Tag Info

New answers tagged

2

Schrödinger's Wave Equation is an application of Hamiltonian Mechanics. Unlike Newtonian Mechanics, Hamiltonian Mechanics relies on knowing about the things that contribute to the energy of the system. If you know the things which contribute to the energy of a system, then you can determine things like forces, accelerations, and positions. (All through the ...


1

The Hamiltonian for a particle with charge $q$ is $$H = -\frac{\hbar^2}{2m} (\nabla - q\vec A)^2 + q\phi$$ where $\vec A$ is the vector potential and $\phi$ the scalar potential. So the charge appears explicitly.


2

The hamiltonian involves potential energy! So the effect of charge is inherently involved in the Schrodinger equation. If you write down the time dependent Schrodinger Equation, you will see that.


0

As you said, you want that your wave function is normalized. So, it is not correct that the solution outside the wall is $$\psi=Cexp(\mu x)+Dexp(-\mu x).$$ In fact this function diverges for $x\rightarrow\pm\infty.$ In order to keep your wave function normalized, you must use one solution for $x<-a$ and another one for $x>a.$ For $x<-a$, you ...


1

You should show your work, but my guess is that you have to notice the change of variables: $$\frac{d\chi}{dx}=\frac{d\chi}{d\xi}\frac{d\xi}{dx}$$ You need to do this a second time (using the derivate of a product. See if you can continue from there.


0

If $E< V(x) $ everywhere, and if we assume that the kinetic energy operator $T=\frac{p^{\dagger}p}{2m}$ is a (semi)positive operator, then the TISE implies $$ 0 ~\leq~ \langle \psi | T | \psi \rangle ~=~ \langle \psi | (E-V) | \psi \rangle~<~ 0, $$ which is impossible. Here $H=T+V$ is the Hamiltonian operator.


0

This particle with have an unphysical wave function which blows up (as can be quite easily derived). Therefore, in quantum mechanics, we do not have any particles with $E<V_\text{min}$.


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 ...


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. ...


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 ...


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 ...


0

Tong's lecture notes cover this in detail. In particular, your lagrangian is easy to obtain from a complex scalar field that obeys the Klein-Gordon equation in the non-relativistic limit. This lagrangian does lead to a conserved Nöther current of the same form as that of Schrödinger's, but this does not have the interpretation of conserved probability, ...


1

The lecturer's answer assumes $p$ can be positive or negative, and your answer assumes $p=\sqrt{2mE}$ is positive. The factor of $\exp{(-iEt/\hbar)}$ is just the time-dependent part of the separated solution.


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 ...


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 ...


0

The terms in $A$ and $B$ represent waves traveling in opposite directions. Their values are set by the initial conditions of the problem. Some problems will involve waves traveling to the right, others to the left, others a combination of the two. Evidently your book chose to focus on a wave traveling to the right for simplicity. Note that this does ...


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 ...


0

Solution of an initial value problem can be written as integral of the initial function $\psi_0$ multiplied by the propagator of the Schr. equation. Depending on the function $\psi_0$, the integral may or may not be calculable in terms of simple functions. I do not know of any initial function $\psi_0$ and potential $A(t)$ that would admit simple exact ...


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 ...



Top 50 recent answers are included