Calculating the Integral of Electric Field Wavefunctions

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The discussion focuses on calculating the integral of electric field wavefunctions defined as E(t)=E_0(t) e^{-i \omega_0 t}, where E_0(t)=A for -t1 ≤ t ≤ t1 and zero otherwise. The integral G(t) is evaluated using the limit as T approaches infinity, leading to a conclusion that the final result is zero. The participants identify errors in the integration steps, particularly in the treatment of the wavefunction across different time intervals, emphasizing the need to evaluate the integral only within the bounds of -t1 and t1.

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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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