# Expansion of a wavefunction of a two-particle system in one dimension in an arbitrary basis without operations associated with the tensor product

In Shankar's Principle of Quantum Mechanics, Section 10.1, part The Direct Product Revisited (he calls tensor products direct products), he attempts to show that a two-particle state space is the tensor product of two one-particle state space. He begins by letting $$\Omega^{(1)}_1$$ be an operator on the state space $$\Bbb V_1$$ of a particle in one dimension, whose nondegenerate eigenfunctions $$\psi_{\omega_1}(x_1)$$ form a complete basis, and similarly, letting $$\psi_{\omega_2}(x_2)$$ form a basis for the state space $$\Bbb V_2$$ of a second particle. He then states that if a function $$\psi(x_1,x_2)$$ that represents the abstract vector $$|\psi\rangle$$ from the state space $$\Bbb V_{1\otimes2}$$ of a system consisting of both particles has $$x_1$$ fixed at some value $$\bar x_1$$, then $$\psi$$ becomes a function of $$x_2$$ alone and may be expanded as $$\psi(\bar x_1,x_2)=\sum_{\omega_2}C_{\omega_2}(\bar x_1)\psi_{\omega_2}(x_2)\tag{1}\label{e1}$$ where $$C_{\omega_2}(\bar x_1)=\sum_{\omega_1}C_{\omega_1\omega_2}\psi_{\omega_1}(\bar x_1)\tag{2}\label{e2}$$ Substituting \eqref{e2} into \eqref{e1} and dropping the bar on $$\bar x_1$$, the author states that the resulting expansion $$\psi(x_1,x_2)=\sum_{\omega_1}\sum_{\omega_2}C_{\omega_1\omega_2}\psi_{\omega_1}(x_1)\psi_{\omega_2}(x_2)\tag{3}\label{e3}$$ imply that $$\Bbb V_{1\otimes2}=\Bbb V_1\otimes\Bbb V_2$$ for $$\psi_{\omega_1}(x_1)\times\psi_{\omega_2}(x_2)$$ is the same as the inner product between $$|x_1\rangle\otimes|x_2\rangle$$ ($$|x_1\rangle$$ and $$|x_2\rangle$$ are the position basis vectors of $$\Bbb V_1$$ and $$\Bbb V_2$$ respectively) and $$|\omega_1\rangle\otimes|\omega_2\rangle$$ ($$|\omega_1\rangle$$ and $$|\omega_2\rangle$$ are the basis eigenvectors of $$\Omega_1$$ on $$\Bbb V_1$$ and $$\Omega_2$$ on $$\Bbb V_2$$ respectively).

Question: How does \eqref{e1} follow from fixing $$x_1$$ in $$\psi(x_1,x_2)$$ as $$\bar x_1$$? Using the simultaneous eigenbasis $$|\omega_1\omega_2\rangle$$ of the operators $$\Omega_1$$ and $$\Omega_2$$ on $$\Bbb V_{1\otimes2}$$, $$\psi(\bar x_1,x_2)=\langle\bar x_1x_2|\psi\rangle=\sum_{\omega_1}\sum_{\omega_2}\langle\omega_1\omega_2|\psi\rangle\langle\bar x_1x_2|\omega_1\omega_2\rangle=\sum_{\omega_1}\sum_{\omega_2}\langle\omega_1\omega_2|\psi\rangle\psi_{\omega_1\omega_2}(\bar x_1,x_2).\tag{4}\label{e4}$$If the author intends $$C_{\omega_1\omega_2}$$ to mean $$\langle\omega_1\omega_2|\psi\rangle$$, what is the reason (besides the tensor product since he is trying to show that $$\Bbb V_{1\otimes2}=\Bbb V_1\otimes\Bbb V_2$$) for $$\psi_{\omega_1\omega_2}(\bar x_1,x_2)=\psi_{\omega_1}(\bar x_1)\psi_{\omega_2}(x_2)$$?

If we fix $$x_1 = \bar{x}_1$$, the function $$\psi(\bar{x}_1, x_2)$$ describes a valid state of the second particle, so can be decomposed with respect to the eigenbasis $$\psi_{\omega_2}(x_2)$$. The coefficients in this decomposition depend on the value $$\bar{x}_1$$, and denoting these coefficients by $$C_{\omega_2}(\bar{x}_1)$$ we get (1).
• Because $C_{\omega_2}(x_1)$ are now functions in the state space of the first variable and can be decomposed with respect to the eigenbasis $\psi_{\omega_1}(x_1)$. Commented Feb 25 at 13:28
• Why should $C_{\omega_2}(x_1)$ be in the state space of the first variable? Aren't the coefficients, when absolute squared, just the probabilities of a measurement of the observable corresponding to $\Omega_1^{(1)}$ yielding each eigenvalue? Commented Feb 25 at 13:46
• $\psi_{\omega_1}$ is a basis, we can decompose any function with respect to it. Commented Feb 25 at 14:13
• You can also think like this: for every fixed $\omega_2$ we have $\int \psi_{\omega_2}^*(x_2)\psi(x_1, \bar{x}_2) d\bar{x}_2 = \sum_{\omega'_2} C_{\omega'_2}(x_1) \int \psi_{\omega_2}^*(\bar{x}_2)\psi_{\omega'_2}(\bar{x}_2) d\bar{x}_2 = C_{\omega_2}(x_1)$, so $C_{\omega_2}(x_1)$ is a combination of all $\psi(x_1, \bar{x}_2)$, which are states of the first particle, so lies in the state space $\mathbb{V}_1$ Commented Feb 25 at 14:16