The Uncertainty in momentum and the de Broglie wavelength

Question

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 $v = \Delta v$ ?

Answer 1

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.

Answer 2

With $L= 1$ angstrom, I’d say:

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

So

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

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.

Answer 3

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 $\lambda$.