New answers tagged

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. ...


0

There are few choice here. One length scale is the Bohr radius $a_0$. By using this scale, the corresponding frequency scale and energy scale can be defined as $a_0 = \sqrt{\hbar/(2m \omega_0)}$ and $E_0 = \hbar \omega_0$ respectively. Substitute these back, you will get the dimensionless equation. Suppose you are trying to find the ground state, you can ...


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You can substitute: $$u(r) = f(\lambda r)$$ in the differential equation and choose $\lambda$ so that the left hand side becomes dimensionless up to some overall factor (you can then absorb what's left into E and f). You can also reason as follows. You know that $\lambda= \frac{\hbar}{m c}$ has the dimensions of a length, it is the Compton wavelength of ...


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 $v\ll c$. ...


0

$ψ$ and $ψ^*$ are not the same but are mutually connected.Taking a complex conjugate means finding the reflection of a point in the Argand plane about the real axis. So you cannot simply put $ψ$ as a substitute for $ψ^*$ and vice versa. So, if you need to replace $ψ$ by $ψ^*$, you need to put a negative sign for where you see the operator $i$ as an image ...


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 ...


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 ...


0

I figured it out. The problem was that I was simply adding the two phase shifts, ignoring the periodicity in the tangent and arctangent functions. In order to get the proper result/plot, one can resort to the addition formula for tangents $$\delta=\delta_1+\delta_2=\arctan ...


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 ...


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 ...


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. ...


0

You can also use that the shift in the energy eigenvalue due to a perturbation to first order is the expectation value of the perturbation in the unperturbed state. If you multiply the kinetic energy term by $\lambda$ and the potential energy term by $\mu$, you have essentially the same Hamiltonian, so you can write down the ground state energy without much ...


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 ...


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 ...


1

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

$V(x)$ 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., ...


0

The potential $V(x)$ used in Schrodinger's equation is used to either model the motion of particles in real systems (in which case the parameters of the potential depend on the particle properties, from Coulomb forces where the property of interest is electric charge and spin or nucleon-nucleon interactions where isospin and spin is important) through to ...


0

The potential energy definitely depends on the properties of the particle. For example, if you were using electric fields to contain the particle, the potential energy would depend on the charge of the particle; if you were using gravity to contain the particle, the potential energy would depend on the mass of the particle. When there is just one particle ...


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 ...


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What we want us not just $$ \int|\Psi (x)|^2\;dx $$ For example $$ p=\int Ψ^*(x)(- i \hbar)\; \text{grad} Ψ(x)\;dx $$ If the wave function is not differeciable $\text{grad}$ will come into a terrible result



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