Creation/Annihilation operator help

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SUMMARY

The discussion revolves around the application of the differentiation theorem in calculus to understand the relationship between the integral of a function involving an annihilation operator and its derivative. Specifically, the integral expression involves the annihilation operator \(\hat{b}(u)\) and the exponential term \(e^{i \omega(t-u)}\). The key conclusion is that by applying the theorem correctly, the derivative of the integral can be expressed as \(H'(t) = f(t)\), where \(f(u) = e^{i \omega(t-u)}\hat{b}(u)\) and \(g(t) = t\).

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This discussion is beneficial for physics students, particularly those studying quantum mechanics, as well as mathematicians interested in the application of calculus to physical theories. It is also relevant for anyone looking to deepen their understanding of operator theory in quantum physics.

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Homework Statement


I'm trying to go through something shown in a lecture and the part I'm stuck on is shown here.
[tex]\frac{\partial}{\partial t}\int^t_0 \mbox{du }e^{i \omega(t-u)}\hat{b}(u)=\hat{b}(t)[/tex]

[tex]\hat{b}(u)[/tex] is an annihilation operator

The Attempt at a Solution


Can someone explain how this step is made? Obviously a delta function has to come in somewhere but I don't know how to do that with a derivative (only another integral).
 
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this follows from a theorem in calculus

Theorem: Suppose that a function [itex]g:[c,d]\rightarrow [a,b][/itex] (not necessarily a
onto function) is differentiable, and a function [itex]f:[a,b]\rightarrow \mathcal{R}[/itex]
is continuous. If

[tex]H(t)=\int_a^{g(t)} f(u)\, du[/tex]

with [itex]t\in [c,d][/itex], then H is differentiable and

[tex]\large H'=(f(g(t))g'(t)[/tex]

in your example

[tex]f(u)=e^{i \omega(t-u)}\hat{b}(u)[/tex]

and [itex]g(t)=t[/itex]

so

[tex]H'(t)=(f(g(t))g'(t)[/tex]

[tex]g'(t)=1[/tex]

so

[tex]H'(t)=f(t)[/tex]

which gives what you are seeking
 

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