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| student... |
 Posted: Fri Jul 17, 2009 7:08 pm |
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Guest
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On Jul 18, 3:51 am, Lou Pagnucco <pagnu... at (no spam) htdconnect.com> wrote:
[quote:919050c6b6]Interesting. You may be correct, but I am still uncertain.
After all, we could simply add a constant multiple of the
identity matrix to H(t) to offset all of the eigenvalues.
Then, it seems, we could arbitrarily change the accrued
phase.- Hide quoted text -
[/quote:919050c6b6]
Good point. I guess (but would have to check) that the
Berry phase is only of interest for when it can be
compared for two states, as then the relative Berry
phase can be observed via a superposition of the states.
This would mean that the Berry phase is simply not
observable for an energy eigenstate (it is just an
overall phase change as above, and, as you point out,
can be varied simply by varying the choice of
groundstate energy.
However, if you have two energy eigenstates, then
the relative phase from transporting the system
does become observable. Eg, for eigenvalues
E and E', and eigenvectors v and v', one has
v(1) = exp[-iE / hbar] v(0) ,
v'(1) = exp[-iE' / hbar] v'(0) ,
and hence any superposition of these states
will pick up a relative phase
phi = (E-E') /hbar
around the path you defined. This relative
phase is independent of energy shifts.
Not sure if this answers your original
question though? |
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| Lou Pagnucco... |
| Posted: Sun Jul 19, 2009 9:57 am |
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Guest
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On Jul 18, 1:08 am, student <of_1001_nig... at (no spam) hotmail.com> wrote:
[quote:33a0db24a2]On Jul 18, 3:51 am, Lou Pagnucco <pagnu... at (no spam) htdconnect.com> wrote:
Interesting. You may be correct, but I am still uncertain.
After all, we could simply add a constant multiple of the
identity matrix to H(t) to offset all of the eigenvalues.
Then, it seems, we could arbitrarily change the accrued
phase.- Hide quoted text -
Good point. I guess (but would have to check) that the
Berry phase is only of interest for when it can be
compared for two states, as then the relative Berry
phase can be observed via a superposition of the states.
This would mean that the Berry phase is simply not
observable for an energy eigenstate (it is just an
overall phase change as above, and, as you point out,
can be varied simply by varying the choice of
groundstate energy.
However, if you have two energy eigenstates, then
the relative phase from transporting the system
does become observable. Eg, for eigenvalues
E and E', and eigenvectors v and v', one has
v(1) = exp[-iE / hbar] v(0) ,
v'(1) = exp[-iE' / hbar] v'(0) ,
and hence any superposition of these states
will pick up a relative phase
phi = (E-E') /hbar
around the path you defined. This relative
phase is independent of energy shifts.
Not sure if this answers your original
question though?
[/quote:33a0db24a2]
Thanks again, but I do not think that is necessary.
Splitting a single eigenvector (say if its a photon),
then transporting part and recombining it (after it has
been transported) with the other should produce
interference consistent with the Berry phase. |
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