Recent content by Physics Footnotes

Firstly, the discreteness or continuity of an observable generally depends on the system in question. The total energy of some quantum systems, for example, is discrete, while in others is continuous, and in yet other systems is a mix of the two. So the question 'Is the such-and-such observable...

The problem with your expression has nothing to do with Quantum Mechanics specifically, but rather results from a confusion between mathematical objects, on the one hand, and labels for mathematical objects, on the other. Your expression is analogous to the following classical expression...

Because our direct empirical experience of the world is ultimately spatio-temporal in character (dots, flashes, tracks, etc.). Physical quantities like energy, momentum, and spin, are conceptual tools to organize that experience. Take a Stern-Gerlach experiment, for example. You don't actually...

I would put it like this: The outcomes of quantum mechanical experiments are individually unpredictable yet collectively synchronized. Whilst there is nothing inconsistent about this state of affairs at the level of probability theory (the quantum formalism getting the statistics right every...

This is a very elegant articulation of the so-called Measurement Problem, and makes very clear why it is called a 'problem', namely: the experiments used to justify quantum mechanics are, by that very theory, not dynamically possible! Despite what you may read to the contrary (here or...

Ummm... is this addressed to me? If so, where did I mention the word 'Hamiltonian'?? I was addressing a common confusion over the status of the operator ##\mathrm{i}\hbar \partial_t##.

@bhobba (my fellow Brisbaneite!), nothing I am saying contradicts standard textbook QM. I am simply choosing a more mathematically careful way of using our ubiquitous ##\psi## friend in two different ways: first in a concrete version of Schrodinger's Equation, considered as just a partial...

@vanhees71 you are completely missing the point of my notation. Of course for fixed ##t=t_0## the function ##\Psi (t_0,x)## is a square-integrable function, considered now as a function over ##x\in \mathbb R##. However, as I made very clear, I use the notation ##\Psi## to denote the entire...

In my notation, ##\Psi:=\Psi (x,t):{\mathbb R}^2 \to \mathbb C## is not a wavefunction in the quantum-mechanical sense, but rather a function capturing the entire space-time dependence of a system. To see that these functions do not form (or generally even belong to) ##L^2(\mathbb R^2,\mathbb...

I find it best when discussing Schrodinger's Equation to use uppercase ##\Psi=\Psi(x,t)## to capture the entire space-time behavior of a system, and lowercase ##\psi=\psi(x)## to denote the spatial behavior of a system at a particular time (possibly with a subscript, ##\psi_t##, to incorporate a...

Making the claim over and over again that the Measurement Problem and/or the Classical/Quantum Cut are issues that have been resolved by techniques such as decoherence is simply factually false. You can ignore the problems for most practical purposes if they don't interest you, but you are...

Thanks for the stimulating read. I too have come to the conclusion in recent years that anyone serious about the Foundations of Physics must thoroughly acquaint themselves with Bohmian Mechanics; not necessarily because it will turn out to be correct, but because it provides the most coherent...

It is best to think of a quantum state not as something that is observed, but rather as something that is inferred. That is, you observe objective events, and then you infer what kind of state is consistent with those events; there is no concept of 'observing the state'. This remains true even...

Invoking the so-called 'ensemble interpretation' doesn't solve the measurement problem; it simply expresses your optimism for the direction in which you hope the solution will be found.

@David Olivier The reason I left out the details is that you had indicated that you didn't want to deal with the subtleties in this thread. So here's a bit more: Yes, ##\mathcal B (\mathbb R)## is the Borel Algebra of measurable subsets of ##\mathbb R##. You can set up a PVM over any measurable...