Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,054 questions , 2,207 unanswered
5,347 answers , 22,728 comments
1,470 users with positive rep
818 active unimported users
More ...

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

+ 2 like - 0 dislike
493 views

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 ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOv$\varnothing$rflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...