I'd like to pose a straightforward question by providing a brief example to determine whether my current approach is correct or incorrect.

Imagine I have an electron confined within a box measuring $1\times 10^{-10}$ meters in width. If we consider this width as the uncertainty in the electron's position, we can use the equation $\Delta p = \frac{\hbar}{2\Delta x}$ to calculate the minimum uncertainty in the electron's momentum. Now, since momentum is directly proportional to velocity, expressed as $\Delta p = m_e \Delta v$, we can also determine the uncertainty in the particle's velocity.

$$\Delta v = \frac{\Delta p}{m_e}$$

Is it then possible to calculate the de Broglie wavelength of this electron by substituting this value into de Broglie's relation $\lambda = \frac{h}{m_e\Delta v}$?

If not, kindly explain why and provide insights into the implications of such a calculation. Would it this way just imply that we're calculating the wavelength of a particle with a velocity of $v = \Delta v$ ?

  • $\begingroup$ Just a reminder that it is good to accept an answer if you found your problem was resolved so that anyone looking up your question later can see which answer helped you most! $\endgroup$ Commented Oct 27, 2023 at 2:02
  • $\begingroup$ None of them did help. I already knew the information that had been provided. $\endgroup$ Commented Oct 27, 2023 at 11:21

3 Answers 3


This is not, in general, the appropriate way to propagate the error in some measurement; namely, you cannot just plug in the error in a measurement back into the formula for the quantity you desire. First of all, we should clarify that the "error" you are referring to is properly interpreted as a standard deviation, in general. In other words, $\Delta p = \sigma_p$. For a multi-variable function $f(\{x_j\})$ where $x_j$ stands for one for one of the $N$ variables that $f$ depends on, we can find the standard deviation $\sigma_f$ for $f$ in the limit that its variables are independent using, \begin{gather*} \sigma_f \approx \sqrt{\sum_j ^N \left( \frac{\partial f}{\partial x_j} \right)^2 \sigma_{x_j}^2 } \end{gather*} where $\sigma_{x_j}$ is the standard deviation in the value for $x_j$. In your case, you want the standard deviation in the value of $\lambda$ from the De Broglie relation, \begin{align*} \lambda &= \frac{h}{p} \\ \implies \frac{\partial \lambda}{\partial p} &= - \frac{h}{p^2} \end{align*} which we can find assuming that the values of the fundamental constants are well-known. Then, using the propagated error formula, \begin{gather*} \sigma_\lambda \approx \sqrt{\left( \frac{\partial \lambda}{ \partial p} \right)^2 \sigma_p ^2 } = \sqrt{ \left( - \frac{h}{p^2} \right)^2 \sigma_p ^2 } = \frac{h \sigma_p}{p^2} \end{gather*} Note again that what you have called $\Delta p$ is really just $\sigma_p$, since this is what is derived in the generalized uncertainty relation.

  • $\begingroup$ Thanks! And yes, that's in general the case. However, since $\Delta v$ is the minimum uncertainty of the particle in the width $\Delta x$, i.e. the amount of velocity the particle cannot avoid (intrinsic in a sense), then would $\lambda = h/(m_e \Delta v)$ also be the maximum wavelength the particle can have in that width? It is correct this way, right? $\endgroup$ Commented Oct 26, 2023 at 11:32
  • $\begingroup$ Your expression is, unfortunately, just incorrect. To see why, plug in some numbers. Assuming the deviation of the wavelengths is far smaller than the wavelength itself, then your estimate for the error would blow up to a value far larger than the value of the wavelength itself! This is obviously unphysical and just an incorrect estimate. $\endgroup$ Commented Oct 26, 2023 at 11:44
  • $\begingroup$ Say $\Delta v = v_{min}$, the amount of velocity the particle cannot avoid, then $\Delta \lambda = \frac{h mv_{min}}{(mv_{min})^2}$, so $\Delta \lambda = \frac{h}{mv_{min}}$. The uncertainty in the wavelength cannot get any larger because $v$ is already the minimum. Now if this $\Delta \lambda \ge \Delta x$ we can say that this particle is behaving very heavily quantum mechanically. $\endgroup$ Commented Oct 26, 2023 at 11:59

With $L= 1$ angstrom, I’d say:

$$ \sigma_x = L/\sqrt{12}$$


$$ \sigma_{px}=\frac 1 2 \hbar / \sigma_x $$

Now add all three components in quadrature:

$$ \sigma_p = \frac{\sqrt 3} 2 \hbar \sqrt{12}/L=3\hbar/L$$

As a lower bound.

As long as $L$ is much greater than the Compton wavelengh, you can use the nonrelativistic formula for velocity.

But after that, I am not liking the three, and would just say:

$$ \frac v c = \frac{\lambda_C}L$$

And be done with it.


No, you can't substitute electron's speed uncertainty $\Delta v$ (or $\Delta v/2$, or anything similar) into some speed value like $v_{min}, v_{max} ~\text{or}~\langle v \rangle$. Similarly, you can't use any uncertainties as a substitute for a dynamic variables expected value $(x, p, E, \ldots)$. To understand the reasons why,- look at most general form of Heisenberg uncertainty, which is :

$$ \sigma _{x}\sigma _{p}\geq {\frac {\hbar }{2}} \tag 1 $$

So at the core, uncertainty principle just dictates how measurement errors are distributed around their expected values, that's the point of standard deviation metrics ! So, like you can't say from your ruler systematic error how tall your room is,- analogically you can't say from the position or momentum standard deviation, what $x$ or $p$ expected values of particle are.

These must be measured or modeled separately. To calculate electron de Broglie wavelength, you need to know it's average speed in a box, which you can infer by 1D particle in the box model, namely by formula :

$$ v_n = \frac{1}{\sqrt{m}} \frac{nh}{2L} \tag 2 $$

However, as particle in a box has many energy levels and you don't know in what state particle "lives in", probably you have to do some additional guesswork. For starters, I would propose to take base level $n=1$, as most expected is to have particle in the ground energy level. However, keep in mind, that electron can jump between energy levels and so does it's kinetic energy and accordingly - average speed (2) and average de Broglie wavelength $\lambda$.

  • $\begingroup$ My idea is: say $\Delta v = v_{min}$, the amount of velocity the particle cannot avoid, then $\Delta \lambda = \frac{h mv_{min}}{(mv_{min})^2}$, so $\Delta \lambda = \frac{h}{mv_{min}}$. Now if this $\Delta \lambda \ge \Delta x$ we can say that this particle is behaving very heavily quantum mechanically. $\endgroup$ Commented Oct 26, 2023 at 12:02
  • $\begingroup$ No. Minimum particle speed is associated with it's base energy level and is described by my given (1) equation with $n=1$, i.e. minimum speed is $v_1$. So in the base level particle speed is $v_1 \pm \Delta v$, where $\Delta v$ is speed uncertainty. $\endgroup$ Commented Oct 26, 2023 at 12:10
  • $\begingroup$ I am not sure of that. But thanks anyway. $\endgroup$ Commented Oct 26, 2023 at 12:14
  • $\begingroup$ Your answer is blatantly incorrect and should be removed. The OP’s original formula is NOT an approximation to the error in the De Broglie wavelength. $\endgroup$ Commented Oct 26, 2023 at 12:24
  • $\begingroup$ @AnkyPhysics You are welcome. In general uncertainty principle gives you just error of variable when you compare it with expected value. So you can't directly substitute uncertainty into $x_{min}, x_{max}$ or $\overline x$, because it gives none of these, but just a variable fluctuation range. Also, as (1) shows when $n \to \infty$, then it can be that $v_n \gg \Delta v$. $\endgroup$ Commented Oct 26, 2023 at 12:28

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