## Posts

### 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 $$E_{\text{ground}}$$,

### 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