Why does a free electron with definite energy not exist in the quantum mechanical sense?

Question

I'm reading Introduction to Quantum Mechanics by Griffiths and there is a part that I don't understand. He says that a free electron with a well defined energy cannot exist (bottom of pg. 60, 2nd edition). I don't understand why though.

Let's say that I make an experiment. On the left side I have a device which will eject a single electron towards the right. On the right I have a wall that will notice exactly when a free electron hits it. I run the experiment: the electron is emitted and the wall receives it. The time taken to reach the wall is recorded. I now know the energy of the electron using $E=\frac{mv^2}{2}$, where $v=d/t$, $d=$ the distance that the electron travelled and $t$ is the time it took for the electron to travel the gap. This experiment used a free particle in the sense that it traveled a vacuum gap without interacting with anything before it hit the wall.

Where is my understanding flawed?

Answer 1

If a free particle has a definite energy, then it has a definite momentum by $p=\sqrt{2mK}$ ($p$ is the particle's momentum, $m$ is the particle's mass, and $K$ is the particle's kinetic energy). The Heisenberg Uncertainty Principle states that the uncertainty of the position of a particle ($\Delta x$) multiplied by the uncertainty of the momentum of a particle ($\Delta p$) has a minimum of $$\Delta x \Delta p \geq \frac{h}{2\pi},$$ where $h$ is Planck's constant. If $\Delta p = 0,$ then $\Delta x = (h/2\pi)/0 = \infty,$ meaning the electron could hit the screen after an arbitrary amount of time since it starts from a completely uncertain location. This makes your energy measurement impossible since $d$ is completely unknown. If you prepare your electron so that it starts from a definite position, reducing $\Delta d$, then it's momentum must be more uncertain, meaning your timing measurements will have more uncertainty ($\Delta t$).