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?


2 Answers 2


You might as well ask why classically the phase space of some system is not always the full product of the free phase spaces of its constituents. The answer is: Because physically, constraints can arise. There is nothing peculiar to quantum theory here, and nothing mathematical going on.

Consider two particles, connected by a rod. We often model the rod as massless and basically non-existent, and the only thing it changes in the physical model is that it constrains the difference of the two particles to be constant. This leads to the actual phase space accessible to this two-particle system being much smaller than the full product of the two phase spaces of the individual particles.

Mathematically, you cannot detect those constraints on the level of the state spaces. Someone needs to hand you the constraints, in one form or another. In many cases, it is just so obvious what the actual state space is that we skip the constraint procedure and just directly write down the theory on the proper subspace of the full product space.

  • $\begingroup$ I understand your point about the constraints. But IMO in classical mechanics the constraints are usually easier to see. The rod example is quite easy to see what's happening. But what about that thing of identical particles and bosons and fermions? It doesn't seem at first to be a question of constraints. Is there some "hidden" constraint in this case? $\endgroup$
    – Gold
    Commented Feb 26, 2016 at 1:18
  • 1
    $\begingroup$ @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. $\endgroup$
    – ACuriousMind
    Commented Feb 26, 2016 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.

  • $\begingroup$ Thanks for the answer. Indeed I've once heard about this in a Statistical Mechanics course I took. I've already seem cases however where we don't have information about this. I've added an edit to explain my point, could you take a look? Thanks again. $\endgroup$
    – Gold
    Commented Feb 22, 2016 at 3:03

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