Chapter 5

Take the spin Hamiltonian for the hydrogen atom in an external magnetic field \(B_{0}\) in the \(z\) direction to be $$ \hat{H}=\frac{2 A}{\hbar^{2}} \hat{\mathbf{S}}_{1} \cdot \hat{\mathbf{S}}_{2}+\omega_{0} \hat{S}_{1 z} $$ where \(\omega_{0}=g e B_{0} / 2 m c\), with \(m\) the mass of the electron. The contribution \(-\hat{\mu}_{2} \cdot \mathbf{B}_{0}\) of the proton has been neglected because the mass of the proton is roughly 2000 times larger than the mass of the electron. Determine the energies of this system. Examine your results in the limiting cases \(A \gg \hbar \omega_{0}\) and \(A \ll \hbar \omega_{0}\) by expanding the energy eigenvalues in a Taylor series or binomial expansion through first nonvanishing order.

In an EPR-type experiment, two spin- \(\frac{1}{2}\) particles are emitted in the state $$ |1,1\rangle=|+\mathbf{z},+\mathbf{z}\rangle $$ \(\mathrm{A}\) and \(\mathrm{B}\) have their SG devices oriented along the \(x\) axis. Determine the probabilities that the resulting measurements find the two particles in the states \(|1,1\rangle_{x},|1,0\rangle_{x}\), and \(|1,-1\rangle_{x}\).

At time \(t=0\), an electron and a positron are formed in a state with total spin angular momentum equal to zero, perhaps from the decay of a spinless particle. The particles are situated in a uniform magnetic field \(B_{0}\) in the \(z\) direction. (a) If interaction between the electron and the positron may be neglected, show that the spin Hamiltonian of the system may be written as $$ \hat{H}=\omega_{0}\left(\hat{S}_{1 z}-\hat{S}_{2 z}\right) $$ where \(\hat{\mathbf{S}}_{1}\) is the spin operator of the electron, \(\hat{\mathbf{S}}_{2}\) is the spin operator of the positron, and \(\omega_{0}\) is a constant. (b) What is the spin state of the system at time \(t\) ? Show that the state of the system oscillates between a spin- 0 and a spin- 1 state. Determine the period of oscillation. (c) At time \(t\), measurements are made of \(S_{1 x}\) and \(S_{2 x}\). Calculate the probability that both of these measurements yield the value \(\hbar / 2\).

Determine the four states with \(s=\frac{3}{2}\) that can be formed by three spin- \(\frac{1}{2}\) particles. Suggestion: Start with the state \(\left|\frac{3}{2}, \frac{3}{2}\right\rangle\) and apply the lowering operator as in (5.36).

The annihilation of positronium in its ground state produces two photons that travel back to back in the positronium rest frame along an axis taken to be the \(z\) axis. The polarization state of the two-photon system is given by $$ |\psi\rangle=\frac{1}{\sqrt{2}}|R, R\rangle-\frac{1}{\sqrt{2}}|L, L\rangle $$(c) Compare the probability for the two photons to be in the state \(|x, x\rangle\) or in the state \(|y, y\rangle\) with what you would obtain if the two-photon state were either \(|R, R\rangle\) or \(|L, L\rangle\) rather than the superposition \(|\psi\rangle\). Note: Since the photons are traveling back to back along the \(z\) axis, if photon 1 is traveling in the positive \(z\) direction, then photon 2 is traveling in the negative \(z\) direction. Consequently, $$ |R\rangle_{2}=\frac{1}{\sqrt{2}}|x\rangle_{2}-\frac{i}{\sqrt{2}}|y\rangle_{2} \quad \text { and } \quad|L\rangle_{2}=\frac{1}{\sqrt{2}}|x\rangle_{2}+\frac{i}{\sqrt{2}}|y\rangle_{2} $$ (a) What is the probability that a measurement of the circular polarization state of the two photons will find them both right-handed? Both left-handed? (b) What is the probability that photon 1 will be found to be \(x\) polarized and photon 2 will be found to be \(y\) polarized, that is, that the system is in the state \(|x, y\rangle\) ? What is the probability that the system is in the state \(|y, x\rangle\) ?

A spin- \(\frac{1}{2}\) particle is in the pure state \(|\psi\rangle=a|+\mathbf{z}\rangle+b|-\mathbf{z}\rangle\). (a) Construct the density matrix in the \(S_{z}\) basis for this state. (b) Starting with your result in (a), determine the density matrix in the \(S_{x}\) basis, where $$ |+\mathbf{x}\rangle=\frac{1}{\sqrt{2}}|+\mathbf{z}\rangle+\frac{1}{\sqrt{2}}|-\mathbf{z}\rangle \quad|-\mathbf{x}\rangle=\frac{1}{\sqrt{2}}|+\mathbf{z}\rangle-\frac{1}{\sqrt{2}}|-\mathbf{z}\rangle $$ (c) Use your result for the density matrix in (b) to determine the probability that a measurement of \(S_{x}\) yields \(\hbar / 2\) for the state \(|\psi\rangle\).

Given the density operator $$ \hat{\rho}=\frac{1}{2}(|+\mathbf{z}\rangle\langle+\mathbf{z}|+|-\mathbf{z}\rangle\langle-\mathbf{z}|-|-\mathbf{z}\rangle\langle+\mathbf{z}|-|+\mathbf{z}\rangle\langle-\mathbf{z}|) $$ construct the density matrix. Use the density operator formalism to calculate \(\left\langle S_{x}\right\rangle\) for this state. Is this the density operator for a pure state? Justify your answer in two

Show that $$ \hat{\rho}=\frac{1}{2}|+\mathbf{n}\rangle\left\langle+\mathbf{n}\left|+\frac{1}{2}\right|-\mathbf{n}\right\rangle\left\langle-\mathbf{n}\left|=\frac{1}{2}\right|+\mathbf{z}\right\rangle\left\langle+\mathbf{z}\left|+\frac{1}{2}\right|-\mathbf{z}\right\rangle\langle-\mathbf{z}| $$ where $$ \begin{aligned} &|+\mathbf{n}\rangle=\cos \frac{\theta}{2}|+\mathbf{z}\rangle+e^{i \phi} \sin \frac{\theta}{2}|-\mathbf{z}\rangle \\ &|-\mathbf{n}\rangle=\sin \frac{\theta}{2}|+\mathbf{z}\rangle-e^{i \phi} \cos \frac{\theta}{2}|-\mathbf{z}\rangle \end{aligned} $$

Show that the Curie constant for an ensemble of \(N\) spin-l particles of mass \(m\) and charge \(q=-e\) immersed in a uniform magnetic field \(\mathbf{B}=B \mathbf{k}\) is given by $$ C=\frac{2 N \mu^{2}}{3 k_{\mathrm{B}}} $$ where \(\mu=g e \hbar / 2 m c\). Compare this value for \(C\) with that for an ensemble of spin- \(\frac{1}{2}\) particles, as determined in Example \(5.6 .\)

An attempt to perform a Bell-state measurement on two photons produces a mixed state, one in which the two photons are in the entangled state $$ \frac{1}{\sqrt{2}}|x, x\rangle+\frac{1}{\sqrt{2}}|y, y\rangle $$ with probability \(p\) and with probability \((1-p) / 2\) in each of the states \(|x, x\rangle\) and \(|y, y\rangle\). Determine the density matrix for this ensemble using the linear polarization states of the photons as basis states.

Show for the density operator for a mixed state $$ \hat{\rho}=\sum_{k} p_{k}\left|\psi^{(k)}\right\rangle\left\langle\psi^{(k)}\right| $$ that the probability of obtaining the state \(|\phi\rangle\) as a result of a measurement is given by \(\operatorname{tr}\left(\hat{P}_{|\phi\rangle} \hat{\rho}\right)\), where \(\hat{P}_{|\phi|}=|\phi\rangle\langle\phi|\).

Use the density operator formalism to show that the probability that a measurement finds two spin- \(\frac{1}{2}\) particles in the state \(|+\mathbf{x},+\mathbf{x}\rangle\) differs for the pure Bell state $$ \left|\Phi^{(+)}\right\rangle=\frac{1}{\sqrt{2}}|+\mathbf{z},+\mathbf{z}\rangle+\frac{1}{\sqrt{2}}|-\mathbf{z},-\mathbf{z}\rangle $$ for which $$ \hat{\rho}=\left|\Phi^{(+)}\right\rangle\left\langle\Phi^{(+)}\right| $$ and for the mixed state $$ \hat{\rho}=\frac{1}{2}|+\mathbf{z},+\mathbf{z}\rangle\left\langle+\mathbf{z},+\mathbf{z}\left|+\frac{1}{2}\right|-\mathbf{z},-\mathbf{z}\right\rangle\langle-\mathbf{z},-\mathbf{z}| $$ Thus, the disagreement between the predictions of quantum mechanics for the entangled state \(\left|\Phi^{(+)}\right\rangle\)and those consistent with the views of a local realist are apparent without having to resort to Bell inequalities.

Show that the equation governing time evolution of the density operator for a mixed state is given by $$ i \hbar \frac{d}{d t} \hat{\rho}(t)=[\hat{H}, \hat{\rho}(t)] $$

(a) Show that the time evolution of the density operator is given by $$ \hat{\rho}(t)=\hat{U}(t) \hat{\rho}(0) \hat{U}^{\dagger}(t) $$ where \(\hat{U}(t)\) is the time-evolution operator, namely $$ \hat{U}(t)|\psi(0)\rangle=|\psi(t)\rangle $$ (b) Suppose that an ensemble of particles is in a pure state at \(t=0\). Show the ensemble cannot evolve into a mixed state as long as time evolution is governed by the Schrödinger equation.