a (nonunique) product of transpositions. Even though this product is not unique, what is unique is the parity of the number of transpositions needed to express a permutation. That is, a permutation that can be written as an even number of transpositions can only be written as an even number of transpositions, and similarly, a permutation that can be written as an odd number of transpositions can only be written as an odd number of transpositions.
The sign of a permutation $\sigma$, denoted $\operatorname{sgn}(\sigma)$, is defined as $1$ if it can be written with an even number of transpositions, and $1$ if it can be written with an odd number of transpositions. With this definition, we are nearly done. Suppose that $\alpha_{(i~j)}=1$ for all transpositions. Then we have, for any permutation $\sigma$, that
\[T_\sigma = T_{\tau_1\cdots\tau_m} = T_{\tau_1}\cdots T_{\tau_m} = 1\cdots 1 I = I,\]where $\tau_1,\dots\tau_m$ are transpositions that compose $\sigma$. This is what we call the trivial representation of $\mathfrak{S}_k$. In this case, we require the canonical action of permutations to be the trivial representation:
\[T_\sigma = I.\]However, we fall into the same trap as in the previous section. By definition, this is simply not true for products of linearly independent vectors, so we must restrict our scope to a subspace of $\otimes^{k}\mathcal{H}$ where there is this kind of permutation invariance. We define the $k$th symmetric product of $\mathcal{H}$ as
\[\mathcal{S}^k\mathcal{H} := \left\{\left\vert \psi \right\rangle\in \otimes^k\mathcal{H}~:~T_\sigma \left\vert \psi \right\rangle = \left\vert \psi \right\rangle \text{ for all }\sigma\in \mathfrak{S}\_k\right\}.\]And we call it the boson state space of $k$ particles. An example of such a state is
\[\left\vert \psi \right\rangle = \frac{1}{\sqrt{3}}\left(\left\vert \psi_1\psi_2\psi_2 \right\rangle + \left\vert \psi_2\psi_1\psi_2 \right\rangle + \left\vert \psi_2\psi_2\psi_1 \right\rangle \right)\in\mathcal{H}\otimes\mathcal{H}\otimes\mathcal{H}.\]But now we have the other choice, which is that $T_{(i~j)}=1$ for all transpositions. In this case, the symmetry is not as simple. If $\sigma$ is a permutation, then we decompose it into transpositions, $\sigma = \tau_1\cdots \tau_m$. The action of $T_\sigma$ is, then
\[T_{\sigma} = T_{\tau_1\cdots \tau_m}=T_{\tau_1}\cdots T_{\tau_m} = (1)^mI = \operatorname{sgn}(\sigma)I.\]This follows since $(1)^m$ is precisely $1$ if $m$ is even and $1$ if $m$ is odd. Then our other choice for permutation symmetry is the requirement that \(T_{\sigma} = \operatorname{sgn}(\sigma)I,\) which we call the alternating representation of $\mathfrak{S}_k$. But again! The same trap as before! This relationship does not hold for many vectors in $\otimes^k\mathcal{H}$, so we restrict to a subspace where this relationship does hold, which is the $k$th alternating (or exterior) product of $\mathcal{H}$:
\[\Lambda^k\mathcal{H} := \left\{\left\vert \psi \right\rangle \in \otimes^k\mathcal{H}~:~T_\sigma \left\vert \psi \right\rangle =\operatorname{sgn}(\sigma) \left\vert \psi \right\rangle \text{ for all }\sigma\in \mathfrak{S}\_k\right\}.\]And this space is called the fermion state space of $k$ particles. Again, it can be seen that there are no vectors in which there are two particles in the same state (for if there were, a permutation of such two particles would need to yield a $1$… which it doesn’t). This means that two fermionic particles cannot be in the same state, something that is called Pauli’s exclusion principle.
These spaces will be the setting for doing quantum mechanics of many particles. They seem pretty similar, but in reality, they are extremely different.
Our objective for next week will be writing down an orthonormal basis for these spaces, given a basis for $\mathcal{H}$. Then we will consider states spaces where there can be arbitrarily many particles, and hopefully find a way into field theory.
The takeaway
The assumption of indistinguishability implies that the Hilbert space that represents multiparticle states (particles of the same kind) must not distinguish between “particle $a$ is in state $x$ and particle $b$ is in state $y$” from “particle $a$ is in state $y$ and particle $b$ is in state $x$”. Coming up with such a Hilbert space from scratch is not trivial, so what we do instead is construct a Hilbert space whose elements represent multiparticle states, but also in which the particles are distinguishable. If one particle is represented by the Hilbert space $\mathcal{H}$, then such a “distinguishable” multiparticle space is $\otimes^k\mathcal{H}$. Then, in order to pass to indistinguishability, we consider subspaces of $\otimes^k\mathcal{H}$ in which permutation is an absolute symmetry. There are only two of those, namely the bosonic space which is the symmetric product $\mathcal{S}^k\mathcal{H}$, and the fermionic space which is the antisymmetric product $\Lambda^k\mathcal{H}$. In particular, the fermionic space does not have any elements in which two particles are in the same state. This is Pauli’s Exclusion Principle.
References

Folland, G. B. (2008). Quantum Field Theory: A Tourist Guide for Mathematicians, Chapters 3 and 4.

Sakurai, J. J. (1995). Modern Quantum Mechanics, Revised Edition, Chapters 1, 4, and 6.

Beck, M. (2012). Quantum Mechanics: Theory and Experiment, Chapter 8.

Weinberg, S. (2002). The Quantum Theory of Fields I: Foundations, Chapter 2, Appendix A.
Appendix: $T_\sigma$ is the same for all $\sigma$
The quick way to do this, which requires a bit of machinery from algebra, is as follows (a more downtoearth proof, I think, will follow soon). Note that the representation $T:\mathfrak{S}_k\to \left\{I,I\right\}\cong \mathbb{Z}_2$ is actually a onedimensional (since it maps to multiples of the identity), and therefore $\operatorname{im}(T)$ is abelian. This means that $T$ must factorize through the abelianization $\mathfrak{S}_k^{ab}=\mathfrak{S}_k/[\mathfrak{S}_k,\mathfrak{S}_k]$, where $[\mathfrak{S}_k,\mathfrak{S}_k]$ is generated by the commutators $ghg^{1}h^{1}$. However, the map $\operatorname{sgn}:\mathfrak{S}_k\to \left\{1,1\right\}\cong \mathbb{Z}_2$ is surjective whose kernel is precisely^{10} $[\mathfrak{S}_k,\mathfrak{S}_k]$, so $\mathfrak{S}_k^{ab}\cong \mathbb{Z}_2$. This means that $T=\tilde{T}\circ \pi$, where $\tilde{T}:\mathbb{Z}_2\to\mathbb{Z}_2$ is a homomorphism and $\pi=\mathrm{sgn}$ is the projection to the abelianization. There are only two possible endomorphisms of $\mathbb{Z}_2$, namely the trivial one and $\operatorname{id}$. So if $\tilde{T}$ is trivial, then $T$ is trivial too. If $\tilde{T}=\operatorname{id}$, then $T=\operatorname{sgn}$.

Maybe the hypothesis is justified in that otherwise the standard definition of entropy would not be extensive (Gibbs’ paradox)… But maybe that is a problem with the definition of entropy, not the distinguishability (oof long word!) of particles. Again, I’m not opening that can of worms. Yet. Stay tuned! ↩

Of course, there is the subtlety of the case where $\hat{A}$ has a continuous spectrum without eigenstates, but this can be formally extended as probability densities. I think. Don’t think about it too much. ↩

Don’t quote me on this, I just made this up. ↩

If this is not convincing, then ask yourself “how you I tell two such states apart?” There is no observable that can distinguish between them. ↩

The pragmatic justification for this is that this is the simplest nontrivial way to make a Hilbert space out of $\mathcal{H}_1$ and $\mathcal{H}_2$, but I am sure that this can be justified using the spectra of observables of the joint system and a little bit of probability theory. Stay tuned! ↩

This notation is not standard. It is usually called $p$ or $P$ or something like that or maybe something entirely different because nobody wants confusion with the momentum operator. ↩

“Why “boson”? What does this have to do with spin? Aaaah!” Well, again. A good question… for another time. ↩

Recall that this means that for each $i$ there is one and only one number $j$ such that $\sigma(j)=i$. This implies that when you write down $\sigma(1),\dots,\sigma(k)$, you obtain exactly $1,\dots,k$ but in disorder. ↩

that is, a post in the future ↩

One of the inclusions is trivial to prove, the other requires noting that the product of two transpositions is a product of $3$cycles, and that $3$cycles are always commutators. This is not hard, but not obvious! ↩