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I am a novice on QM and until now i have allways been using sinusoidal form of wave equation: $$A = A_0 \sin(kx - \omega t)$$

Well in QM everyone uses complex exponential form of wave equation: $$A = A_0\, e^{i(kx - \omega t)}$$

QUESTION: How do i mathematically derive exponential equation out of sinusoidal one? Are there any caches? I did read Wikipedia article where there is no derivation.

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If you want more ways to prove Euler Identity, this is a very good reading:…. – Physicist137 Jul 25 at 9:31
up vote 4 down vote accepted

You asked about the second equation. See below:

$e^{ix}{}= 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \frac{(ix)^6}{6!} + \frac{(ix)^7}{7!} + \frac{(ix)^8}{8!} + \cdots \\[8pt] {}= 1 + ix - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \frac{ix^5}{5!} - \frac{x^6}{6!} - \frac{ix^7}{7!} + \frac{x^8}{8!} + \cdots \\[8pt] {}= \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \cdots \right) + i\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \right) \\[8pt] {}= \cos x + i\sin x \ .$

To calculate the expansions I have used in the above equation, you need to understand the procedure for finding Taylor expansions of functions. This youtube video teaches the procedure:

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Where can i seeh for myself proof that $e^ix$, $\sin x$ and $\cos x$ are trully the sequences that you said they are? Please post some great youtube videos or articles. – 71GA Feb 4 '13 at 15:15
If you google Taylor expansion it is the procedure to prove the sequences are as I say. I will update my answer with these proofs soon. – Mew Feb 4 '13 at 23:17 – Mew Feb 5 '13 at 11:51
One more thing. Isn't $e^{i(kx-\omega t)} = \cos (kx - \omega t) + i \sin(kx \omega t)$? So it should than hold that $\sin(kx - \omega t) = e^{i(kx-\omega t)} -\frac{\cos (kx - \omega)}{i}$? What gives us right to use only real part? – 71GA Feb 6 '13 at 20:24
I have shown that e^i(kx-wt) is an oscillating function with the same frequency as sin(kx - wt). Whenever sin(kx - wt) is the solution to a differential equation, so will e^i(kx-wt) be. This is because in an equation, the Real part of the left hand side will always equal the Real part of the right hand side. Similarly, the complex part of the left hand side will always equal the complex part of the right hand side. So an equation with a complex number is effectively two equations, one for the real part, and one for the complex part. The equation is more general solution to the DE. – Mew Feb 7 '13 at 0:00

As user1104 commented, you use Euler's identity:

$$ e^{ix} = cos(x) + i \space sin(x) $$


$$ sin(kx-wt) = \frac{ e^{i(kx-wt)} - e^{-i(kx-wt)}}{2i} $$

But we wouldn't normally procede by replacing sin by this expression. Both the sin form and the exponential form are mathematically valid solutions to the wave equation, so the only question is their physical validity. In QM we don't worry about having a complex solution because the observable is the squared modulus, which is always real.

For a guitar string obviously the complex form isn't physically valid, but any sum of solutions to the wave equation is also a solution to the wave equation. That's why we can add (or subtract) the complex solutions to get a real solution.

Response to comment:

$$ e^{ix} = cos(x) + i \space sin(x) $$

so replacing $x$ by $-x$ gives:

$$ \begin{split} e^{-ix} &= cos(-x) + i \space sin(-x)\\ &= cos(x) - i \space sin(x) \end{split} $$

because $cos(x) = cos(-x)$ and $sin(x) = -sin(-x)$. So subtracting $e^{-ix}$ from $e^{ix}$ gives:

$$ \begin{split} e^{ix} - e^{-ix} &= cos(x) + i \space sin(x) - cos(x) + i \space sin(x)\\ &= 2i \space sin(x) \end{split} $$


$$ \frac{e^{ix} - e^{-ix}}{2i} = sin(x) $$

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Could you please do this in a couple more steps? – 71GA Feb 4 '13 at 8:41
Can someone remind me how you get the = signs to line up using MathJax? – John Rennie Feb 4 '13 at 9:07
You should use begin{split} end{split} environment and allign equall sign in it by using &. I did edit your post accordnigly so you can take a look. Thanks for the add. I know how to derive second equation now, but what about first one? – 71GA Feb 4 '13 at 10:11
Oh my god look at this beautifull video: – 71GA Feb 4 '13 at 14:57
@71GA, the music is brilliant isn't it! – Mew Feb 5 '13 at 11:49

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