Different definitions of the uncertainty principle

Question

In the book Introduction to Quantum Mechanics by Griffiths, the mathematical form of the uncertainty principle is stated as $$\sigma_x \sigma_p \ge \frac{\hbar}{2}.$$ However, another book on QM, that by Bransden and Joachain, states the principle as $$\Delta x \Delta p \gtrsim \hbar.$$ What I am confused about them is, what is the mathematical difference between these two representations, since they physically mean the same thing? To be precise, I know that I can calculate $\sigma_x$ as $\sqrt{\langle x^2 \rangle - \langle x \rangle^2}$, and similarly for $\sigma_p$. Thus, for a given wavefunction, I can check myself using the definition by Griffiths that it satisfies are uncertainty relation. Is there a similar way to calculate $\Delta x$ in the second representation? Is $\Delta x$ the same as $\sigma_x$?

Answer 1

The general uncertainty principle for two observables $\hat{A}$ and $\hat{B}$ is given by:

$$ \sigma_{A}\sigma_{B}\geq |\langle[\hat{A},\hat{B}]\rangle|/2 $$

where $\sigma_{A}=\sqrt{\langle\hat{A}^2\rangle-\langle\hat{A}\rangle^2}$. You can find a simple derivation on wikipedia. If $\hat{A}$ and $\hat{B}$ are $\hat{x}$ and $\hat{p}$, then $[\hat{x},\hat{p}]=i\hbar$, which substituting into the expression above gives your first result:

$$ \sigma_{x}\sigma_{p}\geq \hbar/2. $$

What could be going on with the second expression? I do not have access to the book, but my best guess is that they only use an "approximately equal or larger than" symbol. This suggests that their derivation of the uncertainty principle is not an exact derivation, but an approximate one which only allows them to provide an "order of magnitude" estimate to the true relation, which is the first one you write.

Answer 2

As far as I know, the sign ≳ means "is greater than or of the same order as", so you can use the second formula to estimate $\Delta x$ (from below) by the order of magnitude. As this is just about the order of magnitude, it is not important if $\Delta x$ is exactly the same as $\sigma_x$.

Answer 3

In the Wikipedia article the wave mechanical derivation of the Heisenberg uncertainty is given by :

$$\sigma_x \sigma_p \ge \frac{\hbar}{2}$$

I repeat my comment here:

DvijD.C.when I gave a wiki link, there was a special link to be clicked in order to see the derivation of the statement

$$\Delta x \Delta p \gtrsim \hbar.$$

and it was written just so in the referred link.

I now do not find the link there, and unfortunately do not remember the author . (It must have been edited away and I do not have access to earlier editions to check.)

I checked on the net various calculations and did not find the second form derived, there is always a factor of 2 with the usual one. There is this similar question , and I think the answer by Charles Francis is the correct one:

Heisenberg proposed:

$$\Delta x \Delta p \gtrsim \hbar.$$

but he only gave a heuristic argument and did not use a precise definition of uncertainty.

Kennard (1927) defined uncertainty precisely, to coincide with the definition of standard deviation in statistics.

$⟨ΔK⟩2=⟨K2⟩−⟨K⟩2$

He then proved

$$\sigma_x \sigma_p \ge \frac{\hbar}{2}$$

So it seems to be a historical progress in defining the uncertainty mathematically. In any case it is irrelevant because the larger limit includes the smaller one, and as the relationship is an order of magnitude limit , the results will be the same.