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## New answers tagged schroedinger-equation

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So one detail I omitted from the question was that: $$\psi_{sc}(k)=\frac{g+g I}{2\pi(k^2-p^2)}$$ Where: $$I=\int^{\infty}_{-\infty}\psi(q)dq \space\space\space\space (1)$$ (I had used in arbitrary prescription in the original description of the problem, this is what I obtain before solving for $I$)$$\\$$ Using equation (1) we can solve for I, obtaining: ...

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For a connection between Schr. eq. and complex Klein-Gordon eq., see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein. The complex Klein-Gordon eq. in QFT describes both particle and anti-particle excitations. When scaling to the appropriate non-relativistic one-particle sector to derive the Schr. eq. for the complex ...

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Klein-Gordon (and actually also Dirac) equation is usually considered a classical field equation. To obtain a quantum field theory, you have to quantize it to become an operator on the symmetric (for bosons) Fock space. Then you have again a Schrödinger-type equation for the wavefunction, with an Hamiltonian that embodies the Klein-Gordon dynamics. For a ...

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I think this problem is similar to the problem of finding modes of rectangular dielectric waveguide. In this case, you can use the effective-index method for finding the approximated solution (For your problem, we can call it effective-potential method). For more information about effective-index method see the following articles: Effective-index analysis ...

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You can think about "solving X equations in Y unknowns". When $X>Y$, you generally expect infinite solutions, when $X=Y$ you generally expect a unique solution, when $X<Y$ you generally expect no solutions. This kind of statement is not always mathematically rigorous but you can usually argue it rigorously in specific circumstances. Pick an energy $W$ ...

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A more precise mathematical way of asking your question is: given a (usually unbounded) self-adjoint operator $H$ on a Hilbert space $\mathscr{H}$, am I able to characterize its spectrum? Finding "closed" solutions to the equation you are writing, means finding eigenfunctions of your operator $H$, possibly belonging to the Hilbert space $\mathscr{H}$ (since ...

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An electron with total orbital angular momentum of $L^2 = \hbar^2 l(l+1)$ will experience a centrifugal force in addition to the Coulomb force from the nucleus. The result is that, in the frame rotating with the electron (don't read too much into this), the electron will see an effective potential energy given by: V_l(r) = - \frac{e^2}{4 \pi \epsilon_o ... 2 For 1D potentials, the sequence of bound state energy eigenvalues E_n cannot grow faster than what happens in the case of an infinite well, i.e. E_n cannot grow faster than n^2. 0 I think you need to add a postulate to (3) in order to obtain (1). This postulate it's actually assumed in every picture of the dinamic and it is that the evolution operator is a one group parameter. It is such a natural assumption that it comes no harm in taking it for true. Even in classical hamiltonian mechanics the flow in the coordinate space form a ... 4 The result can be proved in a general way using, well, math. In particular the theory of semigroups of linear operators on Banach spaces (I know that seems advanced and maybe not physical, but it is an elegant way of proving the result you seek ;-) ). Define the Banach space \mathscr{L}^1(\mathscr{H}) of trace class operators over a separable Hilbert ... 1 Just calculate (3) for \hat{A}_H=\hat{U}_H. With \hat{U}_H(t)=\hat{U}^{\dagger}(t)\hat{U}(t)\hat{U}(t)=\hat{U}(t) and \hat{H}_H=\hat{H} (as you said) you get: \frac{d\hat{U}_H}{dt}=\frac{\partial \hat{U}}{\partial t}+\frac{1}{i\hbar}\underbrace{[\hat{H},\hat{U}]}_{=0 \text{(as you ... 4 CAUTION - ANSWER INCOMPLETE There is a gap in my argument (see the send); it relies on the claim that \begin{align} - \hat O^\dagger \hat A = \hat A\hat O \end{align} for all hermitian \hat A implies \hat O = 0 which may not be true. Please comment if you know how to prove this or know of a counterexample. Update. Actually the claim above is ... 0 If this isn't at a higher level than the original lectures (which I haven't seen, actually), the correct way to go from \psi (x,0) to \psi (x,t) in Quantum Mechanics, is to employ the time evolution operator \exp(-i{\hat H}t/\hbar), where {\hat H} is the Hamiltonian, on \psi (x,0), i.e.\psi(x,t) = {\hat T} \left(\psi (x,0)\right) = \exp(-i{\hat ...

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By the simple form of the equation (1) you wrote down, Allan really meant $$\frac{\partial}{\partial t} \Psi(x,t)|_{t=0} = \frac{E}{i\hbar}\Psi(x,0)$$ He just used the notation where $t=0$ is substituted from the beginning but he clearly did mean that $\Psi(x)$ is first considered as a general function of $t$, then differentiated, and then we substitute ...

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Going from your (1) to your (2) is bogus as you suspect. Your (1) isn't even really a differential equation: the independent variable $t$ does not appear. (1) is just using the initial condition to find what the time derivative is at $t = 0$. Indeed to directly conclude that the solution is (2), we would need your stronger assumption. One correct way to ...

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Since Hamiltonian is a linear operator, one could always choose real eigenfunctions. This is because if f is a complex eigenfunction of any linear operator O with eigenvalue e, then its both real and imaginary parts are also O's eigenfunction with the same eigenvalue e. It does not require the operator to be Hermitian. Moreover, this is true for all ...

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The energy operator is obtained via the so-called correspondence principle. This means that one considers the classical expression for the total energy $$\frac{p^2}{2m}+V(x)$$ and replaces the momentum and position variables (numbers in classical mechanics) by the momentum and position operators. $p^2/2m$ is the kinetic energy (it's just another way of ...

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The "Energy operator" in a quantum theory obtained by canonical quantization is the Hamiltonian $H = \frac{p^2}{2m} + V(x)$ (with $V(x)$ some potential given by the concrete physical situation) of the classical theory promoted to an operator on the space of states. Since the core of the quantization procedure is promoting the classical phase space ...

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