In QM why can wavefunctions be written as $\psi (x,t)= Ae^{i(kx-\omega t+\phi)}$?

In my textbook, a wave traveling in the positive $x$-direction can be described using $$\psi(x,t)=Ae^{i(kx-\omega t+\phi)}.$$

I understand that the equation for a classical wave can be extrapolated by solving the differential equation associated with simple harmonic motion, giving us $\psi(x,t)=A\cos(kx-\omega t +\phi)$. I know Euler's identity is $e^{ix}=\cos(x)+i\sin(x)$. It would seem that we could simply equate $\psi(x,t)=A\cos(kx-\omega t +\phi)$ with $\Re (Ae^{i(kx-\omega t+\phi)})$ by omitting the $i\sin(kx-\omega t +\phi)$ term. My book seems to equate $\psi(x,t)=A\cos(kx-\omega t +\phi)$ to the entire complex version of the wavefunction $\psi (x,t) = Ae^{i(kx-\omega t+\phi)}$ with the stipulation that you can simply extract the real portion of the function when needed. It seems like these two functions should not be able to be equated for reasons described above. Is this a typo or am I not understanding this correctly?

Also, why in QM do we need to include the imaginary terms associated with the $i\sin(kx-\omega t +\phi)$ to get an accurate picture of the wave? Can the above relation only hold because it is derived from a wave associated with simple harmonic motion with classical not quantum waves?

• The imaginary part is needed to get the probability distribution since $\mathbb{P}[x]=\psi(x)\psi^*(x)$. With regards to equating $e^{i\theta}$ to $\cos(\theta)$ some authors just assume that it is understood that they are always taking $\Re (e^{i\theta})$. – Jepsilon May 28 '18 at 0:28

$\psi (x,t) = Ae^{i(kx-\omega t+\phi)}$ is a solution of the free particle Schrödinger equation, but $\psi(x,t)=Acos(kx-\omega t +\phi)$ is not.
• How can the cosine equation satisfy the wave equation $\frac {\partial^2 \psi(x,t)}{ \partial x^2}$ - $\frac {1}{v^2} \frac {\partial^2 \psi (x,t)}{\partial t^2}=0$ but not the Schrodinger equation if you get the Schrodinger equation using the wave equation with substitutions of the de Broglie equations? – Joe May 28 '18 at 0:42
As stated in the answer by @my2cts, $$\psi(x,t)=Ae^{i(kx-\omega t+\phi)}$$ is the solution of the free particle Schrödinger equation, but $$\psi(x,t)=A\cos(kx-\omega t+\phi)$$ is not, even if it is a solution to wave equation $$\frac{\partial^{2}\psi(x,t)}{\partial x^{2}}-\frac{1}{v^{2}}\frac{\partial^{2}\psi(x,t)}{\partial t^{2}}=0$$ because Schrödinger equation has the form $$\frac{\partial^{2}\psi(x,t)}{\partial x^{2}}\propto\frac{\partial\psi(x,t)}{\partial t}$$. Therefore, the solution to Schrödinger equation is a linear combination of sines and cosines (as you can show), which is represented as the imaginary exponential function by Euler's identity.