I have just begun my Introduction to Quantum Mechanics course in my undergrad and I am trying to understand the uncertainty principle on a fundamental level. I think the best way to understand the fundamentals of this principle is to question the principle and understand why doing so is wrong.

Consider the following thought experiment:

I have a setup consisting of two independent regions. Region A with an electric field along the positive x-axis, lets say, and Region B with a magnetic field, also along the positive x-axis. In Region A I have a cathode and the electric field causes a potential difference thereby accelerating the electrons from rest to a particular velocity (along positive x-axis).

Let us just focus on a single electron. I know the exact velocity (magnitude and direction) to which this electron is accelerated to (energy conservation). Now this electron is going to enter the magnetic field in a direction that is parallel to the field. I know that the electron is going to move in a straight line in the magnetic field, without losing any energy as the magnetic field does zero work. So I know the exact momentum of the electron at all times as this momentum is not going to change as long as the electron is inside the magnetic field. So uncertainty in momentum is zero. Since I know the exact trajectory of the electron, the uncertainty in position is also zero and I know both of these simultaneously.

So it appears like this though experiment violates the Heisenberg uncertainty principle.

What is the flaw in my argument? Why can't this happen?

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    $\begingroup$ The flaw in your argument is that electrons don't "go" in straight lines. That's an entirely classical picture of an electron and real electrons simply don't behave that way. $\endgroup$ – CuriousOne Jan 26 '16 at 6:56
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    $\begingroup$ The whole story is full of many other wrong statements. For example, electrons aren't moving along straight lines in magnetic fields even in a classical theory. They move along circles or spirals, right? Also, all the claims that "the uncertainty is zero" are absolutely unjustified. To say that the uncertainty is zero means to be able to measure it, and know that the measurement is precise etc. In reality, it never is quite precise. $\endgroup$ – Luboš Motl Jan 26 '16 at 7:14
  • $\begingroup$ @LubošMotl, In the classical theory, when an electron enters a magnetic field in a direction parallel to the field, its velocity and the magnetic field will make an angle of zero with each other. So cross product of those two times the charge of the electron is zero. So the magnetic force is zero. So the electron will indeed go in a straight line. $\endgroup$ – Rohit C Jan 26 '16 at 14:18
  • $\begingroup$ Dear Rohit, don't forget that the electron also has a magnetic moment - it's a small magnet oriented along the direction of its spin which generally differs from the direction of the magnetic field. $\endgroup$ – Luboš Motl Jan 26 '16 at 15:22

You say:

Let us just focus on a single electron. I know the exact velocity (magnitude and direction) to which this electron is accelerated to (energy conservation).

but this isn't true. You know the electron energy has increased by $E$ eV, where $E$ is the potential difference you're using but you don't know what its energy was initially i.e. when it left the anode and before being accelerated by your field. The only way you can know the initial momentum of the electron precisely is if it's completely delocalised i.e. you don't know where it is or when it was emitted.

I applaud your attempts at understanding the uncertainty principle, but you are going about it the wrong way. You need to start by writing down the wavefunction for a free particle. The eigenfunctions for a free particle are infinite plane waves, which have a precise momentum but completely unspecified position. Assuming you start with a partially localised particle you construct its wavefunction by using Fourier synthesis i.e. you build up the initial probability distribution by summing (an infinite number of) plane waves. Because this requires combining waves with different momenta that means your partially localised particle has a spread of momenta.

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    $\begingroup$ Just a little nit: I think it's important to avoid language such as "you know" when talking about the uncertainty principle. Folks tend to get this idea that the principle is about what we "can know" about a particle. While that is in a sense correct because wave functions are fundamentally a representation about what can be known about a quantum system, I think this language more often than not leads people to think that the particle has some properties that the uncertainty principle prevents us from knowing, which is not correct as far as we know. $\endgroup$ – DanielSank Jan 26 '16 at 8:26

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