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Why adj( U(t)) * U(t) = I where U(t) is a propagator in QM?

 
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Jul19-07, 11:24 PM   #1
 

Why adj( U(t)) * U(t) = I where U(t) is a propagator in QM?


This is probably really obvious but can someone explain to me why adjiont( U(t)) * U(t) = I where U(t) is a propagator in QM and I is the identity.
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Jul20-07, 12:31 AM   #2
 
Isn't that basically the definition of U?
Jul20-07, 01:17 AM   #3
 
Look at what happens if you time-reverse the Schrödinger equation.
Jul20-07, 02:12 AM   #4
 

Why adj( U(t)) * U(t) = I where U(t) is a propagator in QM?


Quote by ehrenfest View Post
This is probably really obvious but can someone explain to me why adjiont( U(t)) * U(t) = I where U(t) is a propagator in QM and I is the identity.
This is not obvious, but can be proven. The time evolution operator U(t) [also known as propagator] must preserve probabilities. E. P. Wigner proved (I think it was in 1931) that this implies that U(t) is a unitary operator (formally, U(t) can be also antiunitary, but this possibility can be discarded on the basis of continuity of U(t)). This result is called "Wigner theorem". The condition you wrote is equivalent to saying that U(t) is a unitary operator.

Eugene.
Jul20-07, 05:29 PM   #5
 
in the language of operator theory, i believe another proof is via Stone's theorem.

it is always a bit startling to realize that many of the properties of QM follow very naturally from the mathematical properties of the Hilbert space. for example, many people are (for some reason) surprised when i tell them that the resolution of the identity, or complete set of states, [tex]\sum_i |i><i| = 1[/tex] is merely a trivial result of vector calculus, e.g. [tex]\vec{v} = \sum_i \vec{e_i} (\vec{e_i} \cdot \vec{v}) [/tex] with an arbitrary basis
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