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I am required to show that the operator $\partial_t$ is Anti-Hermitian. This operator is defined such that

$$\partial_t: s(t) \rightarrow \partial_t s(t) $$

Where the definition of an Anti-Hermitian operator in terms of the inner product is

$$<s_1, As_2> = - <As_1, s_2>$$

using this definition of the inner product:

$$<s_1, s_2> = \int_{-\infty}^{+\infty} s_1(t)[s_2(t)]^{*}dt$$

where $^{*}$ denotes the conjugate. This is probably quite a simple question (I am told) but I can't quite seem to make it make much sense. Is it a matter of using integration by parts? I tried but couldn't get the negative sign to appear from anywhere. Maybe I'm slightly confused about the purpose of the complex conjugate in these Fourier transforms.

Any help much appreciated

Victoria
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1 Answers1

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It is simply using integration by parts. Note that boundary terms must vanish otherwise the integral wouldn't converge. Then note that the derivative commutes with conjugation.

  • He means $\left(\overline{s(t)}\right)' =\overline{s'(t)}$. – science Mar 14 '15 at 05:14
  • Okay so $u = s_1(t) \rightarrow du = \partial_{t}s_1(t)$ and then $dv = \partial_t[s_2(t)]^{*}dt $. But how do you integrate the $dv$? Are there plus and minus infinity bounds on the integral? – Victoria Mar 14 '15 at 05:25
  • You simply remove the \partial. As I had said, the boundary terms vanish otherwise the integral wouldn't converge. –  Mar 14 '15 at 05:26
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    That's what integration by parts is. What's the anti-derivative of a derivative? –  Mar 14 '15 at 05:28
  • and then $uv - \int vdu = s_1(t)s_2(t)^{} - \int s_2(t)^{}\partial_{t} s_1(t) dt$ – Victoria Mar 14 '15 at 05:33
  • That's almost right, make sure you include the limits of integration! –  Mar 14 '15 at 05:35
  • ... and those limits make the $\int vdu$ part equal to the negative inner product I'm after? But what about the $s_1(t)s_2(t)$ part that's still hanging about? – Victoria Mar 14 '15 at 05:36
  • As I said, make sure to include the limits of integration! That INCLUDES the part that's still hanging about. –  Mar 14 '15 at 05:37
  • I'm sorry, I'm pretty slow... I've wound up getting: $$<s_1(t), \partial_{t}s_2(t)> = s_1(t)s_2(t) - <\partial_{t}s_1(t), s_2(t)$$ – Victoria Mar 14 '15 at 05:42
  • Again what are the limits of integration? You aren't taking an indefinite integral, correct? You are integrating from $-\infty$ to $\infty$ correct? What can $s_1(\infty)$ be? –  Mar 14 '15 at 05:43
  • In order for the original integral to converge (which it does by assumption), what does $s_1(\infty)$ have to be? –  Mar 14 '15 at 05:48
  • That's (one of) the limits of integration, as you had said. –  Mar 14 '15 at 05:51
  • You apply the bounds to the $s_1(t)s_2(t)^{*}$ as well. –  Mar 14 '15 at 05:56
  • If you are integrating with respect to t, how can you end up with leftover t? The statement of ibp includes applying bounds to the extra part. –  Mar 14 '15 at 05:58
  • http://tutorial.math.lamar.edu/Classes/CalcII/IntegrationByParts.aspx look at "Integration by parts, definite integrals" –  Mar 14 '15 at 05:59
  • Of course. :) So using the limits, and noting that $s_{1,2}(\pm \infty)=0$, what do you get? –  Mar 14 '15 at 06:02
  • AHA! I simply never learnt that aspect of by parts. What a fool I've been. I see it all now, thanks so much ! – Victoria Mar 14 '15 at 06:06
  • Hi! The shear number of comments in this thread triggered a system flag. It seems to me that William now understands the argument. Therefore I have mild suggestions/requests: 1) avid19 could now add some of those details to the answer (+1 in advance). I very much approve of the style to first post the punchline only, and add details in the comments, no argument there, but now you may consider cleaning it up. 2) Do consider deleting those comments that have now served their purpose. This is just a suggestion - not an order :-) – Jyrki Lahtonen Mar 14 '15 at 06:32