# Homework Help: Neutron Beam Width after localization via a slit in a sheet (QM)

1. Jul 17, 2013

### LowBathWater

1. The problem statement, all variables and given/known data
A parallel beam of neutrons with speed 200m/s is incident on an absorbing sheet with a slit of width 0.01mm. Calculate the width of the beam 10m behind the slit.
Hint The slit localizes the neutrons transversely (y) to their propagation direction x. The resulting ΔPy gives a range of sideways motions, associated with the y uncertainty, superimposed on the x motion.
neutron mass: 1.67x10-27kg
Momentum: P

2. Relevant equations
1) Tn=(h/2pi)2n2/2ma2 explicit kinetic energy calculation for given excitation state n

2) Δy2=a2[(1/3)-(1/2pi2n2)] mean square of positional variance

3) T=mv2/2 kinetic energy

4) T≥(h/2pi)/2m(Δy)2 kinetic energy of confinement demonstrated by uncertainty principle

5) ΔyΔPy≥(h/2pi)/2 uncertainty principle

where n=excitation state, m=mass, a=width of potential well (slit width), v= velocity, T= kinetic energy, Δy=uncertainty in y (y=position in y dimension)

3. The attempt at a solution
i'm currently working through a quantum mechanics primer text and have come to an exercise where I am a little uncertain (no pun to be inferred) whether my reasoning is leading me anywhere close to the general direction of the solution since no solutions are given in the text or online support materiel. The text so far has taken the wave function approach introducing the time-independent Schrodinger equation in one dimension as a postulate then studying its consequences for an infinite square well potential.

Workings

Ty=(h/2pi)2pi2/2(1.67x10-27)(0.01x10-3)2

Ty=mvy2/2

vy=√2Ty/m
=√(h/2pi)2pi2/(1.67x10-27)2(0.01x10-3)2
=(h/2)√1/(1.67x10-27)2(0.01x10-3)2

t=10/200-1/20

d=vy/20

w=2d+0.01(10-3)
=[vy/10]+(0.01x10-3)
=[(h/20)√1/(1.67x10-27)2(0.01x10-3)2]+(0.01x10-3)
=1.99502994x10-3m
≈2mm

EXPLANATION
My initial approach was to model the slit as an infinite potential localizing the particles in the y plane then calculate the kinetic energy of confinement for the ground state via formula 1 in the y direction (I assumed that the wave function would be in its ground state as no excitation, only localization, had occurred). From this I then calculated the potential speed in the y direction via rearranging formula 3. I then calculated that it would take a neutron 1/20th of a second to travel 10 meters. Using this value I calculated the distance d a neutron could travel in the y direction given the previously calculated speed. I reasoned that a neutron could "exit" the slit at either side owing to the nature of the wave function inside the slit and that the associated uncertainty in momentum meant that said particle could have a positive or negative velocity (travelling left or right) of previously calculated magnitude. Therefore particles could travel d either in the positive or negative y direction giving a beam width at ten meters of 2d+0.01mm. Using this method I got a value of approximately 2mm.
I am very uncomfortable with this method as it felt very botched when I came up with it. Firstly I am unsure what happens when the neutrons enter and exit the slit. I assumed their wave functions acted akin to the wave function in a potential well in the y dimension. The phrase, "The resulting ΔPy gives a range of sideways motions, associated with the uncertainty in y" gives me the feeling that that I was supposed to use equation 2 at some point, especially as it was derived in the immediately preceding exercise, because my method doesn't explicitly involve a range of motions. considering inequalities 4 and 5 I can appreciate that if I made and explicit calculation of Δy2 for n=1 I could have a potential range of motions.

2. Jul 17, 2013

### Staff: Mentor

The neutrons are not trapped in any potential well. Those energy states are not relevant (even if I think you can get a similar numerical value with the ground state).

Just use the slit width (or half of it, usually a factor of 2 does not matter here) as position uncertainty, and calculate the corresponding momentum uncertainty. The remaining calculations are classical physics.
Alternatively, you can use the de-Broglie wavelength of those neutrons.

3. Jul 21, 2013

### LowBathWater

Ok thank you. So I use the uncertainty inequality to calculate the associated uncertainty in momentum then I take that value as an estimate for the momentum calculate from that the velocity then find the distance traveled in 1/20th of a second? I should mention at this point that I'm an AS UK student so may be missing a few things in terms of mechanics. I've self taught the M1 syllabus now covering the M2 materiel, looking over the A2 physics syllabus too. However I have already covered the whole A level maths syllabus so there's a lot of the maths behind the quantum that I'm fine with. Just needed to put my difficulties into context. It's riveting!

4. Jul 21, 2013

### Staff: Mentor

Right.

I have no idea what those abbreviations mean, but I guess it is not important.

5. Jul 21, 2013

### LowBathWater

Sorry, forgot this is not a UK forum. They're just maths modules I've taken M for mechanics etc...
Not really important, just stating I'm a bit Naive about things.
Again cheers.