Hey Everyone,(adsbygoogle = window.adsbygoogle || []).push({});

I'm working on a question and can't quite get the answer out.

QUESTION:

Part (a)

"[itex]\left|\alpha\right\rangle[/itex] and [itex]\left|\beta\right\rangle[/itex] are the eigenfunctions for neutrons polarized respectively along positive and negative z directions. If the neutron, initially in state [itex]\left|\alpha\right\rangle[/itex] at t = 0 is subjected to a constant magnetic field B in the x-direction, show that its wavefunction becomes

[itex]\left|\Psi(t)\right\rangle = \left|\alpha\right\rangle \cos(\omega t) + i\left|\beta\right\rangle \sin(\omega t) [/itex]

where [itex]\omega = \mu B / \hbar[/itex]

Obtain also the expectation values of each of the components of the magnetic moment as a function of time and interpret the results.

Part (b)

In an interferometer, a beam of monoenergetic neutrons initially in state [itex]\left|\alpha\right\rangle[/itex] is split into two separate beams that propagate in different directions, one of which is subjected to a magnetic field as above. The beams travel equal effective path lengths and are then recombined coherently, i.e their wavefunctions are added. Show that the number of neutrons detected in the combined beam is proportional to (1+cos(ωt)).

Attempt at a Solution

The first part is reasonably easy to get out. We can make use of the property:

[itex]

e^{i\theta(\mathbf{n}\cdot \mathbf{\sigma } )} = \mathbf{I} \cos\theta + i (\mathbf{n}\cdot\mathbf{\sigma})\sin\theta [/itex]

The hamiltonian because of a uniform magnetic field has an extra term of the form

[itex]

\widehat{H} = - \mathbf{\mu} \cdot \mathbf{B}

[/itex]

where μ is the magnetic moment which can be related to their definition of ω. We can use the time evolution operator applied on a general state of psi to get the answer they want out. For a magnetic field in the x direction:

[itex]

\mathbf{B} = (B,0,0)

[/itex]

so in the relation we wrote above, we're left with the the time evolution operator U(t) = exp(iω*σ_{x}t), the pauli matrix in the x direction. Theta is Omega. We want a state of the form:

[itex]\left|\Psi(t)\right\rangle = a \left|\alpha\right\rangle + b\left|\beta\right\rangle [/itex]

which can be written as a spinor of the form:

[itex]

\begin{pmatrix}

a\\

b

\end{pmatrix}

[/itex]

and we have the equation:

[itex]

U(t) \left|\Psi(t)\right\rangle = \exp(i\omega\widehat{\sigma_x} t) \begin{pmatrix}

a\\

b

\end{pmatrix}

=

\begin{pmatrix}

\alpha(t)\\

\beta(t)

\end{pmatrix}

[/itex]

Now, solving for alpha and beta, and applying the initial boundary conditions, we get the required result.

[itex]\left|\Psi(t)\right\rangle = \left|\alpha\right\rangle \cos(\omega t) + i\left|\beta\right\rangle \sin(\omega t) [/itex]

I'm struggling with the expected values. If I understand the question properly, we want

<μ> = <ψ|μ|ψ>, and we should have three components, one for each component of μ (the three pauli matrices. So if I want the expected value of μ in the x direction, I would have:

[itex]

\left\langle\psi|\mu_x|\psi\right\rangle = \frac{\gamma \hbar}{2} \begin{pmatrix}

\cos(\omega t) & i\sin(\omega t)

\end{pmatrix}

\begin{pmatrix}

0 & 1 \\

1 & 0

\end{pmatrix}

\begin{pmatrix}

\cos(\omega t)\\

i\sin(\omega t)

\end{pmatrix}

\\

\left\langle\psi|\mu_x|\psi\right\rangle = \frac{\gamma \hbar}{2} (2i\cos(\omega t)\sin(\omega t)) = \frac{i\gamma \hbar}{2} \sin(2\omega t)

[/itex]

and that's an imaginary answer, so it must be wrong.. since an observable has to be real! I'm not quite sure where I've gone wrong and would really appreciate the feedback.

For part (b), this is how far I get:

one half of the beam goes through the same uniform magnetic field as before, so it behaves as we've just derived, and the wavefunction after it has gone through becomes:

[itex]\left|\Psi_B(t)\right\rangle = \left|\alpha\right\rangle \cos(\omega t) + i\left|\beta\right\rangle \sin(\omega t) [/itex]

on the other path, we have a stationary state of the hamiltonian, so the state remains at [itex]\left|\alpha\right\rangle [/itex]. We add the two wavefunctions to get:

[itex]

\left|\chi(t)\right\rangle = \left|\Psi_B(t)\right\rangle + \left|\Psi_0\right\rangle = \left|\alpha\right\rangle \cos(\omega t) + i\left|\beta\right\rangle \sin(\omega t) + \left|\alpha\right\rangle \\

\left|\chi(t)\right\rangle = \left|\alpha\right\rangle(\cos(\omega t)+1) + i\left|\beta\right\rangle \sin(\omega t)

[/itex]

Now, from here I'm not quite sure how to calculate the 'number of neutrons detected'. Does it have anything to do with the flux? Or do we take the probability density? The (1+cosθ) we want is in there, so it seems close.

Would really appreciate anyone's help on this!

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Neutron Spin Time Evolution in Interferometer

Loading...

Similar Threads - Neutron Spin Evolution | Date |
---|---|

B Spin and polarizations in momentum space | Mar 5, 2018 |

I Unnaturalness of Neutron | Dec 11, 2017 |

I Neutron Star Feynman Diagram | Nov 20, 2017 |

I Neutron Star Question | May 20, 2017 |

B Neutron, White Dwarfs and Degeneracy Pressure | Feb 2, 2017 |

**Physics Forums - The Fusion of Science and Community**