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 $\psi_{\omega_1\omega_2}(\bar x_1,x_2)=\psi_{\omega_1}(\bar x_1)\psi_{\omega_2}(x_2)$?


1 Answer 1


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).

  • $\begingroup$ Why are the coefficients equal to the sum in Equation (2)? $\endgroup$
    – Qwertyuiop
    Commented Feb 25 at 13:18
  • $\begingroup$ 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 $\psi_{\omega_1}(x_1)$. $\endgroup$ Commented Feb 25 at 13:28
  • $\begingroup$ 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? $\endgroup$
    – Qwertyuiop
    Commented Feb 25 at 13:46
  • $\begingroup$ $\psi_{\omega_1}$ is a basis, we can decompose any function with respect to it. $\endgroup$ Commented Feb 25 at 14:13
  • $\begingroup$ 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 $\mathbb{V}_1$ $\endgroup$ Commented Feb 25 at 14:16

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