Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

How do we calculate uncertainty in kinetic energy $\Delta E_k$ if we only know that an (a) electron (b) proton is closed in a 1-D box of width $d=10fm$.

I first assumed that uncertainty in position $\Delta x=d$ and then calculated a momentum $\Delta p=\frac{\hbar}{2\Delta x}=9.845MeV/c$?

If I try to calculate uncertainty in kinetic energy relativisticaly I start with a Lorentz invariance:

\begin{align} E^2 &= p^2c^2 + {E_0}^2\\ E &= \sqrt{p^2c^2 + {E_0}^2}\\ E_k + E_0 &= \sqrt{p^2c^2 + {E_0}^2}\\ E_k &= \sqrt{p^2c^2 + {E_0}^2} - E_0\\ &\left\downarrow\substack{\text{Because energy $E_k$ is a function}\\\text{of only one variable $p$, we use}\\\text{standard formula for calculating}\\\text{uncertainty}.}\right. \quad \boxed{\Delta q = \frac{dq}{dp}\Delta p}\\ \Delta E_k &= \frac{d}{dp}\left(\sqrt{p^2c^2 + {E_0}^2} - E_0\right) \cdot \Delta p\\ \Delta E_k &= \frac{d}{dp}\left(\sqrt{p^2c^2 + {E_0}^2} - E_0\right) \cdot \Delta p\\ \Delta E_k &= \frac{1}{2}\frac{1}{\sqrt{p^2c^2+{E_0}^2}}2c^2p\cdot \Delta p\\ \Delta E_k &= \frac{c^2p}{\sqrt{p^2c^2+{E_0}^2}}\cdot \Delta p\\ \end{align}

From this I can't calculate $\Delta E_k$ because I don't know the expectation value for $p$. Should I use $\langle p\rangle$instead of only p? Can anyone please guide me through the procedure to calculate $\Delta E_k$ for (a) electron and (b) proton?


EDIT 1:

I think I have found the solution for the electron:

(a) ELECTRON: If we take $d=\Delta x$ as the absolute uncertainty and calculate $\Delta p=9.845MeV/c$ we notice that $\Delta pc < E_{0e}$ but $\Delta p$ is supposed to be small compared to $p$ which means $p$ must be huge! So i can with a confidence say that $pc\gg E_{0e}$ and this means i can deal with this problem ultrarelativistically!

So i can write the Lorentz invariance: \begin{align} E^2 &= p^2c^2 + {E_{0e}}^2\\ (E_k+E_{0e})^2 &= p^2c^2 + {E_{0e}}^2\longleftarrow \substack{\text{Here i use the ultrarelativistic approach and neglect the $E_{0e}$}}\\ E_k &= pc\\ \end{align} I got a fairly simple relation between momentum and energy and i can use standard formula for propagation of uncertainty on it like this: \begin{align} \Delta E_k &= \frac{d E_k}{dp}\Delta p\\ \Delta E_k &= c \Delta p\\ \Delta E_k &= 9.845MeV\\ \end{align} The result matches with my book's result! So I understand this problem now. Or so I thought untill I try to redo it for the proton.

(b)PROTON: Well if i compare $\Delta p c$ with a rest energy of a proton $E_{0p}=933.41MeV$ i find that $\Delta p c > E_{0p}$ but the $p$ is much larger than than $\Delta p$ and therefore i think i can say that $pc \approx E_{0p}$ which means i have to deal with this problem relativistically which again means i can't solve it (if i dealt with it superrelativistically i would get the same result as for electron):

\begin{align} E^2 &= p^2c^2 + {E_0}^2\\ E &= \sqrt{p^2c^2 + {E_0}^2}\\ E_k + E_0 &= \sqrt{p^2c^2 + {E_0}^2}\\ E_k &= \sqrt{p^2c^2 + {E_0}^2} - E_0\\ &\downarrow\\ \Delta E_k &= \frac{d}{dp}\left(\sqrt{p^2c^2 + {E_0}^2} - E_0\right) \cdot \Delta p\\ \Delta E_k &= \frac{d}{dp}\left(\sqrt{p^2c^2 + {E_0}^2} - E_0\right) \cdot \Delta p\\ \Delta E_k &= \frac{1}{2}\frac{1}{\sqrt{p^2c^2+{E_0}^2}}2c^2p\cdot \Delta p\\ \Delta E_k &= \frac{c^2p}{\sqrt{p^2c^2+{E_0}^2}}\cdot \Delta p\\ \end{align}

Please help me to understand how to solve this for a proton as well. The result for a proton from the book is $\Delta E_k = 53keV$.

Will i have to use the variances and expectation values? I am not sure how to connect expectation values with an absolute uncertainty. Are those the same? So can i write $\Delta p = \langle\Delta p \rangle$? What i know about a particle in a box is its wavefunction $\psi$ from which i can calculate $\langle x\rangle$, $\langle x^2\rangle$, $\langle p\rangle$ and $\langle p^2\rangle$:

\begin{align} &\psi = \sqrt{\frac{2}{d}}\sin\left(\frac{N\pi}{d}x\right) \longleftarrow \substack{\text{I assume that for minimum uncertainty i}\\\text{have to use the ground state for which $N=1$}}\\ \phantom{asd}\\ \phantom{asd}\\ & \left. \begin{aligned} \langle x\rangle&=\frac{d}{2}\\ \langle x^2\rangle&=\frac{1}{3}d^2\left( 1 - \frac{3}{2}\frac{1}{N^2\pi^2} \right)\\ \end{aligned} \right\}~~ \begin{aligned} \Delta x &= \sqrt{\langle x^2\rangle - \langle x \rangle^2}= d\sqrt{\tfrac{1}{12}-\tfrac{1}{2N^2\pi^2}} =\\ &=10fm \sqrt{\tfrac{1}{12}-\tfrac{1}{2\cdot1\cdot\pi^2}} = 1.8fm \end{aligned}\\ \phantom{asd}\\ \phantom{asd}\\ & \left. \begin{aligned} \langle p\rangle&=0\\ \langle p^2\rangle&=\frac{\hbar^2\pi^2}{d^2}N^2\\ \end{aligned} \right\}~~\Delta p = \sqrt{\langle p^2\rangle - \langle p \rangle^2} = \frac{\hbar\pi}{d}N = 3.31\times10^{-20} \frac{kgm}{s}= 61.86MeV/c \end{align}

How can I now calculate $\Delta E_k$ for a proton?

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.