## Posts

### 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

### Schrödinger equation in momentum space

Consider a quantum particle of mass $$m$$ moving in an $$n$$-dimensional space in the presence of an external (scalar) potential $$V(\boldsymbol{x})$$. The Hamiltonian governing the motion is $H = \frac{\boldsymbol{P} \cdot \boldsymbol{P}}{2 m} + V(\boldsymbol{X}) \,,$ where $$\boldsymbol{X}$$ and $$\boldsymbol{P}$$ are the position and momentum operators, respectively. The time-dependent state of the particle $$| \Psi_t \rangle$$ satisfies the Schrödinger equation $i \hbar \frac{\partial | \Psi_t \rangle}{\partial t} = H | \Psi_t \rangle \,.$ In position representation, we have $\boldsymbol{X} = \boldsymbol{x} \,, \qquad \boldsymbol{P} = -i \hbar \frac{\partial}{\partial \boldsymbol{x}} \,,$ and so the Schrödinger equation has the following familiar form: \[i \hbar \frac{\partial \psi(\boldsymbol{x},t)}{\partial t} = -\frac{\hbar^2}{2 m} \frac{\partial}{\partial \boldsymbol{x}} \cdot \frac{\partial \psi(\boldsymbol{x},t)}{\partial \boldsymbol{x}} + V(\boldsymbol{x}) \