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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?


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?

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