In Quantum Mechanics, if a particle's state is a superposition of many states, then we say that its position is well-defined (by the Heisenberg uncertainty principle, because here we have ill-defined momentum). In the addition of each wave of momentum $p$ using Fourier series, where do we involve the initial phase of each wave that we are adding?
For example, say we add those two waves and get the function that I aforementioned. But if we add up those waves with only one of them having an initial phase of $\pi/12$ (making it $\exp[i(k x + \pi/12)]$ ), won't we get a function different from the one from the first addition? So we still add up the same states of definite momentum but get a different function.
Is there something wrong with my reasoning? Where do we involve the phase of each added wave(I believe it is something in complex analysis which I don't know about)? As someone stated at the comments, this at heart is a Fourier series question, so some mathematics indicating where the phase comes into play in Fourier series would be much appreciated.
NOTE: this question was edited so as to delete some things that I wrote which where wrong, so you might see some answers that try to explain something that might now be missing from the question but was present before.
Answer
There are some correct things and some incorrect things in your question. But let me just give the main points.
First of all, if you add two states together with different relative phases, you get different states. For instance,
$$\frac{1}{\sqrt{2}}|\psi_1\rangle + \frac{1}{\sqrt{2}}|\psi_2\rangle$$
is a different quantum-mechanical state than
$$\frac{1}{\sqrt{2}}|\psi_1\rangle + e^{i\pi/12}\frac{1}{\sqrt{2}}|\psi_2\rangle.$$
There are in fact a (continually) infinite number of different states that you can make out of $|\psi_1\rangle$ and $|\psi_2\rangle$,
$$\cos\theta|\psi_1\rangle + e^{i\phi}\sin\theta|\psi_2\rangle,$$
parameterized by the quantities $\theta$ and $\phi$.
For concreteness, and directly related to your question, if we work in the position representation, you can form the equal-amplitudes superposition of plane-wave states and make a delta-function, which can be thought of as a state of well-defined position (although such states do not actually physically exist). One possibility is
$$\delta(x) = \int\frac{dk}{2\pi}e^{ikx}$$
(In the abstract ket-space of position eigenvectors $|x\rangle$, this state would be the state $|0\rangle$.)
However, if you add a relative phase $e^{-ikx_0}$ to each plane-wave, then this becomes
$$\delta(x-x_0) = \int\frac{dk}{2\pi}e^{ikx}e^{-ikx_0},~~~~~~(1)$$
which in the abstract ket space is $|x_0\rangle$. To see this maybe a little more clearly, we can re-write the previous equations in the abstract ket space completely:
$$|0\rangle = \int{dk}|k\rangle.$$ $$|x_0\rangle = \int{dk}e^{-ikx_0}|k\rangle.~~~~~~(2)$$
In other words, we are adding up the momentum states $|k\rangle$ with different relative phases, so we get different states.
Most simply, if you add two plane waves with different relative phases, you obviously get different functions:
$$e^{ikx}+e^{-ikx}=2\cos{kx}$$
but
$$e^{ikx}+e^{i\pi}e^{-ikx}=2i\sin{kx}$$
Cleaning up some misconceptions
In Quantum Mechanics, if a particle's state is a superposition of many states of definite momentum, then we say that it's position is well-defined
This is not in general true. We say that the "position is well-defined" only if the superposition of momentum states is of a particular form (see equations (1) and (2) above). And it is always necessary to add this caveat: there really is no such thing as a state of well-defined position; it a mathematical convenience that, if used carefully, can be used as an approximation of a narrow wave-packet.
So say we add up two states of definite momentum (sine functions) so we get a function in position space.
This might just be an imprecise statement, but it's important to understand that a state is an abstract quantity that can be represented in either the position or momentum representation. The states of definite momentum are the eigenstates of the momentum operator, and when written in the position representation (i.e. as functions of $x$) take the form of plane waves:
$$|k\rangle = \int dx \langle x|\psi_k\rangle |x\rangle,\\ \langle x|\psi_k\rangle=\psi_k(x)=\frac{1}{\sqrt{2\pi}}e^{ikx}.$$
If you add up two momentum eigenstates, you can represent this function in position space, or you can represent this function in momentum space.
For example, say we add those two waves and get the function that I aforementioned. But if we add up those waves with only one of them having an initial phase of $\pi/12$ (making it $\exp[i(kx+\pi/12)]$), won't we get a function different from the one from the first addition?
Yep. See above.
So we still add up the same states of definite momentum but get a different function.
Yep! Nothing to see here really. I feel like this is the crux of the problem, but I can't tell what it is you're actually confused about.
...so some mathematics indicating where the phase comes into play in Fourier series would be much appreciated.
Under suitable niceness conditions of the functions, suitable boundary conditions etc., a complex-valued function $f(x)$ can be expanded in a Fourier series as
$$f(x) = \sum_{n=-\infty}^{\infty} A_n e^{i2\pi nx},$$
where the $A_n$'s are in general complex, so they can be represented in polar form as $re^{i\phi}$. And the point is, if you change those phases $e^{i\phi}$, you get out a different function $f$. That's it.
Final note
An overall complex phase is irrelevant. That is, $|\psi\rangle$ is the same state as $e^{i\phi}|\psi\rangle$, but as soon as you are forming superpositions, then the relative phases matter, as I've discussed.
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