Neutrino Oscillations

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Homework Statement


Considering just the electron and the muon neutrino, The Free hamiltonian is written in this basis as:

H = 1/2 (E + Δ Cos[θ]) |Ve><Ve| + 1/2 Δ Sin[θ] |Vu><Ve| + 1/2 Δ Sin[θ]|Ve><Vu| + 1/2 (E - Δ Cos[θ]) |Vu><Vu|

Since H is not diagonal in this basis, they will exhibit oscillations between them. Find the eigenstates |V1> and |V2> along with the eigenvalues E1 and E2.

Homework Equations


Book gives us:
|Ve> = Cos(θ) |V1> - Sin(θ)

|V2> |Vu> = Sin(θ)|V1> +Cos(θ) |V2>

The Attempt at a Solution


I tried to do:
solve for the Eigenvalues using the eq: H-λI = 0

I arrived at two eigenvalues E1=1/2 (Δ +E) and E2=1/2 (E-Δ).

I put these two eigenvalues back in, and solve for the EigenVectors using H|v1> = E2 |v1>.
Then H|v2> = E2 |v2>...

I arrived at:

|v1> = {{(1+Cos(θ))/Sin(θ)},{1}}

|v2> = {{(1+Cos(θ))/Sin(θ)},{1}}

Now I Don't think I made an error in the math per say, but I do feel as though i should have had to do something to the non diagonal H before i attempted to find the eigenvalues?
 

Answers and Replies

  • #2
Orodruin
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You do not have to do anything to the non-diagonal Hamiltonian apart from solving the eigenvector equation correctly (which you have not done - your eigenvalues seem correct, but you cannot have the same eigenvector for two different eigenvalues).

One thing to check is your Hamiltonian which is not on the standard form used in neutrino oscillations. The angle in the Hamiltonian should be ##2\theta## and the sign of ##\Delta## on the diagonal is switched between electron and muon neutrinos.

The quoted eigenvectors are wrong with this definition of the Hamiltonian.
 
  • #3
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You do not have to do anything to the non-diagonal Hamiltonian apart from solving the eigenvector equation correctly (which you have not done - your eigenvalues seem correct, but you cannot have the same eigenvector for two different eigenvalues).

One thing to check is your Hamiltonian which is not on the standard form used in neutrino oscillations. The angle in the Hamiltonian should be ##2\theta## and the sign of ##\Delta## on the diagonal is switched between electron and muon neutrinos.

The quoted eigenvectors are wrong with this definition of the Hamiltonian.
AHHH I actually typed my eigenstates in wrong from what i have here on mathematica!
V2 is supposed to be a minus instead!
And I am not sure what to do about the Hamiltonian issue, as it was provided to us on the problem statement.
I don't quite understand what Δ is supposed to represent in this case, besides something with units of energy?
 
  • #4
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Homework Statement


Considering just the electron and the muon neutrino, The Free hamiltonian is written in this basis as:

H = 1/2 (E + Δ Cos[θ]) |Ve><Ve| + 1/2 Δ Sin[θ] |Vu><Ve| + 1/2 Δ Sin[θ]|Ve><Vu| + 1/2 (E - Δ Cos[θ]) |Vu><Vu|

Since H is not diagonal in this basis, they will exhibit oscillations between them. Find the eigenstates |V1> and |V2> along with the eigenvalues E1 and E2.

Homework Equations


Book gives us:
|Ve> = Cos(θ) |V1> - Sin(θ)|V2>

|Vu> = Sin(θ)|V1> +Cos(θ) |V2>

The Attempt at a Solution


I tried to do:
solve for the Eigenvalues using the eq: H-λI = 0

I arrived at two eigenvalues E1=1/2 (Δ +E) and E2=1/2 (E-Δ).

I put these two eigenvalues back in, and solve for the EigenVectors using H|v1> = E2 |v1>.
Then H|v2> = E2 |v2>...

I arrived at:

|v1> = {{(1+Cos(θ))/Sin(θ)},{1}}

|v2> = {{(-1+Cos(θ))/Sin(θ)},{1}}

Now I Don't think I made an error in the math per say, but I do feel as though i should have had to do something to the non diagonal H before i attempted to find the eigenvalues?
Edit
 
  • #5
Orodruin
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And I am not sure what to do about the Hamiltonian issue, as it was provided to us on the problem statement.

If the problem statement has exactly that Hamiltonian and quotes those eigenstates, then it is simply wrong. This can be easily checked by simply acting on one of the eigenstates with the Hamiltonian and seeing that what you get out is not proportional to the original state. If you change the ##\theta## in the eigenstates to ##\theta/2## it might work out. I suggest you rewrite your eigenstates in terms of ##\theta/2## by using the trigonometric expressions for the double angle (do not forget to normalise the states).

The quantity ##\Delta## is usually given by
$$
\Delta = \frac{\Delta m^2}{2E},
$$
where ##\Delta m^2 = m_2^2 - m_1^2## and ##m_i## is the mass of the ##i##th neutrino mass eigenstate. Heuristically, you can see this by expanding ##E =\sqrt{p^2 + m^2}## in the ultra-relativistic limit, which gives you ##E = p + m^2/(2p)## - and energy and momentum in this limit are essentially interchangeable. But as I said, this given Hamiltonian has some issues with respect to the standard definitions.
 

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