Past exam of the theoretical physics II course of the University of Cambridge / 2017 / 2

Complete the square in the potential energy. With $x_F=F/(M{\omega }^2)$,The quantum harmonic oscillator is translated without changing its level spacing. If ${\phi }_n(x)$ is its normalized unforced energy eigenstate, thenEquivalently, $|n;F⟩=e^{-i{x}_{F}p/\hslash }|n⟩$. The ground state position is centered at $x_F$, and its position uncertainty remains $\sqrt{\hslash /(2M\omega )}$.

A normalized oscillator coherent state is an eigenstate of the annihilation operator, $a|\alpha ⟩=\alpha |\alpha ⟩$. Expanding it in Fock states givesSince $x=\sqrt{\hslash /(2M\omega )}(a+a^{†})$, $⟨a^2⟩={\alpha }^2$, and $⟨a{a}^{†}⟩=|\alpha {|}^2+1$, the variance is $⟨x^2⟩-⟨x{⟩}^2=\hslash /(2M\omega )$. Henceindependently of the complex amplitude $\alpha$.

For $A=\alpha a^{†}$ and $B=-{\alpha }^{\ast }a$, the commutator is the scalar $[A,B]=|\alpha {|}^{2}I$. The Baker--Campbell--Hausdorff formula therefore terminates and gives the displacement operatorBecause $a|0⟩=0$, applying it to the Fock vacuum yields exactly the Fock state expansion above:For a general complex $\alpha$, its complex conjugate is required in the annihilation term. The expression with $-\alpha a$ printed in the paper is valid when $\alpha$ is real.

The quantum state is continuous through a sudden finite-force quantum quench, so it is still $|0⟩$ at $t=0^{+}$. DefineThe new Hamiltonian operator is $H=\hslash \omega (b^{†}b+1/2)-\hslash \omega {\lambda }^2$. The old Fock vacuum obeys $b|0⟩=-\lambda |0⟩$, so the coherent state after shifting an oscillator equilibrium rotates with amplitude $-\lambda e^{-i\theta }$ relative to the new equilibrium. In the original oscillator basis, the full state isOne way to obtain the phase is to writeand apply the displacement operator composition rule. The coherent-state oscillator expectation values giveThe position uncertainty stays $\sqrt{\hslash /(2M\omega )}$. These mean values obey the forced classical equation, while the nonzero variance retains the quantum width.

Specify the magnetic coupling before fixing its sign: let the magnetic moment be $\mathbf{\mu }=\gamma \mathbf{S}$, where $\gamma$ is a signed gyromagnetic ratio. The Zeeman effect gives $H_Z=-\gamma \mathbf{B}\cdot \mathbf{S}$. For spin one-half, $\mathbf{S}=\hslash \mathbf{\sigma }/2$, so with $B=|\mathbf{B}|$ and $u=\beta \gamma \hslash B/2$,This is the quantum Gibbs state. Its eigenvalues are $(1\pm \tanh u)/2$, so its purity is $(1+{\tanh }^{2}u)/2$. For nonzero level splitting it approaches a pure state as $|u|\to \infty$: positive inverse temperature selects the lower energy, while negative inverse temperature selects the upper one. It is a maximally mixed state at $u=0$, including infinite temperature or zero magnetic splitting. At zero field the Gibbs state remains $I/2$; lowering temperature alone does not select a unique pure state from that degeneracy.

Negative temperature is meaningful for an internally equilibrated spin sector with a bounded energy spectrum and population inversion. It must be sufficiently isolated from unbounded motional degrees of freedom: the oscillator's unbounded spectrum, for instance, makes its partition sum diverge at negative inverse temperature. The two-level spin partition sum itself stays finite for either sign.

For $\mathbf{B}=B{\mathbf{e}}_z$, the quantum Gibbs state is diagonal in the $S_z$ basis. Its two probabilities are proportional to $e^{\pm \beta \gamma \hslash B/2}$. ThusAt small inverse temperature, $⟨S_z⟩\simeq \beta \gamma {\hslash }^{2}B/4$. As a function of $u=\beta \gamma \hslash B/2$, the normalized spin component is the smooth odd curve $\tanh u$: it crosses zero at $u=0$ and saturates at $\pm 1$. As a function of $\beta$ its direction is reversed if $\gamma B<0$. For $\gamma B>0$, positive temperature favors $S_z=+\hslash /2$, while negative temperature gives the inverted population.