Explanation for the EPR-like paradox I am trying to understand the process of Quantum Entanglement for use in Quantum computers. 
The problem I have is this: 
Suppose some nuclear process emits an electron-positron pair. Now after adequate separation, I measure the electron's position and the positron's momentum at time $t$ simultaneously, both with sufficiently high degrees of accuracy. 
By momentum conservation, I should be able to tell both the position and momentum of the electron (or positron) at the measuring time $t$, thus violating Heisenberg's Principle.
What is wrong with this logic? 
Specifically, I cannot understand how will entanglement affect momentum or position?
I am an Electronics student, not Physics so I apologize if this is too simple.
 A: What is wrong with this logic is that you are supposing a particle has simultaneously well-defined position and momentum. This is not true - a state localized in real space is delocalized in momentum space, and vice versa. The classical conservation laws hold on the quantum level as operator laws, not as laws on the states.
A: Since you're an electronics student, I'll speak your language. Think of momentum and position as parameters in time and frequency domain of a signal rather than classical observable that are well defined. If you do so, you can easily realize that your frequency isn't well defined if you don't do an infinitely long measurement. This is simply due to the wave nature of variables that come from a Fourier  transform.
It is not an issue of tricking the system to read a position, but rather it is more about whether what you read from an experiment makes sense as a real, reproducible and reliable physical observable. 
A: I think the key point that you're missing is that as soon as you make a measurement, the entanglement between the two particles is broken.  It should also be noted that the original particle also obeyed the uncertainty principle, and that at a quantum level there is no direct relationship between position, momentum and time.
Another confusing factor is that you haven't specified what is actually causing the entanglement.  One possibility is that it is because the particles can be shot off in any direction, but one particle must go in the opposite direction to the other; that's both boring and difficult to think about, because the entangled part of the uncertainty in the position is perpendicular to the direction of travel.  So I'm going to examine the one-dimensional case; we can create entanglement by making the energy of the original particle uncertain.  In this case, we don't know what momentum the particles have relative to the center of mass, but we do know that they are equal.
If you make the position measurement (on the electron) first:


*

*the uncertainty about the position of the electron becomes arbitrarily small;

*the momentum of the electron changes, with the uncertainty becomes correspondingly large;

*the uncertainty about the position of the positron becomes smaller, but not arbitrarily so;

*the uncertainty about the momentum of the positron also becomes smaller (because it is correlated with the energy of the original particle, and hence with the position of the electron) but not arbitrarily small.
It may be helpful to imagine that the remaining uncertainty in the position of the positron is because of the uncertainty of the position of the original particle, though this is not entirely accurate - the uncertainty of the position of the original particle does affect the uncertainty of the position of the positron after measuring the electron position, but the relationship isn't as straightforward as that.
However, even though we haven't actually calculated exactly what the wavefunction looks like, we can guarantee that the uncertainty principle holds, simply because it is true for any wavefunction, no matter how constructed.
Also, since the entanglement between the particles is broken by the position measurement, when we then measure the momentum of the positron:


*

*nothing happens to the electron or our knowledge of it;

*our uncertainty about the momentum of the positron becomes arbitrary low;

*the position of the positron is changed, and the uncertainty becomes correspondingly high.
If, on the other hand, we made the momentum measurement (on the positron) first:


*

*the uncertainty about the momentum of the positron becomes arbitrarily small;

*the position of the positron changes, with the uncertainty becoming correspondingly large;

*the uncertainty about the momentum of the electron becomes smaller, but not arbitrarily so;

*the uncertainty about the position of the electron becomes smaller (because it is correlated with the energy of the original particle, and hence with the momentum of the positron) but not arbitrarily so.
And, as before, the entanglement is broken, so when we then measure the position of the electron nothing happens to the positron or our knowledge of it.
What if you make the measurements at the same time?  Well, that would be complicated to analyze, but we can cheat by calculating the results in a different frame of reference, one in which the measurements did not occur at the same time.  It so happens that the results of QM never depend on the reference frame, so we can be sure that this produces the right result.
(OK, the fact that in the real world the measurements take a finite time messes this up, unless you do them sufficiently far apart.  At that point you would really have to model the exact behaviour of both measurement devices to figure out exactly what happens.  But the end result will be the same: the particles are no longer entangled, and the wavefunction always obeys the uncertainty principle.)
A: Suppose the particles are initially in the (entangled) state
$$A\otimes B+C\otimes D$$
where $A$ and $C$ are position eigenstates for particle 1 and $B$ and $D$ are position eigenstates for particle 2.  
Note that this state is the same as 
$$X\otimes Y+Z\otimes W$$
where $X=(1/2)(A+C)$ and $Y=(1/2)(A-C)$ are momentum eigenstates for particle 1 and $Z=(1/2)(B+D)$ and $W=(1/2)(B-D)$ are momentum eigenstates for particle 2.  
(I am using "position" and "momentum" here for arbitrary observables with eigenstates related as above.)
Now observe the first particle's position.  Without loss of generality, you get $A$.  Therefore the pair is now in state $$A\otimes B=A\otimes Z+A\otimes W$$
Now observe the second particle's momentum.  It is either $Z$ or $W$, equiprobably.  The pair is now in either state $A\otimes Z$ or $A\otimes W$.
Or, if you insist on treating the two measurements as "simultaneous", note that the initial state is also equal to
$$A\otimes Y+A\otimes W+C\otimes Y+C\otimes W$$
so that a measurement of "particle 1's position and particle 2's momentum" returns 
$(A,Y)$, $(A,W)$, $(C,Y)$ or $(C,W)$ equiprobably.
Where's the problem?
A: The problem is that you propose to make the two measurements "at time $t$ simultaneously".  Measuring the particle's momentum cannot be done instantaneously; the more precisely you want to measure it, the longer the minimum required observation time becomes.  (Rougly speaking that's because knowing the particle's momentum is equivalent to measuring its frequency, but to know its frequency precisely you have to count for a significant fraction of its period. That's an effect you should be familiar with from electronics, where it shows up in signal processing. The more cycles you count for, the more precisely you'll know its frequency and hence its momentum.) But then your knowledge of the other particle's position becomes fuzzy because you're now asking for its location not at a single point in time, but over some small but nonzero time interval.
So, the longer the measurement takes, the better you may be able to know the one particle's momentum, but the worse you'll know the other's position. 
A: Ok. Let's suppose that the initial state of the two particle are an eigenstate of the momentum operator (momentum is well defined). Quantum mechanics tell us that the position of the center of mass is not well defined. If we measure the position of the particle 1 (electron), then we do two thing in the system:


*

*We apply a measurement in a part of the whole system (trace in the hilbert space of the second particle), this means that we don't know precisely what state actually describing the particle, we can only determine the probability of each state be the right state if we know the state of whole system (two particle). 

*We measure the position of the particle: we "collapse" our possible momentum state in a position state.


Now let's see the another particle:


*

*We apply a measurement in a part of the whole system (trace in the hilbert space of the second particle), this means that we don't know precisely what state actually describing the particle, we can only determine the probability of each state be the right state if we know the state of whole system (two particle).

*We measure what momentum state is right for the second particle (positron), and consequently the first particle by conservation law.


The key point is actually in the first procedure made by both measurements. We are ignoring a part of the system. If I exchange some informations after or before the measurements we need to a better description of what actually happens. We need to look for the total state of the particles, and apply the measurements in whole state. Then we see that when we measure the position of some of the particle and the momentum of another particle the total momentum turn to be ill defined (we collapse the total state). Is not so certainly that, when we measure what momentum state is right for the second particle (positron), the momentum of the first particle is achieved by conservation law, because we don't know what happens in the first particle. We can't make assumption of the total state if we are working in a part of this state.
