I appreciate the statement of Heisenberg's Uncertainty Principle. However, I am a bit confused as to how exactly it applies to the quantum mechanical situation of an infinite square well.
I understand how to apply Schrödinger's equation and appreciate that energy Eigenvalues can be deduced to be
$$E_n=\frac{n^2\hbar^2\pi^2}{2mL^2}.$$
However, I have read somewhere that the reason that the quantum particle cannot have $n = 0$—in other words, $E = 0$—is because by having zero energy we also have a definite momentum with no uncertainty, and by the Heisenberg uncertainty principle this should lead to an infinite uncertainty in the position of the particle. However, this cannot the case be in an infinite well, as we know the particle should be somewhere in the box by definition. Therefore $n$ can only be greater than or equal to one.
Surely when $n = 1$ we have the energy as
$$E_1 = \frac{\hbar^2 \pi^2}{2mL^2},$$
which is also a known energy, and so why does this (as well as the other integer values of $n$) does not violate the uncertainty principle?
Best Answer
This started off as comment, but right now I only have the reputation to answer. But this is by no means a rigorous answer.
There are a couple of assumptions your question makes that aren't strictly true. For a start, you seem to say that any definite value of energy would entail a definite value of momentum. This is true for a completely free particle, but this is no longer true for a particle that is undergoing some interaction (where is the interaction, you ask? Well the fact that it is placed in a box, of course!)
There's a simple and (in my opinion) instructive way to see this. If what you said were true, then the states of definite energy would also be the states of definite momentum. In other words, they would satisfy the eigenvalue equation $$\hat{p}\psi_n = p_n\psi_n$$ where $\hat{p}=\frac{\hbar}{i}\frac{d}{dx}$ is the momentum operator and $p_n$ is a constant (which would represent the measured momentum). Let us check if this is the case. The states of definite energy are given by
$$\psi_n = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$$
The action of the momentum operator is thus
$$\hat{p}\psi_n = \frac{\hbar}{i} \sqrt{\frac{2}{L}} \frac{n\pi}{L} \cos\left( \frac{n\pi x}{L}\right) \neq p_n \psi_n$$
In other words, the states of definite energy are not states of definite momentum, since the momentum operator doesn’t yield the original wave function times a constant! Of course, you're right that the magnitude of the momentum would be fixed.
If you're still having trouble with this, here's a sloppy "intuitive" semi-classical example: say I gave you a (1-dimensional!) box with a particle (of unit mass) bumping around constantly inside it. I tell you that I measured the energy of this particle many times and it always came out to be exactly 8. Now I ask you to give me its momentum. "Aha!" you say, "when it is inside the box, there are no forces acting on it, and so the energy is simply given by:"
$$\frac{p^2}{2m} =\frac{p^2}{2} = 8$$
Thus, you find that the 'momentum' is 4! But wait a minute, you don't know if it's bouncing to the left or to the right. In other words, if it's $+4$ or $-4$! The fact that the particle interacts with the wall is responsible for its momentum 'flipping' sign.
In the same way, for the particle in the box, the magnitude of the momentum $|p|$ is given by
$$|p| = \pm \sqrt{2mE}$$
So what's the uncertainty in $p$? Well, it's simply $\Delta p = +|p| - (-|p|) = 2|p| = \frac{2 n\pi \hbar}{L}$. What about the uncertainty in $x$? Well, it could be anywhere in the box, and so $\Delta x = L$, the box's length.
Let us try to find
$$\Delta x \Delta p = 2n\pi \hbar > \frac{\hbar}{2}$$ for all values of $n\geq 1$.
Clearly, when $n=0$ this no longer works. We can understand this in many ways. A simple way would be to realise that when $n=0$, the magnitude of the momentum is $0$, and thus there are no 'positive' and 'negative' values it could take: it most definitely has a momentum of exactly zero, with no uncertainty. This would be allowed, if you were not in a box. However, placing yourself in a box, meaning that $\Delta x < \infty$, means that you necessarily have a minimal non-zero momentum, using the argument you mentioned earlier.
Moreover, the mathematics tells you that a state with $n = 0$ is the trivial state $\psi_0(x) = 0$. The mod-square integral of this function is 0, which can be interpreted as such a particle simply not existing.