- #1

- 326

- 142

- Homework Statement:
- psb

- Relevant Equations:
- psb

a proton is confined to an infinite potential well of width ##a=8fm##. The proton is in the state

$$\psi(x,0)=\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)+\sqrt{\frac{2}{56}}sin\Big(\frac{2\pi x}{8}\Big)+\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big)$$

(a) What are the values of energy that will be measure and with what probabilities? express in units of MeV. (##mc^2=939MeV, \hbar c=197MeVfm##).

(b) What is the expectation value of the energy?

(c) Use the uncertainty principle to estimate the proton's speed as a function of c.

Solution attempt:

We normalize the wavefunction $$\psi(x,0)=\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)+\sqrt{\frac{2}{56}}sin\Big(\frac{2\pi x}{8}\Big)+\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big)$$

by summing on the squares of the coefficients

$$\Big(N\sqrt{\frac{4}{56}}\Big)^2+\Big(N\sqrt{\frac{2}{56}}\Big)^2+\Big(N\sqrt{\frac{8}{56}}\Big)^2=1\Rightarrow N=2$$

The normalized wavefunction is $$\psi(x,0)=2\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)+2\sqrt{\frac{2}{56}}sin\Big(\frac{2\pi x}{8}\Big)+2\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big)$$

##\hat{H}## operates on the unambiguous states in ##\psi(x,0)## to give ##E_n##.

$$\hat{H}\psi(x,0)=\frac{-\hbar^2}{2m}\frac{d}{dx}\psi(x,0)$$

$$\Rightarrow\frac{-\hbar^2}{2m}\frac{d}{dx}2\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)+\frac{-\hbar^2}{2m}\frac{d}{dx}2\sqrt{\frac{2}{56}}sin\Big(\frac{2\pi x}{8}\Big)+\frac{-\hbar^2}{2m}\frac{d}{dx}2\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big))$$

$$\Rightarrow2\sqrt{\frac{4}{56}}\frac{\pi^2\hbar^2}{128m}sin\Big(\frac{\pi x}{8}\Big)+2\sqrt{\frac{2}{56}}\frac{4\pi^2\hbar^2}{128m}sin\Big(\frac{2\pi x}{8}\Big)+2\sqrt{\frac{8}{56}}\frac{9\pi^2\hbar^2}{128m}sin\Big(\frac{3\pi x}{8}\Big)$$

Energies:

$$E_1=\frac{\pi^2\hbar^2}{128m}=\frac{\pi^2(197MeVfm)^2}{(128fm)^2(939MeV)}=2.5\times10^{-2}MeV$$

$$E_2=\frac{4\pi^2\hbar^2}{128m}=\frac{4\pi^2(197MeVfm)^2}{(128fm)^2(939MeV)}=9.9\times10^{-2}MeV$$

$$E_3=\frac{9\pi^2\hbar^2}{128m}=\frac{9\pi^2(197MeVfm)^2}{(128fm)^2(939MeV)}=2.2\times 10^{-1}MeV$$

Probabilities:

$$P_1=\Big(2\sqrt{\frac{4}{56}}\Big)^2=\frac{16}{56}$$

$$P_2=\Big(2\sqrt{\frac{2}{56}}\Big)^2=\frac{8}{56}$$

$$P_3=\Big(2\sqrt{\frac{8}{56}}\Big)^2=\frac{32}{56}$$

(b) The expectation value of the energy is

$$<\hat{H}>=E_1P_1+E_2P_2+E_3P_3=1.5\times10^{-1}MeV$$

(c) The uncertainty principle states ##\Delta x\Delta p>\frac{\hbar}{2}##.

$$\Delta x =8fm$$. The uncertainty in speed is given by

$$\Delta p =m\Delta v >\frac{\hbar}{2\Delta x }\Rightarrow \Delta v>\frac{\hbar}{2m\Delta x}$$

##V(x)=0## inside of the well. so, the momentum of the proton is ##0##.

$$<\hat{H}>=KE=1.5\times10^{-1}MeV$$

$$KE=\frac{1}{2}mv^2\Rightarrow v=\sqrt{\frac{2\times1.5\times10^{-1}MeV}{2m}}\Rightarrow \sqrt{\frac{2\times1.5\times10^{-1}MeV}{2m}}$$

$$\psi(x,0)=\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)+\sqrt{\frac{2}{56}}sin\Big(\frac{2\pi x}{8}\Big)+\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big)$$

(a) What are the values of energy that will be measure and with what probabilities? express in units of MeV. (##mc^2=939MeV, \hbar c=197MeVfm##).

(b) What is the expectation value of the energy?

(c) Use the uncertainty principle to estimate the proton's speed as a function of c.

Solution attempt:

We normalize the wavefunction $$\psi(x,0)=\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)+\sqrt{\frac{2}{56}}sin\Big(\frac{2\pi x}{8}\Big)+\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big)$$

by summing on the squares of the coefficients

$$\Big(N\sqrt{\frac{4}{56}}\Big)^2+\Big(N\sqrt{\frac{2}{56}}\Big)^2+\Big(N\sqrt{\frac{8}{56}}\Big)^2=1\Rightarrow N=2$$

The normalized wavefunction is $$\psi(x,0)=2\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)+2\sqrt{\frac{2}{56}}sin\Big(\frac{2\pi x}{8}\Big)+2\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big)$$

##\hat{H}## operates on the unambiguous states in ##\psi(x,0)## to give ##E_n##.

$$\hat{H}\psi(x,0)=\frac{-\hbar^2}{2m}\frac{d}{dx}\psi(x,0)$$

$$\Rightarrow\frac{-\hbar^2}{2m}\frac{d}{dx}2\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)+\frac{-\hbar^2}{2m}\frac{d}{dx}2\sqrt{\frac{2}{56}}sin\Big(\frac{2\pi x}{8}\Big)+\frac{-\hbar^2}{2m}\frac{d}{dx}2\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big))$$

$$\Rightarrow2\sqrt{\frac{4}{56}}\frac{\pi^2\hbar^2}{128m}sin\Big(\frac{\pi x}{8}\Big)+2\sqrt{\frac{2}{56}}\frac{4\pi^2\hbar^2}{128m}sin\Big(\frac{2\pi x}{8}\Big)+2\sqrt{\frac{8}{56}}\frac{9\pi^2\hbar^2}{128m}sin\Big(\frac{3\pi x}{8}\Big)$$

Energies:

$$E_1=\frac{\pi^2\hbar^2}{128m}=\frac{\pi^2(197MeVfm)^2}{(128fm)^2(939MeV)}=2.5\times10^{-2}MeV$$

$$E_2=\frac{4\pi^2\hbar^2}{128m}=\frac{4\pi^2(197MeVfm)^2}{(128fm)^2(939MeV)}=9.9\times10^{-2}MeV$$

$$E_3=\frac{9\pi^2\hbar^2}{128m}=\frac{9\pi^2(197MeVfm)^2}{(128fm)^2(939MeV)}=2.2\times 10^{-1}MeV$$

Probabilities:

$$P_1=\Big(2\sqrt{\frac{4}{56}}\Big)^2=\frac{16}{56}$$

$$P_2=\Big(2\sqrt{\frac{2}{56}}\Big)^2=\frac{8}{56}$$

$$P_3=\Big(2\sqrt{\frac{8}{56}}\Big)^2=\frac{32}{56}$$

(b) The expectation value of the energy is

$$<\hat{H}>=E_1P_1+E_2P_2+E_3P_3=1.5\times10^{-1}MeV$$

(c) The uncertainty principle states ##\Delta x\Delta p>\frac{\hbar}{2}##.

$$\Delta x =8fm$$. The uncertainty in speed is given by

$$\Delta p =m\Delta v >\frac{\hbar}{2\Delta x }\Rightarrow \Delta v>\frac{\hbar}{2m\Delta x}$$

##V(x)=0## inside of the well. so, the momentum of the proton is ##0##.

$$<\hat{H}>=KE=1.5\times10^{-1}MeV$$

$$KE=\frac{1}{2}mv^2\Rightarrow v=\sqrt{\frac{2\times1.5\times10^{-1}MeV}{2m}}\Rightarrow \sqrt{\frac{2\times1.5\times10^{-1}MeV}{2m}}$$