| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
21 Jul 2005 01:45:58 PM |
| Object: |
Feynman Vol. 3 Chapter 5 Question |
I have a question for anyone who has read Feynman Vol. 3 Chapter 5.
He discusses the base states (+S,0s,-S,+T,0T,-T etc.) He then talked
about
the probability amplitude of a particle going from going from one state
to another in the form of <final state|initial state>. So, he gives the
ampitude of going from say +S to -S as <-S|+S>. He then uses the short
hand of i,k representing the state of the filter refered to such as
1~+S ect. He makes the statement that <j|i>=0. Looking at, for
example, equation 5.15 this appears not to be true. Here the particle
starts in the +S state and goes to the -S state. This could be written
as <-S|+S>. Couldn't this be written as <j|i>=not zero? Does the
notation <j|i> mean going from i to j state without passing through
another filter? In 5.30 the particle goes through appartus A and it is
written <j|A|i>.
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| User: "Jim Black" |
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| Title: Re: Feynman Vol. 3 Chapter 5 Question |
21 Jul 2005 10:56:35 PM |
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wrote:
I have a question for anyone who has read Feynman Vol. 3 Chapter 5.
He discusses the base states (+S,0s,-S,+T,0T,-T etc.) He then talked
about
the probability amplitude of a particle going from going from one state
to another in the form of <final state|initial state>. So, he gives the
ampitude of going from say +S to -S as <-S|+S>. He then uses the short
hand of i,k representing the state of the filter refered to such as
1~+S ect. He makes the statement that <j|i>=0. Looking at, for
example, equation 5.15 this appears not to be true. Here the particle
starts in the +S state and goes to the -S state.
This does not affect the answer to the question, but in my text, the
particle ends up in the 0S state.
This could be written
as <-S|+S>. Couldn't this be written as <j|i>=not zero? Does the
notation <j|i> mean going from i to j state without passing through
another filter? In 5.30 the particle goes through appartus A and it is
written <j|A|i>.
You've pretty much got it. In standard Dirac notation, just writing
<0S|+S> would represent the probability amplitude for the particle to
go from the +S state to the 0S state with nothing significant happening
to it in between, not even the passage of time if it would change the
state of the particle.
In the case of (5.15), the probability we'd want to compute would be:
|<0S|0T><0T|+S>|^2
This probability amplitude can be by from the usual rules; we multiply
the probability amplitude for the first step (going from +S to 0T) with
the probability amplitude for the second step (going from 0T to 0S).
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| User: "Puppet_Sock" |
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| Title: Re: Feynman Vol. 3 Chapter 5 Question |
21 Jul 2005 03:04:45 PM |
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wrote:
I have a question for anyone who has read Feynman Vol. 3 Chapter 5.
He discusses the base states (+S,0s,-S,+T,0T,-T etc.) He then talked
about
the probability amplitude of a particle going from going from one state
to another in the form of <final state|initial state>. So, he gives the
ampitude of going from say +S to -S as <-S|+S>. He then uses the short
hand of i,k representing the state of the filter refered to such as
1~+S ect. He makes the statement that <j|i>=0. Looking at, for
example, equation 5.15 this appears not to be true. Here the particle
starts in the +S state and goes to the -S state. This could be written
as <-S|+S>. Couldn't this be written as <j|i>=not zero? Does the
notation <j|i> mean going from i to j state without passing through
another filter? In 5.30 the particle goes through appartus A and it is
written <j|A|i>.
I don't have the text in front of me. But I suspect what is
going on in the case that -S goes to +S is *not* that <-S|+S>
is non-zero. Rather, it's that <-S|something|+S> is non-zero.
The something is the interaction that allows the change. It
could be the field of some equipment, or a passing particle,
or the nucleus of an atom, etc.
Yes, <j|i> means the overlap of two states without anything
between. The usual case for Eigenvectors is they are orthogonal,
that is, for j not = i, <j|i> = 0. So, in the normal course of
things, a particle can't just spontaneously flip from one state
to another with nothing else happening. That's where the idea
of conserved quantum numbers comes in.
In the case of spin, for example, the something else is that the
particle interacts with the electromagnetic field. An electron,
for example, can emit or absorb a photon to flip spin. And that
is very suggestive of putting <j|A|i> because the EM field is
often written as A. <j|A|i> is the amplitude that interaction A
takes a particle in state i and puts it in state j.
Socks
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