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Every purely attractive potential in 1D has at least one bound state

Every purely attractive one-dimensional potential well, no matter how shallow, has at least one bound state. More precisely, this statement can be formulated as follows. Consider a one-dimensional non-relativistic quantum particle of mass \(m\) inside a purely attractive well potential \(V(x)\): the function \(V(x)\) is such that \(V(x) < 0\) for all \(x\), and \(V(x) \to 0\) as \(x \to \pm \infty\). This system has at least one bound state. That is, the ground state energy \(E_{\text{ground}}\) of the Hamiltonian \[H = -\frac{\hbar^2}{2 m} \frac{d^2}{d x^2} + V(x)\] is strictly negative: \[E_{\text{ground}} < 0 \,.\] Proof: [D. ter Haar, Selected Problems in Quantum Mechanics (Academic, New York, 1964).] According to the variational principle, \[E_{\text{ground}} \le \langle \psi | H | \psi \rangle = \int_{-\infty}^{\infty} dx \, \psi^*(x) H \psi(x)\] for any normalized state \(\psi(x)\). Hence, in order to prove the negativity of \(E_{\text{ground}}\),

Virial theorem

Consider a quantum particle of mass \(m\) moving in a \(D\)-dimensional space under the action of a Hamiltonian \[H = T + V \,,\] where \[T = \frac{\boldsymbol{p} \cdot \boldsymbol{p}}{2 m} = \frac{p_1^2 + p_2^2 + \ldots + p_D^2}{2 m}\] is the kinetic energy, and \[V = V(\boldsymbol{x}) = V(x_1, x_2, \ldots, x_D)\] is the potential energy. Here, \(x_j\) and \(p_j\), with \(j = 1, 2, \ldots, D\), are Cartesian components of the particle's position and momentum vectors, respectively, satisfying the standard commutation relation: \[[x_j, p_k] = i \hbar \delta_{jk} \,.\] Suppose further that the potential \(V\) is confining, and that the particle is in a bound state \(| \psi \rangle\) of energy \(E\), i.e. \[H | \psi \rangle = E | \psi \rangle \,.\] The virial theorem states that the expectation value of the kinetic energy is given by \[\boxed{ \langle T \rangle = \tfrac{1}{2} \langle \boldsymbol{x} \cdot \boldsymbol{\nabla} V \rangle }\] Here, \[\langle \cdot \rangle \equi

Lower bound on survival probability

Let \(| \psi(t) \rangle\) be the time-dependent state of a quantum system evolving under the action of a time-independent Hamiltonian \(H\), i.e. \[i \hbar \frac{d | \psi(t) \rangle}{d t} = H | \psi(t) \rangle \,.\] Suppose that initially, at \(t=0\), the system is in some state \(| \psi(0) \rangle = | \psi_0 \rangle\). The autocorrelation function \[P(t) = \big| \langle \psi(t) | \psi_0 \rangle \big|^2\] quantifies the survival probability: \(P(t)\) is the probability that the system would be found in its original state after time \(t\). The survival probability equals unity at \(t=0\) and, generally, decays as \(t\) increases. How fast can \(P(t)\) decay? In particular, can the decay be exponential, i.e. \[P(t) \stackrel{?}{=} e^{-\gamma t}\] with some decay rate \(\gamma>0\) on a time interval \(0 \le t \le T\)? (The assumption of exponential decay is commonplace in back-of-the-envelope arguments. For instance, the number of atoms in a sample undergoing radioactive d

No-cloning theorem

Consider two quantum systems of the same nature. For concreteness, let us take them to be two hydrogen atoms. Suppose that the first H-atom is in an arbitrary unknown state \(| \alpha \rangle\), while the second H-atom is in the ground state \(| 0 \rangle\). Is it possible to construct a perfect cloning machine operating as follows: The machine changes the state of the second H-atom from \(| 0 \rangle\) to \(| \alpha \rangle\) without altering (or destroying ) the state of the first H-atom? More specifically, the machine takes the initial state of the composite system, \[| \alpha 0 \rangle = | \alpha \rangle \otimes | 0 \rangle \qquad \text{(initial state)}\] where \(\otimes\) denotes the tensor product, and transforms it into \[| \alpha \alpha \rangle = | \alpha \rangle \otimes | \alpha \rangle \qquad \text{(final state)}\] module perhaps some physically irrelevant global phase. The no-cloning theorem states that constructing such a machine is impossible. Proof 1 (usin

Mandelstam-Tamm uncertainty relation

The Mandelstam-Tamm enerty-time uncertainty relation states that \[\Delta E \Delta t_A \ge \frac{\hbar}{2} \,,\] where \(\Delta E\) is the energy uncertainty of a quantum system, and \(\Delta t_A\) is the time required for a significant change of the expectation value of an observable \(A\). Derivation Consider a quantum system with a Hamiltonian \(H\). Let \(| \psi \rangle\) be the time-dependent state of the system, and let \(A\) be some observable. The uncertainty in the system's energy and the uncertainty in \(A\) are defined, respectively, as \[\begin{align} &\Delta E = \sqrt{\langle H^2 \rangle - \langle H \rangle^2} \,, \\ &\Delta A = \sqrt{\langle A^2 \rangle - \langle A \rangle^2} \,, \end{align}\] where \(\langle \cdot \rangle = \langle \psi | \cdot | \psi \rangle\) denotes the expectation value. \(\Delta E\) and \(\Delta A\) satisfy the uncertainty relation \[\Delta E \Delta A \ge \frac{\big| \langle HA - AH \rangle \big|}{2} \,.\] Since the rate

Free-particle Gaussian wave packets

Below are basic properties of quantum Gaussian packets describing the motion of a free particle on a line (taken to be the \(x\) axis). Position representation A time-dependent Gaussian wave function, in its most general form, is given by \[\psi(x,t) = \left( \frac{2 \operatorname{Re} \alpha_t}{\pi} \right)^{1/4} \exp \left( -\alpha_t (x - x_t)^2 + \frac{i}{\hbar} p_0 (x - x_t) + i \frac{p_0^2 t}{2 \hbar m} + \frac{i}{2} \arg \frac{\alpha_t}{\alpha_0} \right) \,,\] where \[x_t = x_0 + \frac{p_0 t}{m} \,, \qquad \alpha_t = \frac{\alpha_0}{1 + i \frac{2 \hbar t}{m} \alpha_0} \,.\] Here, \(m\) is the particle's mass, \(x_0\) and \(p_0\) are real-valued parameters, and \(\alpha_0\) is a complex-valued parameter, such that \(\operatorname{Re} \alpha_0 > 0\). The wave function \(\psi\) is normalized, \[\int_{-\infty}^{+\infty} dx \, |\psi(x,t)|^2 = 1 \,,\] and satisfies the free-particle Schrödinger equation, \[i \hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{

Two-dimensional hydrogen atom as a harmonic oscillator

Bound state energy levels \(E<0\) of a two-dimensional hydrogen atom are determined by the Schrödinger equation \[\left( -\frac{\hbar^2}{2 m} (\partial_x^2 + \partial_y^2) - \frac{q^2}{\sqrt{x^2 + y^2}} \right) \psi = E \psi \,,\] where \(m\) and \(q\) are the mass and charge of the electron, respectively, and \(\psi(x,y)\) is the electronic wave function. This equation can be solved in terms of Kummer's confluent hypergeometric function (see this post for details). Here we show how the two-dimensional hydrogen atom can be mapped onto a two-dimensional harmonic oscillator. This mapping has been discussed, e.g., in Quantum Mechanics of H-Atom from Path Integrals . We begin by making a coordinate transformation from \((x,y)\) to \((u,v)\) defined by \[\begin{align} x &= u^2 - v^2 , \\ y &= 2 u v \,. \end{align}\] From \[\begin{pmatrix} \partial_u \\ \partial_v \end{pmatrix} = J \begin{pmatrix} \partial_x \\ \partial_y \end{pmatrix}\] with \[J = \begin{pmatrix