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Let us first restate the mathematical statement that two operators $\hat A$ and $\hat B$ commute with each other. It means that $$\hat A \hat B - \hat B \hat A = 0,$$ which you can rearrange to $$\hat A \hat B = \hat B \hat A.$$ If you recall that operators act on quantum mechanical states and give you a new state in return, then this means that with $\hat ... 17 There is a fair amount of background mathematics to this question, so it will be a while before the punch line. In quantum mechanics, we aren't working with numbers to represent the state of a system. Instead we use vectors. For the purpose of a simple introduction, you can think of a vector as a list of several numbers. Therefore, a number itself is a ... 14 Are we talking quantum mechanics? Then I'd say that a "measurement" is any operation that entangles orthogonal states of the system under consideration with orthogonal states of the environment. "Measurement" is the important thing in most formulations of QM. Colloquially speaking, an observer is something that performs measurements. The only other place ... 12 There are observables corresponding to the light going through both slits. You can write down a basis: "it went through slit A + slit B", and "it went through slit A - slit B". Although maybe you can't detect these observables easily with an experiment, they're perfectly good observables, they're orthogonal, and a clever enough experiment should be able to ... 12 Time is not a variable in Quantum Mechanics (QM), it's a parameter — much in the same way as it is in Classical (Newtonian) Mechanics. So, if you have a Hamiltonian, e.g., for the harmonic oscillator, you have$\omega$as a parameter, as well as the masses of the particle(s) involved, say$m$, and you also have time — even though it's not something that ... 12 In non-relativistic quantum mechanics the mass can, in principle, be considered an observable and thus described by a self-adjoint operator. In this sense a quantum physical system may have several different values of the mass and a value is fixed as soon as one performs a measurement of the mass observable, exactly as it happens for the momentum for ... 11 I don't have a complete answer, but maybe the following is useful for your purposes: Consider the Laplacian$\Delta$on a circular drum of unit radius. As explained on the wikipedia page, the axially symmetric eigenvectors$\Delta u(r) = -\lambda^2 u(r) $are Bessel functions$u(r)=J_0(\lambda r)$. Obviously, the boundary condition requires ... 9 OP asks: Is there any physical meaning to this? Yes, the Pauli matrix$\sigma_j$represents (up to a proportionality factor) the spin in the$j$th direction of a spin$\frac{1}{2}$system. Such system has only two spin states:$\uparrow$and$\downarrow$, with opposite eigenvalues. The square$\sigma_j^2$can no longer see the sign, so it only has one ... 9 Several reasons: Orthogonal functions arise naturally in the study of Sturm-Liouville theory which includes many classical and quantum system mathematical models; More generally, it is the class of normal operators (and an important special case self adjoint operators) which the spectral theorem most readily works and is most complete for. The eigenvectors ... 8 A simple example of non-commutativity is rotations in 3D, cf. figure. Physically, the rotations around the$x$- and the$y$-axis are generated by angular momentum operators$\hat{L}_x$and$\hat{L}_y$, respectively, which do not commute. From the mathematical expressions for$\hat{L}_x$and$\hat{L}_y$, you may proceed with the mathematical derivation, ... 7 As Lubos has mentioned$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\}$is unitless quantity ) In QM observables are required to be hermitian operators (so that they can have real eigenvalues). In ... 7 It's a very good question but the answer is No, there is nothing such as "it went through both slits" observable (i.e. no linear operator that would correspond to this Yes/No question). The reason is that such "information" cannot be observed, not even in principle and not even statistically. Much more generally, there don't exist any observables that would ... 7 There is quite a lot of very important information hidden in the term hermitian. For an operator$A$on a finite-dimensional Hilbert space$\mathcal H$, one can show that there exists an orthonormal basis for the Hilbert space consisting of eigenvectors of the operator$A$. Moreover, one can show that the eigenvalues corresponding to these eigenvectors ... 7 Every observable in the technical or mathematical sense (linear Hermitian operator on the Hilbert space) is, in principle, observable in the physical operational sense, too. That's why it's called this way. Magnetic fields may be measured, for example, by compasses. Analogous methods exist for electric fields, scalar fields, or any other fields. For ... 7 Mass-squared is a Hermitian linear operator, it's a Casimir operator$\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}$are Hermitian. It commutes with all the generators of the Poincare group and so it's eigenvalues (mass-squared) are constant on each ... 6 Of course, mass is an observable, although in simple models it is constant. This is already the case classically. One cannot determine the path of as rocket that burns fuel (which forms a large fraction of its mass) without taking into account that the mass is variable. The same holds in quantum mechanics, whenever the mass is not fixed by the modeling ... 6 There are many experiments, the most famous of them is probably the Cavendish experiment, done in 18th century by a British scientist. I believe it was the first one to measure the force acting between two masses in a laboratory. He calculated the attraction between two lead spheres. http://en.wikipedia.org/wiki/Cavendish_experiment So, I believe 'yes' is ... 6 The statement is that if$K$and$L$are Hermitian operators – which means $$K = K^\dagger, \quad L = L^\dagger$$ and it implies that the eigenvalues of$K,L$are real and the eigenvectors with different eigenvalues are orthogonal to each other, then$i(KL-LK)$(the same as yours) is also Hermitian. This is easily proved by computing the Hermitian ... 6 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 ... 6 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 ... 6 Assume that you have a maximal set$A,B,C,\ldots$and two states$\phi_1$and$\phi_2$with the same set of eigenvalues in that set. Then construct the operator$Z = |\phi_1\rangle\langle\phi_1|$. Convince yourself that it would distinguish between$\phi_1$and$\phi_2$, and that it would commute with all of$A,B,C,\ldots$--- i.e. your original set was not ... 6 Yes, since it is the maximal set of compatible observables, it includes all observables for which$|a\rangle$,$|b\rangle$,$|c\rangle$, etc. are the eigenvectors (I'll use the notation$|\psi_1\rangle$,$|\psi_2\rangle$,$|\psi_3\rangle$etc instead). Hence this includes the observable$D = \sum_k k |\psi_k\rangle \langle \psi_k|$. However$D$has a unique ... 6 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 ... 5 I'll just supplement Prof. Kalitvianski's answer by adding that the mantra « observables as operators » only applies to the special kind of measurement called a « quantum measurement », which always involve amplification. Other kinds of measurement, such as measuring physical constants, are not modelled by observables either, like the speed of light, the ... 5 There exists a set of eigenstates,$\{ |\psi_\lambda\rangle \}$, such that$L|\psi_\lambda\rangle = \lambda|\psi_\lambda\rangle$where$\lambda$is a real eigenvalue and$|\psi_\lambda\rangle$is an eigenstate of$L$. The$\lambda$represents the value of a physical observable associated with the eigenstate$|\psi_\lambda\rangle$. There is currently no ... 5 If$\psi$is a normalized eigenvector of$L^2$and$\lambda$the corresponding eigenvalue then$\lambda=\psi^*\lambda\psi=\psi^*L^2\psi=(L\psi)^*(L\psi)$. Thus$\lambda$is manifestly real and nonnegative. This even holds if$L$is a vector of noncommuting real (i.e., Hermitian) quantities, such as angular momentum. Then we get in the last step$\sum_i ...

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I would actually expect this to be rare, and only generically true when the state of the system corresponds to an eigenstate. This simply because, for a state $\psi = \sum a_{n}\lvert\phi_{n}\rangle$ with eigenvalues $V_{n}$, you would have $\langle V\rangle = \sum V_{n}\lvert a_{n}\rvert^{2}$, which is not constrained to be equal to one of the $V_{n}$. ...

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I don't think this question has been answered. The original poster asked: why is \begin{align} \int dp |p\rangle \langle p| = 1, \end{align} up to unimportant normalization issues that have been discussed extensively (but are off the point). The above statement is equivalent to the question of completeness of a set of states on an Hilbert space (which ahmed ...

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There is no time operator in quantum mechanics. At least, there's no nontrivial time operator. You could have an operator whose action is just to multiply a function by $t$, but time is a parameter in QM, so the operator will never do anything more complicated than that. Its eigenfunctions wouldn't be terribly useful either because they would just be delta ...

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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 ...

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