# Errors and differentials

$$\Delta p \Delta x \geq \hbar/2$$

we write the inherent errors in measurement as $\Delta p$ and $\Delta x$ .

Does these two errors behave the same as the differentials $dp$ or $dx$ or some other form of differentials?

• No. Differentials are infinitesimal, whereas these are on the left side of a $\ge$ sign with something finite on the right. – Chris Jan 23 '18 at 8:06
• I assume that your point is valid but when we take errors in physics, we represent the relative error in the form $\Delta x / x = \Delta a / a + \Delta b/b$. Almost like a logarithmic differentiation. What do we make of this? – Yuzuriha Inori Jan 23 '18 at 8:09
• Two formulas looking similar doesn't necessarily they are the same thing. (Also, if those are uncorrelated errors, they should be squared. Which incidentally makes it not look like logarithmic differentiation) – Chris Jan 23 '18 at 8:16
• The logarithmic differentiation bit simply arises because the logarithm turns multiplication into summation, so logarithmic derivatives of products combine linearly, exactly as ordinary derivatives of sums combine linearly. But this observation has nothing to do with the HUP. – WetSavannaAnimal Jan 23 '18 at 11:15

The Heisenberg inequality deals with a highly specific measure of "uncertainty", i.e. spread in the probability distributions that govern observed measurement outcomes applied to the same quantum state. Here, the $\Delta$s are the standard deviations of the relevant probability distributions, and have nothing to do with differentials.
More specifically, two quantum observables $\hat{x}$ and $\hat{p}$ are conjugate, i.e. fulfill the canonical commutation relationship $[\hat{x},\,\hat{p}]=i\,\hbar\,\mathrm{id}$, then if we apply $\hat{x}$ and $\hat{p}$ to the same quantum state, the measurements we glean from them have standard deviations related by the Heisenberg inequality. The inequality is saturated (becomes equality) if and only if the probability distributions are both Gaussian.
Furthermore, it can be shown that (1) $\hat{x}$ and $\hat{p}$ have continuous spectrums, and that measurements represented by them can take on any real value and (2) if one writes the quantum state in $\hat{x}$-co-ordinates (i.e. wherein $\hat{x}$ is a simple multiplication operator) so we represent the quantum state by the $\mathbf{L}^2$ function $\psi(x)$, then $\hat{p}=-i\,\hbar\,\mathrm{d}_x$ and so (3) the Fourier transform (a unitary mapping) maps the quantum state and co-ordinates into the Hilbert space wherein the $\hat{p}$ operator becomes the simple multiplication operator.