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Quantum mechanics describes the microscopic properties of nature in a regime where classical mechanics no longer applies. It explains phenomena such as the wave-particle duality, quantization of energy and the uncertainty principle and is generally used in single body systems. Use the quantum-field-theory tag for the theory of many-body quantum-mechanical systems.

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Look up the "Wigner's Friend" thought experiment: your scenario is the one at the point when the friend knows the state of the cat but before Wigner comes back and asks his friend whether his friend's …
answered Dec 2 '13 by WetSavannaAnimal
2
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The answer is no, your result does not follow. To understand why, suppose we have some $N\times N$ projector $\mathbf{P}$, with a range space $\operatorname{Ran}(\mathbf{P})$ (the subspace that the p …
answered Aug 26 '13 by WetSavannaAnimal
2
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Let's begin with knzho's comment: A combination of eigenstates (with different eigenvalues) is a non-eigenstate. But yes, 1/2 is what happens. and take up your further question in comment: …
answered Jul 11 '16 by WetSavannaAnimal
1
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To add a slightly different angle to PhotonicBoom's sound answer, the link between the two entities - discrete sum and integral - is the concept of measure, not of limit. You can think of your sum as …
answered Oct 14 '14 by WetSavannaAnimal
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I'll add to Trimok's answer in the light of your further question as to "what does it look like in higher dimensional bases. The operators going to be higher order tensors now?". Since you're "impleme …
answered Aug 1 '13 by WetSavannaAnimal
1
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I'll just first add to Alfred's answer and also point out that in more generalized discussions (theory of distributions and rigged Hilbert spaces) the notion of adjoint is broadened so that in rigged …
answered Oct 29 '13 by WetSavannaAnimal
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However, there is not a unitary transformation that connecting two different Fock states $|m\rangle$ and $|n\rangle$. That's not true. There is always a unitary transformation that maps between …
answered Jul 16 '15 by WetSavannaAnimal
2
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The key idea here is that, just after measurement, a quantum system is always in an eigenstate of the observable (operator) that represents the measurement. Therefore, if the same measurement is appli …
answered Mar 18 '18 by WetSavannaAnimal
1
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It has been a few years since I've heard that name. As I understand it, his claimed results of his thought experiment are at variance with what actually happens if you do the experiment. Even if his c …
answered Jan 15 '17 by WetSavannaAnimal
0
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The uncertainty principle has a somewhat misleading name. If you're seeking everyday analogies, the UP has less to do with inaccuracy and is much more to do with noncommutativity: a phenomenon illustr …
answered Jan 21 '15 by WetSavannaAnimal
1
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As in the comments, you need to specify the probability distribution for the electron eigenstates. This distribution can depend on the hydrogen atom's environment - if the hydrogen atom is in a resona …
answered Feb 25 '18 by WetSavannaAnimal
3
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To add to Ernie's Answer: Actually, tides on Earth are tantamount to an observation of a Moon-like object. Bernard Schutz in his book "A First Course in General Relativity" imagines an instance of yo …
answered Aug 28 '15 by WetSavannaAnimal
1
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I agree with joshphysic's answer that your reasoning is right, but you need to justify $V(x)$'s being a multiplication operator (in position space, that is). Maybe the way I tend to look at this may …
answered Oct 27 '13 by WetSavannaAnimal
9
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To add to ACuriousMind's answer by bringing my personal experience to the table, I too had the background mathematics for many years and understood perfectly all the algebraic machinations in many tex …
answered Sep 2 '15 by WetSavannaAnimal
1
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The intrinsic difference is that a dimensionless parameter is invariant if we change our units, whereas other parameters do depend on the units. The former arises when we can specify the property in q …
answered Jul 16 '15 by WetSavannaAnimal

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