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What does measurable mean ? It means that one can do an experiment and get a value for a+ib , the complex number. A complex number to be measurable one should be able to measure a value at the same time for a and b and put a point on the complex plane. This means two independent variables, a and b can be measured and a point defined. In quantum mechanics ...


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Angular momentum is that which is conserved under rotations. Equivalently, the angular momentum operators are the generators of rotations. This holds both classically and quantumly by (versions of) Noether's theorem. Defining "angular momentum" as $\vec x \times \vec p$ classically and then showing that it is conserved is doing it the wrong way around from ...


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I) Well, $r=0$ is the boundary of a $d$-dimensional spherical coordinate patch, i.e. an artifact of our choice of coordinate system, but $r=0$ does not correspond to a physical boundary per se, other than what we can deduce from the TISE. The mantra is that a boundary condition (for finite $r$) should only be imposed if it is a consequence of the TISE. See ...


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A slight expansion on danimal's comment: you can generally get the state $\psi(x,t)$ from the $\psi(x,0)$ you provided by operating on it with the unitary time evolution operator $\exp(-i \hat{H} t/\hbar)$. Since you know the eigenstates, you can write the Hamiltonian in a diagonal basis and this operator will appear to multiply $\psi_n$ simply by $e^{-i E_n ...


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First I want to point out that most of these questions do not bring up issues specific to Bohmian mechanics. That's not a criticism, I'm just pointing out that these notations and concepts are already employed in standard quantum mechanics, or even in classical mechanics. I am going to answer this a little casually, and then make my answer "community ...


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Suppose $\psi$ satisfies the (dimensionless) time-dependent Schrödinger equation: $$ i\frac{\partial\psi}{\partial t}=-\frac{\partial^2\psi}{\partial x^2}+V(x)\psi $$ It will also satisfy the conjugate equation: $$ -i\frac{\partial\psi^*}{\partial t}=-\frac{\partial^2\psi^*}{\partial x^2}+V(x)\psi^* $$ Now consider how the normalisation changes over time: ...


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The solution you give is unphysical, because it cannot be normalized (as you noticed). This happens a lot in physics (e.g. try solving the wave equation in cylindrical coordinates: you'll get Bessel functions that rise to infinity at the origin, which you'll also discard if the origin is part of the solution domain). Unphysical solutions are discarded, and ...


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Eigenstates aren't the only allowed physical states. It's a postulate of quantum mechanics that the most general quantum state can be written as a superposition of eigenstates of some operator (the Hamiltonian for instance). For instance $\Psi(x)=\sum_nc_n\psi_n(x)$ is a general quantum state for a particle in a box, where $\psi_n(x)$ are the energy ...


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It actually is the very essence of the QM. In short, when we observe a superposed state, the probability of observing specific eigenvalue is the square of the norm of the corresponding eigenstate in the superposed state. And this is more like a postulate, rather than a mathematical derivation. For example, particle in a box has discrete eigenvalues, bounded ...


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Further to Anna V's answer, in the case of an electron there is an important physical meaning to the "lack" measurability of the phase of the electron's wavefunction. This is because the electron is coupled to the electromagnetic field. And, if one models this by the Minimal Coupling between the electron and the electromagnetic field, one gets the ...


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For any arbitrary collection of such travelling waves will always be a wave envelope that retains the same shape as the collection of waves propagate? No, it will not. For example, a Gaussian wave-packet will spread out in time. Wave packets are used to represent localization of particles in Quantum Mechanics.Group velocity will give the physical velocity of ...


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The wavefunction $\psi(x_0, x_1, t)$ gives the probability of finding one electron between $x_0$ and $x_0+dx_0$ and the other between $x_1+dx_1$: $$P(x_0, x_0+dx_0; x_1, x_1+dx_1; t) = \vert\psi(x_0, x_1, t)\vert^2dx_0dx_1$$ We expect the brightness of the scintillation at $y$ to be given by the probability (density) that either electron is found at $y$: ...


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SUMMARY OF EDITED VERSION: You cannot place any conditions on $V(x)$ and $E$ that guarantee that solutions to the time-independent Schrödinger equation are normalizable, for something of a silly reason. Initial, partial answer: If the potential is bounded below by some value $V_\text{min}$, then a solution to the time-independent Schrödinger equation ...


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Since $V(x)$ is bounded from above we have three possibilities. Either it oscillates at infinity with an upper bound, or it asymptotes to a constant $<E$ or it diverges to $-\infty$. Since we are interested in $x\rightarrow\infty$ we may average the oscillation in the first case to the mean, and if it diverges then we concern our selves with the leading ...


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The two questions are slightly different. Each individual measurement of $L^2$ or $L_z$ will return an eigenvalue. In this case, you have only one possible measurement for $L^2$ (corresponding to $l=1$), but you have two possible measurements for $L_z$; 2/3 of the time you'll get $m=1$, and 1/3 of the time you'll get $m=0$. The expectation value, on the ...


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For a normalised linear combination of (orthogonal) states like this, the probability of measuring one of them is the absolute square of the coefficient in the combination: If $$\Psi = a_1\psi_1+a_2\psi_2+...$$ where $|a_1|^2+|a_2|^2+... = 1$ then $$P(\psi_1) = \left|a_1\right|^2, P(\psi_2) = \left|a_2\right|^2$$etc. Slightly more generally, if you know ...


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It seems to me that you are making some confusion. The problem with the passage [1] (and [2]) that you outline is that you are not allowed to do that (on a rigorous level) if the operator has continuous spectrum, for there are no corresponding eigenvectors on the Hilbert space (and it is wrong also on a non-rigorous level as pointed out by others). Anyways, ...


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I suspect your text is taking $$\hat x=x\times \;,\qquad \text{ and } \qquad \hat p_x=\frac {\hbar}i\frac d{dx},$$ as postulates (it only holds in the Schrödinger Picture and with the Position Representation). And what it is saying is that it expects you to take any other observable $O$ and write it as a function of $t$, $x$, $p_x$, etcetera and replace ...



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