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
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