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- Newsgroups: sci.physics
- Path: sparky!uunet!cs.utexas.edu!convex!news.utdallas.edu!nariani
- From: nariani@utdallas.edu (Sushil Nariani)
- Subject: QM question
- Message-ID: <Bzs9F7.Ion@utdallas.edu>
- Sender: usenet@utdallas.edu
- Nntp-Posting-Host: csclass.utdallas.edu
- Organization: Univ. of Texas at Dallas
- Date: Thu, 24 Dec 1992 21:37:54 GMT
- Lines: 20
-
-
- There's this thing about QM which Liboff writes in his book
- "Introductory QM" :
- "The measurement of an observable of a particle leaves it
- in an eigenstate of the corresponding operator till another
- measurement is made"
- Also, delta(x) is the eigenfunction of the position operator x.
- Now what bothers me is this: Suppose I make an infinite precision
- measurement on the particles position. That would leave it in the
- state delta(x-x') where x' would be the measured value for position.
- Now if i do not make any other measurement, the particle should
- remain in this state. Does'nt this indicate that the particle should
- remain at the same position later on? In which case the momentum
- uncertainty is zero. Now I think I've got something wrong here
- but can't figure it out. The author does'nt help much. Pliss to
- illuminate the ignoranti.
-
- Sushil
-
- ps : needless to say, I'm not a Physics student :)
-