Paper 2, Section II, B

Principles of Quantum Mechanics
Part II, 2019

(a) Let i|i\rangle and j|j\rangle be two eigenstates of a time-independent Hamiltonian H0H_{0}, separated in energy by ωij\hbar \omega_{i j}. At time t=0t=0 the system is perturbed by a small, time independent operator VV. The perturbation is turned off at time t=Tt=T. Show that if the system is initially in state i|i\rangle, the probability of a transition to state j|j\rangle is approximately

Pij=4iVj2sin2(ωijT/2)(ωij)2P_{i j}=4|\langle i|V| j\rangle|^{2} \frac{\sin ^{2}\left(\omega_{i j} T / 2\right)}{\left(\hbar \omega_{i j}\right)^{2}}

(b) An uncharged particle with spin one-half and magnetic moment μ\mu travels at speed vv through a region of uniform magnetic field B=(0,0,B)\mathbf{B}=(0,0, B). Over a length LL of its path, an additional perpendicular magnetic field bb is applied. The spin-dependent part of the Hamiltonian is

H(t)={μ(Bσz+bσx) while 0<t<L/vμBσz otherwise H(t)= \begin{cases}-\mu\left(B \sigma_{z}+b \sigma_{x}\right) & \text { while } 0<t<L / v \\ -\mu B \sigma_{z} & \text { otherwise }\end{cases}

where σz\sigma_{z} and σx\sigma_{x} are Pauli matrices. The particle initially has its spin aligned along the direction of B=(0,0,B)\mathbf{B}=(0,0, B). Find the probability that it makes a transition to the state with opposite spin

(i) by assuming bBb \ll B and using your result from part (a),

(ii) by finding the exact evolution of the state.

[Hint: for any 3-vector a,eiaσ=(cosa)I+(isina)a^σ\mathbf{a}, e^{i \mathbf{a} \cdot \boldsymbol{\sigma}}=(\cos a) I+(i \sin a) \hat{\mathbf{a}} \cdot \boldsymbol{\sigma}, where II is the 2×22 \times 2 unit matrix, σ=(σx,σy,σz),a=a\boldsymbol{\sigma}=\left(\sigma_{x}, \sigma_{y}, \sigma_{z}\right), \quad a=|\mathbf{a}| and a^=a/a.]\left.\hat{\mathbf{a}}=\mathbf{a} /|\mathbf{a}| .\right]