- #1
Normandy
- 2
- 0
Homework Statement
https://dl.dropboxusercontent.com/u/62834965/neutron_interferometer.png
Consider a neutron interferometer (NI), such as the Mach-Zehnder interferometer in the figure.
We send in a beam of neutrons. We assume that the flux of neutrons is so low (neutrons can be very slow) so that only one neutron is present at any time inside the interferometer. The neutrons are initially in an eigenfunction of the momentum with eigenvalue [itex]p = \hbar k[/itex], [itex]\left|\psi\right> = \left|+k\right>[/itex]. The first beamsplitter divides the neutron flux into two parts, that will go into the upper arm or the lower arm with positive or negative momentum. The lower and upper beams are then reflected at the mirrors and recombined at the second beam splitter, after which the neutron flux is measured at one arm. We assume that both beamsplitters work in the same way, delivering an equal flux to each arm (that is, the transmission and reflection are the same).
What is the probability of measuring a neutron with a negative momentum at the location marked 4 in the interferometer?
Homework Equations
If [itex]\left|\psi\right> = a\left|+k\right> + b\left|-k\right>[/itex], then [itex]P(-\hbar k) = \left|b\right|^2[/itex]
The beamsplitter transforms the states by the matrix
[tex]\frac{1}{\sqrt{2}}
\left( \begin{array}{cc}
1 & 1 \\
1 & -1 \end{array} \right)
[/tex]
According to our lecture notes the output at the end of the detector will be [itex]\left|\psi\right> = \cos(\varphi)\left|+k\right> + \sin(\varphi)\left|-k\right>[/itex] (this is a bit sketchy since it depends on
The Attempt at a Solution
I answered 0, since the detector is placed only at the top beam, but the solutions state that the answer is [itex]\sin(\varphi)^2[/itex]. However, this confused me since only neutrons with positive momentum would be able to reach the top detector, even though the positive and negative beams interfere at the second beamsplitter. This differs from the photon spin interferometer experiment since photon spin isn't correlated with the photon beam momentum.
I consulted with my TA but he wasn't able to produce a satisfactory explanation (this is a nuclear engineering class so quantum physics isn't the main focus of the class). I came up with a number of possible scenarios and I was wondering which was the correct resolution to this problem:
- The solutions are correct with no fine print
- The solutions are incorrect and the probability of measuring negative momentum is 0
- The solutions are correct for true momentum eigenstates, since the position uncertainty is infinite and so the location of the detector does not matter, but the solutions are incorrect for "real" states since the position uncertainty is finite and so, far enough away from the second beamsplitter, the beam will split back into a beam of positive momentum neutrons and a beam of negative momentum neutrons.
Last edited by a moderator: