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  How to expand a wavefunction of a two-particle system in one dimension in some basis without operations associated with the tensor product?

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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 spaces. 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 of a second particle. He then states that if a function $\psi(x_1,x_2)$ that represents an 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 $$\psi(\bar x_1,x_2)=\sum_{\omega_2}C_{\omega_2}(\bar x_1)\psi_{\omega_2}(x_2)\tag{1}$$ where $$C_{\omega_2}(\bar x_1)=\sum_{\omega_1}C_{\omega_1\omega_2}\psi_{\omega_1}(\bar x_1)\tag{2}$$ Substituting Equation $(2)$ into Equation $(1)$ and dropping the bar on $\bar x_1$, he states that the resulting expansion $$\psi(x_1,x_2)=\sum_{\omega_1}\sum_{\omega_2}C_{\omega_1\omega_2}\psi_{\omega_1}(x_1)\psi_{\omega_2}(x_2)\tag{3}$$ 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 Equation $(1)$ 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 $\Bbb V_{1\otimes2}$, $$\psi(\bar x_1,x_2)=\langle\bar x_1x_2|\psi\rangle=\sum_{\omega_1}\sum_{\omega_2}\langle\omega_1\omega_2|\psi\rangle\langle\bar x_1x_2|\omega_1\omega_2\rangle=\sum_{\omega_1}\sum_{\omega_2}\langle\omega_1\omega_2|\psi\rangle\psi_{\omega_1\omega_2}(\bar x_1,x_2).\tag{4}$$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 $\psi_{\omega_1\omega_2}(\bar x_1,x_2)=\psi_{\omega_1}(\bar x_1)\psi_{\omega_2}(x_2)$?

asked Feb 25 in Theoretical Physics by anonymous [ no revision ]

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