Calculating the Integral of Electric Field Wavefunctions

[ilvreth]

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?

 

[LeonhardEuler]

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.

 

[ilvreth]

i fixed the typo. The limit from 1 step to 2 is different, yes. Is this step correct?

 

[LeonhardEuler]

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?

 

[ilvreth]

It must be... $$T\rightarrow \infty$$ for all steps?

 

[LeonhardEuler]

That's true, but also, the form of the integral needs to change in the different regions.

 

[ilvreth]

How can be written?

 

[LeonhardEuler]

Since the integral is additive, you can just break the integral up into different regions and add the integrals over different regions up.

 

[ilvreth]

How could you break an integral which is within a limit?

 

[LeonhardEuler]

$$\int_{a}^{b}f(x)dx = \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx$$

 

[ilvreth]

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)?

 

[LeonhardEuler]

No, you should separately do the integral for t<-t1, then for -t1<t<t1, then for t>t1.

 

[ilvreth]

But the integral for t<-t1 and t>t1 gives zero (because the wavefunction is zero). Finally what the result is?

 

[LeonhardEuler]

Yes, it does give 0, so in other words, you only need to evaluate the integral between -t1 and t1.

 

[LeonhardEuler]

And, of course, T still approaches infinity, not t1.

 

[ilvreth]

So the final result is zero?

 

[LeonhardEuler]

You got it.