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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 ...


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 ...


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 ...


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 $\psi$: ...


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 ...


0

You ask a couple of different questions: 1) You say "by Heisenberg's uncertainty, we cannot measure the exact momentum or position of a particle/wave ever". No, Heisenberg's uncertainty principle doesn't say that. It says that IF you measure the position of a QUANTUM particle with precision Δx, i.e. you localize the particle within an interval Δx, then the ...


0

Obviously it is discussed the phase of a quasi-classical solution to this exact equation. The quasi-classical solution is approximate; it is an analytical formula for the wave function. It is compared to the exact solution to this equation.


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.


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.


0

I) Ignoring the metaplectic correction/Maslov index, the Bohr-Sommerfeld quantization rule reads $$\tag{1} N ~\approx~\int_a^b \!\frac{\mathrm{d}x}{\pi\hbar} p(x) ~=~ \int_a^b \! \frac{\mathrm{d}x}{\pi\hbar} \sqrt{2m(E-V(x))}, $$ so that $$ \tag{2} \frac{dN}{dE} ~\approx~\int_a^b \! \frac{\mathrm{d}x}{\pi\hbar} {\sqrt{\frac{m}{2(E-V(x))}}}~=~\int_a^b ...


1

Comment to the question (v2): The question formulation conflates on one hand a non-relativistic particle in an infinite potential well/box, which has a quadratic dispersion relation $E \propto p^2$; and on another hand a massless relativistic dispersion relation $E \propto p$. See this Phys.SE question and links therein for a similar misunderstanding.


2

Wave equations have a long history in physics, they are usually equations involving second derivativs, the solutions are sinusoidal ( sines and cosines) and have been used to model classical waves, starting from water, sound, pressure waves, and finally light classically, with Maxwell's equations. When Schrodinger's equation was able to reproduce the Bohr ...


5

The quantum wavefunction is not a wave in the classical sense. In particular, it is not a wave obeying $E = \hbar \omega$, as electromagnetic waves do. The term wavefunction is, in this sense, just a bad naming decision. It is no wave, there is nothing oscillating, and it has no connection whatsoever except the mathematical form to physical waves.


4

Here's a paper with a proof that the ground state must be l=0 for spherically symmetric potentials for a single particle, assuming there's a bound state. Abstract: The variational principle is used to show that the ground-state wave function of a one-body Schrödinger equation with a real potential is real, does not change sign, and is nondegenerate. As a ...


0

Consider (at least perturbative) contribution of the additional effective "potential" in the radial equation when $l\ne 0$, analyze its sign and judge correspondingly. The physical explanation is simple - it is an additional kinetic energy of a rotational motion.


-1

You have two questions which are not quite the same. One is Is there only radial motion in the hydrogen ground state? and the other is If so, that would be a rather odd result. Let us say the electron is at position $(x, 0, 0)$. Then the kinetic energy would be the result of motion either away from the nucleus (direction $+x$) or towards the ...


1

Unfortunately I cannot tell you what went wrong on your first try, since I don't exactly know what you did. However, I sat down and tried to solve the system you describe: We are looking for Eigenstates of the Hamiltonian $$ H = \frac{p^2}{2} + V(x),\qquad \text{where }V(x)=\left\{\begin{array}{ll}\infty & \text{if } x<0 \\ x & ...


0

On average there is no motion at all, i.e., there is no systematic displacements. But there are "fluctuations" with non zero squares averaged. Classically speaking, it is like a Brownian motion in a limited space. But let us set aside a classical picture. Apart from momentum representation of the wave function, there is a simple proof that the electron may ...


2

Yes, you need to put in an extra factor of $\hbar$ to get the units to work out. This is easily seen since, if an operator $T$ generates evolution in a parameter $\tau$, the corresponding evolution operator is $e^{iT\tau}$, so $T\tau$ must be unitless. When $\tau$ is time then $T$ has to be $H/\hbar$. This is a quirk of language usage that you'll have to ...


2

The Schodering equation is a wave equation, not a diffusion equation. While the equations look similar, the $i$ in Schrodinger equation differentiates them; that allows non-decaying oscillatory solutions, which diffusion equations do not allow. That said there are certainly relations between the two. The Schrodinger equation is analogous to the ...


4

They are both densities, that is they have the form $$ \frac{\text{base quantity}}{\text{"volume"}}$$ The base quantity has whatever units it normally has. Amperes for current, or (dimensionless) for probability. The "volume" (which is in quotes because (a) it's not necessarily 3d as in a linear charge density and (b) can exist in abstract space with ...


9

Probability, as such, has no units — it is simply a dimensionless number. A probability density, however, measures probability over a unit of space (or time, or phase space, or whatever), and thus its unit is the inverse of the unit you're using to measure the space the density is distributed over. For example, if you have a probability density over ...


1

I don't know whether Schrodinger proved or guessed the equation with his name, but this equation CAN be derived similarly with the diffusion equation - see Gordon Baym, "Quantum Mechanics" (I will give a more detailed reference when the book is again in my hands). HOWEVER: to the difference from the diffusion equation, the diffusion coefficient in the ...


1

Say we have two different coordinate systems $F=(x,y,z)$ and $F'=(x',y',z')$. Consider the basis spanned by the eigenvectors of the $L_z$ operator for a given $l$, $\{|l \; m\rangle\; ; \;m=-l,\dots,l\}$. Now, one can find the eigenvectors of the $L_{z'}$ on this basis. In general, the eigenvectors of $L_{z'}$ will be linear combinations of the ...


1

I'm not familiar with quantum mechanics yet, but I have taken a course on partial differential equations where we did look at Ficke's law. The form of the equations do seem to be quite similar - the first time derivative is proportional to the second spatial derivative. This implies solutions that end up settling down over time (i.e. steady-state ...


0

I want to write a simple answer to my question based on the knowledge of one-dimensional system; which can qualitatively answer why 2D is similar to 1D while 3D is not. (1). For one dimensional system, there is at least one bound state for pure attractive potential. (Proof can be done by using a Gaussian trail wave function for variational principal) ...


2

The $\phi_i(r)$ form an orthonormal basis of (square-integrable) functions on $\mathbb{R}$, i.e. you should have a relation like $$ \int dr\,\phi_i(r)\phi_j^*(r)=\delta_{ij}. $$ This is what you need in order to expand $\psi(r,t)$ and $\psi^\dagger(r,t)$ the way you did above. You can use this to write the $b_i(t)$ in terms of the $\psi(r,t)$ in the ...


2

To study bound states, we have to find solutions to the Schrödinger time-independent equation $$-\frac{\hbar^2}{2m}\nabla^2\psi+V\psi=E\psi$$ Using separation of variables, in spherical coordinates $$\psi(r,\theta,\phi)=Y^m_l(\theta,\phi)\frac{u(r)}{r},$$ where $Y^m_l(\theta,\phi)$ are the spherical harmonics, the radial part can be shown after substitution ...


3

Why does the statement "any negative potential supports a bound state" hold in 1D, but not in 3D? In short, this is because for a bound state to occur, any positive kinetic energy needs to be fully offset by a negative potential energy. Achieving a large negative potential energy requires the particle to be localized in the volume where the potential is ...


2

The precise theorem is the following, cf. e.g. Ref. 1. Theorem 1: Given a non-positive (=attractive) potential $V\leq 0$ with negative spatial integral $$\tag{1} \int_{\mathbb{R}^n}\! d^n r~V({\bf r}) ~<~0 ,$$ then there exists a bound state$^1$ with energy $E<0$ for the Hamiltonian $$\tag{2} H~=~K+V, \qquad K~=~ -\frac{\hbar^2}{2m}{\bf ...



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