Neutron Spin Time Evolution in Interferometer

[hpar]

Hey Everyone,

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_xt), 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!

 

[fzero]

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.

First, this post should probably be in the HW forum. You've done a great job showing your work, though.

Second, when you compute ##\langle\psi|## you have to take both the transpose and the complex conjugate. When you fix the appropriate sign, you will find that the imaginary part is zero.

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.

Suppose you were asked a different question: how many neutrons are detected in the ##|\alpha\rangle## state? How would you compute that, supposing that the total number of neutrons in the beam was ##N##? Is there another quantity that we could add to this to get the total number of neutrons detected?

 

[hpar]

ok, I get it now I think, Thanks! And should have posted in the HW section.. sorry about that

So for part (a), I just forgot to take the complex conjugate, so our expected value ends up being zero.. Physically how do we interpret this? I mean we have a magnetic field in the x direction, but our magnetic moment's expected value in that direction is zero..

for part (b), if I was to look for the number of neutrons in state alpha, I would take the coefficient squared to get the probability of being in state alpha, so effectively, for all neutrons in all states, we just take ||ψ>|² = (1+cos(ωt))² + sin²(ωt) to get the required answer. That gives us the probability, that we then multiply by N, the total number of neutrons.

Thanks again for your help!

 

[fzero]

So for part (a), I just forgot to take the complex conjugate, so our expected value ends up being zero.. Physically how do we interpret this? I mean we have a magnetic field in the x direction, but our magnetic moment's expected value in that direction is zero..

Classically, the torque on a magnetic dipole is perpendicular to both the magnetic moment and external field.