# Two particle system state space

I'm trying to understand how the state space of a bigger system composed of smaller subsystems relates to the state spaces of the individual subsystems.

To get started I'm currently trying to understand how the state space of a two-particle system relates to the state space of a single particle. In the book I'm reading (Cohen-Tannoudji's Quantum Mechanics) the author directly talks about the $|\mathbf{r}_1, \mathbf{r}_2\rangle$ representation.

After thinking a little it makes sense that $|\mathbf{r}_1,\mathbf{r}_2\rangle = |\mathbf{r}_1\rangle \otimes |\mathbf{r}_2\rangle$. Indeed if we consider $\mathcal{E}_i$ the state space of particle $i$ alone, and consider $X^i_j$ the $i$-th coordinate observable of the $j$-th particle, that is $X^i_j \in \mathcal{L}(\mathcal{E}_j)$, these operators could be extended to the tensor product $\mathcal{E}_1\otimes \mathcal{E}_2$ as $X^i_1\otimes \mathbf{1}$ and $\mathbf{1}\otimes X^i_2$.

With this we see, for example, that

$$X^i_1 |\mathbf{r}_1,\mathbf{r}_2\rangle = (X^i_1\otimes \mathbf{1})|\mathbf{r_1}\rangle\otimes |\mathbf{r}_2\rangle = x^i_1|\mathbf{r}_1\rangle\otimes |\mathbf{r}_2\rangle = x^i_1|\mathbf{r}_1,\mathbf{r}_2\rangle,$$

and everything works fine: $|\mathbf{r}_1,\mathbf{r}_2\rangle$ is eigenstate of $X^i_j$ with eigenvalue $x^i_j$ as expected. All of this, together with the fact that being $|\mathbf{r}_i\rangle$ basis of $\mathcal{E}_i$ we have that $|\mathbf{r}_1\rangle\otimes |\mathbf{r}_2\rangle$ is basis of $\mathcal{E}_1\otimes \mathcal{E}_2$ seems to imply that the state space of the combined system is $\mathcal{E}_1\otimes \mathcal{E}_2$.

On the other hand, I've read sometimes (though I don't remember where), that the state space of a combined system is not the tensor product of the subsystems. It's just a subspace of the tensor product. That is, instead of having $\mathcal{E}=\mathcal{E}_1\otimes \mathcal{E}_2$ we have $\mathcal{E}\subset \mathcal{E}_1\otimes \mathcal{E}_2$.

Why is that? Why the combined state space is just a subset and not the whole tensor product? If it's really just a subset, how do we determine which subset it is?

EDIT: I've seem that idea of bosons and fermions before, but what about cases where we are not considering this? For example, when dealing with two particle systems in his book, Cohen changes from the $|\mathbf{r}_1,\mathbf{r}_2\rangle$ representation to the $|\mathbf{r}_g, \mathbf{r}\rangle$ representation where $\mathbf{r}_g$ is the center of mass position and $\mathbf{r} = \mathbf{r}_1-\mathbf{r}_2$ is the separation.

With this we have two "ficticious particles". Quoting the book:

The state space $\mathcal{E}$ of the system can then be considered to be the tensor product $\mathcal{E}_{\mathbf{r}_G}\otimes \mathcal{E}_\mathbf{r}$ of the state space $\mathcal{E}_{\mathbf{r}_G}$ associated with the observable $\mathbf{R}_G$ and the space $\mathcal{E}_{\mathbf{r}}$ associated with $\mathbf{R}$.

In this case we can't classify them as bosons nor fermions, after all they are not real particles. In this case it seems that we use the whole tensor product. Now, why for this case we use the whole tensor product and why in other cases we don't?

• @user1620696: The "hidden" constraint is that, if the particles are identical, you cannot actually distinguish particle "1" being at A and particle "2" being at B from particle "2" being at A and particle "1" being at B. The two states are physically equivalent for all purposes (it's a discrete $\mathbb{Z}/2\mathbb{Z}$ gauge symmetry, if you will), so you have to quotient this interchange symmetry out of your phase space, just like one quotients out the gauge orbits constraints generate. – ACuriousMind Feb 26 '16 at 22:12
If you have identical particles, the system state space is given by taking an appropriately symmetrized tensor product: alternating for identical fermions, symmetric for identical bosons. In the simplest case, with two identical particles $$|r_1,r_2\rangle = 2^{-1/2} ( |r_1\rangle\otimes |r_2\rangle \pm |r_2\rangle\otimes |r_1\rangle )$$ with the plus sign for bosons and the minus sign for fermions.