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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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

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  • $\begingroup$ 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. $\endgroup$ – 71GA Feb 4 '13 at 15:15
  • $\begingroup$ 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. $\endgroup$ – Kenshin Feb 4 '13 at 23:17
  • $\begingroup$ youtube.com/watch?v=GUtLtRDox3c $\endgroup$ – Kenshin Feb 5 '13 at 11:51
  • $\begingroup$ 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? $\endgroup$ – 71GA Feb 6 '13 at 20:24
  • $\begingroup$ 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. $\endgroup$ – Kenshin Feb 7 '13 at 0:00

As user1104 commented, you use Euler's identity:

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


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

But we wouldn't normally proceed 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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Consider the following derivative: $[\cos{x} +i\sin{x}]' = i \sin{x} - i\cos{x} = i(\cos{x} +i\sin{x})$. That sure looks like $[e^{ix}]' = ie^{ix}$. So the question from a physics point of view is why is the oscillatory behaviour of $\sin$ and $\cos$ so fundamentally connected to the behaviour of $\exp$ governing growth and decay?

One answer may be self-similarity--$\exp$ is self-similar, so for instance in radioactive decay, the number of decays is always proportional to the number of atoms present. Compare that with an pendulum, where the acceleration (velocity change) is proportional to the displacement and the displacement change is proportional to the velocity.

Both these ideas are combined in the damped oscillator-where a single complex frequency's real part describes the oscillation and the imaginary part describes the damping.

When applied to unstable particles one considers the Breit Wigner resonance formula, so when one makes Delta baryons the mass is on average 1232 MeV--but not always. The lifetime is so short ($5 \times 10^{-24}\ $ s) that we speak of the width of the resonance (~114 MeV)--with the two being connected by the Heisenberg Uncertainty Principle. (The mass drives the oscillator part of the wave function, while the width drives the decay--so that a complex frequency unifies the two phenomena).

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