I'm studying identical particles in Quantum Mechanics and I'm having a hard time to understand the idea of permutations of particles from a mathematical standpoint.

From one intuitive point of view it is quite simples: we have two identical particles and we label them arbitrarily as $1,2$. Permuting the particles them means permuting the labels, so that the particle once labeled $1$ is now $2$ and the particle once labeled $2$ is now $1$.

Now, mathematically things are more complicated. If the description of each particle alone is given by the state space $\mathcal{E}$, it seems at first, that for the two particle system the state space should be $\mathcal{E}\otimes \mathcal{E}$.

I know that later on we see that it is a subspace of that, but just to make my point clear, what is important is, it seems that the first $\mathcal{E}$ that appears is for particle $1$ and the second $\mathcal{E}$ that appears is for particle $2$.

Now, I'm reading Cohen's book and regarding that the author states the following:

Consider a system composed of two particles with the same spin $s$. Here it is not necessary for these two particles to be identical: it is sufficient that their individual state spaces be isomorphic. Therefore, to avoid problems which arise when the two particles are identical, we shall assume that they are not: the numbers (1) and (2) with which they are labeled indicate their natures. For example, (1) will denote a proton and (2), an electron.

We choose a basis, $\{|u_i\rangle\}$, in the state space $\mathcal{E}(1)$ of particle (1). Since the two particles have the same spin, $\mathcal{E}(2)$ is isomorphic to $\mathcal{E}(1)$, and it can be spanned by the same basis. By taking the tensor product, we construct, in the state space $\mathcal{E}$ of the system, the basis:

$$\{|1: u_i; 2: u_j\}$$

Since the order of the vectors is of no importance in a tensor product, we have

$$|2:u_j;1:u_i\rangle = |1:u_i;2:u_j\rangle.$$

However note that:

$$|1:u_j; 2:u_i\rangle \neq |1:u_i; 2:u_j\rangle, \quad \text{if} \ i\neq j.$$

The permutation operator $P_{21}$ is then defined as the linear operator whose action on the basis vectors is given by:

$$P_{21}|1:u_i;2:u_j\rangle = |2:u_i;1:u_j\rangle = |1:u_j;2:u_i\rangle.$$

Now I must confess this doesn't make any sense to me. What this notation $|1:u_i;2:u_j\rangle$ means? To say particle $1$ is at $|u_i\rangle$ and particle $2$ is at $|u_j\rangle$ is the same as saying that the system is at the state $|u_i\rangle\otimes |u_j\rangle$. But I can't have any idea on what this notation he uses means.

So, how to understand this piece of text the author says? How to understand his notation, and specially, how $P_{21}$ is rigorously defined. I really can't understand how:

$$|1:u_i;2:u_j\rangle \to |1:u_j;2:u_i\rangle$$

is any different than

$$|u_i\rangle \otimes |u_j\rangle \to |u_j\rangle \otimes |u_i\rangle.$$

So how do we understand this notation and the action of this operator from a mathematical standpoint?


Let us label the state spaces clearly as $\mathcal{H}_1$ and $\mathcal{H}_2$ for the first and second particle respectively and denote the canonical isomorphism sending a state in $\mathcal{H}_1$ of the first particle to the exact same state of the second particle by $\phi : \mathcal{H}_1\to\mathcal{H}_2$. Let us further denote the canonical "flip isomorphism" of the tensor product as $\mathrm{flip}: \mathcal{H}_1\otimes\mathcal{H}_2\to\mathcal{H}_2\otimes\mathcal{H}_1, v\otimes w\mapsto w\otimes v$.

Then $\lvert 1:u_i;2:u_j\rangle$ is the element $u_i\otimes u_j\in\mathcal{H}_1\otimes\mathcal{H}_2$ and $\lvert 2:u_j;1:u_i\rangle$ is its image $u_j\otimes u_i$ under $\mathrm{flip}$ in $\mathcal{H}_2\otimes \mathcal{H}_1$.

In contrast, $\lvert 1:u_j;2:u_i\rangle$ is the image under the exchange map $$P : \mathcal{H}_1\otimes \mathcal{H}_2\to\mathcal{H}_1\otimes\mathcal{H}_2, v\otimes w\mapsto \phi^{-1}(w)\otimes \phi(v).$$ Unlike $\mathrm{flip}$, which is a map between two different (but isomorphic) spaces, $P$ is an automorphism of $\mathcal{H}_1\otimes\mathcal{H}_2$. It has eigenvalues 1 and -1, whose eigenspaces are the spaces of symmetric and antisymmetric tensors, respectively.


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