Can the uncertainty principle be derived in quantum field theory? If yes, does is have a different interpretation than quantum mechanics because the coordinates $x_i$ are now parameters and not operators?

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    $\begingroup$ When saying "the uncertainty principle" it is worthwhile to realize that the famous $\sigma_x \sigma_p \ge \frac{\hbar}{2}$ is just one instantiation of the general relation for operators, and no notion of "coordinates" is needed to get this principle. $\endgroup$
    – ACuriousMind
    Jun 24, 2015 at 20:45

2 Answers 2


There are some possibilities to produce a statement in QFT similar to the one valid for QM. In this case $X$ and $P$ must be replaced by the analogous objects in QFT, the field operator and its conjugate momentum. Consider a quantum scalar field $\phi$ and the equal time CCR: $$[\phi(t, \vec{x}), \pi(t, \vec{y})] = i\hbar \delta(\vec{x}-\vec{y}) I \:.$$ The rigorous version is: $$[\phi(t, f), \pi(t, g)] = i\hbar (f|g) I$$ where $f,g : \mathbb R^3 \to \mathbb R$ are spatial smearing test functions and $$(f|g)= \int _{\mathbb R^3} \overline{f(\vec{x})} g(\vec{x}) d^3x\:.$$ With the same procedure as for standard CCR you easily get $$\Delta \phi(t, f)_\Psi \:\Delta \pi(t, g)_\Psi \geq \frac{\hbar}{2} |(f|g)|$$ for every normalized vector state $\Psi$ which belongs to the domain of $\phi(t, f), \pi(t, g)$ and their second order powers. In particular you see that if $f$ and $g$ have disjoint supports, $|(f|g)|=0$, so that $\Delta \phi(t, f)_\Psi \:\Delta \pi(t, g)_\Psi \geq 0$, in accordance with the fact that $\phi(t, f)$ and $\pi(t, g)$ commute in that case...

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    $\begingroup$ Hello. May I ask: Can I understand $\Delta \phi$ or $\Delta \pi$ as fluctuations of the quantum fields? Thanks. $\endgroup$ Mar 15, 2017 at 17:36
  • $\begingroup$ Yes, that is the most direct interpretation. $\endgroup$ Mar 15, 2017 at 19:39
  • $\begingroup$ But those relationships given in your answer, would still limit the precision with which the linear momentum of a particle and its position can be determined? $\endgroup$
    – Davius
    Aug 26, 2023 at 18:53
  • $\begingroup$ I do not think so. In particular, no notion of position is discussed here. $\endgroup$ Aug 26, 2023 at 22:53

Quantum Field Theory is essentially modelled on top of the theory of Quantum Mechanics for finitely many degrees of freedom. With the creation and annihilation operators one can define the analogue of the position and momentum operators $q$ and $p$ as the closures of $$q_0(x) = \frac1{\sqrt2}[a(x) - a(x)^*],\qquad p_0(x) = \frac i{\sqrt2}[a(x)+a(x)^*]$$ respectively. The Heisenberg relations are then $$[q(x),p(y)] = i (x,y)I,$$ for any pair of vectors $x,y$ in the single particle Hilbert space.

The same relations come directly from the canonical fields $\phi$ and $\pi$, which satisfy the Heisenberg relations at a certain time, say $t=0$. By choosing an orthonormal basis $\{e_n\}$ of the single particle Hilbert space one can set $$q_k = \phi(e_k),\qquad p_k=\pi(e_k),$$ whence $$[q_i,q_k]=[p_i,p_k] = 0,\qquad [q_i,p_k] = i\delta_{ik}I.$$

As the uncertainty relations stem directly from the Heisenberg relations, one have them for the degrees of freedom of the theory, but there is no link between them and the coordinates of space-time.


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