How does Heisenberg's uncertainty principle work with more than one quantum field? I am specifically asking about the time-energy uncertainty: $$\Delta E \Delta t \ge \frac {\hbar}{2}\tag{1}$$ Imagine we're sampling a volume of space in vacuum. The uncertainty in energy is: $$\Delta E \ge \frac{\hbar}{2\Delta t}\tag{2}$$ I want to know if this uncertainty is accounting for all the quantum fields in that volume. And if we are constraining ourselves to just one quantum field, say the photon field, how would the energy uncertainty look? Would it be $$\Delta E \ge \frac{\hbar}{2n\Delta t}\tag{3}$$ where $n$ is the total count of quantum fields? Would it still be $$\Delta E \ge \frac{\hbar}{2\Delta t}\tag{4}$$ Or something else?

If it is equation 4 wouldn't the uncertainties in all quantum fields add up so the total uncertainty for the energy for the volume to be $$\Delta E \ge n\frac{\hbar}{2\Delta t}\tag{5}$$ where $n$ is the total count of quantum fields?

If any of the equations, containing $n$ is true, how do we know what $n$ is? We can't say for sure that our particle model is complete.

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    $\begingroup$ Besides the fact that the energy-time uncertainty principle is really vague to begin with, that's not how uncertainty relations work. The other fields are independent. You don't get a factor of $n$, for the same reason that in $\Delta p_x \Delta x \gtrsim \hbar/2$ you don't get a factor of the number of dimensions. The other dimensions have their own independent uncertainty principles. $\endgroup$
    – knzhou
    Commented Jan 10, 2019 at 15:16
  • $\begingroup$ The energy-time uncertainty relation is of a different character than the Heisenberg's uncertainty principle, where $x$ and $p$ are canonical conjugate pair of operators, unlike energy and time. $\endgroup$
    – Avantgarde
    Commented Jan 11, 2019 at 19:20


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