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In a one-dimensional infinite potential well of width $l$, suppose the wave function is given by

$$\psi_{n}(x) = \sqrt{\frac{2}{l}} \sin\left(\frac{n\pi x}{l}\right).$$

The uncertainty between the position and the momentum is calculated to be

$$\Delta(x)\Delta(p) = \frac{nh}{4}\sqrt{\frac{1}{3}-\frac{2}{n^2\pi^2}}.$$

With the increase in values of '$n$', the uncertainty or value $\Delta(x)\Delta(p)$ increases.

I understand that the increase in the amplitude of the wave function makes the position more uncertain, but how does this affect the measurement in potential?

Also, according to the correspondence principle, increase in the value of the quantum number transitions the quantum well problem to the classical domain.

If that is so, shouldn't the uncertainty decrease but not become less than: $$\frac{h}{4\pi}$$ with the increase in '$n$'?

What is the physical significance behind this relation between quantum number and the uncertainty?

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In a classical sense, when going for a very large $n$, the particle can be anywhere in the well with about equal probability, meaning a uniform distribution with length $L$, so $\Delta x \to \frac{L}{\sqrt{12}} $ .

Now the momentum on the other hand can be $p = \pm \frac{h}{\lambda} = \pm \hbar k = \pm \hbar \frac{\pi n}{L}$. The standard deviation of a 50-50 chance to get $\pm 1$ is just $1$, and so we expect $\Delta p$ to approach $\hbar \frac{\pi n}{L}$, and their multiplication to approach (for large $n$): $$\hbar \frac{\pi n}{\sqrt{12}} = \frac{hn}{2\sqrt{12}}=\frac{hn}{4} \sqrt{\frac{1}{3}}$$ which is exactly what you got!

So one can see that the classical interpretation is consistent with the results. Why did you assume it needs to go to minimal unceratinty?

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    $\begingroup$ Is there a name for dx*dp? In signals, it's called the "time bandwidth product", or "the sophistication". Anyway, whatever it is called, I don't think there is a potential in which it is independent of $n$, and if it goes as $n^a$, what is the smallest $a$ you can get? $\endgroup$ Commented May 16 at 22:44
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From a classical point of view the maximum magnitude of momentum increases as the total energy of the particle increases. The same is true from a quantum point of view. The expectation value of the momentum will be zero becaue the particle has a net zero velocity averaged over time. If, however, we look at the expectation value of the momentum squared that will increase as the expectation value of the 'kinetic energy' increases. Thus as $n$ increases $p$ values will increase. As you noted in your question the range of motion in $x$ also increases a bit as $n$ increases, but this is not a large effect because it is a one dimensional infinite well with a fixed length.

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"I understand that the increase in the amplitude of the wave function makes the position more uncertain, but how does this affect the measurement in potential?"

This statement is not right. The maximum amplitude of the eigenstates are independent of $n$ and are always $\sqrt{2/l}$.

But in general: if the amplitude does increase, say, as $\psi(x)$ approaches $\delta(x)$, the position uncertainty decreases.

However, if in the infinte well, $\psi(x)$ looks like:

$$\psi(x) \propto \delta(x-\frac l 2+\epsilon) + \delta(x +\frac l 2-\epsilon) $$

which is two equal peaks as far away as possible (with $\epsilon\rightarrow 0$ for boundary conditions), that should be the maximally uncertain position state. I didn't calculate it, so maybe it's not.

The position uncertainty is really bound by $l$, dead or alive: the particle is in the box. There is a finite domain for $\psi(x)$; however, the momentum is not.

Since the particle is stuck in a box, you're going to have equal and opposite weight at positive and negative momentum (or else it will propagate out of the box). The momentum is related to $\sqrt{2ME_n}$, and since $E_n$ is unbound and goes up as $n^2$, one would expect $\Delta p$ to scale as $n$ at large $n$.

Again: I have calculated nothing, just speculating with the ol' quantum intuition, for what it is worth.

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