# Is the QFT vacuum $T$-invariant?

A quantum vacuum corresponds to closed loops of particle-antiparticle pairs, in the language of Feynman diagrams. There is no relation between $$E$$ and $$p$$ for these particles, i.e., they are not on their mass shell. It is suggested that the QV is insensitive to the direction of time (seen as an irreversable thermodynamic process), but I'm not sure what this means. On top of that it is said that we can't make a mental picture of a QV, and it's purely a mathematical construct (which means we can make a time reversal operation on the QV). But it has to correspond something physical. So suppose we could make a film of the process (which involves time), could we see a difference between the film running forwards or backwards. That is, is the QFT vacuum $$T$$-invariant?

• This is a perfectly good question and completely clear. I don't understand the down votes or close vote. May 23 '18 at 13:37
• Virtual particles don't "pop" into existence. To exist means to move in time. Virtual particles don't move in time. So reversing time would not change the picture. May 23 '18 at 14:46
• @Nathaniel Because "virtual particles popping in and out of existence" has no basis in actual QFT, it's a crude picture used in pop science. If the question starts from an incorrect premise, it can't be answered. May 23 '18 at 14:59
• Also, OP has about 25 questions about this exact same subject, and the answer to every single one of them is "don't take popsci too literally" or perhaps "read a textbook". May 23 '18 at 15:00
• @knzhou-I've read Lewis H. Ryder's book on QFT together with a piece of work written by the professor (and many articles on the net after that) for my exam on QFT and strange enough I did well! To say I only read popular science books is simply not true. Who are you to say so? You don't know anything about me! I don't make any difference between so-called experts and laymen. And where are all the questions I asked about this? This is the first time I asked this! May 24 '18 at 7:40

A quantum vacuum is an eigenstate of the Hamiltonian, so the energy is certain and the occupation numbers are certain too. "Fluctuations" of some variables (like $x$ and $p$) do not change this fact. So it is a T-invariant state.