| Topic: |
Science > Physics |
| User: |
"Natski" |
| Date: |
19 Oct 2004 10:50:33 AM |
| Object: |
Hermitian conjugates |
Having trouble proving that for two non commutative operators written
PQ (in that order) has the hermitian conjugate Q*P*; where the *
denotes a hermitian conjugate since I can't find the correct symbol on
this keyboard. Thanks for any help.
Ben
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| User: "George Jones" |
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| Title: Re: Hermitian conjugates |
19 Oct 2004 11:16:41 AM |
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"Natski" <diesel8684@hotmail.com> wrote in message
news:bd655f2b.0410190750.560404a3@posting.google.com...
Having trouble proving that for two non commutative operators written
PQ (in that order) has the hermitian conjugate Q*P*; where the *
denotes a hermitian conjugate since I can't find the correct symbol on
this keyboard. Thanks for any help.
Before I try and help you, I'd like to know what notation you're using,
so as not to confuse you anymore than necessary :-).
Let A be an operator, and phi and psi be states. How is A* defined?
By one of the following?
1) Dirac notation:
<psi|A*|phi> = <phi|A|psi>* (where the 2nd * is conjugate of a number).
2) Inner product notation:
<A* phi, psi> = <phi, A psi>.
Or are you using some other (equivalent) notation and/or definition.
Hint: Try C = PQ in whatever definition you use.
Regards,
George
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| User: "Bruce Scott TOK" |
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| Title: Re: Hermitian conjugates |
19 Oct 2004 11:20:51 AM |
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Natski asked:
|> Having trouble proving that for two non commutative operators written
|> PQ (in that order) has the hermitian conjugate Q*P*; where the *
|> denotes a hermitian conjugate since I can't find the correct symbol on
|> this keyboard. Thanks for any help.
Same as for matrix multiplication and the inverse thereof: it is the
order. With two operators, working on a function f in some space
(doesn't have to be specifically the Hilbert space from QM), you write
A B f
for the combination, with B working on f and then A working on the
result of that, giving you a new result. Hermitianity refers to scalar
products, which are defined as
(g,f) = \int dV g f
so that you would also have
(g, A B f) = \int g A B f
The Hermitian conjugate of the A which acts upon B f is A\dagger, but
now it acts in the other ``direction,'' upon g. Then you turn B around
in similar fashion, obtaining
(g, A B f) = \int f B\dagger A\dagger g = (f, B\dagger A\dagger g)
Hence, the Hermitian conjugate of A B is B\dagger A\dagger.
--
cu,
Bruce
drift wave turbulence: http://www.rzg.mpg.de/~bds/
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| User: "Bjoern Feuerbacher" |
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| Title: Re: Hermitian conjugates |
19 Oct 2004 11:13:54 AM |
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Natski wrote:
Having trouble proving that for two non commutative operators written
PQ (in that order) has the hermitian conjugate Q*P*; where the *
denotes a hermitian conjugate since I can't find the correct symbol on
this keyboard. Thanks for any help.
First, for hermitation conjugate one usually writes a "dagger". That
sign is not on keyboards, agreed - but you could e.g. use a "t".
Second, consider the definition of "hermitean conjugate":
An operator A is said to be the hermitean conjugate of an operator
B if for all vectors phi, psi, the scalar product of phi with B*psi is
equal to the scalar product of A*phi with psi.
Using that definition, it is quite easy to prove that the hermitean
conjugate of PQ is Q^t P^t ...
Bye,
Bjoern
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