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Accidently
Sep8-11, 12:30 AM
I have a puzzle when I study the hybrid inflation model.

Suppose we have two scalar fields, \phi_1 and \phi_2
first, lets consider the situation where they are in their independent potentials
V(\phi_i)=m_i^2\phi_i^2, i = 1,2
with initial value
\phi_i^{ini}
We can solve the scalar dynamic equations for them. And they are both in harmonic oscillation. This is Okay.

But when a 'mixing term' \lambda^2 \phi_1\phi_2 is introduced, \phi_1 and \phi_2 get infinite values, if \lambda is large. This can be showed numerically. What I thought is the large mixing term would lead to \phi_1 = \phi_2. So why it goes to infinite?

And we can rotate \phi_1 and \phi_2 to a basis where there is no mixing term. In this basis, we would not get infinite values for \phi_1 or \phi_2. So it seems I get a different result working in different basis. What is the problem

Chronos
Sep8-11, 01:00 AM
You are confusing scalar quantities with vector quantities.

Chalnoth
Sep9-11, 04:19 AM
I have a puzzle when I study the hybrid inflation model.

Suppose we have two scalar fields, \phi_1 and \phi_2
first, lets consider the situation where they are in their independent potentials
V(\phi_i)=m_i^2\phi_i^2, i = 1,2
with initial value
\phi_i^{ini}
We can solve the scalar dynamic equations for them. And they are both in harmonic oscillation. This is Okay.

But when a 'mixing term' \lambda^2 \phi_1\phi_2 is introduced, \phi_1 and \phi_2 get infinite values, if \lambda is large. This can be showed numerically. What I thought is the large mixing term would lead to \phi_1 = \phi_2. So why it goes to infinite?

And we can rotate \phi_1 and \phi_2 to a basis where there is no mixing term. In this basis, we would not get infinite values for \phi_1 or \phi_2. So it seems I get a different result working in different basis. What is the problem
How large are we talking? I don't think you can go above \lambda^2 = m_1^2 + m_2^2 and have sensible results.

Accidently
Sep14-11, 08:51 AM
You are confusing scalar quantities with vector quantities.

do you mean scalars can not mix? I thought about that. But my understanding is two fields can mix if they have exactly the same quantum number.

Accidently
Sep14-11, 08:59 AM
How large are we talking? I don't think you can go above \lambda^2 = m_1^2 + m_2^2 and have sensible results.

The limit sounds reasonable. But why do we have this limit? Unfortunately, I am consider some process which can go beyond this limit (for example, a fast scattering between the two scalars, bringing the two fields to equilibrium.)

Chalnoth
Sep14-11, 10:12 AM
The limit sounds reasonable. But why do we have this limit? Unfortunately, I am consider some process which can go beyond this limit (for example, a fast scattering between the two scalars, bringing the two fields to equilibrium.)
Well, one way to think about this is that the fundamental particles are different from the particles we observe, and that the fundamental particles are mixed, through virtue of some matrix, into the particles we observe. This mixing matrix gives rise to the cross-term interaction.

If your cross term is zero, then the mixing matrix is diagonal, and the particles we observe are the fundamental particles. If, however, the mixing term is at the limit \lambda^2 = m_1^2 + m_2^2, then the mixing matrix is saying that there are is in actuality only one fundamental particle that is mixed into these two, and the behavior of the system is fully-specified by the behavior of one of the particles. If you try to get larger off-diagonal terms, the mixing matrix ceases to make any sort of physical sense.