- #1
muppet
- 602
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Hi all,
I'm being a bit thick in following some introductory lectures on the Standard Model: http://hepwww.rl.ac.uk/hepsummerschool/Teubner-%20Standard%20Model%202008.pdf
The thing I'm struggling to understand is the treatment of how "box" diagrams contribute to B-Bbar mixing (see page 70 of that link). My attempt to understand it would be the following. The box diagram adds terms that couple B, Bbar to the effective Hamiltonian, so that the mass matrix becomes of the form
\left(\begin{array} \<br /> M_{B} & \Delta M \\<br /> (\Delta M )^* & M_{B} \end{array} \right)
where \Delta M \sim (V_{tb}V_{td}^{*})^2 is the complex contribution from the box diagram. The eigenvalues of this matrix sum to 2 M_B and their product is M_{B}^2-|\Delta M |^2=(M_B - |\Delta M|)(M_B + |\Delta M|). So define \Delta m =2 |\Delta M|, and the eigenvalues of this matrix must be M_B \pm \frac{1}{2} \Delta m.
where the minus sign corresponds to the heavy mass eigenstate and the plus sign to the lighter eigenstate (because (A-\lambda I)v=0).
The eigenvectors of this matrix should then lie in the kernel of
\left(\begin{array} \<br /> \mp |\Delta M| & \Delta M \\<br /> (\Delta M )^* & \mp |\Delta M| \end{array} \right)
If, for a moment, we take \Delta M to be real, I then think that the heavy normalised eigenstate is (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}) and the lighter one is (\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}), which is the other way round to what the notes suggest. This is problem number one that I have. (Does [\itex] \Delta M [/itex] turn out to be negative or something?)
Problem number two is that if I write \Delta M= |\Delta M|e^{i \alpha}, so that \alpha is the phase of the box diagram, then I think my mass eigenstates are solutions of
|\Delta M| \left(\begin{array} \<br /> \mp 1 & e^{i \alpha} \\<br /> e^{-i \alpha} & \mp 1\end{array} \right) \left(\begin{array} \ p \\ q \end{array} \right)=0
so that normalised eigenvectors will look like \frac{1}{\sqrt{2}}(1,\pm e^{i \alpha})
-i.e. that the phase of the diagram is the ratio p/q. But according to the notes, p/q is the square root of the phase of the diagram. What am I doing wrong?
Thanks in advance.
I'm being a bit thick in following some introductory lectures on the Standard Model: http://hepwww.rl.ac.uk/hepsummerschool/Teubner-%20Standard%20Model%202008.pdf
The thing I'm struggling to understand is the treatment of how "box" diagrams contribute to B-Bbar mixing (see page 70 of that link). My attempt to understand it would be the following. The box diagram adds terms that couple B, Bbar to the effective Hamiltonian, so that the mass matrix becomes of the form
\left(\begin{array} \<br /> M_{B} & \Delta M \\<br /> (\Delta M )^* & M_{B} \end{array} \right)
where \Delta M \sim (V_{tb}V_{td}^{*})^2 is the complex contribution from the box diagram. The eigenvalues of this matrix sum to 2 M_B and their product is M_{B}^2-|\Delta M |^2=(M_B - |\Delta M|)(M_B + |\Delta M|). So define \Delta m =2 |\Delta M|, and the eigenvalues of this matrix must be M_B \pm \frac{1}{2} \Delta m.
where the minus sign corresponds to the heavy mass eigenstate and the plus sign to the lighter eigenstate (because (A-\lambda I)v=0).
The eigenvectors of this matrix should then lie in the kernel of
\left(\begin{array} \<br /> \mp |\Delta M| & \Delta M \\<br /> (\Delta M )^* & \mp |\Delta M| \end{array} \right)
If, for a moment, we take \Delta M to be real, I then think that the heavy normalised eigenstate is (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}) and the lighter one is (\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}), which is the other way round to what the notes suggest. This is problem number one that I have. (Does [\itex] \Delta M [/itex] turn out to be negative or something?)
Problem number two is that if I write \Delta M= |\Delta M|e^{i \alpha}, so that \alpha is the phase of the box diagram, then I think my mass eigenstates are solutions of
|\Delta M| \left(\begin{array} \<br /> \mp 1 & e^{i \alpha} \\<br /> e^{-i \alpha} & \mp 1\end{array} \right) \left(\begin{array} \ p \\ q \end{array} \right)=0
so that normalised eigenvectors will look like \frac{1}{\sqrt{2}}(1,\pm e^{i \alpha})
-i.e. that the phase of the diagram is the ratio p/q. But according to the notes, p/q is the square root of the phase of the diagram. What am I doing wrong?
Thanks in advance.