In the textbook ''Concepts of Modern Physics -Arthur Beiser'' in chapter 3 section 3.7 where the book talks about the uncertainty princple

While illustrating the physics behind the uncertainty principle ''I will put an image from the book'' he began talking about the $\Delta x$ and defined it as the width of wave group associated with the particle,, and this is logical as the width of wave group increase the less precise we know it's position

but while deriving the famous formula which is :

\begin{gather*} \Delta x\Delta p\ge \frac{h}{4\pi} \end{gather*}

he said that $\Delta x $ is the standard deviation of the ${\Psi}(x)$

this is what my confusion about ,,is the $\Delta x $ the width of wave group represented by ${\Psi}(x)$ or the standard deviation ${\Psi}(x)$? and why we began by talking about it by the width of wave group of the wave function ${\Psi}(x)$ and then we said it represent the standard deviation of ${\Psi}(x)$? and what does mean by saying that the $\Delta x$ is the uncertainty what do mean by uncertainty?

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1 Answer 1


Formally, $\Delta x$ here means the standard deviation in $x$, which can be computed given the wavefunction $\Psi(x)$ as (I'll assume we're in one spatial dimension) \begin{equation} \Delta x^2 = \langle x^2\rangle - \langle x \rangle^2 = \left[\int dx \Psi^\star x^2 \Psi \right] - \left[\int dx \Psi^\star x \Psi\right]^2 \end{equation}

However, intuitively, given a wavepacket, $\Delta x$ indeed corresponds to the spatial width of the wavepacket. It is not too difficult to check this analytically for a Gaussian wavepacket \begin{equation} \Psi(x) = \frac{1}{\left(2\pi \sigma^2\right)^{1/4}}e^{-(x-x_0)^2/(4\sigma^2)} e^{i k x} \end{equation} using the formula for $\Delta x^2$ given above.

  • $\begingroup$ If $\Delta x$ correspond to the width of wave packet ,why we say it's the standard deviation of $x$ ? $\endgroup$
    – Mans
    Jan 6, 2023 at 2:58
  • $\begingroup$ @Mans $\Delta x$ is defined to be the standard deviation of $x$. In the case when the wavefunction has the form of a wavepacket, the standard deviation of the wavepacket is a measure of the width of the wavepacket. $\endgroup$
    – Andrew
    Jan 6, 2023 at 2:59
  • $\begingroup$ But this is not always the case if the the wave packet has a gaussian shape the width of wave packet will be greater than standard deviation $\endgroup$
    – Mans
    Jan 6, 2023 at 3:01
  • $\begingroup$ @Mans By "width of a Gaussian wavepacket," physicists generally mean "one standard deviation" -- in other words, a region which contains most of the probability, not all of the probability. (Indeed using the latter definition I suppose a Gaussian wavepacket would have an infinite width). $\endgroup$
    – Andrew
    Jan 6, 2023 at 3:10
  • 2
    $\begingroup$ @Mans Probably because the textbook author was trying to give an intuitive, not overly-mathematical description of the physics. For a Gaussian wavepacket, the standard deviation is the width. But, the width of a wavepacket is generally an easier concept for students. So (I assume, I haven't read the book) that the author used the "width" to sweep the mathematical details under the rug to be able to talk about the physics of the uncertainty principle without defining standard deviation. I assure you, the fully correct statement uses the standard deviation. $\endgroup$
    – Andrew
    Jan 6, 2023 at 3:43

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