Past exam of the theoretical physics II course of the University of Cambridge / 2018 / 4

Using $[x,p]=i\hslash$, the Heisenberg picture equations for the quantum harmonic oscillator areSolving these coupled linear equations givesThe oscillator annihilation operator and creation operator areThey obey $[a,a^{†}]=1$ and $H_0=\hslash \omega (a^{†}a+1/2)$. Substitution of the position and momentum solutions, or the corresponding commutators with $H_0$, yields

Complete the square, keeping the equilibrium displacement signed:Write $X=x-x_0$ for position relative to the new minimum. By the sudden approximation, the state immediately after the quantum quench is still the old vacuum. Its initial means areThe old ground state has $⟨H_0⟩=\hslash \omega /2$ and $⟨x⟩=0$. Thus $⟨H⟩=\hslash \omega /2$, but the energy measured above the new potential minimum isThe extra term is the excitation energy above the new ground state; the zero-point term remains part of the energy above the potential minimum.

The variables $X,p$ obey the oscillator equations in part (a). Using the above initial means givesThe wavepacket's centre therefore oscillates about the new equilibrium, beginning at the old origin with zero mean momentum.

Define the signed dimensionless displacementThe spatial translation operator is $T(x_0)=e^{-ip{x}_0/\hslash }$, and the phases of the new energy eigenstates are fixed by $|n^{\prime }⟩=T(x_0)|n⟩$. Their expansion coefficients are thereforeSince $p=i\sqrt{m\hslash \omega /2}(a^{†}-a)$,Set $A=-\lambda a^{†}$ and $B=\lambda a$. Their commutator is $[A,B]={\lambda }^{2}I$, which commutes with both operators. Thus the stated special case of the Baker--Campbell--Hausdorff formula is applicable and givesBecause $a|0⟩=0$,Taking the overlaps with $⟨n|$ therefore provesThis is the coherent state after shifting an oscillator equilibrium. Its normalization follows from $e^{-{\lambda }^2}{\sum }_n{\lambda }^{2n}/n!=1$.

There is a sign qualification in the printed expansion: $(m\omega x_0^2/(2\hslash ){)}^{n/2}=|\lambda {|}^n$ agrees with $(-\lambda {)}^n$ when $x_0\le 0$, equivalently $\gamma \ge 0$. For a general signed displacement $x_0>0$, the odd coefficients have the opposite sign in the specified translated basis. The squared displacement alone fixes the occupation probabilities, not these coherent relative phases. The two-exponential identity also requires the central-commutator condition used above; it is not valid in that simple form for arbitrary operators.

The annihilation operator associated with the new coordinate $X=x-x_0$ isAs the initial state is the old vacuum,Equivalently, applying $a^{\prime }$ to the number-state expansion in part (c) and shifting its summation index returns $-\lambda$ times the same state. It is therefore a coherent state of the displaced oscillator.

The new Hamiltonian is $H=U_{\min }+\hslash \omega (a^{\prime †}{a}^{\prime }+1/2)$. In the Schrödinger picture its evolution givesConsequentlyThe displayed prefactor is only a global phase. Equivalently, in the Heisenberg picture $a^{\prime }(t)=a^{\prime }(0)e^{-i\omega t}$. The mean occupation $|\alpha (t){|}^2={\lambda }^2$ is constant, and $\hslash \omega ({\lambda }^2+1/2)$ reproduces the energy above the new minimum found in part (b). Its mean position and momentum likewise reproduce that part's oscillating wavepacket centre.