Time evolution of a state. (a missing t)

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SUMMARY

The discussion focuses on the computation of the term in equation 1068 related to time evolution in quantum mechanics, specifically addressing the absence of a "t" in the sine function. The user derives an expression for the integral involving an exponential function and identifies that the sine function should include a "t" in the denominator to match the expected sinc function form. The resolution comes from recognizing the correct form of the coefficient in the time evolution formula, specifically ##c_n(t) = it/\hbar...##, which clarifies the confusion regarding the missing "t".

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naima
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Hi PF there is one thing that i cannot understand here.
Please look at eqn 1068
I try to compute the first term (without ##V^\dagger##)
I get something like
##c_f (t) =-i/\hbar exp [i(\omega + \omega_{fi})t/2] \frac{sin(\omega + \omega_{fi})t/2}{(\omega + \omega_{fi})/2} \}##

Unlike eqn 1071 a "t" is lacking under the sine and i cannot write it as a sinc function.

Could you help me? I do not see my mistake.
(I can write my intermediate formulas)
 
Last edited:
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We begin with a ##\int_0^t xxx... dt' term##
With a change of variables t"" = t' -t/2 we get
##\int_{-t/2}^{t/2} exp[i \Omega (t"" + t/2)] dt""##
##= exp[i \Omega t/2]\int_{-t/2}^{t/2} exp[i \Omega t""] dt""##
In the integral the odd i sin function gives no contribution so we have
##= exp[i \Omega t/2]\int_{-t/2}^{t/2}cos (\Omega t"") dt""##
so we finish with a ##sin (\Omega t/2) / (\Omega/2) ##
But the real formula has a ##sin (\Omega t/2) / (\Omega t/2) ##
Where does this t come from?
 
Last edited:
Sorry I did not see that in the link we had ##c_n(t) = it/\hbar ...##
I have no more problem with the formula :smile:
 

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