Calculating the Integral of Electric Field Wavefunctions

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Homework Help Overview

The discussion revolves around the calculation of an integral involving electric field wavefunctions, specifically the expression G(t) defined in terms of the electric field E(t). The problem involves evaluating the integral under certain conditions and limits.

Discussion Character

  • Mathematical reasoning, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the correctness of the steps in the integration process, particularly the transition from one step to another and the implications of the defined intervals for E(t). Questions arise about the treatment of the integral across different regions and the limits involved.

Discussion Status

There is an ongoing examination of the steps taken in the integration process, with some participants pointing out potential errors in the assumptions made about the wavefunction's definition across different intervals. Guidance is offered regarding the need to evaluate the integral separately for different regions, and the discussion reflects a mix of interpretations regarding the limits and conditions of the integral.

Contextual Notes

Participants note that the wavefunction is defined as zero outside the interval -t1 to t1, which affects the evaluation of the integral. There is also mention of the confusion surrounding the limit as T approaches infinity in relation to the integral's bounds.

ilvreth
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The question is about the integration bellow.

We assume the the electric field wavefunction is E(t)=E_0(t) e^{-i \omega_0 t} and E_0(t)=A for t_1 \leq t \leq t_1 and zero when t<-t_1 , t>t_1.

Calculate the integral G(t)= \lim_{T\rightarrow \inf} \frac{1}{2T} \int_{-T}^{T} E(t)^{\ast} E(t+\tau) dt where \tau is a constant.
My solution is the following

Step 1 G(t)= \lim_{T\rightarrow \infty} \frac{1}{2T} \int_{-T}^{T} E(t)^{\ast} E(t+\tau) dt =

Step 2 \lim_{T\rightarrow t_1} \frac{1}{2T} \int_{-T}^{T} A^{\ast} e^{i\omega_0 t} A e^{-i\omega_0 t } e^{-i\omega_0 \tau} dt =

Step 3 \lim_{T\rightarrow t_1} \frac{1}{2T} \int_{-T}^{T} |A|^2 e^{-i\omega_0 \tau} dt =

Step 4 \lim_{T\rightarrow t_1} \frac{1}{2T} \left[ |A|^2 e^{-i\omega_0 \tau} t \right]^{T}_{-T} =

Step 5 \lim_{T\rightarrow t_1}\frac{1}{2T} |A|^2 e^{-i\omega_0 \tau} 2T =|A|^2 e^{-i\omega_0 \tau}The questions

1) Is the the step 1 to 2 correct?
2) Is the result correct?
 
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There seems to be a problem going from 1 to 2. E(t) is defined differently for the interval -t1<t<t1 (I assume that's what it's supposed to be and t<t<t1 is a typo) compared to outside this interval, but you substitute one formula for all cases.
 
i fixed the typo. The limit from 1 step to 2 is different, yes. Is this step correct?
 
Ok, except it should be -t1<t<t1 not t1<t<t1. Anyway, what about the mistake in going to step 2? Do you see the error there?
 
It must be... T\rightarrow \infty for all steps?
 
That's true, but also, the form of the integral needs to change in the different regions.
 
How can be written?
 
Since the integral is additive, you can just break the integral up into different regions and add the integrals over different regions up.
 
How could you break an integral which is within a limit?
 
  • #10
\int_{a}^{b}f(x)dx = \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx
 
  • #11
I know that. But the limit makes a confusion in my mind because T runs to infinity. Anyways. In my country is 4.48am and i am sleepless. Is result correct (the last step)?
 
  • #12
No, you should separately do the integral for t<-t1, then for -t1<t<t1, then for t>t1.
 
  • #13
But the integral for t<-t1 and t>t1 gives zero (because the wavefunction is zero). Finally what the result is?
 
  • #14
Yes, it does give 0, so in other words, you only need to evaluate the integral between -t1 and t1.
 
  • #15
And, of course, T still approaches infinity, not t1.
 
  • #16
So the final result is zero?
 
  • #17
You got it.
 

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