Unphysical nature of ##\phi^{3}## interaction

In summary: The energy of the photons doesn't really matter. But if you include radiation, then the excited states quickly decay back to the ground state. But nonrelativistically, excited states are stable, if you ignore radiation. The energy of the photons doesn't really matter. But if you include radiation, then the excited states quickly decay back to the ground state.
  • #1
spaghetti3451
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In page 77 of Peskin and Schroeder, it's mentioned that for a ##\phi^{3}## interaction, the energy is not positive-definite unless we add a higher even power of ##\phi##.

Can someone please prove this statement?
 
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  • #2
One reason is that ##V=\frac{1}{2}m^2\phi^2+\lambda \phi^3## is unbounded below (##V\to-\infty## as ##\phi \to-\infty##).

On the other hand, ##V=\frac{1}{2}m^2\phi^2+\lambda\phi^3+\sigma \phi^4## is bounded below.

Does that help in any way?
 
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  • #3
Yes, that's the reason. As long as V is bounded below, we can add a constant to V that makes the minimum value of V equal zero. Then each term in the hamiltonian is positive. But if V is unbounded below, this cannot be done.
 
  • #4
Here's a related question: What goes wrong in QM if the energy levels are not bounded below? We could reason that particles tend to radiate, and drop to lower energy levels. So if there is no lower bound to the energy levels, then the particle will just keep falling to lower and lower energy levels, and will radiate an infinite amount of energy. This would be a catastrophe, and would end up destroying the universe, I suppose. But would it be actually inconsistent with QM?
 
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  • #5
stevendaryl said:
Here's a related question: What goes wrong in QM if the energy levels are not bounded below? We could reason that particles tend to radiate, and drop to lower energy levels. So if there is no lower bound to the energy levels, then the particle will just keep falling to lower and lower energy levels, and will radiate an infinite amount of energy. This would be a catastrophe, and would end up destroying the universe, I suppose. But would it be actually inconsistent with QM?
I think by QM you actually mean QFT, because AFAIK we can't have creation of photons in QM! But maybe you mean classical radiation from quantum mechanical matter? Is it reasonable?
 
  • #6
Shyan said:
I think by QM you actually mean QFT, because AFAIK we can't have creation of photons in QM! But maybe you mean classical radiation from quantum mechanical matter? Is it reasonable?

Well, prior to actually developing QED, physicists had a simplistic model of radiation: that a particle makes a transition from energy level [itex]E_n[/itex] to [itex]E_m[/itex] and releases a photon of frequency [itex]\nu = (E_n - E_m)/h[/itex]. But I could rephrase the question: Does QM without radiation run into inconsistencies if the energy levels are not bounded from below?
 
  • #7
stevendaryl said:
Well, prior to actually developing QED, physicists had a simplistic model of radiation: that a particle makes a transition from energy level [itex]E_n[/itex] to [itex]E_m[/itex] and releases a photon of frequency [itex]\nu = (E_n - E_m)/h[/itex]. But I could rephrase the question: Does QM without radiation run into inconsistencies if the energy levels are not bounded from below?

What immediately comes to mind is that we wouldn't have stable states like the Hydrogen ground. But I feel like there would be a more general issue with the formalism?
 
  • #8
HomogenousCow said:
What immediately comes to mind is that we wouldn't have stable states like the Hydrogen ground. But I feel like there would be a more general issue with the formalism?

But nonrelativistically, excited states are stable, if you ignore radiation.
 

1. What is the unphysical nature of ##\phi^{3}## interaction?

The unphysical nature of ##\phi^{3}## interaction refers to the fact that it violates the principle of unitarity, which states that the total probability of all outcomes in a physical process must equal 1. In other words, the interaction leads to probabilities that are greater than 1, which is not physically possible.

2. Why is the unphysical nature of ##\phi^{3}## interaction a problem?

The unphysical nature of ##\phi^{3}## interaction is a problem because it leads to inconsistencies in calculations and predictions in quantum field theory. It also raises questions about the validity of the theory and its ability to accurately describe physical phenomena.

3. What are the implications of the unphysical nature of ##\phi^{3}## interaction?

The implications of the unphysical nature of ##\phi^{3}## interaction are still being studied and debated. Some theories suggest that this issue can be resolved through renormalization techniques, while others propose modifications to the theory itself. It also raises questions about the interpretation of physical results and the need for further research.

4. Can the unphysical nature of ##\phi^{3}## interaction be observed in experiments?

No, the unphysical nature of ##\phi^{3}## interaction cannot be directly observed in experiments. This is because it is a theoretical issue that arises when making calculations and predictions using the theory. However, its implications on physical results and the need for further research can be observed.

5. Are there any other interactions in physics that have unphysical properties?

Yes, the ##\phi^{3}## interaction is not the only interaction in physics that has unphysical properties. For example, the ##\phi^{4}## interaction also violates unitarity and leads to similar issues in quantum field theory. Other interactions, such as the weak nuclear interaction, also have unique properties that are not fully understood and continue to be studied by scientists.

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