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11

I) The substitution $f=r\psi$ is the standard substitution to get a radial 3D problem to resemble a 1D problem, see e.g. Ref. 1. II) From the perspective of the normalization of the wavefunction $\psi(r)$, a $1/r$ singularity of $\psi(r)$ at $r=0$ is fine because $|\psi(r)|^2$ is suppressed by a Jacobian factor $r^2$ coming from the measure in 3D spherical ...


5

Yes. Yes, it is true. More generally, for any state $\lvert \psi \rangle$, the probability to find it in the state $\lvert\phi\rangle$ is $\lvert \langle \phi \vert \psi \rangle \rvert^2$, and so, since $\lvert \mathrm{e}^{\mathrm{i}\phi_n}\rvert^2 = 1$, phases do not influence probability in this case.


5

This is the same notation that you'll find in Weinberg's books. $$(\psi, \chi)$$ is the inner product of the two states $\psi$ and $\chi$, and corresponds to $$\langle \psi \mid \chi \rangle$$. So, the above corresponds literally to $$ \frac{1}{\sqrt{\langle \psi_k \mid \psi_k \rangle}} \left| \psi_k \right>$$ This new object is just the normalized ...


4

It should be understood that physics - at least in it's current form - does not provide answers to "Why these laws?" questions. It can only describe an emergent law from a deeper and more fundamental one. Quantum theory is so far the most fundamental framework we have, so there is no more fundamental "reason" to describe it's structure aside from finding ...


3

It's not really a product of functions, it's a tensor product. In a sense, you have two systems: The spatial system and the spin system. The combined system is thus the tensor product of the two spaces by the fundamental postulates and an eigenfunction is a (tensor!) product of eigenfunctions. Not that an addition of two wave functions wouldn't make ...


2

The general form, equivalent to eqn (5) but for outside the well, is $\Psi_o = A_o \dfrac{e^{-k_o r}}{\sqrt{r}} + B_o \dfrac{e^{k_o r}}{\sqrt{r}}$ But just like we know $\Psi_i$ is not infinite at the origin, we also know that $\Psi_o$ doesn't go to infinity at large values of r. So we know $B_o$ must be 0.


2

The "wavefunction" way of talking about things is a special case of the more abstract "Hilbert space" formulation. The abstract formulation says that states live in a Hilbert space, that is a complex vector space with an inner product (plus some technical assumption about completeness). The Hamiltonian is then a linear operator on that space. The way you ...


1

A free quark wavefunction does not exist. Instead, inside nucleons quarks are relativistic and asymptotically free, which means that they only behave like individual particles for sufficiently large momentum/energy transfer. Imagine a classical solid state analog: if you apply a very small, slowly acting force to a single atom of a crystal, the force will ...


1

First of all, not every Hamiltonian admits a basis of eigenvectors. An operator has to be either compact or with compact resolvent to admit a basis of eigenvectors. With those type of operators, eigenvalues can accumulate (i.e. being distinct, but getting arbitrarily close) only at zero (compact) or infinity (compact resolvent). A discrete eigenvalue can ...


1

Evidently this is a homework problem. One way to approach the integral is to look at the symmetry of the integrand. To get started: the integrand has two factors. one of them, $y$, is odd under inversion of the axis. Once you have the symmetry of the integrand, you can make quick progress with the integral. Make a sketch of the integrand. That will ...


1

I am not aware of the exact context of your work, but anyway, only case that I recall having read on "coherent wave functions", was in the quantum mechanical study of harmonic oscillators, where the states (wavefunctions) that minimized the uncertainty between position and momentum of the oscillator, without undermining Heisenberg's uncertainty principle, ...


1

'Coherent wave function of macroscopic body' usually refers to wave functions such as found in superconductors, superfluids and Bose-Einstein condensates. Usually, owing to numerous sources of decoherence caused by coupling to environment, it is formidably difficult to maintain the coherence of a wave function describing a macroscopic body. By 'coherence', ...


1

The simplest answer to the question is that the particles of concern (electrons or atoms) are all in the same single-particle state with the same phase. Of course this is not possible for electrons themselves, but they can form Cooper pairs and then are able to, in a sense, occupy the same single-particle state. The explanation I have given here is a little ...


1

See for example Topics in Koopman-von Neumann Theory by D. Mauro. This should be one of the most extensive overviews of KvN Theory, it also contains some examples of applying this theory to some well known problems such as Aharonov-Bohm Effect.


1

It is better seen in the Heisenberg representation. Physical quantities, Observables, are represented by hermitian linear operators. Equation of movement is then (for a non-relativistic massive particle) : $$ m \dfrac{d^2\hat X(t)}{dt^2} = - \dfrac{\partial V(\hat X)}{\partial \hat X}(t) \tag{1}$$ with the quantization conditions : $[\hat X(t),m ...



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