Can It Be Proven That Optics Field Amplitudes Satisfy Negative Frequencies?

AI Thread Summary
In quantum optics, the relationship E(-ω) = E*(ω) is often stated, but its proof is questioned. The discussion highlights that this relationship may not be specific to quantum cases and can be derived from definitions related to oscillating electric fields. The participants agree that the property seems trivial and stems from the way field amplitudes are expressed mathematically. The conversation concludes with a realization that the concept is indeed straightforward. Overall, the proof of this relationship remains largely based on definitions rather than a rigorous derivation.
Niles
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Hi

In (quantum) optics, many authors state that the field amplitudes satisfy

<br /> E\left( { - \omega } \right) = E^* \left( \omega \right)<br />

But how is it that one can prove that this is correct? I have never seen any book do this,
 
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Isn't this trivial and not specific to quantum cases anyways? Suppose you have an oscillating E-field (of course, the same works for the B-field component),
E(\vec{r},\omega,t)=E_0(\vec{r},t)e^{-i \omega t}

Then the property you mention is trivial.
 
You might be right; it probably isn't related to quantum cases. Here is how I have understood it: We can generally write

<br /> E\left( {r,t} \right) = \sum\limits_{n &gt; 0} {\left( {E\left( r \right)e^{ - i\omega t} + c.c.} \right) = \sum\limits_{n,\,\,all} {E\left( \omega \right)e^{ - i\omega t} } } <br />

The last equality follows it we define

<br /> \begin{array}{l}<br /> E\left( r \right) \equiv E\left( \omega \right) \\ <br /> E\left( \omega \right)^* = E\left( { - \omega } \right) \\ <br /> \end{array}<br />

But these are just definitions. So I don't see how we can really prove that <br /> <br /> E\left( { - \omega } \right) = E^* \left( \omega \right)<br /> <br />Niles.
 
Ok, it's pretty obvious now. Thanks.
 
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