How to get “complex exponential” form of wave equation out of “sinusoidal form”?

Question

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.

Answer 1

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: http://www.youtube.com/watch?v=GUtLtRDox3c

Answer 2

As user1104 commented, you use Euler's identity:

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

so:

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

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:

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

therefore:

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