# Does vacuum excitation violate the conservation of energy?

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## Summary:

Does vacuum excitation violate the conservation of energy?
Hi, there. I am reading the article Relativistic quantum optics: The relativistic invariance of the light-matter interaction models by Eduardo Martin-Martinez el al (2018).

Here he calculate the transition probability of a vacuum excitation for a detector.

Suppose there is a lab where the electricmagnetic field is quantizedthe, and a detector atom is travelling relative to the lab. Assume that the initial state of the detector atom and the field is the ground state ##\left | g,0 \right >##. Then the transition probability of the vacuum excitation is given by
##p\left ( \Omega \right)=\sum_{out} \left | \left < {e}, {out} \right | U \left | {g}, 0 \right > \right | ^2##
where the sum over states ##\left | {out} \right >## represent a sum over an orthonormal basis of possible final states of the field.

At last, he derived a expression for ##P\left ( \Omega \right )## which is not zero.

But if the detector atom was at the ground state and the field was at the vacuum state initially, and then the atom was excited and the field could be some state other than the vaccum state, the law of conservation of energy seems to be violated.

How could that be possible?

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vanhees71
Gold Member
2019 Award
Well, since there is a "detector atom" it's not vacuum anymore. I think the idea of "vacuum fluctuations" as used in hand-waving arguments (usually in popular-science textbooks but unfortunately sometimes even in textbooks in an attempt of some authors wanting to be "didactical") is the most confusing misguidance of students ever! Don't take it too literarly.

• bhobba
Well, since there is a "detector atom" it's not vacuum anymore. I think the idea of "vacuum fluctuations" as used in hand-waving arguments (usually in popular-science textbooks but unfortunately sometimes even in textbooks in an attempt of some authors wanting to be "didactical") is the most confusing misguidance of students ever! Don't take it too literarly.
So the state of the field is not really the vacuum state? Wait a munite. I think I have read something similar in the quantum field theory.

I remember that the vacuum states are different in the free field theory and the interacting field theory. Is this relavant?

vanhees71
Gold Member
2019 Award
Yes, it is. The vacuum state is by definition the ground state of your quantum field theory and thus there are no energy fluctuations, because it's an energy eigenstate by definition.

• bhobba and Haorong Wu
Yes, it is. The vacuum state is by definition the ground state of your quantum field theory and thus there are no energy fluctuations, because it's an energy eigenstate by definition.
Thanks! I got it. Thus initially, the field is at the vacuum state for the interacting theory since it is coupled to the atom. After interacting with the atom, the field becomes the vacuum state for the free theory.

vanhees71
Gold Member
2019 Award
No, it's even simpler: If there is an atom, there's no vacuum anymore, because there's a nucleus and electrons.

• bhobba
No, it's even simpler: If there is an atom, there's no vacuum anymore, because there's a nucleus and electrons.
Well, I am confused again.

If the vacuum state is the state with the lowest energy, then after the atom is excited, the field should lose some energy and drop to a lower state, but there is no lower state.

If the vacuum state is not the lowest state, then the name "vacuum state" is quite odd.

vanhees71
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