0

I'm not a physicist. My understanding of Heisenberg's uncertainty principle and its proof (that is given by an imaginary microscope) is that for example: at a specified time determining the exact energy of a system is not possible. But I do not understand that if we can't experimentally exactly determine the energy of a vacuum at a specified time, this does't mean that it's energy can not be zero and there must be fluctuations in vacuum.

Qmechanic
  • 201,751
Alberto
  • 177
  • 1
    You might want to read this answer of mine. – AccidentalFourierTransform May 09 '16 at 22:19
  • That quantum mechanics is the correct theory follows from a wealth of experimental data, all of which agree with the predictions of the theory. The uncertainty relations are a simple corollary to the structure of the theory. That measurements produce uncertainty is a consequence of us requiring that the measurement leave a lasting record. This, in a thermodynamic sense, can only be done if we open up the quantum system to the environment, which we don't know perfectly. Without the measurement itself, the "fluctuations" wouldn't happen, so one shouldn't imagine them to be part of the system. – CuriousOne May 09 '16 at 22:27
  • @AccidentalFourierTransform I couldn't understand "A virtual particle is just a contraction of fields that depend on variables over which you integrate, when using Dyson's perturbative series for the S-matrix." You said before that "the vacuum is the same at every point in space." But if it is the same in every point in space how can be contraction of field in some point? – Alberto May 09 '16 at 22:55
  • How do you get a "point in space" to hold still long enough to examine it? – Howard Miller May 09 '16 at 23:20
  • @HowardMiller one should separate the mathematical model from the reality which is modeled. In the mathematical model all space points exist as the variables of the model. The feynman diagram spatial description can have loop integrations that lead to zero over the HUP bounds. Loops acquire reality when there are external lines to the loops, i.e. energy is supplied. – anna v May 10 '16 at 06:55
  • @Alberto because fields depend on position. The vacuum does not. Therefore, for example $\langle \Omega|\hat Q(x)|\Omega\rangle$ does not depend on $x$ but $\langle \Omega|\hat Q(x)\hat W(y)|\Omega\rangle$ does. – AccidentalFourierTransform May 10 '16 at 11:19
  • Can anyone explain this without math? – Alberto May 10 '16 at 11:38

0 Answers0