Why is it impossible to measure position and momentum at the same time with arbitrary precision? I'm aware of the uncertainty principle that doesn't allow $\Delta x$ and $\Delta p$ to be both arbitrarily close to zero. I understand this by looking at the wave function and seeing that if one is sharply peeked its fourier transform will be wide.
But how does this stop one from measuring both position and momentum at the same time?
I've googled this question, but all I found were explantions using the 'Observer effect'. I'm not sure, but I think this effect is very different from the intrinsic uncertainty principle.
So what stops us from measuring both position and momentum with arbitrairy precision?
Does a quantum system always have to change when observerd? Or does it have to do with the uncertainty principle?
Thank you in advance.
 A: It is possible to measure both the position and the momentum of a particle to arbitrary position "at the same time", if you take that phrase to mean "within such quick succession that you can be confident that the probability distribution for the first measured quantity has not changed via Schrodinger evolution between the two measurements".
But doing so isn't very useful, because there will always be some infinitesimal delay between the two measurements, and whichever one comes second will effectively erase the information gained from the first measurement. For example, if you measure position and then momentum immediately afterwards, then you can get a very precise value for both measurements, but the process of getting a precise momentum reading will change the wavefunction such that it's position after the momentum measurement now has large uncertainty with respect to a subsequent measurement. So the momentum measurement "nullifies" the information from the prior position measurement, in the sense of rendering it unrepeatable.
So it's better to talk about the inability to "know" the position and momentum at the time than about the inability to "measure" both (which actually is possible). Fully understanding why requires understanding both the "state collapse" behavior of measurements, and the "wide <-> narrow" relation between non-commuting observables (e.g. via the Fourier transform) that you mention.
That's for measurements in extremely quick succession. You could ask about measurements that take place at exactly the same time, but that gets into philosophical waters as to whether two events ever occur at exactly the same time even in classical physics. In practice, if you try to do both measurements at once, then you'll always find that the particle comes out with either very tightly bounded position or momentum, and with large uncertainty in the other quantity.
A: That is a very deep question and it takes years to understand it. I try my best to answer it.
"what stops us from measuring both position and momentum with arbitrary precision?"
level 1: nature. That is how the nature works.
level 2: particles are neither wave nor particle. They behave differently from everything that we see in everyday life. You can watch Feynmann lecture on youtube where he explains this concept. In some experiment they behave like a particle and in some experiment they behave like a wave.
level 3: To our best understanding they behave like a field which exist everywhere. How can you measure the position and momentum of a field with infinite certainty? This way of looking at the particles explains everything except gravity.
"Does a quantum system always have to change when observerd?"
The quantum system is observed in one of the states. Before the observation the system is all the possibilities at the same time with different weight. You can learn more about this when you study path integral.
A: Others have already said this, but here is the succinct version:  You cannot determine the electron's position and momentum at the same time for the exact same reason that you cannot determine the electron's favorite flavor of ice cream, namely this:  The electron does not have a favorite flavor of ice cream.  Likewise, most electrons, most of the time, do not have a definite position or a definite momentum.  You can force it to have one of those (or at least a very good approximation thereto), but then it surely does not have the other.
A: The correct form of the uncertainty principle is that the product of delta x and delta p is always greater than or equal to (hbar)/2. Among other things, this means that the more precisely we know x, the less precisely we can know p.
This has nothing to do with the so-called observer effect; it has to do with the wave-like behavior of quantum particles.
A: To measure speed, you measure time between two positions.  Once you have a speed determined, which of the two positions would you associate with it simultaneously?  You can't rightly do it with either position.  To associate it with the average of the two positions would require that you assume constant velocity but you only measured the average velocity between the two positions and have no way to know if it was constant between them.
A: Your statement that the uncertainty relation comes from the Fourier transform is quite glib.  Physics is not just a collection of math results.  QM developed as a theory to account for experimental observations that did not respect classical mechanics, Newtonian or Relativistic.  Really they pointed to the fact that our paradigm regarding the nature of matter was wrong.  In QM every observable quantity is represented by an operation acting on a linear function space.  The eigenvalues of those operators represent the only allowed measurements of that quantity that can be observed.  For example, position ($x$), momentum ($p_x$), energy ($E$), etc are all operators.  The eigenfunctions of these operators represent the "state" that the system will be prepared in once a measurement is made.
When one measures $x$, which perhaps one can do with arbitrary precision and accuracy, and gets a specific value, the particle is left in an eigenstate of the operator $x$, which is a Dirac delta function.  Now if one tries to measure $p_x$ immediately after there is an equal probability to get any value of $p_x$.   Once you measure $p_x$ your previous measurement of $x$ is completely ruined.  You are not at all justified in making the claim that you know the value of $x$.  If you try to measure $x$ again you will get a different answer.  This is what the uncertainty relation is describing.  To say that it has to do with measurement precision is a red herring.
A: The OP wrote:

I understand that for a given wavefunction that if $\Delta x$ is small, $\Delta p$ will be big and how this arises from fourier transformations. But I fail to see how this prevents anyone from doing a simultaneous measurement of both $x$ and $p$ with infinite precision.

This seems to boil down to the issue of what "simultaneous measurement" means. What it means to simultaneously measure two observables $A, B$ is to perform a single measurement on the system, obtaining the values $a$ and $b$, such that, immediately after the measurement, the system is in a state where the value of $A$ is certainly $a$ and the value of $B$ is certainly $b$.
In other words, the result of the measurement is that the system is in a simultaneous eigenstate of $A$ and $B$. Since there are no simultaneous eigenstates of $x$ and $p$ (as the OP already understands), this isn't possible for this particular pair of observables.
A: The uncertainty principle is not a limitation on "measurements", but rather expresses a fundamental limitation of the Universe regarding its information capacity. In a rough sense, the Universe only "allocates", so to speak, so many bits to each particle, and thus there are only so many bits available that can, at any given time, realize its actual position and momentum together. This is most evident when it is written in the - and this is a more accurate form! - form involving the informational entropy of the position and momentum:
$$H_x + H_p \ge \lg(e\pi \hbar)$$
which quite simply says that there is always going to be entropy - a lack of information, compared to their classical counterpart - in either one, the other, or both.
There isn't a lot of magic about that. The Universe simply is economical and doesn't splurge an infinite number of bits to detail its particles' parameters.
That's why that, as the other answer mentions, if you do now bring in measurement, which most properly understood is the transaction of information between a system and an agent, and try to measure them both to higher precision than the amount above (about 170.18 bits jointly, if taken relative to a scale of 1 m and 1 N·s), you will not be able to repeat the measurement immediately afterward and obtain the same values. Obtaining the same value would require that information be in the particle so it could be retrieved again, but there isn't storage room for that. Hence what you get is junk.
A: You can't measure precise values at the same time because precise values for both don't exist at the same time.
All the properties of, say, an electron and be inferred from the electron's wave function, $\Psi(\vec x)$. The wave function is a mathematical object that covers all of space. It has a complex value at each point.
The electron doesn't have a precise position. Instead, it has a probability of being found at each point, $\vec x$, in space on being measured. That probability is $\Psi(\vec x)^*\Psi(\vec x)$. (That is a little loose. Really the probability of being found in a small region $d \vec x$ is $\int \Psi(\vec x)^*\Psi(\vec x) d \vec x$.)
The probability of being found somewhere is $1$, and so $\int\Psi(\vec x)^*\Psi(\vec x)dx = 1$. A function like this must approach $0$ everywhere except in some finite region.
There is a limiting case where it is $0$ everywhere except at one point, where it is infinite. In that case, it has a definite position.
You can also get the momentum from $\Psi(\vec x)$. Again, a definite momentum doesn't exist, except in a limiting case.
In general, $p = h\lambda$. That means an electron with a definite momentum would have a constant amplitude sinusoidal wave function with a definite wavelength. Such a wave function would cover all of space. $\Psi(\vec x) = A e^{i \vec p \cdot \vec x}$. This isn't possible, except as a limiting case where the amplitude approaches $0$. But in this limiting case, the wave function has the same (infinitesimal) amplitude everywhere. The electron has no location at all. It is spread over all space.
These limiting cases are at opposite ends of a range of possibilities. Most wave functions are non-zero over some finite region. Or at least, given any small number $\epsilon$, $|\Psi(\vec x)| > \epsilon$ only over a finite region.
The electron will be found in that finite region, but it doesn't have a precise location. Just a region where it will be found.
Likewise it doesn't have a definite momentum. You can use Fourier analysis to break a function up into a sum of functions of the form $A e^{i \vec p \cdot \vec x}$. $\Psi(\vec x) = \sum A(\vec p) e^{i \vec p \cdot \vec x}$. In the case of a non-periodic function like we have here, it is an infinite sum of infinitesimal functions. It is expressed as an integral rather than a sum. $\Psi(\vec x) = \int A(\vec p) e^{i \vec p \cdot \vec x} d \vec p$
You can think of $A(\vec p)$ as another way of expressing the wave function. This is another mathematical function, defined over that set of all possible momenta. It is useful for describing the momentum of the electron.
It can be shown that $A(\vec p)$ has lots of the same kinds of properties that $\Psi(\vec x)$ does. For example, the probability of finding the electron has momentum $\vec p$ is (again loosely) $A(\vec p)^*A(\vec p)$.
It can be shown $\int A(\vec p)^*A(\vec p)d\vec p = 1$. That is, the probability of finding the electron with some momentum is $1$. It can be shown the function can only be non-zero for a finite range of $\vec p$'s.
There is a limiting case where where $A(\vec p)$ is $0$ everywhere except for one value of $\vec p$. In this limiting case, the electron has a definite $\vec p$.
But the usual case is that the electron has neither a definite $\vec x$, nor a definite $\vec p$. That is, when the wave function is expressed as $\Psi(\vec x)$, it has a finite region where $\Psi(\vec x) > 0$. In this case, it turns out that when the wave function is expressed as $A(\vec p)$, there is a finite range of $\vec p$'s where $A(\vec p) > 0$.
The Uncertainty Principle is an important relation between the size of these two finite regions. $\Delta \vec x \Delta \vec p > \hbar/2$.
This video from 3blue1brown illustrates the idea. In particular, it shows how the Uncertainty Principle comes from wave properties.

Addendum - I didn't address an area where pglpm's answer really shines. I thought I would add my 2 cents.
Suppose you have an electron prepared in a state given by a particular wave function, $\Psi(\vec x)$. The position and momentum can be calculated to be particular values $\vec x$ and $\vec p$, with uncertainties $\Delta \vec x$ and $\Delta \vec p$. Note that uncertainties are often expressed as standard deviations of expected outcomes. This means position and momentum can be predicted to be $\vec x \pm \Delta x$ and $\vec p \pm \Delta p$.
Suppose the electron is just arriving at a free standing thin film surface containing many atoms.
If  $\Delta \vec x$ is large, it is not possible to predict which atom the electron will hit in advance. Nevertheless, the electron will hit a particular atom. It may the atom is affected in some permanent way, say by being ejected and leaving a hole. In that case, it is possible to go back afterward and find out very precisely what the position of the electron was.
If $\Delta \vec p$ is large, it is not possible to predict in advance what the electron's momentum will be measured to be. But if it ejects an atom, it may be possible to measure time of flight of the scattered electron and atom to detectors with high spatial resolution and get a very precise value for what the electron's initial momentum turned out to be.
The Uncertainty Principle does not limit how precisely we can determine the outcomes of these measurements. It limits how precisely we can predict them in advance. If you have many electrons in the same state, it limits how repeatable multiple measurements will be.
Immediately after the collision, the electron and atom will be in new states. Both states will have a $\Delta \vec x$ and $\Delta \vec p$. It is not possible to predict in advance when and where either will hit their detectors. But it is possible to say that the combined outcomes of the position and momentum measurements of the scattered electron and atom will add up to a momentum consistent with the electron's initial momentum and uncertainty.
A: The Uncertainty Principle has almost nothing to do with measurement. It's intrinsic to wave phenomena that if a composition of waves has a definite frequency, it has great uncertainty in duration and vice versa. Technically, a pure sine wave of a single frequency that lasts only temporarily isn't so pure. When expanded into its Fourier representation, one finds a density distribution of multiple frequencies.
The frequency and duration are conjugate quantities of the wave. So are wave number and wavelength.
De Broglie's Principle allows us to associate a wavelength to a material particle based on its momentum. Classical Mechanics establishes a relationship between wavelength and momentum for light waves.
Together we have that "Matter Waves" have momentum and position as conjugate quantities. A particle with a narrow band of momenta must have a broad distribution in space. The Born interpretation of the particle wave associates the modulus of its value with the probability of being located in that position.
Uncertainty in position is the standard deviation of the position given by its wave function. The Inverse Fourier Transform is the wave function in momentum space.
It can then be proven that having a definite position, i.e. a high concentration of likelihood to be located about a specific point, requires that we have a broad concentration of momenta, that is, the particle has a probability of being in multiple momentum states that are far apart.
It all comes down to the intrinsic nature of waves and waves being associated with momentum and position.
This manifests itself through measurement in various ways.
As it happens, The Uncertainty Principe tells us we can't simultaneously know with arbitrary accuracy any two components of a quantum particle's Spin Angular Momentum, having spin 1/2.
Measure $L_x$ and we might get $\hbar/2$. Measure it a bunch more times, and you get the same answer, repeating. Now measure $L_x$ and suppose you get $\hbar/2$, then measure $L_y$. There's only a 50/50 chance of getting either of the two allowed values. Measure $L_x$ again, you have only a 50/50 chance of once again obtaining $\hbar/2$.
Quantum Mechanically, to be in a specific state of $L_x$ is necessarily to be in multiple states of $L_y$ or $L_z$. One gets a definite answer only if a particle is in a pure state of a quantum observable.
In more complicated quantum system one finds that there are only certain allowed quantum states. Measurement will yield only certain specific results regardless of how the kinematic value is measured. We only observe the Uncertainty Principle in play when we make observations, i.e. execute measurements, but the behavior is indicative of fundamental attributes of the system and not of the measurement apparatus.
A: For wave $\Delta \nu \Delta t$ and  $\Delta k \Delta x$ are limited to be larger than some function of the amplitude. This holds for sound, light, water waves etc. In the case of a Schrödinger wave function the amplitude cannot be arbitrarily small as the wave must be normalised to 1. Because of this there is a minimum value of $\hbar/2$.
Once you understand this, the question becomes 'why should we use quantum mechanics?'. To this the answer is that it is 100% accurate, so far. Why this is no one knows.
A: As a laymen, I think my way through this one without invoking the quantum world, but simple the definitions of "position" and "velocity".

*

*If you want to measure momentum, you need to measure velocity.

*If you want to measure velocity, you need to measure distance and the time taken across that distance.

*To measure distance, you need to measure 2 positions.

*If you measure 2 positions, then which is "the" position of the
object?

I imagine this problem the same as looking at a perpendicularly travelling car through binoculars.  If you've ever tried this, you find yo have to keep moving your binos over, to follow moving car. Then someone asks you "Where is it now?"
A: I guess if you can gauge the precise format of the wavefunction, in either $x$ or $p$ domain (you can certainly calculate it), and consider that quantum state to be the real description of its position and momentum, then you know both. It is just that to say you "know the particle position precisely" usually means the wavefunction is forced (i.e. collapsed) into a very narrow peak in the $x$ domain, which implies a wide wavefunction in the $p$ domain (or vice-versa). So it boils down to the semantics of "knowing precisely".
