# Solution of equation for decaying real scalar field

1. May 6, 2015

### karlzr

Suppose there is a real scalar field $\phi$ with some decay width $\Gamma$ to some fermion. The quantum equation of motion after one-loop correction takes the form
$\ddot{\phi}+(m^2+im\Gamma)\phi=0$
where $m$ is the renormalized mass.
The solution can be obtained as $\phi=\phi_0 e^{imt}e^{-\Gamma t/2}$. So how do we use this complex solution? since the solution we need for application must be real. Usually we can take the real part when we have a complex solution. But In this case the real part of this solution does not solve the full quantum equation of motion due to the imaginary component.

2. May 8, 2015

### Brage

Surely the equation $\ddot{\phi}+(m^2+im\Gamma)\phi=0$ has the solution

$$\phi=Ae^{-im^2t}e^{\Gamma m t/2}+Be^{im^2t}e^{-\Gamma m t/2}$$, where A and B are complex constants to be determined from initial and boundary conditions. Unless I've missed something fundamental here?

3. May 8, 2015

### karlzr

But $\phi$ starts as a real field, how can it have complex solution?
In quantum mechanics, an imaginary part in Hamiltonian of some system indicates decaying amplitude. But isn't the total Hamiltonian of the whole system always real in order to obey unitarity? I don't know how to draw an analogy in this case.

4. May 8, 2015

### bhobba

Some solutions to equations are not necessarily physically sensible eg the requirement for the solution to be real will put constraints on A and B if you expand things out.

Thanks
Bill

Last edited: May 8, 2015
5. May 8, 2015

### karlzr

When $\Gamma << m$, the full solution is $\phi(t)= A e^{imt-\Gamma t/2}+B e^{-imt+\Gamma t/2}$. It doesn't seem to be possible to make it real by constraining the two constant coefficients $A$ and $B$.
Actually, this question is from Phys.Lett. B117 (1982) 29

6. May 8, 2015

### bhobba

Well that simply means it has no real solution. So?

Thanks
Bill

7. May 8, 2015

### karlzr

I need to find the evolution of its energy density $\rho=\dot{\phi}^2/2+m^2\phi^2/2$. With complex solution for $\phi$, I get complex energy density which doesn't make sense.

8. May 8, 2015

### bhobba

Maybe the answer is to apply Noethers theorem to get the stress energy tensor:
http://www.itp.phys.ethz.ch/research/qftstrings/archive/12HSQFT1/Chapter04.pdf [Broken]

Notice the Lagrangian in 4.1 while containing a complex field is real.

Thanks
Bill

Last edited by a moderator: May 7, 2017
9. May 8, 2015

### Brage

Well the real part of $\phi(t)= A e^{imt-\Gamma t/2}+B e^{-imt+\Gamma t/2}$ would be $$Re(\phi(t))[Re(A)cos(mt)-Im(A)sin(mt)]e^{-\Gamma t/2}+[Re(B)cos(mt)+Im(B)sin(mt)]e^{\Gamma t/2}$$ just by use of Euler's formula and taking the resulting real part.

10. May 8, 2015

### karlzr

But the real part doesn't solve the equation.

11. May 8, 2015

### bhobba

It has no real solution.

If you want to find the Hamiltonian use the methods in the link I gave and apply Noether. I am pretty sure a simple modification of the Lagrangian in 4.1 of my link will give your equation.

The problem as you stated it at the start is inconsistent - there is no real scalar field that is the solution to that equation. I am pretty sure its Lagrangian contains a complex field so its obvious why that's so. However when you work out the Hamilitonian it will give real values of energy - as it must.

Thanks
Bill

12. May 8, 2015

### karlzr

My equation is indeed from the quantum action. A fermion loop ($m_f<m_\phi/2$) will contribute a complex term to the effective potential, which means the free scalar particle is not the eigenstate of Hamiltonian. It will decay. I believe the Hamiltonian is also complex just like in quantum mechanics.
When I say the scalar field is real, I mean there is only one degree of freedom. I don't know whether it is inconsistent for such a real scalar field to have a complex solution. But my problem is to find the evolution of energy density from my solution to the equation and I expect it to be real. Actually I expect the solution to oscillate as in free field theory and also the energy density to decrease since the decay channel is open in our case.

13. May 8, 2015

### bhobba

A complex field is simply a tricky way of elegantly writing the equation of two real fields. You cant have a single real field that is complex. It makes no sense.

What I think you should do is post the full details of what you are trying to do. We have some quite high powered theorists that post here (I an not one - my knowledge of QFT is not as good as I would like) and hopeful they can sort it out.

Thanks
Bill

14. May 9, 2015

### karlzr

Thanks.
This question is from Phys.Lett. B117 (1982) 29
This is a 5-page paper and my question is from page two. The scalar is the inflaton which reheats the universe at the end of inflation by perturbative decaying. We can assume the potential energy of inflaton has only quadratic term. That's pretty much all the background of my question.

15. May 9, 2015

### bhobba

16. May 9, 2015

### fzero

If we look back at the equation of motion, we've replaced the mass by a complex number. As you've found, the solutions have a complex energy and are no longer real. Sometimes in the literature, these solutions are called Gamow states. The rationalization is that these states represent a resonance that decays after a short time. Usually, the solution of an equation of motion represents an equilibrium configuration (stationary state), but we've cooked up our system in such a way that it does not have equilibrium solutions but quasistationary ones.

As you can see from the paper in question, there is a physical interpretation of the solutions that has a predictive value for certain problems so the utility of the method is evident. But you will not be able to demand that you find a real solution to an equation that has been analytically continued to complex parameters.

17. May 10, 2015

### karlzr

Thanks. Since we are on this paper, I have a related question.
I am a little confused about how energy goes from inflaton to fermions. So during reheating, this paper says we can find the amplitude of fermion production from $<f_i,\bar{f}_i|\phi \bar{\psi}_i\psi_i|0>$. Here $\phi$ is purely classical field which should be replaced by the complex solution of its equation, right? It seems this amplitude depends only on the value of $\phi$ and has nothing to do with the energy density which doesn't make sense to me. Also, in my opinion, a classical field $\phi$ in this case serves only to give a oscillating mass to the fermion (preheating, non-perturbative decay), and I don't know why we can get information of its perturbative decay. Is it the inflaton field or inflaton particle that decays to fermion?

18. May 10, 2015

### fzero

As you say, we have a classical value of $\phi$ so we are talking about the field and not a particle state. The physical situation is that we start with the scalar somewhere away from the minimum of its potential, as it would be at high temperature. This is modeled by putting in a source term that is suddenly turned off at $t=0$. The field will naturally settle into a minimum of the potential, but the energy that was stored in the field has to go somewhere. The paper considers the case where all of this energy goes to fermions via the Yukawa-type coupling.

Given an amplitude $\mathcal{A}_{fi}$ from some initial state $i$ to a final state $j$, Fermi's Golden rule:

$$\Gamma_{i\rightarrow f} \sim | \mathcal{A}_{fi} |^2$$

gives the probability per unit time to create the final state $f$. I think you can convert this into an energy released per unit time by considering the expression for the Hamiltonian evaluated on the classical solution for $\psi$. When you integrate that over time, you will find the energy density of the fermions.

Last edited: May 10, 2015