News neutrino discovery

Published on October 2nd, 2011 | by Carl Mundy

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Neutrino Oscillations: A Mathematical Introduction

Neutrinos have recently flooded the headlines with a result that may be a hint of hidden extra dimensions, a hint that we may have the wrong view of our universe or even a hint that we don’t know the distance between France and Italy! Today we discuss neutrino oscillation and the history behind it…

These last weeks have forced neutrinos into the spotlight with the suggestion that they could travel faster than the speed of light. If such a suggestion gains any momentum it could overturn our current way of thinking about the universe. On the other hand, it could simply be a miscalculation of the distance between source and detector – that’s science! What the story did do though is widen knowledge about particle physics and in particular the mysterious particles known as neutrinos.

Neutrinos were first theorised by Wolfgang Pauli as a means to conserve several quantities including energy and momentum in beta decay. Since Pauli, two more types of neutrino were theorised and by the very beginning of the twenty-first century all three types of neutrino had been experimentally discovered. From these discoveries, we know that neutrinos are produced in particular types of radioactive decay and nuclear reactions like those that take place inside stars such as our own Sun.

The Solar Neutrino Problem

It was known that nuclear processes inside the Sun produce electron neutrinos and in the late 1960s an experiment was devised to detect and count these neutrinos coming from our Sun. Housed in a gold mine nearly 1.5km underground, a large tank of what was essentially cleaning fluid (chlorine) was used to detect these ghostly particles. The experiment relied on the reaction between an electron neutrino and a chlorine atom which resulted in an electron and an argon atom being produced.

    \[  {\nu}_{e} + {^{37}_{17}Cl} \longrightarrow e^{-} + {^{37}_{18}Ar} \]

The tank was filled with chlorine and left for a few weeks after which the physicists would count how many chlorine atoms had turned into argon. The result was a third of the amount they had predicted they were going to get. At first it was thought there may have been errors in the team’s prediction or indeed a problem with the experiment itself but further experiments and scrutiny revealed that their result was indeed true – neutrinos were missing! This became known as the solar neutrino problem.

One experiment that was sensitive to all three types of neutrino (electron, muon and tau neutrinos) was the Sudbury Neutrino Observatory (SNO) which used heavy water (^{2}H_{2}O) instead of chlorine. The original experiment ran from 1996 until late 2006 but it was results published in 2001 that finally settled the problem of the mysteriously absent solar neutrinos. Their results showed that if they took into account all three types of neutrino then the number they detected agreed with prediction – this meant that the electron neutrinos that originated in the Sun must have somehow changed into one of the other two types on their way to the detector here on Earth. This change is known as neutrino oscillation and the SNO provided the first direct evidence that solar neutrinos oscillated.

A Mathematical Introduction

Using a result from quantum mechanics, some matrices and employing algebra to rearrange things, we can look at how two of the three neutrino states oscillate between each other. In this case we will look at how the electron and muon neutrino can be described, but you could do the following with any two from the three we all know and love.

Firstly, we consider two Hamiltonian eigenstates, \nu_{1} and \nu_{2}, that are a superposition of the two flavour eigenstates (electron and muon flavours) \nu_{e} and \nu_{\mu}. In matrix form, these can be represented as shown below.

    \[  {\left( {\begin{array}{c} \nu_{1}(t) \\ \nu_{2}(t) \\ \end{array} } \right)} = {\left( {\begin{array}{cc} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{array}}\right)}{\left( {\begin{array}{c} \nu_{e}(t) \\ \nu_{\mu}(t) \\ \end{array} } \right)} \]

The above is just another way of writing two equations describing how the Hamiltonian eigenstates oscillate between the electron and muon neutrino states. These involve the parameter \theta which is known as the mixing angle. One other thing we know from quantum mechanics is how an eigenstate develops over time, so we can write down an expression for our two Hamiltonian eigenstates with respect to time.

    \[  \nu_{1}(t) = \nu_{1}(0)e^{-iE_{1}t} \]

    \[  \nu_{2}(t) = \nu_{2}(0)e^{-iE_{2}t} \]

The state at any time t is determined by the initial state (at t = 0) and the state’s energy E_{1,2}. These above expressions can be expressed in matrix form so that we can work with matrices later on.

    \[   {\left( \begin{array}{c} \nu_{1}(t) \\ \nu_{2}(t) \\ \end{array} \right)} = {\left( \begin{array}{cc} e^{-iE_{1}t} & 0 \\ 0 & e^{-iE_{2}t} \\ \end{array} \right)}{\left( \begin{array}{c} \nu_{1}(0) \\ \nu_{2}(0) \\ \end{array} \right)} \]

Now we have these expressions we can begin the long process of substituting, rearranging and rewriting to find an expression for the probability of finding either an electron or muon neutrino. Using these relationships we can say the following:

    \[   {\left( \begin{array}{cc} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{array} \right)}{\left( \begin{array}{c} \nu_{e}(t) \\ \nu_{\mu}(t) \\ \end{array} \right)} = {\left( \begin{array}{cc} e^{-iE_{1}t} & 0 \\ 0 & e^{-iE_{2}t} \\ \end{array} \right)}{\left( \begin{array}{cc} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{array} \right)}{\left( \begin{array}{c} \nu_{e}(0) \\ \nu_{\mu}(0) \\ \end{array} \right)} \]

Next, the inverse of the far left matrix is multiplied from the left on both sides. It must be noted that multiplying any square matrix by it’s inverse gives the identity matrix which, for our purposes here, is essentially the matrix form of the number one – UU^{-1} = U^{-1}U = I. Using this, the following is obtained:

    \[  {\left( \begin{array}{c} \nu_{e}(t) \\ \nu_{\mu}(t) \\ \end{array} \right)} = {\left( \begin{array}{cc} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \\ \end{array} \right)}{\left( \begin{array}{cc} e^{-iE_{1}t} & 0 \\ 0 & e^{-iE_{2}t} \\ \end{array} \right)}{\left( \begin{array}{cc} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{array} \right)}{\left( \begin{array}{c} \nu_{e}(0) \\ \nu_{\mu}(0) \\ \end{array} \right)} \]

Applying our work to the solar neutrino problem, we know that the neutrinos originating from the Sun start off as electron neutrinos. This gives us boundary conditions we can insert into the expression above; namely that \nu_{e}(0) = 1 and \nu_{\mu}(0) = 0 so that we get:

    \[  {\left( \begin{array}{c} \nu_{e}(t) \\ \nu_{\mu}(t) \\ \end{array} \right)} = {\left( \begin{array}{cc} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \\ \end{array} \right)}{\left( \begin{array}{cc} e^{-iE_{1}t} & 0 \\ 0 & e^{-iE_{2}t} \\ \end{array} \right)}{\left( \begin{array}{cc} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{array} \right)}{\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right)} \]

Now taking on the right hand side of the equation, we will attempt to simplify the expression two matrices at a time from the right of the right hand side. See if you can follow.

    \[  {\left( \begin{array}{c} \nu_{e}(t) \\ \nu_{\mu}(t) \\ \end{array} \right)} = {\left( \begin{array}{cc} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \\ \end{array} \right)}{\left( \begin{array}{cc} e^{-iE_{1}t} & 0 \\ 0 & e^{-iE_{2}t} \\ \end{array} \right)}{\left( \begin{array}{cc} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{array} \right)}{\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right)} \]

    \[  {\left( \begin{array}{c} \nu_{e}(t) \\ \nu_{\mu}(t) \\ \end{array} \right)} = {\left( \begin{array}{cc} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \\ \end{array} \right)}{\left( \begin{array}{cc} e^{-iE_{1}t} & 0 \\ 0 & e^{-iE_{2}t} \\ \end{array} \right)}{\left( \begin{array}{c} \cos\theta \\ \sin\theta \\ \end{array} \right)} \]

    \[  {\left( \begin{array}{c} \nu_{e}(t) \\ \nu_{\mu}(t) \\ \end{array} \right)} = {\left( \begin{array}{cc} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \\ \end{array} \right)}{\left( \begin{array}{c} {\cos\theta e^{-iE_{1}t}} \\ {\sin\theta e^{-iE_{2}t}} \\ \end{array} \right)} \]

    \[  {\left( \begin{array}{c} \nu_{e}(t) \\ \nu_{\mu}(t) \\ \end{array} \right)} = {\left( \begin{array}{c} {\cos^{2}\theta e^{-iE_{1}t} + \sin^{2}\theta e^{-iE_{2}t}} \\ {\sin\theta\cos\theta e^{-iE_{2}t} - \sin\theta\cos\theta e^{-iE_{1}t}} \end{array} \right)} \]

Now all that is left are two equations describing the electron and muon neutrino states at any time t. We will take the equation describing the muon neutrino, \nu_{\mu} and replace the exponentials with a sine function as well as using a trigonometric identity to replace the cosines with sine functions.

    \[  \nu_{\mu}(t) = \sin\theta\cos\theta e^{-iE_{2}t} - \sin\theta\cos\theta e^{-iE_{1}t} \]

    \[  \nu_{\mu}(t) = \sin\theta\cos\theta {\left( e^{-iE_{1}t} - e^{-iE_{2}t} \right)} \]

    \[  \nu_{\mu}(t) = {\frac{1}{2} {\sin2\theta}} {\left( {2 \sin \left[ {\frac{(E_{2} - E_{1})t}{2}} \right]} \right)} \]

    \[  \nu_{\mu}(t) = \sin(2\theta) \sin{\left( {\frac{(E_{2} - E_{1})t}{2}} \right)} \]

Now we are in a position to consider the probability that the neutrino is in either the electron or muon neutrino state. The probability is defined as the modulus squared and in the case of our expression for the muon neutrino state is given by |\nu_{\mu}(t)|^{2}.

    \[  {P_{\nu_{\mu}}} = {|\nu_{\mu}(t)|^{2}} = {\sin^{2}{(2\theta)}{\sin^{2}{\left( {\frac{(E_{2} - E_{1})t}{2}} \right)}}} \]

With an expression for the probability, we can begin to make some approximations due to neutrinos being highly relativistic and the fact that they are light compared to their energy. To this end we can say that |\underline{p}| \sim E and t \sim x – in particle physics we deal with natural units where the speed of light is set to 1 (c = 1) as well as other numbers such as the reduced Planck’s constant \hbar. These units mean that the famous equation E^{2} = \underline{p}^{2}c^{2} + m^{2}c^{4}becomes E^{2} = \underline{p}^{2} + m^{2}. From this we can say that:

    \[   E_{2} - E_{1} = \sqrt{\underline{p}^{2} + m_{2}^{2}} - \sqrt{\underline{p}^{2} + m_{1}^{2}} \]

Performing an expansion of the square roots we find this becomes:

    \[  E_{2} - E_{1} \approx {|\underline{p}|(1 + \frac{1}{2}\frac{m_{2}^{2}}{|\underline{p}|^{2}})} - {|\underline{p}|(1 + \frac{1}{2}\frac{m_{1}^{2}}{|\underline{p}|^{2}})} \]

    \[  E_{2} - E_{1} \approx {\frac{m_{2}^{2} - m_{1}^{2}}{2|\underline{p}|}} \]

And using the other approximations mentioned above as well as saying \Delta m^{2} = m_{2}^{2} - m_{1}^{2}, the expression for the muon neutrino probability can be approximated in the form shown below.

    \[  P_{\nu_{\mu}} \approx \sin^{2}(2\theta) \sin^{2}{\left( {\frac{x \Delta m^{2}}{4E}} \right)} \]

As the stuff inside the second sine squared function must be dimensionless there must be another length, which we can call L, that is defined by all the stuff in the sine squared that isn’t the length x we already have. Let us define this as the following:

    \[  P_{\nu_{\mu}} \approx \sin^{2}(2\theta) \sin^{2}{\left( {\frac{\pi x}{L}} \right)} \]

    \[  L = \frac{4 \pi E}{\Delta m^{2}} \]

This length gives us an idea of the kind of distance over which a neutrino oscillates into one of the other types. Note though that as we are dealing with natural units we have to put the speed of light and the reduced Planck’s constant back into our expression for the distance L. To do this we perform dimensional analysis and see where these constants fit in. After doing so we see that the expression for the oscillation scale distance is given by:

    \[  L = \frac{4 \pi E \hbar}{c^{3} \Delta m^{2}} \]

The only step left to do is to input some numbers into this equation and find out the sort of distances we are dealing with.

Test Your Knowledge

Q.: Experiments have shown the square mass difference \Delta m^{2} between the electron and muon neutrino to be on the order of 10-76 kg. If we have an experiment that deals with 8 GeV neutrinos, what is the oscillation scale length in kilometres? How does this relate to the findings of the experiments that lead to the solar neutrino problem?

A.: Plug in the values into the equation for the scale distance!

    \[  L = \frac{(4 \pi)(8 \times 10^{9} \times 1.6 \times 10^{-19})(1.05 \times 10^{-34})}{(3\times 10^{8})^{3} (10^{-76})} \]

    \[  L \sim 6 \times 10^{5} \mathrm{km} \]

Comparing this with the Earth-Sun distance (1.5 x 108 km) we find that the oscillation scale length is much smaller and so the neutrinos that originate from the Sun will be well mixed and so you would expect to find ~1/3 of each flavour, as the first experiment did!

Header image courtesy Argonne National Laboratory.

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About the Author

Astronomy PhD student from the UK with a passion for astronomy and science outreach projects. Involved with weekly science-based radio programme The Science Show on University Radio Nottingham (URN).



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