# Tag Info

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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 ...

3

Here is the Heisenberg uncertainty principle: Here is h_bar 1.054571726(47)×10^−34 joulesecond All it needs is some algebra to see that a kilogram ball moving at a micron per second and measurement accuracies of the order of a micron will still fulfill the HUP constraint as h_bar is a very small number. For classical dimensions h_bar is essentially zero ...

2

I would boldly claim that this thought experiment (also known as the Heisenberg microscope) is simply the wrong picture to understand the origin of uncertainty principle. The reason why it is so is because it mixes up between uncertainty due to measurement and uncertainty due to quantum state; nonetheless it had made its way into numerous textbooks and ...

2

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 $$q_0(x) = \frac1{\sqrt2}[a(x) - a(x)^*],\qquad p_0(x) = \frac i{\sqrt2}[a(x)+a(x)^*]$$ ...

1

In quantum mechanics observables are represented by (some classes) of self-adjoint operators on some Hilbert space. Saying that you can precisely measure a quantity given by the operator $A$ means that your state can be one of the eigenstates of that operator. Likewise, if you want to precisely measure two quantities $A, B$ together your state needs to be an ...

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When defining the uncertainty principle one has first of all to be very careful with the domain of definitions the operators have: in particular, if $A, B$ are the observables whose uncertainties we want to measure together, what needs to be calculated is the commutator $\left[A,B\right]$. In order this commutator to be well defined we must have that the ...

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