Re: quantum amplitude
- From: principalbundles@xxxxxxxx
- Date: Wed, 16 Aug 2006 21:35:16 +0000 (UTC)
Thank you for your kind answer. Some comments follow:
Oh No wrote:
Better to translate <q',t'| as the statement ?a particle has position
|q'> at time t'? and to translate <q',t'|q,t> as the probability
amplitude for the statement ?a particle with position |q> at time t
would have position |q'> if we were to do a measurement at time t'?
I suppose you meant the ket |q',t'>, not the bra. Anyway, it seems to
me that your rephrasing adds nothing to mine, since «a particle with
position |q> at time t» means nothing but «some device measures the
particle having position |q> at time t». Doesn't it?
This isn't quite right. If |q',t'> is a position eigenstate at time t'
then |q',T+a> isn't a position eigenstate at time T+a
Of course. But I never wrote that! I simply take |q',T+a> to be an
eigenstate. Then obviously |q',t'> is not an eigenstate, although is a
time-evoluted state in my picture.
Not sure why you want to do this. Generally we know the state at t,
(which is to say we know the measurement result at t). There isn't a
measurement at time T=(t+t')/2, so we don't know the state at that time.
I was just driven by the formal expression <q',t'|q,t> to attach to it
a meaning in terms of time-evoluted bras and kets. Surely <q',t'|q,t>
can be obtained, as you rightly say, letting evolve eigenstate |q> from
time t to time t', then sandwiching the resulted state with eigenstate
|q'>. And this has a sound physical meaning. But the formal expression
<q',t'|q,t> suggests a different meaning to me, precisely that of two
time-evoluted states sandwiched together. Am I the only one?
.
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