# The Klein-Gordon equation with a potential

1. Nov 13, 2012

### Sekonda

Hello,

My question concerns the Klein-Gordon Equation under some potential of the form (and refers to a higgs-like interaction, i assume as that's what we're researching):

$$\delta V= \lambda \Psi^{*}\Psi$$

For substitution into the Klein-Gordon equation:

$$(\frac{\partial^2 }{\partial t^2}-\bigtriangledown ^2+m^2)\Psi=\lambda \Psi^{*}\Psi$$

This is all in reference to the feynman diagram of:

The internal line is supposed to represent some scalar mediator, and we also assume the $$\Psi$$ are 'free' states. Also there are $$\lambda$$ factors at the vertices of the feynman diagram, I believe this to be just some scaling factor of the perturbation - however I'm not sure. Also from what I've wrote the Psi's in the potential may have subscripts i' and f' but I'm not really sure why this is (they may supposed to be i and f?)

I don't understand much of this - beginning with my main issue of the form of the perturbation being what seems to be a probability density? Also I'm not really sure what interaction is exactly happening in this feyman diagram.

If what I have wrote is nonsensical, please tell me why so and I'll see if I can clear anything up. Any explanations will be appreciated!

Thanks guys,
SK.

2. Nov 14, 2012

### Sekonda

My professor has told me that the use of the Psi's in the potential is 'not really a 'potential' but a non-linear term we add to the wave equation that induces self interactions of the wave'

Does anybody know why self-interactions are caused by terms like ψ*ψ?

Thanks

3. Nov 16, 2012

### andrien

How $$\delta V= \lambda \Psi^{*}\Psi$$ is an interaction term and what is that λ.I can not see any kind of interaction associated with this term.

4. Nov 16, 2012

### jfy4

lambda is a coupling constant. perturbative expansions are taken around lambda small. As to why certain terms intuitively lead to vertices, and the powers of fields directly correspond to the number of legs entering the vertices, I would love to know. You can find many derivations starting from a scalar action that work out the perturbative expansions in $\lambda$ and $J$ for some source. But as far as why nonlinearity equates to interaction I'm still not sure. If you look through some of the past threads here I can guaranteed the question has been answered though, it just might be buried somewhere deep.

5. Nov 16, 2012

### Sekonda

Thanks guys for the information, I'll have a look around.

Lambda is the coupling constant - is this basically the coupling of the particle to the field? What does it mean?

Thanks

6. Nov 16, 2012

### the_pulp

Look at Leonard Susskind QFT courses in YouTube. That is explained in the first or the second course (each one have 10 lectures) in a very intuitive manner.
More or less, the idea is that ψ, $\phi$ or A are creation / destruction operators so if you write, for example, ψ(x)*ψ(x)*$\phi$(x) you are saying that you are creating 2 ψ particles and 1 $\phi$ in position x from nothing (or destructing 1 and creating 2, or creating 1 and destructing 2 and so on).
So, in that way one can interpretate that, for example in the 2 desctruction-1 creation example, a ψ particle interacts with a $\phi$ particle, it sort of eats it and as a result we have a ψ particle. All this being said in a veeeeryyyy intuitive way.

Hope it helps

7. Nov 16, 2012

### Sekonda

Thank you for the suggestion of the youtube videos, I can tell this will help me understand this whole subject a lot better. Though there are quite a few so I better get watching and making notes!

Thanks again the_pulp!

8. Nov 18, 2012

### andrien

I still don't get it,if λ is a coupling constant then where is the coupling term.I mean let's take qed for example in which coupling term is of form jμAμ,where jμ is current associated with particles and Aμ which represents potentials.only this term corresponds to interaction in qed.but I can not see anything similar with the given interaction because to me it does not seem any interaction.

9. Nov 18, 2012

### jfy4

It's coupling to itself. here are some example couplings: $(\lambda / 4!) \phi^4$ is common example. Also $(\lambda /4!)(\phi \phi^{\dagger})^2$ is another example where there are anti-particles, the phi and phi dagger.

10. Nov 18, 2012

### andrien

yes,it is a kind of self interaction.I was just not thinking that both ψ* and ψ should be associated to same spacetime point to cause interaction.It was generating the anharmonic term which gives interaction

11. Nov 18, 2012

### Sekonda

jfy4 what do you mean when you say we take perbutative expansions around λ?

Thanks!

12. Nov 18, 2012

### jfy4

I mean, starting with the partition function, one can expand the exponential of the potential term in a series expansion in $\lambda$ (the coupling constant, theres nothing special about the letter lambda). Then, in nice cases you can write the poential as variations in a source $J$, $V(\delta / \delta J)$. One can solve the free field partition function in terms of the propagator and sources, and then pull down green's functions with the variations in $J$. These will be related to whatever order you decide to take the series expansion to in $\lambda$.

13. Nov 20, 2012

### andrien

But there is no source term J here which has really caused me trouble something like J∅,in qed jμ is generally ieψ*γ0γμψ but here there is no such thing.

14. Nov 20, 2012

### jfy4

for the complex klein gordon equation you can write one as $J^{\dagger}\phi + J\phi^{\dagger}$ and for the real one you can write $J\phi$. You can set $J$ equal to zero when you are done with the perterbative expansion.

15. Nov 21, 2012

### andrien

There is no necessity of it,one can directly go on with iL(lagrangian) and put the planewave form and separate all the unnecessary terms(factors taken care by external line etc.) to get the vertex factor.one can put the interaction term equal to delta function and go to momentum space to find the propagator.But so far there is no internal line corresponding to ay virtual quanta as is drawn because there is no source term and it's interaction with any field.

16. Nov 21, 2012

### Dickfore

The r.h.s. of KGE should read:
$$\frac{\lambda}{2} \left(\Psi^{\ast} \, \Psi \right) \Psi$$
for a self-interaction of the type the Higgs boson has:
$$\mathcal{L}_{\mathrm{int}} = -\frac{\lambda}{2! 2!} \, \left( \Psi^{\ast} \, \Psi \right)^2$$

17. Nov 21, 2012

### andrien

Second one makes sense,first one does not.

18. Nov 21, 2012

### Dickfore

The KGE is the Euler-Lagrange eqn. for the action defined as $S = \int{d^d x \, \mathcal{L}}$, where the Lagrangian (density) is:
$$\mathcal{L} = \mathcal{L_0} + \mathcal{L}_{\mathrm{int}}$$
Here $\mathcal{L_0} = \partial^{\mu} \Psi^{\ast} \, \partial_{\mu} \Psi - m^2 \Psi^{\ast} \, \Psi$ is the Lagrangian for the free complex scalar field. Varying w.r.t. $\Psi^{\ast}$, one obtains:
$$\frac{\delta S}{\delta \Psi^{\ast}(x)} = -\left\lbrace \partial^2 + m^2 \right\rbrace \Psi(x) - \frac{\lambda}{2! 2!} 2 (\Psi^{\ast} \Psi) \, \Psi$$
Equate this variation to zero and you get:
$$\left\lbrace \partial^2 + m^2 \right\rbrace \Psi = - \frac{\lambda}{2} (\Psi^{\ast} \Psi) \, \Psi$$
Apart from a wrong sign, the r.h.s. has the form I gave in my previous post.

19. Nov 21, 2012

### andrien

I thought your first term also some sort of a lagrangian which is not possible.

20. Nov 21, 2012

### Dickfore

How can a Lagrangian term be present in a Klein-Gordon equation?!