Tag Info

Hot answers tagged

13

An explanation for the motivation of Schrödinger's that I have heard is similar and is an extension of your own. Recall energy conservation, where the total energy $E$ is the sum of kinetic $K$ and potential energy $V$. $$K+V=E$$ We then multiply both sides by $\Psi$ and assume it has the wave form $\Psi \sim e^{i(kx-\omega t)}$. Where $k$ is the wave ...


10

"Where did we get that (equation) from? Nowhere. It is not possible to derive it from anything you know. It came out of the mind of Schrödinger. ” —Richard Feynman Now, if you want to see how Schrödinger got to this equation, I would suggest to read his own paper on the subject. In my opinion, and I am only an amateur in science history, the cleanest ...


6

The very minimum that a wavefunction needs to satisfy to be physically acceptable is that it be square-integrable; that is, that its $L_2$ norm, $$ \int |\psi(x)|^2\mathrm d x, $$ be finite. This rules out functions like $\sin(x)$, which have nonzero amplitude all the way into infinity, and functions like $1/x$ and $\tan(x)$, which have non-integrable ...


4

The Schrödinger picture and the Heisenberg picture are unitarily equivalent. None is more accurate, more fundamental, or in any other objective sense "better" than the other.


3

This question is about Guassian wave-packet propagation and the corresponding Green's function in ordinary quantum mechanics. Assuming $\hbar=m=1$ for simplicity, consider the solution (with its initial condition) to the following Schrodinger equation: $$i\partial_t G=-\partial^2_x G \\ G(t=0,x)=\delta(x) $$ Now assume the Fourier ansatz for $\psi$: ...


3

If you are talking about the time independent Schrödinger equation, it's not a trivial question as it may seem, as the comments suggest. I will restrict the answer to the one-dimensional case, since multiply connected domains in higher dimensions give some additional problems. Not all functions $\psi$ that are solutions of the equation ...


3

I) Let us for simplicity put the physical constants $\hbar=1=m$ to one. OP is considering the usual transcription $u(r)\equiv rR(r)$ of the 3D radial TISE into a 1D TISE, $$\tag{A} - \frac{1}{2} u^{\prime\prime}(r)+U_{\ell}(r)u(r) ~=~E u(r),$$ where the total potential energy $$\tag{49.8b} U_{\ell}(r)~:=~ U(r) + \frac{C_{\ell}}{2r^2} $$ is a sum of a ...


3

The origin of the eigenvalue equation $H\phi=E\phi$ is the separation ansatz $$\psi(x,t)=\exp{\left(-i\frac{E}{\hbar}t\right)}\phi(x)$$ If you conjugate this, this will obviously change the sign of the exponent and therefore you will the same eigenvalue. What you are trying to state would be something like "if $\lambda$ is an eigenvalue of $H$, so is ...


2

As you said, a coherent state is defined by the equation $$a \Psi_{\alpha} = \alpha \Psi_{\alpha} \, .$$ Therefore, to check whether a particular expression is a coherent state, just act $a$ on it and see if you get the same thing back multiplied by a complex number. Let's try it $$ \begin{align} a \Psi_{\alpha} &= a \left[ e^{-|\alpha|^2/2} ...


2

Without magnetic field, spin is conserved. That means, if at the moment you had state with arbitrary spin direction, it will remain the same, apart from phase factor (which does not affect probabilities). Introduction of magnetic fields (for example, Zeeman term $- \mu \vec \sigma \vec B$) breaks symmetry -- only spin along z-axis is conserved. For example, ...


2

In your derivation you've implicitly assumed that the wavefunction does not change its values when you go to the Galilean boosted frame. In other words, you've assumed $\psi'(x', t') = \psi \bigl(x(x',t'), t(x',t') \bigr)$. However, this isn't right. The wavefunction encodes information about a particle's momentum, so when you go to a different frame the ...


2

Consider the following proof by contradiction: Suppose you find a state with $E < -V_0$, then its kinetic energy becomes negative at every point $x$ (classically the velocity is imaginary at every point), which means the whole wavefunction (at all $x$) is an evanescent (exponentially decaying) wave, but such a solution is not a physically stable solution ...


1

They are referring to the Schrodinger and Heisenberg "pictures." In the Schrodinger picture states change over time and operators remain constant over time. In the Heisenberg picture operators change over time, and states do not change. Both pictures give the same numerical answers to any given problem, because they are unitarily equivalent.


1

I will answer this part In addition, we know that the Hamiltonian represents the sum of kinetic and potential energy in a system.However, I'm not quite sure why, intuitively, the time dependent version of the Schrodinger equation becomes Hψ=iℏ ∂/∂t ψ(r,t). Quantum mechanics was developed slowly, because experiments showed that light came in quanta ...


1

Two points relevant here. First, of course, we can only put two electrons in any orbital - one with spin up and one with spin down because of the Pauli exclusion principle. Secondly, When all other things are equal the state with highest multiplicity gives the lowest energy.... For example, in the carbon atom there are two electrons in the 3 2p orbitals. ...


1

This seems like a reasonable homework-like question, so I'll provide a hint. Realize that, for $x>0$, both the simple and semi oscillators have the same potential $V(x)$. For the simple oscillator, draw out the first few of the energy eigenfunctions $\psi_n(x)$. Now for the semi-harmonic oscillator, think about what the boundary condition on any ...


1

The wavefunction depends on the environment - e.g. width of a square well in one dimension - so that you should not be looking for a particle, but an environment with a particle in it. Given that you are interested in this as a theoretical project/thought experiment then depending on what you want in your wave function it may be possible to find a potential ...



Only top voted, non community-wiki answers of a minimum length are eligible