Tensor product in quantum mechanics

Question

In Cohen-Tannoudji's Quantum Mechanics book the tensor product of two two Hilbert spaces $(\mathcal H = \mathcal H_1 \otimes \mathcal H_2)$ was introduced in (2.312) by saying that to every pair of vectors $$|\phi(1)\rangle \in \mathcal H_1, |\chi(2)\rangle \in \mathcal H_2$$ there belongs a vector $$|\phi(1)\rangle \otimes |\chi(2)\rangle \in \mathcal H$$ In a footnote it stated that the order doesn't matter and that we could also call it $$|\chi(2)\rangle \otimes |\phi(1)\rangle$$ I'm a bit confused, since I though that the order of the tensor product generally matters. What would that expression look like if we picked a basis, say: $$|\phi(1)\rangle = a_1|u_1\rangle + a_2|u_2\rangle + \dotsc$$ $$|\chi(2)\rangle = b_1|v_1\rangle + b_2|v_2\rangle + \dotsc$$

Any help will be appreciated!

Answer 1

$|\phi(1)\rangle \otimes |\chi(2)\rangle $ is a cumbersome notation to write ket corresponding to $\psi$ function $\phi(\mathbf r_1)\chi(\mathbf r_2)$, where $\mathbf r_i$ refers to coordinates of the $i$-th subsystem. That's why the order of factors in $\otimes$ product does not matter; the resulting ket corresponds to the same $\psi$ function and is thus the same ket.

On the other hand, $|\phi\rangle \otimes |\chi\rangle $ (without labels) is meant to be read according to different convention; here it is commonly understood that the order of factor signifies the sub-system it refers to. So

$|\phi\rangle \otimes |\chi\rangle $ denotes ket corresponding to $\phi(\mathbf r_1)\chi(\mathbf r_2)$ just as $|\phi(1)\rangle \otimes |\chi(2)\rangle $ does, but:

$|\chi\rangle \otimes |\phi\rangle $ denotes ket corresponding to $\chi(\mathbf r_1)\phi(\mathbf r_2)$ which is not the same. This is because different meaning of the $\otimes$ notation is used.

Answer 2

You are correct in that the tensor product does not commute in general. The ordering of vectors in some tensor product, say \begin{equation*} |\psi\rangle\otimes|\phi\rangle\otimes|\xi\rangle \equiv |u\rangle \end{equation*} implicitly refers to how the resulting Hilbert space is defined via the tensor product. To the above vectors, let us associate $|\psi\rangle\in H_{\psi}$, $|\phi\rangle\in H_{\phi}$ and $|\xi\rangle\in H_{\xi}$. In this case, we have implicity said that the Hilbert space for our vector $|u\rangle$ is \begin{equation*} |u\rangle\in U\equiv H_{\psi}\otimes H_{\phi}\otimes H_{\xi} \end{equation*} One just need to keep in mind the order of how the underlying Hilbert space is formed. As for a physical description in quantum theory the order is of no importance.

Specifically, in your case when you say \begin{equation*} |\chi(2)\rangle\otimes |\phi(1)\rangle \end{equation*} it just means that the vector belongs to $H_{2}\otimes H_{1}$, rather than $H_{1}\otimes H_{2}$.

Answer 3

Imagine two different particles, e.g. a proton and an electron, the former described by functions in the Hilbert space $\mathcal H_p$, and the latter by functions in the Hilbert space $\mathcal H_e$. Now assume a base states $\phi_1, \phi_2,...$ in $\mathcal H_p$ and a base $\chi_1, \chi_2,...$ in $\mathcal H_e$. Your pair of particles is (proton, electron), or is (electron, proton)? It doesn't matter, does it? The same with the description of the state of the pir,

$$\Psi = \sum_{i,j} C_{i,j} \ \phi_i \otimes \chi_j = \sum_{i,j} C_{i,j} \ \chi_i \otimes \phi_j . \tag {i}$$

Now, there exists a situation when the order is important: when the functions $\chi$ and $\phi$ look the same, e.g. the eigenfunctions of the spin-$z$ projection operator, $|\uparrow\rangle, |\downarrow\rangle$. In this case instead of mentioning by subscripts to which particle we refer, we assume an order, e.g. in each product of states we write first the state of the electron, and second of the proton: $|\uparrow\rangle \otimes |\downarrow\rangle$ means that the electron has spin-up and the proton spin-down.