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6

This probably isn't exactly what you're looking for, but if you're looking for the time-independent bound states of a system, the Fourier grid Hamiltonian method may be applicable. Here is an application of it to the following strange-looking potential well: Here are a few low-energy bound states: And here are some of the high-energy ...


4

This calculation agrees with experimentally measured spectral lines, but why would we expect it to be true, even if we accept that the electron moves according to the Schrodinger equation? After all, there's no particular reason for an electron to be in an eigenstate. Good question! The function $\psi$ does not need to be Hamiltonian eigenfunction. ...


4

The idea here is increasingly complex depending on how deep into modern physics you want to delve, but also key to understanding quantum mechanics. So, I'll give a bit deeper explanation than it seems you've seen, but there's plenty more. It's understood that a photon acts both as a particle and a wave. As a particle it has an amount of energy associated ...


3

This calculation agrees with experimentally measured spectral lines, but why would we expect it to be true, even if we accept that the electron moves according to the Schrodinger equation? Your puzzlement arises because you are putting the cart in-front of the horse. The cart is the theoretical model of quantum mechanics and the horse is the data. As ...


3

The general solution is the first one you gave, $$\psi(x,t) =Ae^{i(kx-\omega t)} + Be^{-i(kx+\omega t)}.$$ To get to the second one, you need to impose appropriate boundary conditions, which essentially means having no incoming particle flux from $+\infty$. Alternatively, you can see the solution $$\psi_k(x,t) =Ae^{i(kx-\omega t)}$$ as general enough to ...


2

To have emission (or absorption) of photons you must have a Hamiltonian that includes those degrees of freedom also. If your system consists of (a) the electromagnetic field and (b) a hydrogen atom, you can specify the state with (a) for each frequency, the number of photons with that frequency and (b) the state of the hydrogen atom, in your favorite way, ...


2

Conservation of energy. If we measure the energy of an atom, we will always report an eigenvalue, because we are forcing it into an eigenstate (this is something like the quantum mechanical definition of measurement). Now suppose that we measure the energy of an atom twice, before and after it emits a photon. For conservation of energy to hold, the energy ...


2

There are some simulation tools available online, but whether they are useful to you depends on the details of your requirements. Check out the list of quantum simulators here. Or this one.


1

For any function of $x$ and $t$ that depends on the combination $x\pm vt$ (for constant $v$ represents a wave with a fixed shape that travels in the $\mp x$ direction with speed $v$. That is to say, $$ x\pm vt={\rm constant} $$ In your wave function, $$ \psi(x)=Ae^{i(kx-\omega t)}+Be^{-i(kx+\omega t)}\tag{0}, $$ the first term represents a right-moving wave ...


1

Here is our interpretation of OP's question: We are essentially talking about the asymptotic form of positive energy scattering states to the time-independent Schrödinger equation (TISE) for the two free regions $x \to \pm \infty$. (As OP notes, scattering states are not normalizable, and therefore do not belong to the Hilbert space. Nevertheless they can be ...


1

In general, both the positive and negative real exponentials are needed, but there are special cases where only one is needed. If the barrier is "one-sided", that is it looks like a step function, then the region of space where the classical energy would be negative extends to infinity. One of those exponentials will approach zero as distance increases ...



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