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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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Others have explained why those textbook examples are bad (which they are, but are historically accurate I guess). Let me try to explain the problem itself by an analogy and without just saying that the quantum state character itself leads to ill-defined positions and momentums. Suppose you are a farmer on a flat infinite Earth (pre-medieval times?), and ...

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Here are a few things that you may need clarification on: What is uncertainty? What am I trying to achieve through a measurement? Firstly, as you pointed out, we DO NOT need a measurement to be made for the uncertainty relation to hold good. Going by the statistical interpretation, uncertainty is defined as the standard deviation of the ...

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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 ...

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

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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 ...

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A quantum particle can't simply be a classical particle with unknown but well-defined values for position and momentum. Quantum uncertainty really is a different sort of thing from simple lack of knowledge about the state of an object. One can see this pretty clearly by considering the double-slit experiment. If you send electrons (or any kind of particle) ...

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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 ...

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There is no uncertainty for velocity because velocity is not a well-defined quantum observable. Quantum observables are obtained through canonical quantization from classical observables, which are functions on the classical phase space of Hamiltonian mechanics, whose coordinates are generalized positions and momenta. In particular, $v = ... 0 The uncertaintity principle is a consequence of more fundamental principles and it doesn't trigger anything. In particular, systems evolve according to evolution equations such as wave equations. In particular measurements correspond to operators. In particular (strong) measurement outcomes correspond to eigenvalues of the corresponding operators. In ... 2 As far as I understand, it is not. Entanglement is a consquence of the Hilbert space structure of composite systems in quantum mechanics. When you postulate that if you have system A (with Hilbert space$H_A$) and system B (with Hilbert sapce$H_B$) then the Hilbert space of both systems together (i.e., when they interact) is$H_A \bigotimes H_B\$, then ...

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