# Why can we leave off half of the general solution?

In these pdf notes, it says at the bottom of the first page and beginning of the second:

[...] whose solution is: $$\Psi(\theta) = c_1 e^{i\omega\theta} + c_2 e^{-i\omega\theta}$$ Since we are interested in the real solution only it is sufficient to let: $$\Psi(\theta) = c_1 e^{i\omega\theta}.$$

I don't understand why can we just throw out the $c_2 e^{-i\omega\theta}$. What does the second fundamental solution have to do with our being interested in the real solution only? And what does this person mean by "interested in the real solution only"? After all $c_1 e^{i\omega\theta}$ is not necessarily real.

• I read the pdf and that is the standard horrible "pedagogy" that plagues so much of our field (and others). I struggled with this kind of thing too, Bob Dylan. My advice to you is to ignore the way that pdf does the problem. Keep both terms and do the problem yourself, looking to the pdf for check points if possible. I'm sorry people produce such awful explanations. – DanielSank Oct 1 '14 at 0:46
• a little bit below on the second page the pdf ays it takes the "conjugate" of $\Psi$, interesting since $\Psi$ on top of the 2nd page is taken to be real. Well the point is that the coefficients $c_1$, $c_2$ are in general complex. Note that (in QM) the wavefunction $\Psi$ is necesarily complex (cannot be real since it would make the Schroediger equation inconsistent). Maybe the author means sth else. – Nikos M. Oct 1 '14 at 0:55

Good question. You're right, you can't really throw out one of the solutions. The author of this PDF is using a poorly explained shortcut, which goes like this: if you expand out the full complex solution, you get

\begin{align} \Psi(\theta) &= c_1 e^{i\omega t} + c_2 e^{-i\omega t} \\ &= (a_1 + b_1 i)(\cos\omega t + i\sin\omega t) + (a_2 + b_2 i)(\cos\omega t - i\sin\omega t) \\ &= (a_1 + a_2)\cos\omega t + (b_2 - b_1)\sin\omega t + i[(a_1 - a_2)\sin\omega t + (b_1 + b_2)\cos\omega t] \end{align}

If you know, for reasons I won't get into here, that you only have to consider the real solution, you can assume "without loss of generality" (as they say) that the imaginary part is zero:

\begin{align}a_1 &= a_2 & b_1 &= -b_2\end{align}

That allows you to eliminate half the coefficients; in other words, two of $a_1,a_2,b_1,b_2$ are not independent, but are related to the remaining two.

You can now combine the previous two equations to get

$$\Psi(\theta) = 2a_1\cos\omega t - 2b_1\sin\omega t$$

Notice that the solution now has only two real coefficients, $a_1$ and $b_1$, or equivalently, one complex coefficient, $c_1$. The catch is that $c_1$ isn't really being used as a complex number; rather, it's being picked apart and its components used separately, which is why I don't think the way it's explained in the PDF is very useful.