# Tag Info

3

Complex conjugation has nothing to do with charge conjugation. Charge conjugation flips quantum numbers, which don't appear at all in the standard Schrodinger equation. The actual symmetry related to complex conjugation is time reversal. However, to actually perform time reversal, you must also replace $i$ with $-i$, so the time reversed Schrodinger ...

1

Essentially, separation of variables in the time-independent Schroedinger equation amounts to diagonalizing the Hamiltonian. One can see this easily by considering the case where the Hilbert space is finite-dimensional, and the Hamiltonian is a Hermitian matrix. In case of a partially continuous spectrum one gets the same, except that the sum must be ...

2

The spacetime geometry may be indirectly read from the differential operators in the equations controlling other objects – particles and (matter) fields. Schrödinger's equation contains the differential operator $$\Delta = \nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$$ which is the operator ...

1

You're looking for a set of wavefunctions $\psi_n(x)$ that are solutions to $H\psi_n=E\psi_n$. The index $n$ does not refer to a discretized position. For numerical solutions, you will have to discretize $x$, for example with constant steps $h$ (you call them $\Delta x$); then the second derivative will be approximated as $${\partial^2\psi\over\partial x^2} ... 0 It's true. The Schrodinger equation is a differential equation whose solutions are wavefunctions of spin-1/2 particles - and spin-1/2 particles only. The question is quite deep. The Schrodinger equation for a zero-spin particle (which models energy itself), or a spin-one particle (i.e. a photon) is similar but does not have the 1/2 factor in the 'diffusion ... 0 You have the wavefunction \Psi(x) of the particle given in position representation. In this representation, \int_a^b |\Psi(x)|^2 dx gives you the probability to find the particle somewhere in the interval [a, b]. You can convert this function into the equivalent momentum representation (by taking the Fourier transform of \Psi(x)), in which the ... 1 In one of your preceding statements$$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}$$will not be true for all ... 5 A nice overview of the problem is given in arXiv:1205.3740. I'll summarise the most important points here. Let d be the number of space dimensions. Then the Laplace operator is given by$$ \Delta=\frac{\partial^2}{\partial r^2}+\frac{d-1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\Delta_S\tag{1} $$where \Delta_S is the Laplace operator on the d-1 ... 2 Yes, the gaussian wavepacket can get narrower as the time passes indeed. It's a matter of phases. You know that a gaussian wavepacket is the superposition of plane waves, each one having a precise wavevector. So it really depends on how you "prepare" this superposition, i.e. on how you set the phase of each chromatic component. If at t=0 all the plane ... 1 Differentiating$$\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}$$2 The lower limit should be negative:$$ \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))} ...

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 ...

0

I'll assume the question refers to calculating coefficients $c_i(t)$, not transition probabilities under external interaction. Since in Heisenberg representation all relevant quantities are calculated as observable/operator averages, the problem becomes one of choosing the correct observable to represent the final state. The simplest option is its ...

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 ...

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. ...

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