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21

That the eigenfunctions of the free Hamiltonian $H\propto p^2$ are not actually normalizable due to its completely continuous spectrum and therefore cannot be actual quantum states is well-known, although rarely suitably emphasized. (See e.g. Why are eigenfunctions which correspond to discrete/continuous eigenvalue spectra guaranteed to be ...


10

This is indeed possible only in some situations, e.g. when the continuous spectrum is absent (it may also consist of a single point, see Valter Moretti's comment below). A sufficient condition for that to be true is that either the Hamiltonian is compact or it has compact resolvent. Sadly, very few interesting Hamiltonians satisfy that property (an example ...


5

A nice overview of the problem is given in arXiv:1205.3740. I'll summarise the most important points here. Let $d$ be the number of space dimensions. Then the Laplace operator is given by $$ \Delta=\frac{\partial^2}{\partial r^2}+\frac{d-1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\Delta_S\tag{1} $$ where $\Delta_S$ is the Laplace operator on the $d-1$ ...


4

Yes. For example, consider the harmonic oscillator potential $V(x)$, where the ground state has energy $\hbar \omega / 2$. Then the ground state of the potential $V(x) - \hbar \omega / 2$ has zero energy. This works because in quantum mechanics, like in classical mechanics, absolute energies don't matter. You can always add or subtract constants. ...


2

$|q\phi|\ll mc^2$ means that the rest mass of the particle is way larger than the electrostatic potential energy. This is the usual assumption of non-relativistic mechanics, which is equivalent to $v\ll c$. You can see this with the virial theorem: $$ \frac{1}{2}mv^2\sim q\phi $$ which means that $q\phi\ll mc^2$ iff $v\ll c$. ...


2

The lower limit should be negative: $$ \int_{\color{red}-\sqrt{2E/k}}^0 \sqrt{2m\left(E-\frac{1}{2}kx^2\right)} dx + \int _0^{E/mg} \sqrt{2m(E-mgx)} dx =\left(n-\frac{1}{2}\right) \pi \hbar $$ but the answer is yes: this is the correct expression. The general expression for the WKB approximation is $$ \int_{x^1}^{x^2}\sqrt{2m(E-V(x))} ...


2

Yes, the gaussian wavepacket can get narrower as the time passes indeed. It's a matter of phases. You know that a gaussian wavepacket is the superposition of plane waves, each one having a precise wavevector. So it really depends on how you "prepare" this superposition, i.e. on how you set the phase of each chromatic component. If at $t=0$ all the plane ...


1

In one of your preceding statements $$E\hat{x}=\hat{H}\hat{x}$$ you forgot an important caveat: this is only true for eigenfunctions of both $\hat{H}$ and $\hat{x}$. Schrodinger's equation guarantees our wavefunction $\psi$ to be an eigenfunction of $\hat{H}$, but only that function is guaranteed. So $$E\hat{u}=\hat{H}\hat{u}$$ will not be true for all ...


1

Differentiating $$\sigma=\sqrt{a^2+\left({\hbar t \over 2 m a}\right)^2}$$ you get - $$ \frac{\partial \sigma}{\partial t} = {\hbar^2t \over 4 a^2 m^2\sqrt{a^2+\left({\hbar t \over 2 m a}\right)^2}}$$ In the limit of $t \rightarrow\infty $ you get $$ \left(\frac{\partial \sigma}{\partial t}\right) _{t\rightarrow \infty} = {\hbar \over 2 m a}$$


1

There is yet another solution (maybe more elementary)$^1$, with some components of the answers from Qmechanic and JoshPhysics (Currently I'm taking my first QM course and I don't quite understand the solution of Qmechanic, and this answer complement JoshPhysics's answer) the solution uses the Heisenberg Equation: The time evolution of an operator $\hat{A}$ ...



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