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.
W3Counter Web Stats

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 β tools

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

Attributions

(propose a free ad)

Site Statistics

206 submissions , 164 unreviewed
5,106 questions , 2,251 unanswered
5,412 answers , 23,072 comments
1,470 users with positive rep
822 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
650 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 Ω(1)1 be an operator on the state space V1 of a particle in one dimension, whose nondegenerate eigenfunctions ψω1(x1) form a complete basis, and similarly, letting ψω2(x2) form a basis for the state space of a second particle. He then states that if a function ψ(x1,x2) that represents an abstract vector |ψ from the state space V12 of a system consisting of both particles has x1 fixed at some value ˉx1, then ψ becomes a function of x2 alone and may be expanded as ψ(ˉx1,x2)=ω2Cω2(ˉx1)ψω2(x2)

where Cω2(ˉx1)=ω1Cω1ω2ψω1(ˉx1)
Substituting Equation (2) into Equation (1) and dropping the bar on ˉx1, he states that the resulting expansion ψ(x1,x2)=ω1ω2Cω1ω2ψω1(x1)ψω2(x2)
imply that V12=V1V2 for ψω1(x1)×ψω2(x2) is the same as the inner product between |x1|x2 (|x1 and |x2 are the position basis vectors of V1 and V2 respectively) and |ω1|ω2 (|ω1 and |ω2 are the basis eigenvectors of Ω1 on V1 and Ω2 on V2 respectively).

Question: How does Equation (1) follow from fixing x1 in ψ(x1,x2) as ˉx1? Using the simultaneous eigenbasis |ω1ω2 of the operators Ω1 and Ω2 on V12, ψ(ˉx1,x2)=ˉx1x2|ψ=ω1ω2ω1ω2|ψˉx1x2|ω1ω2=ω1ω2ω1ω2|ψψω1ω2(ˉx1,x2).

If the author intends Cω1ω2 to mean ω1ω2|ψ, what is the reason (besides the tensor product since he is trying to show that V12=V1V2) for ψω1ω2(ˉx1,x2)=ψω1(ˉx1)ψω2(x2)?

asked Feb 25, 2024 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):
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol in the following word:
pyscsOverflow
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
...