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45

Observables don't commute if they can't be simultaneously diagonalized, i.e. if they don't share an eigenvector basis. If you look at this condition the right way, the resulting uncertainty principle becomes very intuitive. As an example, consider the two-dimensional Hilbert space describing the polarization of a photon moving along the $z$ axis. Its ...

28

One problem with the given $3\times 3$ matrix example is that the eigenspaces are not orthogonal. Thus it doesn't make sense to say that one has with 100% certainty measured the system to be in some eigenspace but not in the others, because there may be a non-zero overlap to a different eigenspace. One may prove$^{1}$ that an operator is Hermitian if and ...

25

UPDATE: the insights and conceptual meanings advocated below in this answer are detailed and technically developed, for example, in these wonderful articles: Fuchs; Peres - Quantum Mechanics Needs No Interpretation, Physics Today (2000), vol. 53, issue 3, p. 70. Englert - On Quantum Theory, Eur. Phys. J. D (2013) 67: 238. Duvenhage - The Nature of ...

25

If by "daily life" you mean things we experience on a day to day basis at the macroscopic level (as opposed to the microscopic level), while the principle applies it does so insignificantly. The principle essentially states that you can never simultaneously know the exact position and speed (momentum) of an object because all objects behave like both a ...

23

The relation $$\langle Ay | x \rangle = \langle y | A x \rangle \text{ for all } x \in \text{Domain of } A\tag1$$ makes no reference to any basis at all, so it is indeed basis-independent. In fact, this definition, which seems pretty strange when you first meet it, arises precisely out of a desire to make things basis-independent. The particular ...

22

In non-relativistic quantum mechanics the mass can, in principle, be considered an observable and thus described by a self-adjoint operator. In this sense a quantum physical system may have several different values of the mass and a value is fixed as soon as one performs a measurement of the mass observable, exactly as it happens for the momentum for ...

21

While I agree with tparker saying "If simple examples of the Heisenberg uncertainty principle occurred in everyday life, then it would have been discovered before 1927.", your question reminded me of a problem from Sakurai's Modern Quantum Mechanics: 1.22 Estimate the rough order of magnitude of the length of time that an ice pick can be balanced on its ...

19

The appendix of Mackey talks about superselection rules, and indeed superselection is the phenomenon where there are self-adjoint operators that are not observables. Whether this is obvious or not depends on how one defines "superselection". The standard definition would be that the Hilbert space $H$ splits into the direct sum $H_1\oplus H_2$ such that for ...

19

To answer the question, you have to specify what observable’s eigenstates the “superpositions” you’re talking about are superpositions of. If you measure the energy, you always get one single energy eigenstate, not a superposition of energy eigenstates. The same applies for any other observable quantity. But that one single energy eigenstate might be a ...

15

Is the momentum operator $P=-i\hbar \frac{\mathrm d}{\mathrm dx}$ symmetric when restricted to the compact interval of the well? Are there any subtleties in its definition, via its domain or similar, that are not present for the real-line version? (I henceforth assume the Hilbert space is $L^2([0,1],dx)$.) It depends on the precise definition of the ...

15

In general, for physical reasons not all self-adjoint operators are observables. I will discuss the problem at the level of von Neumann algebras that are (perhaps unawares) more familiar to physicists. There are two further possibilities leading to a bit differently articulated answers: $C^*$-algebras and lattice of elementary propositions theory. Let us ...

15

In practice, you can't actually get yourself into a quantum superposition of dead and alive. Exposing a quantum system to a thermal environment causes the state to lose coherence and collapse into one state or another. In the double slit experiment, you are the thermal environment that causes the collapse. "Observation" in quantum mechanics has nothing to ...

15

Maybe I missed it, but somehow no-one seems to have mentioned the Bandwidth theorem. The Heisenberg Uncertainty Principle is simply one manifestation of it. Basically if you have a function $\psi\left(x\right)$, and its Fourier transform $\tilde{\psi}\left(\nu\right)$, then you can have $\psi\left(x\right)$ being zero everywhere except a small region of $x$-...

14

Color charge in the sense of "being blue, red, green" is not a quantum mechanical observable because the $\mathrm{SU}(3)$ gauge transformations mix the colors. This means it is meaningless to say "We have a blue particle", because we can perform a gauge transformation and then we "have a red particle". Since physical descriptions related by gauge ...

13

Of course, mass is an observable, although in simple models it is constant. This is already the case classically. One cannot determine the path of as rocket that burns fuel (which forms a large fraction of its mass) without taking into account that the mass is variable. The same holds in quantum mechanics, whenever the mass is not fixed by the modeling ...

13

If I try to measure the field at one point in spacetime, I should get a real value which should be an eigenvalue of the quantum field, right? I guess the eigenvectors of the quantum field also live in Fock space? Yes, that's basically correct. If the value of the field at a point is observable, the eigenvalues of the operator representing it are the ...

13

The wave function isn't an operator; the word "operator" in quantum mechanics means something more precise than "function". You might say that the momentum operator is "something to put in an integral to get the expectation value of momentum"; let me explain. We know that for a particle in a state $\psi$,  \langle x \rangle = \int_{-\infty}^{+\infty} x ...

13

In classical physics, the definition of an observable is that it is a (reasonably smooth) function of position and momentum. Therefore, quantum systems obtained purely by quantization of a classical system also have that all their observables are functions of position and momentum. In quantum physics, an observable is just an operator designate to belong to ...

12

This is a rough version of the summary of the complete answer I posted on math.stackexchange, where more details are discussed in a long digression, in particular mathematical motivations for your points 1., 3. and 4. (Any reader interested in more explanations and a longer updated list of references should check out that other answer in Math.SE). I eagerly ...

12

Quantum mechanics is indeed a probability theory, but it is a non-commutative probability theory. So it is not just a matter of having signed/complex measures, but really of having a non-commutative probabilistic framework. Quantum mechanics was developed, historically, before non-commutative probability theories and I think that people in probability ...

12

The Wikipedia quote appears to be lifted from this Solid State Physics text by CTI Reviews, and then plastered all over the web. The text does not give any citation of Lanczos, however. Here is the only passage in Lanczos's 300 pages long Variational Principles of Mechanics that contains the word "self-adjoint": "Schroedinger, on the other hand, ...

12

is any basis a set of eigenfunctions of some observable? This depends on what exactly you want to count as an 'observable' and what you want to rule out. The first example you need to consider is simply the identity operator $\mathbb I$, for which any nonzero vector is an eigenvector with eigenvalue $1$. (Physically, this corresponds to the number $1$ $-$ ...

11

Mass-squared is a Hermitian linear operator, it's a Casimir operator $\hat{C}_{1}=\hat{P}_{0}\hat{P}_{0}-\hat{P}_{i}\hat{P}_{i}$ for the Poincare group. It's Hermitian because the translation generators $\hat{P}_{\mu}$ are Hermitian. It commutes with all the generators of the Poincare group and so it's eigenvalues (mass-squared) are constant on each ...

11

As Lubos has mentioned $QP-PQ=i\hbar$ is one of the basic requirements of quantum mechanics. Classically observables are functions of variables $q$, and $p$ and Poisson bracket relation read $\{q,p\}=1$ (note that $\{q,p\}$ is unitless quantity ) In QM observables are required to be hermitian operators (so that they can have real eigenvalues). In ...

11

Every observable in the technical or mathematical sense (linear Hermitian operator on the Hilbert space) is, in principle, observable in the physical operational sense, too. That's why it's called this way. Magnetic fields may be measured, for example, by compasses. Analogous methods exist for electric fields, scalar fields, or any other fields. For example,...

11

Given a quantum system with associated Hilbert space $\mathcal H$, the set of all self-adjoint bounded operators is $\newcommand{\bh}{\mathcal B(\mathcal H)_\text{sa}}\bh$. In general, only a small subset of $\bh$ will represent physically observable operators. For infinite-dimensional systems, $\bh$ is huge and there's no hope ever finding experiments for ...

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Just open any string text which has a discussion of the relativistic point particle. http://arxiv.org/abs/0908.0333 - Section 1 for example or Green, Schwartz, Witten Volume 1 Punchlines: 1) Time can be introduced as an operator but you need to introduce a 'proper time' parameter for which the system evolves with. In doing this you introduce a gauge ...

11

Hermitian operators (or more correctly in the infinite dimensional case, self-adjoint operators) are used not because measurements must use real numbers, but rather because we almost always decide to use real numbers. As the OP mentions at one point, you might choose to use complex numbers to label a two-dimensional screen, and in that case you'll be able ...

10

Several reasons: Orthogonal functions arise naturally in the study of Sturm-Liouville theory which includes many classical and quantum system mathematical models; More generally, it is the class of normal operators (and an important special case self adjoint operators) which the spectral theorem most readily works and is most complete for. The eigenvectors ...

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