Quantum Physics Corner

A weekly problem from Issue #1 - Correspondence principle of Quantum Newsletter . A full solution can be found here . Subscribe to Quantum Newsletter to receive new issues in your inbox.

Consider a one-dimensional quantum particle, and let \(X\) and \(P\) represent its position and momentum operators, respectively. \(X\) and \(P\) are assumed to satisfy the canonical commutation relation, \[[X,P] = i \hbar\] with \(\hbar\) being the reduced Planck constant. For the following discussion, it is convenient to introduce dimensionless position and momentum as \[x = \frac{X}{L} \qquad \text{and} \qquad p = \frac{L P}{\hbar} \,,\] where \(L\) is some (chosen for convenience, but otherwise arbitrary) length scale. In terms of \(x\) and \(p\), the canonical commutation relation reads \[[x, p] = i \,.\] The operators \(x\) and \(p\) fulfill the Heisenberg uncertainty relation: \[\Delta x \Delta p \ge \frac{1}{2} \,,\] where \(\Delta x \ge 0\) and \(\Delta p \ge 0\) are uncertainties in the particle's position and momentum, respectively. (The Heisenberg uncertainty relation is a straightforward consequence of a more general Robertson-Schrödinger uncertainty rel

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 \(E_{\text{ground}}\),

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

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

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