QM exercise problem

1. Aug 14, 2011

saim_

A beam of neutrons with energy E runs horizontally into a crystal. The crystal transmits half the neutrons and deflects the other half vertically upwards. After climbing to height H these neutrons are deflected through 90 degrees onto a horizontal path parallel to the originally transmitted beam. The two horizontal beams now move a distance $L$ down the laboratory, one distance $H$ above the other. After going distance $L$, the lower beam is deflected vertically upwards and is finally deflected into the path of the upper beam such that the two beams are co-spatial as they enter the detector. Given that particles in both the lower and upper beams are in states of well-defined momentum, show that the wavenumbers $k$, $k^{′}$ of the lower and upper beams are related by

${\large k \simeq k' \left( 1- \frac{m_{{\small N}}gH}{2E} \right) }$

Attempted Solution:
Let E be the kinetic energy of the neutrons in lower beam.

${\large k = \frac{\sqrt{2mE}}{h}}$

${\large k' \simeq \frac{\sqrt{2m(E - mgh)}}{h}}$

$\Rightarrow {\large \frac{k}{k'} \simeq \sqrt{\frac{2mE}{2m(E-mgh)}} = \sqrt{\frac{E}{E-mgh}} }$

I have no idea how to go beyond this.

2. Aug 14, 2011

xts

Is it a yet another most stupidly unrealistic school excercise?

Could you tell me what is the crystal having magical power of reflecting half of the incoming neutrons in perfectly known direction perpendicular to the beam?

And could you tell me what machine you have able to deflect all neutrons by 90° ?

3. Aug 14, 2011

saim_

Its a problem from a textbook; "The Physics of Quantum Mechanics" by James Binney and David Skinner. I think we are supposed to take some careful assumptions and approximations in arriving at the result instead of actually going into the details of a machine that can do the job.

4. Aug 14, 2011

xts

I don't know that textbook (and it makes me happy!)
I just hate such excercises. Primary school excercise about two balls 1kg each elastically colliding at 1000 m/s. That's why kids hate physics at school. You cannot abstract from mechanisms. Especially in QM - are the "crystals" and "deflectors" point-like or have spanning spatially? It changes the analysis dramatically.

Anyway - the excercise seems to be wrong even under such "primary school simplifications", as there is no reason why two beams have different k's. Unless, of course, "the crystal" executes some magic lifting particles without any effort to a 2nd floor.

5. Aug 14, 2011

saim_

6. Aug 15, 2011

mathfeel

I suppose the only point here is that the upper beam has smaller momentum, to be computed using classical formula, that it has a smaller k also. I think one also assumes nonrelativistic neutron.

7. Aug 15, 2011

xts

But how can they have different momenta as beams finally reach the same place just using different paths in gravitation (potential) field and not being subjected to other forces?
The only explanation for that is magic acting in the crystal.

8. Aug 15, 2011

vela

Staff Emeritus
Hint:$$\sqrt{\frac{E}{E-mgh}} = \sqrt{\frac{1}{1-mgh/E}} = \left(1-\frac{mgh}{E}\right)^{-1/2}$$I think there's a sign error in the result you're asked to find.

9. Aug 15, 2011

saim_

Those assumptions are considered in my attempted solution above. What should I do next?
The final result in the book is neither inverted, nor in the square root like my result (or as you suggest). Also, there is a "2" in the denominator, before E (as given in my first post), of the final result in book but there isn't one in my solution.

I don't think all of these are just mistakes in the book. I haven't encountered a single mistake in the book before this one so it seems highly unlikely it will have so many typos or conceptual errors piled up in a single question all of a sudden.

10. Aug 15, 2011

vela

Staff Emeritus
My post was a hint, not the solution. You still have work to do.
Clearly k'<k since E-mgh<E, yet k=k'[1-mgh/(2E)] implies k'>k.

11. Aug 16, 2011

saim_

Oh, I missed that. Thanks, however, I'm still inclined to believe the book; that k must be less than k' then :D

But assuming there is a sign error and we have a plus sign between the two terms, then what? There is still a no square root in the final term and a "2" in the denominator of one of the terms. Do you see a way we can get any of those in place?

If it helps the book states that the problem is a variant of the setup in "R. Colella et al., 1975, Phys. Rev. Let., 34, 1472" though I didn't get much help skimming through the paper. Here is the link: http://www.atomwave.org/rmparticle/ao%20refs/aifm%20refs%20sorted%20by%20topic/inertial%20sensing%20refs/gravity/COW75%20neutron%20gravity.pdf [Broken]

Last edited by a moderator: May 5, 2017
12. Aug 16, 2011

mathfeel

With gravitational potential, $k_y$ is no longer a good quantum number. So the only reasonable $k$ to compute at all is $k_x$.

13. Aug 16, 2011

vela

Staff Emeritus
Use the binomial expansion.

14. Aug 16, 2011

saim_

Oh! nice.... Now I'm pretty sure that minus sign is a mistake :D Thanks a lot.