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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.

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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 ...

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The superposition principle of quantum mechanics is not destroyed by quantum (hamiltonian) unitary evolution operator $U(t_0,t) = \mathrm{e}^{\mathrm{i}H (t - t_0)}$ as per @ACuriousMind's answer. Event if you dont know of the evolution operator in terms of the hamiltonian (which can be derived easily from the Schrodiger equation), still the fact that the ...

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The time evolution operator of a quantum system is (in units with $\hbar = 1$) $$U(t_0,t) = \mathrm{e}^{\mathrm{i}H (t - t_0)}$$ and the "stationary states" are the eigenstates of this operator, i.e. eigenstates of the Hamiltonian. If you are given a collection of stationary states (not a basis of the space, mind you) $\{\lvert \psi_E \rangle\}$ with $H ... 0 I want to ask if it is reasonable that I use the Dirac-Delta function as an intial state ($\psi(x,0)$) for the free particle wavefunction and interpret it such that I say that the particle is exactly at$x=0$during time$t=0$? No, because the delta function is not compliant with the Born interpretation of the function$\psi$. Evolving function that is ... 1 That is indeed how you would go about it. Note, however, that there is nothing to guarantee that the solution is going to be reasonable, or that the integral even exists. In fact, because the Schrödinger equation is time reversible to a large extent, you are essentially guaranteed to not end up in physical states. One thing to note is that the frequency ... 5 Consider evolution of gaussian wave packet. Its wave function in position representation looks like: $$\Psi(\vec r,t)=\left(\frac a{a+i\hbar t/m}\right)^{3/2}\exp\left(-\frac{\vec r\cdot \vec r}{2(a+i\hbar t/m)}\right).\tag1$$ Corresponding relative probability density is$$P(r)=|\Psi|^2=\left(\frac a{\sqrt{a^2+(\hbar ... 2 As per Rob's suggestion, I decided to make this an answer. The Answer The "vanilla" Schrodinger's equation (from non-relativistic QM) does not describe a spin-1/2 particle. The plain, old Schrodinger's equation describes a non-relativistic spin-0 field. Case Studies If we pretend the wave function is a classical field (which happens all the time during ... 0 Schrödinger simply does not account for spin. For spin 1/2, you need the Pauli equation or Dirac. A spinless particle, meaning spin zero such as the Higgs boson or pions (ignoring their internal quark/gluon structure) is described by the Klein-Gordon wave equation. I've read a claim that, to the extent that it does describe a particle with spin, ... 8 The relationship that Sakurai seems to be talking about is a mathematical one: both the Schr:odinger equation in wave mechanics and models of other physical systems, like small vibrations of a stretched elastic membrane, reduce to solving eigenvalue problems with elliptic differential operators. In the quantum mechanical setting, the operator is the ... 1 By observations we know that only whole electrons are distributed in specific hydrogen and helium areas. These areas have certain probabilities stay. For description of these areas there are a lot of rules (Hund, ...) and principles (Pauli, ...) and quantum formulas. That allows to predict more precise statements. But that can not hide the fact that the ... 1 John Rennie gave a nice answer based on the De Broglie hypothesis, however he didn't try the hard part: "Why do only a certain number of electrons occupy each shell?" so let me try! In quantum mechanics particles are described by wave functions. All the observable properties of a particle (like its position) are related to the square of the wave function, ... 7 First of all, strictly speaking, electron shells (as well as atomic orbitals) do not exist in atoms with more than one electron. Such physical model of an atom is simplified (and often oversimplified), it arises from a mathematical approximation, which physically corresponds to the situation when electrons do not instantaneously interact with each other, but ... 2 A big part of it can be explained by combining the constraints of quantum mechanics with the geometry of angular momentum. For the special case of the hydrogen atom, it turns out that when you solve the equations of motion for an electron near a proton, you can't give the electron any old energy. There's a set of energies that are allowed; all others are ... 15 Any answer based on analogies rather than mathematics is going to be misleading, so please bear this in mind when you read this. Most of us will have discovered that if you tie one end of a rope to a wall and wave the other you can get standing waves on it like this: Depending on how fast you wave the end of the rope you can get half a wave (A), one wave ... 0 I can sniff a lot of familiarity with the ''HOW'' answers from what I interpret about you from your post, so I'll only focus on the objective point - ''WHY''. It turns out that it is possible to meaningfully describe nature by postulating that any object would tend to be in the minimum energy state possible under a given set of physical conditions. So, ... 2 This wavefunction is an idealization of a wave function that has very very steep, but not discontinuous, behavior at$0$and$a/2\$. You are right, the wavefunction as written is not a proper wave function. That's a good observation. Often in physics the math is much easier if a real situation is modeled by one that is close to it, but mathematically more ...

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