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QM Fourier Transform and Integral rules

  • Thread starter katkatkat
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Show G(k)=[itex]\sqrt{2π}[/itex]g1(k)g2(k)

Given that G(k) is the fourier transform of F(x), g1(k) is fourier trans of f1(x), g2(k) is fourier trans of f2(X) and

F(x)=[itex]^{+∞}_{-∞}[/itex]∫dyf1(y)f2(x-y)

SO FAR

G(k)=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫F(x)e-ikxdx <-def'n of fourier transform

=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫[itex]^{+∞}_{-∞}[/itex]∫dyf1(y)f2(x-y)e-ikxdx

mult by e-ikyeiky

=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫dyf1(y)e-iky[itex]^{+∞}_{-∞}[/itex]∫f2(x-y)e-ik(x-y)dx

=g1(k)[itex]^{+∞}_{-∞}[/itex]∫f2(x-y)e-ik(x-y)dx

Problem starts here... I have arg of x-y in the remaining part, with dx...
Thought was that sub dx=dx-dy+dx=d(x-y)+dx <-Pretty sure I'm allowed to do this

=g1(k)[itex]^{+∞}_{-∞}[/itex]∫f2(x-y)e-ik(x-y)d(x-y)+g1(k)[itex]^{+∞}_{-∞}[/itex]∫f2(x-y)e-ik(x-y)d(y)

=[itex]\sqrt{2π}[/itex]g1(k)g2(k)+g1(k)[itex]^{+∞}_{-∞}[/itex]∫f2(x-y)e-ik(x-y)d(y)

I'm definitely on the right track up to the problems part, do I say that ∫arg*dy is equal to zero because F(x) is a function of x... or a different reasoning... not sure... If I can make the right half of the equation = 0, I'll have the desired result

=[itex]\sqrt{2π}[/itex]g1(k)g2(k)

=G(k)
 
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Answers and Replies

  • #2
gabbagabbahey
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=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫dyf1(y)e-iky[itex]^{+∞}_{-∞}[/itex]∫f2(x-y)e-ik(x-y)dx

=g1(k)[itex]^{+∞}_{-∞}[/itex]∫f2(x-y)e-ik(x-y)dx
This is not correct, you cannot claim that

[tex]\int_{ -\infty }^{ \infty }\int_{ -\infty }^{ \infty }dydx f_1(y)f_2(x-y) e^{-iky} e^{-ik(x-y)} = \left( \int_{ -\infty }^{ \infty }dx f_2(x-y) e^{-ik(x-y)} \right) \left( \int_{ -\infty }^{ \infty }dy f_1(y)e^{-iky} \right)[/tex]

The first integral contains the variable [itex]y[/itex], so is not constant when integrating over [itex]y[/itex] and cannot be pulled out of the [itex]y[/itex] integral. Instead, try integrating over [itex]x[/itex] first while treating [itex]y[/itex] as a constant (which you can do since they are independent variables over the entire region of integration).
 
  • #3
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I can't actually solve the integral as I'm supposed to be subbing in the symbol for the fourier transform. I have no choice but to separate them somehow.

Here's the prof's solution. I'm not sure how he got to the 3rd step though. How did he get d(x-y) out of dx.

G(k)=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫F(x)eikxdx

=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫dx*eikx[itex]^{+∞}_{-∞}[/itex]∫f1(y)f2(x-y)dy

=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫d(x-y)f2(x-y)eik(x-y)[itex]^{+∞}_{-∞}[/itex]∫f1(y)eikydy

=[itex]\sqrt{2π}[/itex]g1(k)g2(k)

I know his solution is right, this is coming from someone who's specialty is fourier transforms. I'm just not sure how he did the d(x-y) step from the dx step (line 2 to line 3)
 
  • #4
gabbagabbahey
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Just use a substitution: [itex]u(x)=x-y[/itex], when integrating over x, y is constant, so [itex]du=?[/itex]...
 
  • #5
rude man
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Show G(k)=[itex]\sqrt{2π}[/itex]g1(k)g2(k)

Given that G(k) is the fourier transform of F(x), g1(k) is fourier trans of f1(x), g2(k) is fourier trans of f2(X) and

F(x)=[itex]^{+∞}_{-∞}[/itex]∫dyf1(y)f2(x-y)

SO FAR

G(k)=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫F(x)e-ikxdx <-def'n of fourier transform

=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫[itex]^{+∞}_{-∞}[/itex]∫dyf1(y)f2(x-y)e-ikxdx

mult by e-ikyeiky

=[itex]1/\sqrt{2π}[/itex][itex]^{+∞}_{-∞}[/itex]∫dyf1(y)e-iky[itex]^{+∞}_{-∞}[/itex]∫f2(x-y)e-ik(x-y)dx
No, the problem starts here.

The second integral, which, when evaluated (integrated over x), being a function of y, therefore belongs under the 1st integral. You put the second integral outside the first.

For the last step, hint: ∫f2(x-y)exp{-ik(x-y)}dx = same expression with dx → d(x-y), since y is constant in the x integration.
 
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  • #6
Hi,

I just wanted to note that you are essentially trying to prove the convolution theorem for the Fourier transform. Knowing that may help you out by looking at its proof.
 
  • #7
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The second integral, which, when evaluated (integrated over x), being a function of y, therefore belongs under the 1st integral. You put the second integral outside the first.
I don't think that matters. It's a function of the variable (x-y) not y. Can you not sub in z=x-y? Then you no longer have that problem. I don't think it'll fall under the dy integral for that reason since it's not a function of the y variable.
 
  • #8
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I'm going to sub in z(x)=x-y. Then dx=dz and f(x-y)=f(z) etc. Thanks for the help guys :) This looks on the right track according to some of the proofs I've seen.
 
  • #9
vela
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You could use the change of variables u=y, v=x-y. It's straightforward to show that dx dy = du dv so that
$$\int_{-\infty}^\infty \int_{-\infty}^\infty dy\,dx\,f_1(y)\,f_2(x-y) e^{-iky} e^{-ik(x-y)} = \int_{-\infty}^\infty \int_{-\infty}^\infty du\,dv\,f_1(u)\,f_2(v) e^{-iku} e^{-ikv}.$$
 
  • #10
rude man
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I don't think that matters. It's a function of the variable (x-y) not y. Can you not sub in z=x-y? Then you no longer have that problem. I don't think it'll fall under the dy integral for that reason since it's not a function of the y variable.
Ahem - if it' a function of (x-y) it's ipso facto a function of x ... the second integral, when integrated wrt x, is now a function only of y.

You don't need to substitute anything, just realize that dx and d(x-y) are the same thing if y is a constant, which it is in the second integral (the one ending in dx). And you can't just say, "well, I'll get rid of y by substituting z = x-y." That's actually quite amusing ...
 

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