# I Matrix diagonalisation

1. Feb 6, 2017

### spaghetti3451

I want to find the orthogonal matrix $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ which diagonalises the matrix $\begin{pmatrix} 0 & m\\ m & M \end{pmatrix}$.

The eigenvalues are easily found to be $\lambda = \frac{M}{2} \pm \frac{1}{2}\sqrt{M^{2}+4m^{2}}.$

However, I am having trouble finding the eigenvectors. I have the eigenvector equation $\begin{pmatrix} 0 & m\\ m & M \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \lambda \begin{pmatrix} a \\ b \end{pmatrix},$

which gives me $mb = \left( \frac{M}{2} \pm \frac{1}{2}\sqrt{M^{2}+4m^{2}} \right)a$ and $ma+Mb = \left( \frac{M}{2} \pm \frac{1}{2}\sqrt{M^{2}+4m^{2}} \right)b.$

Could you help me out here? The answer's supposed to be $\cos \theta = \frac{1}{2} \arctan \frac{2m}{M}$.

Last edited: Feb 6, 2017
2. Feb 6, 2017

### Orodruin

Staff Emeritus
You do not need to find the eigenvectors. There is only one off-diagonal element in $\mathcal M_d = U\mathcal M U$. Equating it to zero will give you the mixing directly.

3. Feb 6, 2017

### spaghetti3451

How do you know that there is only one off-diagonal element?

4. Feb 6, 2017

### Orodruin

Staff Emeritus
Your matrix is a symmetric 2x2 matrix ...

5. Feb 6, 2017

### spaghetti3451

Okay, so I have

$\mathcal{M}_{d} = \mathcal{U}\mathcal{M}\mathcal{U}$

$= \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} 0 & m\\ m & M \end{pmatrix} \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$

$= \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} m \sin\theta & m \cos\theta \\ m \cos \theta + M \sin \theta & -m \sin\theta + M \cos\theta \end{pmatrix}$

$= \begin{pmatrix} -M\sin^{2}\theta & m - M \sin\theta\cos\theta \\ m + M \sin\theta\cos\theta & M\sin^{2}\theta \end{pmatrix} .$

This is where my problem lies.

6. Feb 6, 2017

### Orodruin

Staff Emeritus
Sorry, it is $U\mathcal M U^T$, not $U\mathcal M U$. Writing on a phone has its drawbacks.

7. Feb 6, 2017

### spaghetti3451

Ah! Of course! But let me give you a bit of a background to this.

Consider the following scalar quantity: $\begin{pmatrix} \chi_{2}^{\dagger} & \chi_{1}^{\dagger} \end{pmatrix} \begin{pmatrix} 0 & m \\ m & M \end{pmatrix} \begin{pmatrix} \chi_{1} \\ \chi_{2} \end{pmatrix}.$

My goal is to diagonalise the middle matrix by a change of basis to $\begin{pmatrix} \psi_{1} \\ \psi_{2} \end{pmatrix}$ such that $\begin{pmatrix} \chi_{1} \\ \chi_{2} \end{pmatrix} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} \psi_{1} \\ \psi_{2} \end{pmatrix}$.

Now,

$\begin{pmatrix} \chi_{1} \\ \chi_{2} \end{pmatrix}^{\dagger} = \begin{pmatrix} \psi_{1} \\ \psi_{2} \end{pmatrix}^{\dagger} \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}^{\dagger}$ so that $\begin{pmatrix} \chi_{1}^{\dagger} & \chi_{2}^{\dagger} \end{pmatrix} = \begin{pmatrix} \psi_{1}^{\dagger} & \psi_{2}^{\dagger} \end{pmatrix}\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$

Therefore,

$\begin{pmatrix} \chi_{2}^{\dagger} & \chi_{1}^{\dagger} \end{pmatrix} \begin{pmatrix} 0 & m \\ m & M \end{pmatrix} \begin{pmatrix} \chi_{1} \\ \chi_{2} \end{pmatrix} = \begin{pmatrix} \psi_{2}^{\dagger} & \psi_{1}^{\dagger} \end{pmatrix}\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} 0 & m \\ m & M \end{pmatrix} \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} \psi_{1} \\ \psi_{2} \end{pmatrix}$

Therefore, I end up finding $\mathcal{U}\mathcal{M}\mathcal{U}$.

What should I do?

8. Feb 6, 2017

### Orodruin

Staff Emeritus
I do not think I need the background story. I am a neutrino physicist after all.

Where did you get your initial equation? It is not how it is supposed to look. The Dirac mass terms should be the ones connecting different fields and in your case they are not.

9. Feb 6, 2017

### spaghetti3451

I guess this thread ought to be moved to the Standard Model forum.

This is in page 105 of Cliff Burgess's textbook on the Standard Model.

You can ignore the following parts in italics, if you'd like to. The crux of what I have written below is in my next post.

The idea is to suppose that a right-handed neutrino for each generation (invariant under $SU_{c}(3) \times SU_{L}(2) \times U_{Y}(1)$) is added to the standard model.

Then the only new renormalizable terms that can appear in the Lagrangian are (also rewriting the kinetic term for the left-handed leptons):

$\mathcal{L} = - \frac{1}{2}\bar{L}_{m}\gamma^{\mu}D_{\mu}L_{m} - \frac{1}{2}\bar{N}_{m}\gamma^{\mu}\partial_{\mu}N_{m} - \frac{1}{2}M_{m}\bar{N}_{m}N_{m} - (k_{mn}\bar{L}_{m}P_{R}N_{n}\tilde{\phi} + \text{h.c.})$

where $N_m$ is the Majorana spinor whose right-handed piece is the right-handed neutrino and $L_m$ is the usual lepton doublet. $M_m$ is a real mass parameter and $k_{mn}$ are Yukawa coupling constants.

Subsituting $\tilde{\phi} \rightarrow \frac{1}{\sqrt{2}}\begin{pmatrix} v \\ 0 \end{pmatrix}$ into the Yukawa interaction, we can show that the neutrino mass terms are:

$- \frac{1}{2}M_{m}\bar{N}_{m}N_{m} - \frac{v}{\sqrt{2}}k_{mn}\left(\bar{\nu}_{m} P_{R}N_{n} + \bar{N}_{n}P_{L}\nu_{m}\right)$.

Rewriting $\displaystyle{\frac{v}{\sqrt{2}}k_{mn}=m_{mn}}$, we get for the neutrino mass terms

$- \frac{1}{2}\left( M_{m}\bar{N}_{m}N_{m} + 2m_{mn}\bar{\nu}_{m} P_{R}N_{n} + 2m_{mn}\bar{N}_{n}P_{L}\nu_{m}\right)$

$- \frac{1}{2}\left( M_{m}\bar{N}_{m}N_{m} + m_{mn}\bar{\nu}_{m}N_{n} + m_{mn}\bar{N}_{n}\nu_{m}\right)$.

In the last line, I used the identities $\bar{\nu}_{m}P_{R}N_{n} = \bar{N}_{n}P_{R}\nu_{m}$ and $\bar{N}_{n}P_{L}\nu_{m} = \bar{\nu}_{m}P_{L}N_{n}$.

This is how I arrived at the expression in my previous post.

Last edited: Feb 6, 2017
10. Feb 6, 2017

### spaghetti3451

The crux of the previous post is that I have something like

$- \frac{1}{2} \begin{pmatrix} \bar{\nu} & \bar{N} \end{pmatrix} \begin{pmatrix} 0 & m \\ m & M \end{pmatrix} \begin{pmatrix} \nu \\ N \end{pmatrix}$

$=- \frac{1}{2} \begin{pmatrix} \nu^{\dagger} & N^{\dagger} \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & m \\ m & M \end{pmatrix} \begin{pmatrix} \nu \\ N \end{pmatrix}$

$=- \frac{1}{2} \begin{pmatrix} N^{\dagger} & \nu^{\dagger} \end{pmatrix} \begin{pmatrix} 0 & m \\ m & M \end{pmatrix} \begin{pmatrix} \nu \\ N \end{pmatrix}$

I am pretty sure my first line is correct, but my third line is wrong.

Where do you think is my mistake?

11. Feb 6, 2017

### Orodruin

Staff Emeritus
Why did you switch the order of $\nu$ and $N$? They are not the parts of a single Dirac spinor, they are separate fields. You need to be careful which indices different operators act on.

12. Feb 6, 2017

### spaghetti3451

Okay, so with $\begin{pmatrix} \nu \\ N \end{pmatrix} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} \psi_{1} \\ \psi_{2} \end{pmatrix}$,

I need to figure out why

$\begin{pmatrix} \bar{\nu} & \bar{N} \end{pmatrix} = \begin{pmatrix} \bar{\psi}_{1} & \bar{\psi}_{2} \end{pmatrix} \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$.

I guess the reason is that the gamma matrices act on the internal spinor indices and not on the indices that define the column or row vector. Is this why we can treat $\bar{\nu}$ as $\nu^{\dagger}$ (and $\bar{N}$ as $N^{\dagger}$) for the purpose of moving the rotation matrix past $\begin{pmatrix} \bar{\nu} & \bar{N} \end{pmatrix}$?

13. Feb 6, 2017

### Orodruin

Staff Emeritus
Yes.

14. Feb 6, 2017

### spaghetti3451

Ah! I see!

Is the following then correct?

$\begin{pmatrix} \bar{\nu} & \bar{N} \end{pmatrix}$

$= \begin{pmatrix} \nu^{\dagger} & N^{\dagger} \end{pmatrix} \begin{pmatrix} \gamma^{0} & 0 \\ 0 & \gamma^{0} \end{pmatrix}$

$= \begin{pmatrix} \nu \\ N \end{pmatrix}^{\dagger} \begin{pmatrix} \gamma^{0} & 0 \\ 0 & \gamma^{0} \end{pmatrix}$

$= \left[ \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} \psi_{1} \\ \psi_{2} \end{pmatrix} \right] ^{\dagger} \begin{pmatrix} \gamma^{0} & 0 \\ 0 & \gamma^{0} \end{pmatrix}$

$= \begin{pmatrix} \psi_{1} \\ \psi_{2} \end{pmatrix}^{\dagger} \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}^{\dagger} \begin{pmatrix} \gamma^{0} & 0 \\ 0 & \gamma^{0} \end{pmatrix}$

$= \begin{pmatrix} \psi_{1}^{\dagger} & \psi_{2}^{\dagger} \end{pmatrix} \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} \gamma^{0} & 0 \\ 0 & \gamma^{0} \end{pmatrix}$

$= \begin{pmatrix} \psi_{1}^{\dagger} & \psi_{2}^{\dagger} \end{pmatrix} \begin{pmatrix} \gamma^{0} & 0 \\ 0 & \gamma^{0} \end{pmatrix} \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$

$= \begin{pmatrix} \bar{\psi_{1}} & \bar{\psi_{2}} \end{pmatrix} \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}.$

15. Feb 6, 2017

### Orodruin

Staff Emeritus
Yes, although there really is no need to write out the gammas. I would go directly to the final expression. The argument that the gammas act on different indicesis sufficient.

16. Feb 6, 2017

### spaghetti3451

Okay, so I have

$\mathcal{M}_{d} = \mathcal{U}^{T}\mathcal{M}\mathcal{U}$

$= \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} 0 & m\\ m & M \end{pmatrix} \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$

$= \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} m \sin\theta & m \cos\theta \\ m \cos \theta + M \sin \theta & -m \sin\theta + M \cos\theta \end{pmatrix}$

$= \begin{pmatrix} 2m\sin\theta\cos\theta + M\sin^{2}\theta & m\cos^{2}\theta - m\sin^{2}\theta + M \sin\theta\cos\theta \\ m\cos^{2}\theta - m\sin^{2}\theta + M \sin\theta\cos\theta & -2m\sin\theta\cos\theta + M\cos^{2}\theta \end{pmatrix}.$

So,

$m\cos^{2}\theta - m\sin^{2}\theta + M \sin\theta\cos\theta = 0$

$2m\cos(2\theta) + M \sin(2\theta) = 0$

$\displaystyle{\tan(2\theta) = - \frac{2m}{M}}$

which contradicts with what I expect to have: $\displaystyle{\tan{\cos2\theta}}=\frac{2m}{M}$.

Have I made a mistake somewhere?

17. Feb 6, 2017

### Orodruin

Staff Emeritus
Yes, you should not be expecting that result.

18. Feb 6, 2017

### spaghetti3451

Thank you!

19. Feb 6, 2017

### spaghetti3451

One final question.

If you have a term like $\bar{N}\gamma^{\mu}\partial_{\mu}N$, how do you apply the same above unitary transformation with the same condition $\tan2\theta = - \frac{2m}{M}$ to get $\bar{\psi}_{1}\gamma^{\mu}\partial_{\mu}\psi_{1}+\bar{\psi}_{2}\gamma^{\mu}\partial_{\mu}\psi_{2}$?

I have the following:

$- \frac{1}{2}\bar{N}\gamma^{\mu}{\partial_{\mu}}N$

$= - \frac{1}{2} \begin{pmatrix} \bar{\nu} & \bar{N} \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & \gamma^{\mu}{\partial_{\mu}} \end{pmatrix} \begin{pmatrix} \nu \\ N \end{pmatrix}$

$= - \frac{1}{2} \begin{pmatrix} \bar{\psi_{1}} & \bar{\psi_{2}} \end{pmatrix} \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & \gamma^{\mu}{\partial_{\mu}} \end{pmatrix} \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} \psi_{1} \\ \psi_{2} \end{pmatrix}$

But this gives me a cross term in $\psi_{1}$ and $\psi_{2}$ and I don't quite obtain canonical kinetic terms in $\psi_{1}$ and $\psi_{2}$.

20. Feb 6, 2017

### Orodruin

Staff Emeritus
You don't. You need to have the kinetic term for the light neutrino as well.