Oct 12, 2022 · If they are, how do you show that $\frac{\delta}{\delta \psi(x)}$, acting on a functional, is indeed the adjoint operator to $\psi(x)$?

In quantum mechanics, the position operator is the operator that corresponds to the position observable of a particle. When the position operator is considered with a wide enough domain (e.g. the space of tempered distributions), its eigenvalues …

Sep 24, 2018 · A simple generalization of this result yields \[\label{e3.69} \int_{-\infty}^\infty f(x)\,\delta(x-x_0)\,dx = f(x_0),\] which can also be thought of as an alternative definition of a delta-function. Suppose that \(\psi(x) = \delta(x-x_0)\).

In quantum mechanics the delta potential is a potential well mathematically described by the Dirac delta function - a generalized function. Qualitatively, it corresponds to a potential which is zero everywhere, except at a single point, where it takes an infinite value.

Jul 11, 2020 · The $\Delta x,\Delta p$ in the Heisenberg uncertainty principle refer to the shape of the states, not to measurements. Although this shape tells you what the outcomes of the measurements can be and how likely each outcome is.

For x \ne 0 x = 0, we simply get the same behavior as for a piecewise constant potential. The interesting behavior is at the delta function. Usually we require \psi' ψ′ to be continuous, but since V (0) V (0) is infinite, \psi' ψ′ can't be continuous, otherwise we would have infinite energy.

Jan 27, 2019 · The state $\psi(x) = \delta(x)$ is a perfectly valid state for the harmonic oscillator to occupy. (With caveats, though: it is not normalizable, so it's not a physically-accessible state. Still, it's a perfectly reasonable thing for the mathematical formalism to handle.)

Dec 1, 2008 · So psi is your particle's state, and x is a dirac delta function-like thingy which is infinite at x0 and zero everywhere else. The dot product is then [tex]<x|\Psi> = \int x \Psi = \int \delta_{x0} \Psi = \Psi[/tex].

Jan 22, 2021 · Let $\psi$ be a test function, i.e. $\psi \in \mathcal{C}^{\infty} _{c}(\mathbb{R})$, then: $$ \int \delta(x^3)\psi (x) dx = \int \delta(y) \psi (y^{\frac{1}{3}}) \frac{1}{3y^{\frac{2}{3}}} dy $$ Note that our new test function $\psi (y^{\frac{1}{3}}) \frac{1}{3y^{2/3}}$ is not smooth, hence we cannot define $\delta(x^3)$ as a distribution.

Mar 9, 2025 · Expand on the introduction of Heisenberg's Uncertainty Principle by calculating the Δx Δ x or Δp Δ p directly from the wavefunction. As will be discussed in Section 4.6, the operators x^ x ^ and p^ p ^ are not compatible and there is no measurement that can precisely determine the coresponding observables (x x and p p) simultaneously.