# Tag Info

7

TL;DR: The supersymmetric partner potential to OP's potential is the constant potential, which is clearly reflectionless. Define for later convenience the constant $\kappa:=\hbar/\sqrt{2m}$. The constant potential and OP's potential are just the two first cases ($\ell=0$ and $\ell=1$) in an infinite sequence of reflectionless attractive$^1$ potentials ...

7

In the examples you've given, the boundary conditions simply say "don't have infinite energy" and "don't be non-normalizable". These don't really have a physical interpretation. Moreover, no boundary conditions ever have an interpretation like "initial position and velocity" because the time-independent Schrodinger equation describes stationary states. ...

4

The choice of boundary conditions fixes the domain and thus a self-adjoint extension of your Schroedinger operator $$-\frac{\hbar^2}{2m} \Delta + V\:.$$ In turn, this choice determines the spectrum of that operator, i.e., the eigenvalues $E$. (There is no analogy with initial conditions here.) No circumstances! Boundary conditions are uncorrelated with ...

4

I) Yes, the time independent Schrödinger equation (TISE) of the form $$\left(-\frac{\hbar^2}{2m} {\bf \nabla}^2 +V({\bf r}) -E \right) \psi({\bf r}) ~=~0$$ is $\mathbb{C}$-linear and invariant$^1$ under complex conjugation. So if the wave function $\psi$ is a solution with finite square-norm, then so will $\psi^{\ast}$, ...

3

You can always change the symbol that stands for a dummy variable, but you can't change its interpretation. Your mistake is tantamount to starting from the equation $$x + 1 = 2$$ which has solution $x = 1$, then declaring $x$ is a dummy variable and replacing it with $-x$, for $$-x + 1 = 2$$ which has solution $x = -1$. This is totally valid, but these two ...

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

Generally speaking the complex conjugate of an operator is not a standard notion of operator theory, though it can be defined after having introduced some general notions. Definition. A conjugation $C$ in a Hilbert space $\cal H$ is an antilinear map $C : \cal H \to \cal H$ such that is isometric ($||Cx||=||x||$ if $x\in \cal H$) and involutive ($CC=I$). ...

2

The Schrödinger equation is $$\left[\frac{1}{2m}(-\Delta)+V(r)\right]\psi=E\psi$$ If we let $\psi=R(r)Y_\ell^m(\theta,\phi)$, we find $$\left[\frac{-1}{2m}\frac{\mathrm d}{\mathrm dr}r^2\frac{\mathrm d}{\mathrm dr}+\frac{\ell(\ell+1)}{2mr^2}+V(r)\right]R(r)=E R(r)$$ Finally, let $R(r)=u(r)/r$, so that $$-\frac{1}{2m} u''(r)+\hat V(r)u(r)=Eu(r)$$ ...

2

To find the instantaneous energy eigenstates, you need to treat $t$ as a parameter and solve the problem for a time independent Hamiltonian depending on the extra parameter $t$. The best way to do that is to complete the square and write the Hamiltonian as: $$H = \hbar \omega (A^{\dagger} A + \frac{1}{2}) - \frac{ f(t)^2}{\hbar \omega}$$ where $$A = a - ... 2 As far as I know, and I'm only an undergraduate student, the boundary conditions in Schrodinger equations are there to hold some special subspace of the Hilbert space of the system or the Hilbert space as a whole. Bound states, for example, form a subspace on the Hilbert space. The boundary condition to that is that \psi\sim e^{-r} at infinity, for every ... 2 As a complement to another answer, I'll demonstrate moving from the coordinate representation (wave function) to the momentum representation. Recalling that$$\Psi(x,t) = \int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t)$$and putting this expression into the (coordinate representation of the) TDSE, we have$$i\hbar\frac{\partial}{\partial ...

2

Generically, any square-integrable function is an admissible wave function, and the space of square-integrable complex functions indeed has uncountable dimension as a vector space over $\mathbb{C}$. And it is also true that the eigenstates of the Hamiltonian span the space of states, and that they are countably many. This is the content of the spectral ...

1

Yes, it is of course possible. Is can be done rigorously, but if you like to have it simple, I will sketch a very nice and intuitive way. It is convenient to understand Schroedinger's equation as an operator equation: $$\frac{p^2}{2m} = E_\mathrm{kin} = E_\mathrm{total} - V$$ Here $p^2$ is proportional to -$\frac{\mathrm d^2}{\mathrm dx^2}$, $V$ might ...

1

Yes, the Probability current is defined as $$J(x,t)=\frac{i\hbar}{2m}\bigg(\Psi\frac{\partial \Psi^*}{\partial x}-\frac{\partial \Psi}{\partial x}\Psi^*\bigg)$$ If the scattering matrix is defined as $$\begin{bmatrix} S_{11}&S_{12}\\ S_{21}&S_{22}\\ \end{bmatrix}$$ The transmission coefficient $T$ is given by $(S_{21})^2$, this is only true in ...

1

As far as I know, the potential energy depends on the properties of the particle itself (its mass, its charge) No, not really. Think of a classical mass-spring system. The potential energy is $$V(x)=\frac{1}{2}kx^2$$ which is independent of the properties of the mass. In some other cases, such as a charged point particle, the potential energy could be ...

1

$V(x)$ is a potential energy function for the system of a particle or particles interacting with a set of constraints. These constraints can be thought of as fields which produce a force on the particle(s) of interest. In the infinite square well (ISW), we examine a particle which has no interaction at all until it gets to some impenetrable constaint, i.e., ...

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The safest way to start with is the representation-free Schrodinger equation, $$i\hbar\frac{\partial}{\partial t}|\Psi(t)⟩= \hat{H}|\Psi(t)⟩.$$ Referring to your case, we take the separable Hamiltonian: $H=\frac{p^2}{2m}+V$ so that $$i\hbar\frac{\partial}{\partial t}|\Psi(t)⟩= \left(\frac{\hat{p}^2}{2m}+V\right)|\Psi(t)⟩.$$ Now is the time the ...

1

You actually want the transpose-conjugate, which applies to everything: $$(i\partial_t |\psi\rangle )^\dagger = (H\|psi\rangle)^\dagger$$ This gives: $$-i\partial_t |\psi\rangle^\dagger = |\psi\rangle^\dagger H^\dagger$$ which reduces to: $$-i\partial_t \langle\psi| = \langle\psi|H$$ The differential operator remains unchanged. Note if your units ...

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