# Proton in a 1D Box: Energy, Probability, Speed

• docnet
In summary, by normalizing the given wavefunction and operating the Hamiltonian on it, we can find the energies and probabilities associated with the system. The energies are found to be 0.025 MeV, 0.099 MeV, and 0.22 MeV, with corresponding probabilities of 16/56, 8/56, and 32/56. The expectation value of the energy is 0.15 MeV. Using the uncertainty principle, we can estimate the proton's speed to be approximately two orders of magnitude slower than the speed of light, with a minimum speed of 1.11x10^-2c.

#### docnet

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##.
$$<\hat{H}>=KE=1.5\times10^{-1}MeV$$
$$KE=\frac{1}{2}mv^2\Rightarrow v=\sqrt{\frac{2\times1.5\times10^{-1}MeV}{2m}}\Rightarrow \sqrt{\frac{2\times1.5\times10^{-1}MeV}{2m}}$$

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.

rewriting the last part

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 $$\Delta x\Delta p>\frac{\hbar}{2}$$
The well has a dimension of ##8fm##, so $$\Delta x=8fm$$
We define $$\Delta p = m \Delta v$$
So $$\Delta v=\frac{\hbar}{16fm}=1.35\times10^{-2}c$$
Because ##V(x)=0## inside the well, the kinetic energy of the neutron is given by ##KE=<\hat{H}>##. Solving this for ##v##, we have $$v=\sqrt{\frac{2<\hat{H}>}{m}}\Rightarrow .01787c =v_0$$
If we assume that ##\Delta v## is symmetrically distributed around ##v_0##, the minimum speed is given by
$$.01787c-\frac{.0135c}{2}=1.11\times10^{-2}c$$
The speed of the neutron is approximately two orders of magnitude slower than c.

docnet said:
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 $$\psi(x,0)=2\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)+2\sqrt{\frac{2}{56}}sin\Big(\frac{2\pi x}{8}\Big)+2\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big)$$
This is incorrect. (This is not how you normalize a wave function.)

docnet said:
$$\hat{H}\psi(x,0)=\frac{-\hbar^2}{2m}\frac{d}{dx}\psi(x,0)$$
The operator is not correct.

docnet said:
Energies:
$$E_1=\frac{\pi^2\hbar^2}{128m}=\frac{\pi^2(197MeVfm)^2}{(128fm)^2(939MeV)}=2.5\times10^{-2}MeV$$
$$E_2=\frac{4\pi^2\hbar^2}{128m}=\frac{4\pi^2(197MeVfm)^2}{(128fm)^2(939MeV)}=9.9\times10^{-2}MeV$$
$$E_3=\frac{9\pi^2\hbar^2}{128m}=\frac{9\pi^2(197MeVfm)^2}{(128fm)^2(939MeV)}=2.2\times 10^{-1}MeV$$
Why are you suddenly getting ##128^2##?

docnet said:
Probabilities:
$$P_1=\Big(2\sqrt{\frac{4}{56}}\Big)^2=\frac{16}{56}$$
$$P_2=\Big(2\sqrt{\frac{2}{56}}\Big)^2=\frac{8}{56}$$
$$P_3=\Big(2\sqrt{\frac{8}{56}}\Big)^2=\frac{32}{56}$$
I think you approach here is. bit backwards. What are the eigenfunctions of the Hamiltonian? What are the corresponding eigenvalues? What is the probability of finding the system in a given eigenstate?

I haven't checked the rest.

DrClaude said:
I haven't checked the rest.

thank you! please don't do so (at least until I have had a chance to re-work the problem).

We compute ##\int|\psi|^2dx=1## and find the wavefunction $$\psi(x,0)=\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)+\sqrt{\frac{2}{56}}sin\Big(\frac{\pi x}{4}\Big)+\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big)$$
##\hat{H}## operates on the eigenstates of ##\psi(x,0)## to give ##E_n##.
$$\hat{H}\psi(x,0)=\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\psi(x,0)$$
$$\Rightarrow\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\Big[\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)+\sqrt{\frac{2}{56}}sin\Big(\frac{\pi x}{4}\Big)+\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big)\Big]$$
$$\Rightarrow\frac{\pi^2\hbar^2}{128m}\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)+\frac{4\pi^2\hbar^2}{128m}\sqrt{\frac{2}{56}}sin\Big(\frac{\pi x}{4}\Big)+\frac{9\pi^2\hbar^2}{128m}\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big)$$
Energies:
$$E_1=\frac{\pi^2\hbar^2}{128m}=\frac{\pi^2(197MeVfm)^2}{(128fm)(939MeV)}=3.18MeV$$
$$E_2=\frac{4\pi^2\hbar^2}{128m}=\frac{4\pi^2(197MeVfm)^2}{(128fm)(939MeV)}=12.75MeV$$
$$E_3=\frac{9\pi^2\hbar^2}{128m}=\frac{9\pi^2(197MeVfm)^2}{(128fm)(939MeV)}=28.68MeV$$
Probabilities:
$$P_1=\int_0^8\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)\Big[\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)+\sqrt{\frac{2}{56}}sin\Big(\frac{\pi x}{4}\Big)+\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big)\Big]dx=\frac{2}{7}$$
$$P_2=\int_0^8\sqrt{\frac{2}{56}}sin\Big(\frac{\pi x}{4}\Big)\Big[\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)+\sqrt{\frac{2}{56}}sin\Big(\frac{\pi x}{4}\Big)+\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big)\Big]dx=\frac{1}{7}$$
$$P_3=\int_0^8\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big)\Big[\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)+\sqrt{\frac{2}{56}}sin\Big(\frac{\pi x}{4}\Big)+\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big)\Big]dx=\frac{4}{7}$$
(b) The expectation value of the energy is
$$<\hat{H}>=E_1P_1+E_2P_2+E_3P_3=19.12MeV$$
(c) The uncertainty principle states ##\Delta x\Delta p>\frac{\hbar}{2}##. The uncertainty in position is
$$\Delta x =a$$ The uncertainty in momentum is given by
$$\Delta p=2\sqrt{2mE}=\frac{2n\pi\hbar}{a}$$
so
$$\Delta x \Delta p=2n\pi\hbar>\frac{\hbar}{2}$$
we re-write $$\Delta p =m\Delta v =\frac{2n\pi\hbar}{\Delta x }\Rightarrow \Delta v=\frac{2n\pi\hbar}{m\Delta x }$$
The uncertainties in the speed corresponding to each eigenstate are
$$\Delta v_1=.165c$$
$$\Delta v_2=.330c$$
$$\Delta v_3=.495c$$
and overall
$$\Delta v=P_1\Delta v_1+P_2\Delta v_2+P_3\Delta v_3=.380c$$
Since ##V(x)=0## inside of the well, $$<\hat{H}>=KE=19.12MeV$$
$$KE=\frac{1}{2}mv^2\Rightarrow v=\sqrt{\frac{2\times 19.12MeV}{m}}\Rightarrow v=.201c$$
The speed of the neutron is 1 order of magnitude less than C.

docnet said:
We compute ##\int|\psi|^2dx=1## and find the wavefunction $$\psi(x,0)=\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)+\sqrt{\frac{2}{56}}sin\Big(\frac{\pi x}{4}\Big)+\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big)$$
I think you did this before (in a previous homework thread). You should express ##\psi(x, 0)## in terms of the eigenfunctions and you can see more easily that it's normalised: $$\psi(x,0)=\sqrt{\frac{2}{7}} \psi_1(x) +\sqrt{\frac{1}{7}} \psi_2(x) +\sqrt{\frac{4}{7}}\psi_3(x)$$
And you can read off the probabilities as the square of the coefficients.

docnet said:
Energies:
$$E_1=\frac{\pi^2\hbar^2}{128m}=\frac{\pi^2(197MeVfm)^2}{(128fm)(939MeV)}=3.18MeV$$
$$E_2=\frac{4\pi^2\hbar^2}{128m}=\frac{4\pi^2(197MeVfm)^2}{(128fm)(939MeV)}=12.75MeV$$
$$E_3=\frac{9\pi^2\hbar^2}{128m}=\frac{9\pi^2(197MeVfm)^2}{(128fm)(939MeV)}=28.68MeV$$
Looks right.

docnet said:
Probabilities:
(b) The expectation value of the energy is
$$<\hat{H}>=E_1P_1+E_2P_2+E_3P_3=19.12MeV$$
Looks okay. Be careful about rounding errors.

docnet said:
$$KE=\frac{1}{2}mv^2\Rightarrow v=\sqrt{\frac{2\times 19.12MeV}{m}}\Rightarrow v=.201c$$
The speed of the neutron is 1 order of magnitude less than C.
I might have calculated the speed for each energy level and then taken the expected value of that, which comes to about the same. As ##\sqrt{E(v^2)} \ne E(v)##

I'm not sure about using the UP. Note that the uncertainty in momentum for each energy eigentsate is only the direction of the momentum: the magnitude of momentum is determined by the energy.

docnet
docnet said:
We compute ##\int|\psi|^2dx=1## and find the wavefunction $$\psi(x,0)=\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)+\sqrt{\frac{2}{56}}sin\Big(\frac{\pi x}{4}\Big)+\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big)$$
Correct.

docnet said:
Energies:
$$E_1=\frac{\pi^2\hbar^2}{128m}=\frac{\pi^2(197MeVfm)^2}{(128fm)(939MeV)}=3.18MeV$$
$$E_2=\frac{4\pi^2\hbar^2}{128m}=\frac{4\pi^2(197MeVfm)^2}{(128fm)(939MeV)}=12.75MeV$$
$$E_3=\frac{9\pi^2\hbar^2}{128m}=\frac{9\pi^2(197MeVfm)^2}{(128fm)(939MeV)}=28.68MeV$$
This is a bit sloppy, and it is not entirely your fault. The factor 1/8 in the sine functions is actually ##1/a##, so it has units of fm. Note that in what you write for the energies would still have a 1/fm. Technically, the factor 128 is actually ##2 (8a)^2##.

docnet said:
Probabilities:
$$P_1=\int_0^8\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)\Big[\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)+\sqrt{\frac{2}{56}}sin\Big(\frac{\pi x}{4}\Big)+\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big)\Big]dx=\frac{2}{7}$$
$$P_2=\int_0^8\sqrt{\frac{2}{56}}sin\Big(\frac{\pi x}{4}\Big)\Big[\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)+\sqrt{\frac{2}{56}}sin\Big(\frac{\pi x}{4}\Big)+\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big)\Big]dx=\frac{1}{7}$$
$$P_3=\int_0^8\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big)\Big[\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)+\sqrt{\frac{2}{56}}sin\Big(\frac{\pi x}{4}\Big)+\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big)\Big]dx=\frac{4}{7}$$
Probability of being in state ##\phi_n## is not
$$\int \phi_n^*(x) \psi(x) dx$$
but rather the absolute value square of that. Also, you do not have the right wave functions for the eigenstates.

Note that your original approach was not incorrect, as if
$$\psi = c_1 \phi_1 + c_2 \phi_2 + c_3 \phi_3$$
then ##P_1 = | c_1 |^2##. But you had also then used the wrong wave function as an eigenstate.

docnet said:
(b) The expectation value of the energy is
$$<\hat{H}>=E_1P_1+E_2P_2+E_3P_3=19.12MeV$$
Turns out to be correct, even though your equations for the ##P##'s was incorrect.

(c) The uncertainty in position is
$$\Delta x =a$$
[/QUOTE]
I am not sure at what level the question is. This is a crude estimate of ##\Delta x##, which might be ok, but there is a way to calculate ##\Delta x## exactly (although the integrals are not trivial here).

docnet said:
The uncertainty in momentum is given by
$$\Delta p=2\sqrt{2mE}=\frac{2n\pi\hbar}{a}$$
Why does ##E## or ##n## come in?

docnet said:
so
$$\Delta x \Delta p=2n\pi\hbar>\frac{\hbar}{2}$$
we re-write $$\Delta p =m\Delta v =\frac{2n\pi\hbar}{\Delta x }\Rightarrow \Delta v=\frac{2n\pi\hbar}{m\Delta x }$$
The uncertainties in the speed corresponding to each eigenstate are
$$\Delta v_1=.165c$$
$$\Delta v_2=.330c$$
$$\Delta v_3=.495c$$
and overall
$$\Delta v=P_1\Delta v_1+P_2\Delta v_2+P_3\Delta v_3=.380c$$
Since ##V(x)=0## inside of the well, $$<\hat{H}>=KE=19.12MeV$$
$$KE=\frac{1}{2}mv^2\Rightarrow v=\sqrt{\frac{2\times 19.12MeV}{m}}\Rightarrow v=.201c$$
The speed of the neutron is 1 order of magnitude less than C.
I don't get this.

At the basic level where ##\Delta x = a## is acceptable, this was a better answer:
docnet said:
(c) The uncertainty principle states $$\Delta x\Delta p>\frac{\hbar}{2}$$
The well has a dimension of ##8fm##, so $$\Delta x=8fm$$
We define $$\Delta p = m \Delta v$$
So $$\Delta v=\frac{\hbar}{16fm}=1.35\times10^{-2}c$$

docnet
PeroK said:
I think you did this before (in a previous homework thread). You should express ψ(x,0) in terms of the eigenfunctions and you can see more easily that it's normalised: ψ(x,0)=27ψ1(x)+17ψ2(x)+47ψ3(x)
And you can read off the probabilities as the square of the coefficients.
DrClaude said:
you do not have the right wave functions for the eigenstates.

Note that your original approach was not incorrect, as if
ψ=c1ϕ1+c2ϕ2+c3ϕ3
then P1=|c1|2. But you had also then used the wrong wave function as an eigenstate.
Sir, I'm sorry to have to ask what trig identity could one possibly use to manipulate
$$\psi(x,0)=\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)+\sqrt{\frac{2}{56}}sin\Big(\frac{\pi x}{4}\Big)+\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big)$$ into another wave function?
Actually, I thought the above terms are already the energy eigenstates of ##\hat{H}## because
$$\hat{H}\psi_i=E\psi_i$$

edited to add indices on ##\psi##

It doesn't need any trig manipulations, it just needs you to use the known normalisation factor, ##\sqrt{\frac 2 a}##, and save all the unnecessary work to calculate the probabilities.

docnet
docnet said:
Sir, I'm sorry to have to ask what trig identity could one possibly use to manipulate
$$\psi(x,0)=\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)+\sqrt{\frac{2}{56}}sin\Big(\frac{\pi x}{4}\Big)+\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big)$$ into another wave function?
The eigenfunctions in this problem are of the form
$$\phi_n(x) = \sqrt{\frac 2L} \sin \left(\frac{n\pi x}{L}\right)$$ where ##L## is the width of the well. So the second term in the wave function you were given, for example, is really
$$\sqrt{\frac{2}{56}} \sin \left( \frac{\pi x}{4} \right) \to \sqrt{\frac 2{14}} \underbrace{\sqrt{\frac{2}{8~\rm fm}}\sin \left( \frac{2 \pi x}{8~\rm fm} \right)}_{\phi_2(x)}.$$ You can do the same thing with the other terms and express the wave function as a linear combination of the ##\phi_n##'s, which can make the calculations a bit simpler.

docnet and PeroK
docnet said:
Sir, I'm sorry to have to ask what trig identity could one possibly use to manipulate
$$\psi(x,0)=\sqrt{\frac{4}{56}}sin\Big(\frac{\pi x}{8}\Big)+\sqrt{\frac{2}{56}}sin\Big(\frac{\pi x}{4}\Big)+\sqrt{\frac{8}{56}}sin\Big(\frac{3\pi x}{8}\Big)$$ into another wave function?
Actually, I thought the above terms are already the energy eigenstates of ##\hat{H}## because
$$\hat{H}\psi_i=E\psi_i$$

edited to add indices on ##\psi##
In your expression for the wavefunction, let a = 8 units of length. To write the coefficients in the form of the normalization constant ##\sqrt{\frac{2}{a}}##, you have to say $$\sqrt{\frac{4}{56}}=\sqrt{\frac{2}{7}\cdot\frac{2}{8}}=\sqrt{\frac{2}{7}}\sqrt{\frac{2}{a}}.$$Then $$\sqrt{\frac{4}{56}}\sin\Big(\frac{\pi x}{8}\Big)=\sqrt{\frac{2}{7}}\sqrt{\frac{2}{a}}\sin\Big(\frac{\pi x}{a}\Big)=\sqrt{\frac{2}{7}}\psi_1(x).$$Similarly for the other two wavefunctions.

docnet and PeroK
Thank you for explaining how to find the energy eigenstates. Just to wrap up this problem, here is my final try.

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.

We compute ##\int|\psi|^2dx=1## and find the wavefunction $$\psi(x,0)=\sqrt{\frac{1}{4}}\sqrt{\frac{2}{7}}sin\Big(\frac{\pi x}{8}\Big)+\sqrt{\frac{1}{8}}\sqrt{\frac{2}{7}}sin\Big(\frac{\pi x}{4}\Big)+\sqrt{\frac{1}{2}}\sqrt{\frac{2}{7}}sin\Big(\frac{3\pi x}{8}\Big)$$
##\hat{H}## operates on the eigenstates of ##\psi(x,0)## to give ##E_n##

Energies:
$$\hat{H}\psi_1(x,0)\Rightarrow E_1=\frac{\pi^2\hbar^2}{128m}=\frac{\pi^2(197MeVfm)^2}{(128fm^2)(939MeV)}=3.18MeV$$
$$\hat{H}\psi_2(x,0)\Rightarrow E_2=\frac{4\pi^2\hbar^2}{128m}=\frac{4\pi^2(197MeVfm)^2}{(128fm^2)(939MeV)}=12.75MeV$$
$$\hat{H}\psi_3(x,0)\Rightarrow E_3=\frac{9\pi^2\hbar^2}{128m}=\frac{9\pi^2(197MeVfm)^2}{(128fm^2)(939MeV)}=28.68MeV$$
Probabilities:
$$P_1=\frac{2}{7}$$
$$P_2=\frac{1}{7}$$
$$P_3=\frac{4}{7}$$

(b) The expectation value of the energy is
$$<\hat{H}>=E_1P_1+E_2P_2+E_3P_3=19.12MeV$$

(c) The uncertainty principle states $$\Delta x\Delta p>\frac{\hbar}{2}$$
The well has a dimension of ##8fm##, so $$\Delta x=8fm$$
We define $$\Delta p = m \Delta v$$
So $$\Delta v=\frac{\hbar}{16fm}=1.35\times10^{-2}c$$
Since ##V(x)=0## inside of the well, $$<\hat{H}>=KE=19.12MeV$$
$$KE=\frac{1}{2}mv^2\Rightarrow v=\sqrt{\frac{2\times 19.12MeV}{m}}\Rightarrow v=.201c$$

The speed of the neutron is approximately one order of magnitude slower than c.

As above, I don't understand what the UP has to do with this. You know the superposition of energy eigenstates, hence you know the possible energies, momenta and speeds you may measure. "The" speed of the particle is not well-defined beyond that.

Also, as pointed out, the expected value of the speed cannot be calculated the way you have calculated it.
PeroK said:
##\sqrt{E(v^2)} \ne E(v)##

docnet
PeroK said:
As above, I don't understand what the UP has to do with this. You know the superposition of energy eigenstates, hence you know the possible energies, momenta and speeds you may measure. "The" speed of the particle is not well-defined beyond that.

Also, as pointed out, the expected value of the speed cannot be calculated the way you have calculated it.
Part (c) asks “Use the uncertainty principle to estimate the proton’s speed as a fraction of c.”

docnet
Thank you for the pointers, for the last time.

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.

We compute ##\int|\psi|^2dx=1## and find the wavefunction $$\psi(x,0)=\sqrt{\frac{1}{4}}\sqrt{\frac{2}{7}}sin\Big(\frac{\pi x}{8}\Big)+\sqrt{\frac{1}{8}}\sqrt{\frac{2}{7}}sin\Big(\frac{\pi x}{4}\Big)+\sqrt{\frac{1}{2}}\sqrt{\frac{2}{7}}sin\Big(\frac{3\pi x}{8}\Big)$$
##\hat{H}## operates on the eigenstates of ##\psi(x,0)## to give ##E_n##

Energies:
$$\hat{H}\psi_1(x,0)\Rightarrow E_1=\frac{\pi^2\hbar^2}{128m}=\frac{\pi^2(197MeVfm)^2}{(128fm^2)(939MeV)}=3.18MeV$$
$$\hat{H}\psi_2(x,0)\Rightarrow E_2=\frac{4\pi^2\hbar^2}{128m}=\frac{4\pi^2(197MeVfm)^2}{(128fm^2)(939MeV)}=12.75MeV$$
$$\hat{H}\psi_3(x,0)\Rightarrow E_3=\frac{9\pi^2\hbar^2}{128m}=\frac{9\pi^2(197MeVfm)^2}{(128fm^2)(939MeV)}=28.68MeV$$
Probabilities:
$$P_1=\frac{2}{7}$$
$$P_2=\frac{1}{7}$$
$$P_3=\frac{4}{7}$$

(b) The expectation value of the energy is
$$<\hat{H}>=E_1P_1+E_2P_2+E_3P_3=19.12MeV$$

(c) The uncertainty principle states $$\Delta x\Delta p>\frac{\hbar}{2}$$
The well has a dimension of ##8fm##, so $$\Delta x=8fm$$
We define $$\Delta p = m \Delta v$$
So $$\Delta v=\frac{\hbar}{16fm}=1.35\times10^{-2}c$$
Since ##V(x)=0## inside of the well, $$<\hat{H}>=KE=19.12MeV$$
$$KE=\frac{1}{2}mv^2\Rightarrow P_1\sqrt{\frac{2\times 3.18MeV}{m}}+P_2\sqrt{\frac{2\times 12.75MeV}{m}}+P_3\sqrt{\frac{2\times 28.68 MeV}{m}} = v\approx .201c$$

The speed of the neutron is approximately one order of magnitude slower than c.

kuruman said:
Part (c) asks “Use the uncertainty principle to estimate the proton’s speed as a fraction of c.”
The proton has no well-defined "speed". It's difficult to make sense of that quantum mechanically. The three energy levels involved are widely spaced.

docnet
In connection to this, it's also a good exercise to think about, whether a particle in the infinite potential well has a well-defined momentum operator (it's more subtle than you think at the first glance!).

docnet
vanhees71 said:
In connection to this, it's also a good exercise to think about, whether a particle in the infinite potential well has a well-defined momentum operator (it's more subtle than you think at the first glance!).

The momentum operator is ##-i\hbar \frac{d}{dx}## which gives a clear and unambiguous answer for the momentum ##p##. The uncertainty principle does not prohibit me from measuring the momentum ##p## in a deterministic manner. So my question is, what is the meaning of "well-defined"?

That's what you think! As I said, the answer is more subtle. Of course, since the momentum operator is defined as the generator of spatial translations, it should be the operator you wrote down, but it's also a question, whether the operator is well-defined on a dense subspace of the Hilbert space of wave functions appropriate to your situation and that it is also a self-adjoint operator. Think about the domain and co-domain of your assumed momentum operator!

docnet
vanhees71 said:
That's what you think! As I said, the answer is more subtle. Of course, since the momentum operator is defined as the generator of spatial translations, it should be the operator you wrote down, but it's also a question, whether the operator is well-defined on a dense subspace of the Hilbert space of wave functions appropriate to your situation and that it is also a self-adjoint operator. Think about the domain and co-domain of your assumed momentum operator!

Can you please explain the statement "the momentum operator generator of spatial translations"? - Does this statement refer to the "space of states" and not x-y-z space?

What makes a subspace "dense" if i may ask?

From wiki:

If I found the correct definition of self-adjoint, then the momentum operator ##\hat{P}## is not a self-adjoint operator, as each operation of ##\hat{P}## on ##\psi = Nsin\Big(\frac{n\pi x}{a}\Big)## multiplies a factor of ##\frac{n\pi}{a}## so ##\hat{P}\hat{P}\neq I##. It might be worth noting that operating with ##\hat{P}## changes the sine to a cosine, so ##\psi## is not an eigenvalue eigenvector eigenfunction of ##\hat{P}##?

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You are on the right track! To see that there's trouble with the momentum operator, it's already enough to study its action on a complete set of orthonormal functions (CONS), and your energy eigenstates are such a CONS. Of course these are not eigenstates of ##\hat{P}##, but that's not so important. Rather, think about, whether ##\hat{P} \psi_n## is an "allowed" wave function for your particle in the box!

docnet
vanhees71 said:
You are on the right track! To see that there's trouble with the momentum operator, it's already enough to study its action on a complete set of orthonormal functions (CONS), and your energy eigenstates are such a CONS. Of course these are not eigenstates of ##\hat{P}##, but that's not so important. Rather, think about, whether ##\hat{P} \psi_n## is an "allowed" wave function for your particle in the box!
If
$$\psi_n=Nsin\Big(\frac{\pi x}{a}\Big)+Nsin\Big(\frac{2\pi x}{a}\Big)+...Nsin\Big(\frac{n\pi x}{a}\Big)$$
Then
$$\hat{P} \psi_n=-i\hbar\Big[C_1cos\Big(\frac{\pi x}{a}\Big)+C_2cos\Big(\frac{2\pi x}{a}\Big)+...C_ncos\Big(\frac{n\pi x}{a}\Big)\Big]$$ is not an allowed wave function in the particle in the box because cosine functions do not satisfy the boundary condition ##\psi=0## at ##x=0##

vanhees71
@vanhees71,

Thank you so much for the help learning these concepts.

If you have time, could you please explain briefly how the cosine result violates spatial translations, and also why momentum operators are expected to be generators of spatial translations to begin with?

regarding spatial translations, I'm assuming because cosine functions are not included in the complete set of orthonormal functions which are only sines.

are there any 1d potential well operators are generators of spatial translations?

Also, how is self adjoint related to being well-defined in the space of states?

Thanks again

That's a lot of stuff to be discussed. Perhaps, it's better to first look at some nice pedagogical papers on this issue, which is rarely discussed in standard textbooks on QM but I think can be important to see where the pitfalls with the somewhat sloppy use of mathematics by us theoretical physics are ;-)):

https://arxiv.org/abs/quant-ph/9907069
https://arxiv.org/abs/quant-ph/0103153

docnet
vanhees71 said:
That's a lot of stuff to be discussed. Perhaps, it's better to first look at some nice pedagogical papers on this issue, which is rarely discussed in standard textbooks on QM but I think can be important to see where the pitfalls with the somewhat sloppy use of mathematics by us theoretical physics are ;-)):

https://arxiv.org/abs/quant-ph/9907069
https://arxiv.org/abs/quant-ph/0103153
Have you considered writing an Insight about the problems with the particle in a box?

DrClaude, vanhees71 and docnet