PSet 5 Discussion

Problem 5-1: Generalized Rabi oscillations¶

(a)¶

Start from the lab-frame Schrödinger equation

\[ i\frac{d}{dt}\ket{\psi(t)} = \hat H(t)\ket{\psi(t)}, \]

with

\[ \hat H(t)=\omega_0 \hat S_z+\frac{\Omega_1}{2}\left(e^{-i\omega t}\hat S_+ + e^{i\omega t}\hat S_-\right). \]

Define the rotating-frame unitary and state by

\[ \hat U(t)=e^{-i\omega t \hat S_z}, \qquad \ket{\psi_{\rm rot}(t)}=\hat U^\dagger(t)\ket{\psi(t)}. \]

workout

Differentiate \(\ket{\psi_{\rm rot}}\):

\[ \frac{d}{dt}\ket{\psi_{\rm rot}} =\dot{\hat U}^\dagger\ket{\psi}+\hat U^\dagger\frac{d}{dt}\ket{\psi}. \]

Multiply by \(i\) and use \(i\frac{d}{dt}\ket{\psi}=\hat H\ket{\psi}\):

\[ i\frac{d}{dt}\ket{\psi_{\rm rot}} =i\dot{\hat U}^\dagger\ket{\psi}+\hat U^\dagger \hat H \ket{\psi}. \]

Rewrite everything in terms of \(\ket{\psi_{\rm rot}}\) using \(\ket{\psi}=\hat U\ket{\psi_{\rm rot}}\):

\[ i\frac{d}{dt}\ket{\psi_{\rm rot}} =\left(i\dot{\hat U}^\dagger \hat U+\hat U^\dagger \hat H \hat U\right)\ket{\psi_{\rm rot}}. \]

This has the standard rotating-frame form

\[ i\frac{d}{dt}\ket{\psi_{\rm rot}} =\hat H_{\rm rot}(t)\ket{\psi_{\rm rot}}, \qquad \hat H_{\rm rot}(t)=\hat U^\dagger \hat H \hat U - i\hat U^\dagger \dot{\hat U}. \]

Now evaluate the extra term \(-i\hat U^\dagger \dot{\hat U}\). Since

\[ \dot{\hat U}(t)=\frac{d}{dt}e^{-i\omega t \hat S_z}=-i\omega \hat S_z\,\hat U(t), \]

we have \(\hat U^\dagger \dot{\hat U}=-i\omega \hat S_z\), hence

\[ -i\hat U^\dagger \dot{\hat U}=-\omega \hat S_z. \]

So

\[ \hat H_{\rm rot}(t)=\hat U^\dagger \hat H(t)\hat U - \omega \hat S_z. \]

Next compute \(\hat U^\dagger \hat H(t)\hat U\). The \(\hat S_z\) term is unchanged because \([\hat S_z,\hat U]=0\):

\[ \hat U^\dagger \hat S_z \hat U=\hat S_z. \]

For the ladder operators, use \([\hat S_z,\hat S_\pm]=\pm \hat S_\pm\), which implies the standard identity

\[ e^{i\theta \hat S_z}\hat S_\pm e^{-i\theta \hat S_z}=e^{\pm i\theta}\hat S_\pm. \]

workout

With \(\theta=\omega t\) and \(\hat U^\dagger=e^{i\omega t \hat S_z}\), this gives

\[ \hat U^\dagger \hat S_+ \hat U = e^{i\omega t}\hat S_+, \qquad \hat U^\dagger \hat S_- \hat U = e^{-i\omega t}\hat S_-. \]

Therefore the time-dependent phases in the drive cancel:

\[ \hat U^\dagger \hat H(t)\hat U =\omega_0 \hat S_z+\frac{\Omega_1}{2}\left(e^{-i\omega t}e^{i\omega t}\hat S_+ + e^{i\omega t}e^{-i\omega t}\hat S_-\right) =\omega_0 \hat S_z+\frac{\Omega_1}{2}\left(\hat S_+ + \hat S_-\right). \]

Putting everything together,

\[ \hat H_{\rm rot} =(\omega_0-\omega)\hat S_z+\frac{\Omega_1}{2}\left(\hat S_+ + \hat S_-\right). \]

This is time-independent. Defining the detuning \(\delta\equiv \omega_0-\omega\),

\[ \boxed{ \hat H_{\rm rot} =\delta\,\hat S_z+\frac{\Omega_1}{2}\left(\hat S_+ + \hat S_-\right). } \]

Now write \(\hat H_{\rm rot}\) as a \(3\times 3\) matrix in the ordered basis \(\{\ket{+1},\ket{0},\ket{-1}\}\) using the provided representations:

\[ \hat S_z= \begin{pmatrix} 1&0&0\\ 0&0&0\\ 0&0&-1 \end{pmatrix}, \qquad \hat S_+ + \hat S_-= \sqrt{2} \begin{pmatrix} 0&1&0\\ 1&0&1\\ 0&1&0 \end{pmatrix}. \]

workout

Hence

\[ \boxed{ \hat H_{\rm rot} = \begin{pmatrix} \delta & \Omega_1/\sqrt{2} & 0\\ \Omega_1/\sqrt{2} & 0 & \Omega_1/\sqrt{2}\\ 0 & \Omega_1/\sqrt{2} & -\delta \end{pmatrix}. } \]

(b)¶

From part (a), in the ordered basis \(\{\ket{+1},\ket{0},\ket{-1}\}\) the rotating-frame Hamiltonian is

\[ \hat H_{\rm rot} = \begin{pmatrix} \delta & \Omega_1/\sqrt{2} & 0\\ \Omega_1/\sqrt{2} & 0 & \Omega_1/\sqrt{2}\\ 0 & \Omega_1/\sqrt{2} & -\delta \end{pmatrix}. \]

To show there is a zero eigenvalue, it suffices to find a nonzero vector \(\vec v=(a,b,c)^T\) such that \(\hat H_{\rm rot}\vec v=\vec 0\).

workout

Define \(\alpha\equiv \Omega_1/\sqrt{2}\) to simplify notation. The equation \(\hat H_{\rm rot}\vec v=\vec 0\) is the linear system

\[ \begin{aligned} \delta a+\alpha b&=0,\\ \alpha a+\alpha c&=0,\\ \alpha b-\delta c&=0. \end{aligned} \]

From the middle equation, since \(\alpha\neq 0\) for a nontrivial drive, we have

\[ a+c=0 \quad\Rightarrow\quad c=-a. \]

Plugging \(c=-a\) into the first equation gives

\[ b=-\frac{\delta}{\alpha}\,a. \]

The third equation becomes

\[ \alpha b-\delta(-a)=\alpha b+\delta a=0, \]

which is automatically satisfied by the same \(b=-(\delta/\alpha)a\). Hence a nontrivial null vector exists for any detuning \(\delta\), so \(\hat H_{\rm rot}\) has an eigenvalue \(E=0\).

Choosing \(a=1\) gives an (unnormalized) eigenvector

\[ \vec v_0= \begin{pmatrix} 1\\[4pt] -\delta/\alpha\\[4pt] -1 \end{pmatrix} = \begin{pmatrix} 1\\[4pt] -\sqrt{2}\,\delta/\Omega_1\\[4pt] -1 \end{pmatrix}. \]

Interpreting this as the state \(\ket{D}\) in the \(\{\ket{+1},\ket{0},\ket{-1}\}\) basis,

\[ \ket{D}\propto \ket{+1}-\frac{\sqrt{2}\,\delta}{\Omega_1}\ket{0}-\ket{-1}. \]

workout

Normalize it. The squared norm of the coefficient vector is

\[ \|\vec v_0\|^2 =1+\left(\frac{\sqrt{2}\,\delta}{\Omega_1}\right)^2+1 =2\left(1+\frac{\delta^2}{\Omega_1^2}\right). \]

Therefore the normalized dark state is

\[ \boxed{ \ket{D} = \frac{1}{\sqrt{2\left(1+\delta^2/\Omega_1^2\right)}} \left( \ket{+1} -\frac{\sqrt{2}\,\delta}{\Omega_1}\ket{0} -\ket{-1} \right), \qquad \hat H_{\rm rot}\ket{D}=0. } \]

As a check, at resonance \(\delta=0\) this reduces to

\[ \ket{D}\big|_{\delta=0}=\frac{1}{\sqrt{2}}\left(\ket{+1}-\ket{-1}\right), \]

which has no \(\ket{0}\) component.

(c)¶

In part (b) we found an eigenstate \(\ket{D}\) of the rotating-frame Hamiltonian with eigenvalue \(0\):

\[ \hat H_{\rm rot}\ket{D}=0. \]

In the rotating frame, the Schrödinger equation is

\[ i\frac{d}{dt}\ket{\psi_{\rm rot}(t)}=\hat H_{\rm rot}\ket{\psi_{\rm rot}(t)}. \]

If we prepare the system in \(\ket{\psi_{\rm rot}(0)}=\ket{D}\), then the time evolution is trivial because the eigenvalue is zero:

\[ \ket{\psi_{\rm rot}(t)}=e^{-i(0)t}\ket{D}=\ket{D}. \]

reminder

Now relate the lab-frame (Schrödinger-picture) state to the rotating-frame state. By definition in part (a),

\[ \ket{\psi_{\rm rot}(t)}=\hat U^\dagger(t)\ket{\psi(t)}, \qquad \hat U(t)=e^{-i\omega t \hat S_z}, \]

so equivalently

\[ \ket{\psi(t)}=\hat U(t)\ket{\psi_{\rm rot}(t)}=\hat U(t)\ket{D}. \]

workout

Using the action of \(\hat U(t)\) on \(\hat S_z\) eigenstates, \(\hat S_z\ket{m}=m\ket{m}\),

\[ \hat U(t)\ket{m}=e^{-i\omega t m}\ket{m}, \qquad m=+1,0,-1. \]

From part (b), we can write \(\ket{D}\) as

\[ \ket{D} = \frac{1}{\sqrt{2\left(1+\delta^2/\Omega_1^2\right)}} \left( \ket{+1} -\frac{\sqrt{2}\,\delta}{\Omega_1}\ket{0} -\ket{-1} \right). \]

Therefore, in the lab frame,

\[ \boxed{ \ket{\psi(t)} = \frac{1}{\sqrt{2\left(1+\delta^2/\Omega_1^2\right)}} \left( e^{-i\omega t}\ket{+1} -\frac{\sqrt{2}\,\delta}{\Omega_1}\ket{0} -e^{+i\omega t}\ket{-1} \right). } \]

This state is not stationary in the lab frame (the \(\ket{\pm 1}\) components acquire opposite time-dependent phases), but the amplitudes in the \(\hat S_z\) basis have constant magnitudes. Hence the probabilities for measuring each \(m\) are time-independent.

workout

\[ \boxed{ \begin{aligned} P_{+1}(t) &= \left|\braket{+1}{\psi(t)}\right|^2 = \frac{1}{2\left(1+\delta^2/\Omega_1^2\right)},\\[6pt] P_{0}(t) &= \left|\braket{0}{\psi(t)}\right|^2 = \frac{\delta^2/\Omega_1^2}{1+\delta^2/\Omega_1^2},\\[6pt] P_{-1}(t) &= \left|\braket{-1}{\psi(t)}\right|^2 = \frac{1}{2\left(1+\delta^2/\Omega_1^2\right)}. \end{aligned} } \]

So, in the dark state, the populations of the three \(\hat S_z\) eigenstates do not oscillate; only relative phases (in particular between \(\ket{+1}\) and \(\ket{-1}\)) evolve in time.

(d)¶

The key facts from parts (b)–(c) are:

In the rotating frame, \(\hat H_{\rm rot}\) is time-independent.
There exists a nontrivial state \(\ket{D}\) such that \(\hat H_{\rm rot}\ket{D}=0\).
Therefore, if the system is in \(\ket{D}\) in the rotating frame, it does not evolve there (no dynamical phase even), so the rotating-frame amplitudes in the \(\hat S_z\) basis are constant in time.
In the lab frame, the state is \(\ket{\psi(t)}=\hat U(t)\ket{D}\) with \(\hat U(t)=e^{-i\omega t \hat S_z}\), which only multiplies each \(\ket{m}\) component by a phase \(e^{-i\omega t m}\).

This is “surprising” because the lab-frame Hamiltonian explicitly drives transitions (\(\hat S_\pm\) terms), so one might expect populations to slosh around. The intuition for why they do not is that \(\ket{D}\) is a coherent superposition in which the drive-induced transition amplitudes destructively interfere.

additional reading

A clean way to see this is to rewrite the coupling part of the rotating-frame Hamiltonian as proportional to \(\hat S_x\):

\[ \hat H_{\rm rot}=\delta \hat S_z+\Omega_1 \hat S_x, \qquad \hat S_x\equiv\frac{\hat S_+ + \hat S_-}{2}. \]

The “drive” couples \(\ket{0}\) to the symmetric combination of \(\ket{+1}\) and \(\ket{-1}\). Meanwhile, the detuning term \(\delta \hat S_z\) couples the antisymmetric combination of \(\ket{\pm 1}\) to \(\ket{0}\) in just the right way to allow a stationary eigenstate.

More explicitly, in the \(\{\ket{+1},\ket{0},\ket{-1}\}\) basis, the drive term (the off-diagonal couplings) only connects nearest neighbors:

\(\ket{+1}\leftrightarrow \ket{0}\) with matrix element \(\Omega_1/\sqrt{2}\)
\(\ket{0}\leftrightarrow \ket{-1}\) with matrix element \(\Omega_1/\sqrt{2}\)

There is no direct \(\ket{+1}\leftrightarrow\ket{-1}\) coupling.

Now take a superposition with opposite amplitudes on \(\ket{+1}\) and \(\ket{-1}\). For the resonant case \(\delta=0\) the dark state is

\[ \ket{D}=\frac{1}{\sqrt{2}}\left(\ket{+1}-\ket{-1}\right). \]

Acting on this with the drive operator \(\hat S_+ + \hat S_-\) produces cancellation in the \(\ket{0}\) channel:

\[ (\hat S_+ + \hat S_-)\ket{+1}=\sqrt{2}\ket{0}, \qquad (\hat S_+ + \hat S_-)\ket{-1}=\sqrt{2}\ket{0}. \]

Therefore

\[ (\hat S_+ + \hat S_-)\left(\ket{+1}-\ket{-1}\right) =\sqrt{2}\ket{0}-\sqrt{2}\ket{0}=0, \]

so the drive has no effect at all: the transition amplitudes from \(\ket{+1}\) and from \(\ket{-1}\) to \(\ket{0}\) are equal in magnitude but opposite in sign, giving perfect destructive interference.

Away from resonance (\(\delta\neq 0\)), \(\ket{D}\) contains a specific amount of \(\ket{0}\),

\[ \ket{D}\propto \ket{+1}-\frac{\sqrt{2}\delta}{\Omega_1}\ket{0}-\ket{-1}, \]

chosen precisely so that the drive-induced flow into and out of \(\ket{0}\) cancels the detuning-induced phase evolution between \(\ket{+1}\) and \(\ket{-1}\) in the rotating frame. In other words, \(\ket{D}\) is an eigenstate of \(\hat H_{\rm rot}\) with eigenvalue \(0\), and eigenstates do not exchange population with other eigenstates.

Finally, why are the lab-frame probabilities constant?

In the rotating frame, \(\ket{D}\) is time-independent, so the magnitudes of its components in the \(\hat S_z\) basis are time-independent.
Transforming back to the lab frame multiplies each \(\ket{m}\) component by a pure phase \(e^{-i\omega t m}\), which cannot change the magnitude of the coefficient of \(\ket{m}\).

Thus the system can have nontrivial time-dependent phases in the lab frame while all \(\hat S_z\) measurement probabilities remain frozen: the drive is “on,” but it cannot couple the state out of the dark superposition because of destructive interference between the two excitation pathways \(\ket{+1}\to\ket{0}\) and \(\ket{-1}\to\ket{0}\).

(e)¶

From part (a), the rotating-frame Hamiltonian is

\[ \hat H_{\rm rot} = \begin{pmatrix} \delta & \Omega_1/\sqrt{2} & 0\\ \Omega_1/\sqrt{2} & 0 & \Omega_1/\sqrt{2}\\ 0 & \Omega_1/\sqrt{2} & -\delta \end{pmatrix}. \]

We already found one eigenpair: eigenvalue \(E_0=0\) with eigenvector \(\ket{D}\).

To find the remaining two eigenvalues, compute the characteristic polynomial. Let \(\alpha\equiv \Omega_1/\sqrt{2}\) and consider \(\det(\hat H_{\rm rot}-E I)=0\):

\[ \det \begin{pmatrix} \delta-E & \alpha & 0\\ \alpha & -E & \alpha\\ 0 & \alpha & -\delta-E \end{pmatrix} =0. \]

workout

Expand along the first row:

\[ \det(\hat H_{\rm rot}-EI) =(\delta-E)\det \begin{pmatrix} -E & \alpha\\ \alpha & -\delta-E \end{pmatrix} -\alpha \det \begin{pmatrix} \alpha & \alpha\\ 0 & -\delta-E \end{pmatrix}. \]

Compute each \(2\times 2\) determinant:

\[ \det \begin{pmatrix} -E & \alpha\\ \alpha & -\delta-E \end{pmatrix} = (-E)(-\delta-E)-\alpha^2 =E(\delta+E)-\alpha^2, \]

and

\[ \det \begin{pmatrix} \alpha & \alpha\\ 0 & -\delta-E \end{pmatrix} =\alpha(-\delta-E). \]

So

\[ \det(\hat H_{\rm rot}-EI) =(\delta-E)\left(E(\delta+E)-\alpha^2\right)-\alpha^2(-\delta-E). \]

Rewrite the last term: \(-\alpha^2(-\delta-E)=+\alpha^2(\delta+E)\). Thus

\[ \det(\hat H_{\rm rot}-EI) =(\delta-E)\left(E(\delta+E)-\alpha^2\right)+\alpha^2(\delta+E). \]

Now expand the first product:

\[ (\delta-E)\left(E(\delta+E)-\alpha^2\right) =(\delta-E)(E\delta+E^2-\alpha^2) =\delta(E\delta+E^2-\alpha^2)-E(E\delta+E^2-\alpha^2). \]

Compute:

\[ \delta(E\delta+E^2-\alpha^2)=\delta^2E+\delta E^2-\delta\alpha^2, \]

\[ E(E\delta+E^2-\alpha^2)=\delta E^2+E^3-E\alpha^2. \]

Subtracting gives

\[ (\delta-E)\left(E(\delta+E)-\alpha^2\right) =\delta^2E+\delta E^2-\delta\alpha^2-\delta E^2-E^3+E\alpha^2 =\delta^2E-E^3-\delta\alpha^2+E\alpha^2. \]

Add \(\alpha^2(\delta+E)=\alpha^2\delta+\alpha^2E\):

\[ \det(\hat H_{\rm rot}-EI) =\left(\delta^2E-E^3-\delta\alpha^2+E\alpha^2\right)+\left(\alpha^2\delta+\alpha^2E\right) =\delta^2E-E^3+2\alpha^2E. \]

Factor out \(E\):

\[ \det(\hat H_{\rm rot}-EI) =E\left(\delta^2-E^2+2\alpha^2\right) =-E\left(E^2-(\delta^2+2\alpha^2)\right). \]

Thus the eigenvalues are

\[ E_0=0, \qquad E_\pm=\pm\sqrt{\delta^2+2\alpha^2}. \]

Since \(2\alpha^2=2(\Omega_1^2/2)=\Omega_1^2\), define the generalized Rabi frequency

\[ \Omega_R\equiv \sqrt{\delta^2+\Omega_1^2}, \]

so

\[ \boxed{ E_\pm=\pm \Omega_R. } \]

Now find the corresponding eigenvectors. Solve \((\hat H_{\rm rot}-E I)\vec v=\vec 0\) with \(\vec v=(a,b,c)^T\).

The equations are

\[ \begin{aligned} (\delta-E)a+\alpha b&=0,\\ \alpha a-E b+\alpha c&=0,\\ \alpha b+(-\delta-E)c&=0. \end{aligned} \]

workout

From the first and third equations, express \(b\) in two ways:

\[ b=-\frac{\delta-E}{\alpha}\,a, \qquad b=\frac{\delta+E}{\alpha}\,c. \]

Hence

\[ -\left(\delta-E\right)a=(\delta+E)c \quad\Rightarrow\quad c=-\frac{\delta-E}{\delta+E}\,a. \]

Now use the middle equation \(\alpha a-Eb+\alpha c=0\) to check consistency and fix the eigenvalue relation. Substitute \(b=-(\delta-E)a/\alpha\):

\[ \alpha a - E\left(-\frac{\delta-E}{\alpha}a\right)+\alpha c=0 \quad\Rightarrow\quad \alpha a+\frac{E(\delta-E)}{\alpha}a+\alpha c=0. \]

Multiply by \(\alpha\):

\[ \alpha^2 a+E(\delta-E)a+\alpha^2 c=0. \]

Insert \(c=-(\delta-E)a/(\delta+E)\):

\[ \alpha^2 a+E(\delta-E)a-\alpha^2\frac{\delta-E}{\delta+E}a=0. \]

Divide by \(a\) (nonzero):

\[ \alpha^2+E(\delta-E)-\alpha^2\frac{\delta-E}{\delta+E}=0. \]

Combine the \(\alpha^2\) terms:

\[ \alpha^2\left(1-\frac{\delta-E}{\delta+E}\right)+E(\delta-E)=0 \quad\Rightarrow\quad \alpha^2\frac{(\delta+E)-(\delta-E)}{\delta+E}+E(\delta-E)=0 \quad\Rightarrow\quad \alpha^2\frac{2E}{\delta+E}+E(\delta-E)=0. \]

Factor \(E\):

\[ E\left(\frac{2\alpha^2}{\delta+E}+\delta-E\right)=0. \]

For \(E\neq 0\) this gives

\[ \frac{2\alpha^2}{\delta+E}+\delta-E=0 \quad\Rightarrow\quad 2\alpha^2+(\delta-E)(\delta+E)=0 \quad\Rightarrow\quad 2\alpha^2+\delta^2-E^2=0, \]

i.e. \(E^2=\delta^2+2\alpha^2\), which is exactly \(E=\pm\Omega_R\). So the vector solution above is consistent.

We can therefore choose a convenient unnormalized eigenvector by setting \(a=1\) and using

\[ b=-\frac{\delta-E}{\alpha}, \qquad c=-\frac{\delta-E}{\delta+E}. \]

It is often cleaner to eliminate fractions by multiplying the whole vector by \(\delta+E\). Define

\[ \vec v_E \equiv \begin{pmatrix} \delta+E\\[4pt] -(\delta-E)(\delta+E)/\alpha\\[4pt] -(\delta-E) \end{pmatrix} = \begin{pmatrix} \delta+E\\[4pt] -(\delta^2-E^2)/\alpha\\[4pt] -(\delta-E) \end{pmatrix}. \]

But \(\delta^2-E^2=-2\alpha^2\), so the middle component becomes \(+2\alpha^2/\alpha=2\alpha\). Hence

\[ \vec v_E \propto \begin{pmatrix} \delta+E\\[4pt] 2\alpha\\[4pt] -(\delta-E) \end{pmatrix}. \]

Returning to \(\alpha=\Omega_1/\sqrt{2}\) gives \(2\alpha=\sqrt{2}\Omega_1\). Therefore, for \(E_\pm=\pm\Omega_R\) we can take unnormalized eigenvectors

\[ \ket{E_\pm}\propto (\delta\pm\Omega_R)\ket{+1}+\sqrt{2}\Omega_1\ket{0}-(\delta\mp\Omega_R)\ket{-1}. \]

Normalize them. Let

\[ \ket{E_\pm}= \mathcal N_\pm\left[(\delta\pm\Omega_R)\ket{+1}+\sqrt{2}\Omega_1\ket{0}-(\delta\mp\Omega_R)\ket{-1}\right], \]

with \(\mathcal N_\pm\) chosen so that \(\braket{E_\pm}{E_\pm}=1\). The squared norm of the unnormalized coefficient vector is

\[ (\delta\pm\Omega_R)^2+2\Omega_1^2+(\delta\mp\Omega_R)^2. \]

Use the identity \((\delta+\Omega_R)^2+(\delta-\Omega_R)^2=2(\delta^2+\Omega_R^2)\) to write

\[ (\delta\pm\Omega_R)^2+(\delta\mp\Omega_R)^2=2(\delta^2+\Omega_R^2). \]

Thus

\[ \|\cdot\|^2=2(\delta^2+\Omega_R^2)+2\Omega_1^2 =2\left(\delta^2+(\delta^2+\Omega_1^2)\right)+2\Omega_1^2 =4(\delta^2+\Omega_1^2) =4\Omega_R^2. \]

So \(\mathcal N_\pm=1/(2\Omega_R).\) Therefore

\[ \boxed{ \ket{E_\pm} = \frac{1}{2\Omega_R} \left[ (\delta\pm\Omega_R)\ket{+1}+\sqrt{2}\Omega_1\ket{0}-(\delta\mp\Omega_R)\ket{-1} \right], \qquad E_\pm=\pm\Omega_R. } \]

Together with the dark state from part (b),

\[ \ket{D}= \frac{1}{\sqrt{2\left(1+\delta^2/\Omega_1^2\right)}} \left( \ket{+1} -\frac{\sqrt{2}\,\delta}{\Omega_1}\ket{0} -\ket{-1} \right), \qquad E_0=0, \]

we have a complete eigenbasis of \(\hat H_{\rm rot}\).

Now address the oscillation frequencies in \(\langle \hat O(t)\rangle\) for an arbitrary initial state.

Write the initial rotating-frame state in the energy eigenbasis:

\[ \ket{\psi_{\rm rot}(0)}=c_+\ket{E_+}+c_0\ket{D}+c_-\ket{E_-}. \]

workout

Time evolution in the rotating frame is

\[ \ket{\psi_{\rm rot}(t)}=c_+e^{-i\Omega_R t}\ket{E_+}+c_0\ket{D}+c_-e^{+i\Omega_R t}\ket{E_-}. \]

For any observable \(\hat O\) (in the rotating frame), the expectation value will contain terms of the form

\[ c_j^*c_k e^{i(E_j-E_k)t}\bra{E_j}\hat O\ket{E_k}. \]

Thus the only possible nontrivial oscillation frequencies are the energy differences among \(\{0,+\Omega_R,-\Omega_R\}\):

Between \(\ket{E_+}\) and \(\ket{D}\): frequency \(|+\Omega_R-0|=\Omega_R\).
Between \(\ket{E_-}\) and \(\ket{D}\): frequency \(|-\Omega_R-0|=\Omega_R\).
Between \(\ket{E_+}\) and \(\ket{E_-}\): frequency \(|+\Omega_R-(-\Omega_R)|=2\Omega_R\).

Therefore, ignoring the trivial \(\omega\)-dependent phases from undoing the rotating-frame transformation (which only add known \(\omega\)-precession factors depending on the observable), the dynamical oscillation frequencies you can see are

\[ \boxed{ \text{possible oscillation frequencies in }\langle \hat O(t)\rangle \text{ are } 0,\ \Omega_R,\ 2\Omega_R. } \]

Here \(0\) corresponds to constant terms (diagonal contributions in the energy basis), while \(\Omega_R\) and \(2\Omega_R\) come from coherences between different energy eigenstates.

(f)¶

At resonance, \(\delta=0\) and hence (from part (a)) the rotating-frame Hamiltonian is

\[ \hat H_{\rm rot} = \frac{\Omega_1}{2}\left(\hat S_+ + \hat S_-\right) = \frac{\Omega_1}{\sqrt{2}} \begin{pmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{pmatrix} \equiv \frac{\Omega_1}{\sqrt{2}}\,M, \]

in the ordered basis \(\{\ket{+1},\ket{0},\ket{-1}\}\).

We will (i) diagonalize \(\hat H_{\rm rot}\), (ii) time-evolve the initial state \(\ket{\psi(0)}=\ket{-1}\) in the rotating frame, then (iii) transform back to the lab frame and compute probabilities.

Step 1: eigenvalues/eigenvectors at \(\delta=0\).

The matrix \(M\) has eigenvalues \(\mu=0,\pm\sqrt{2}\), so \(\hat H_{\rm rot}\) has energies \(E_0=0\) and \(E_\pm=\pm \Omega_1\).

A convenient normalized eigenbasis is

\[ \ket{E_+} = \frac{1}{2}\ket{+1} +\frac{1}{\sqrt{2}}\ket{0} +\frac{1}{2}\ket{-1}, \qquad E_+=+\Omega_1, \]

\[ \ket{E_0} = \frac{1}{\sqrt{2}}\ket{+1} -\frac{1}{\sqrt{2}}\ket{-1}, \qquad E_0=0, \]

\[ \ket{E_-} = \frac{1}{2}\ket{+1} -\frac{1}{\sqrt{2}}\ket{0} +\frac{1}{2}\ket{-1}, \qquad E_-=-\Omega_1. \]

(Can quickly verify by direct multiplication \(M\ket{E_\pm}=\pm\sqrt{2}\ket{E_\pm}\) and \(M\ket{E_0}=0\).)

Step 2: decompose the initial state \(\ket{-1}\) in the energy eigenbasis.

workout

We solve \(\ket{-1}=a\ket{E_+}+b\ket{E_0}+c\ket{E_-}\) by matching components in the \(\{\ket{+1},\ket{0},\ket{-1}\}\) basis, obtaining

\[ \ket{-1} = \frac{1}{2}\ket{E_+} -\frac{1}{\sqrt{2}}\ket{E_0} +\frac{1}{2}\ket{E_-}. \]

Step 3: time evolution in the rotating frame.

workout

Because the rotating-frame Hamiltonian is time-independent, we have

\[ \ket{\psi_{\rm rot}(t)} = \frac{1}{2}e^{-i\Omega_1 t}\ket{E_+} -\frac{1}{\sqrt{2}}\ket{E_0} +\frac{1}{2}e^{+i\Omega_1 t}\ket{E_-}. \]

Now expand back in the \(\ket{m}\) basis. Writing

\[ \ket{\psi_{\rm rot}(t)} = c_{+1}(t)\ket{+1} + c_0(t)\ket{0} + c_{-1}(t)\ket{-1}, \]

workout

one finds (collecting the coefficients of each \(\ket{m}\)):

For \(\ket{+1}\):

\[ c_{+1}(t) = \frac{1}{2}\cdot\frac{1}{2}e^{-i\Omega_1 t} -\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}} +\frac{1}{2}\cdot\frac{1}{2}e^{+i\Omega_1 t} = \frac{1}{4}\left(e^{-i\Omega_1 t}+e^{i\Omega_1 t}\right)-\frac{1}{2} = \frac{1}{2}\cos(\Omega_1 t)-\frac{1}{2}. \]

Equivalently,

\[ c_{+1}(t) = -\sin^2\!\left(\frac{\Omega_1 t}{2}\right). \]

For \(\ket{0}\):

\[ c_{0}(t) = \frac{1}{2}\cdot\frac{1}{\sqrt{2}}e^{-i\Omega_1 t} +\frac{1}{2}\cdot\left(-\frac{1}{\sqrt{2}}\right)e^{+i\Omega_1 t} = \frac{1}{2\sqrt{2}}\left(e^{-i\Omega_1 t}-e^{i\Omega_1 t}\right) = -\frac{i}{\sqrt{2}}\sin(\Omega_1 t). \]

For \(\ket{-1}\):

\[ c_{-1}(t) = \frac{1}{2}\cdot\frac{1}{2}e^{-i\Omega_1 t} -\frac{1}{\sqrt{2}}\cdot\left(-\frac{1}{\sqrt{2}}\right) +\frac{1}{2}\cdot\frac{1}{2}e^{+i\Omega_1 t} = \frac{1}{4}\left(e^{-i\Omega_1 t}+e^{i\Omega_1 t}\right)+\frac{1}{2} = \frac{1}{2}\cos(\Omega_1 t)+\frac{1}{2}. \]

Equivalently,

\[ c_{-1}(t) = \cos^2\!\left(\frac{\Omega_1 t}{2}\right). \]

So the rotating-frame state is

\[ \ket{\psi_{\rm rot}(t)} = -\sin^2\!\left(\frac{\Omega_1 t}{2}\right)\ket{+1} -\frac{i}{\sqrt{2}}\sin(\Omega_1 t)\ket{0} +\cos^2\!\left(\frac{\Omega_1 t}{2}\right)\ket{-1}. \]

Step 4: transform back to the lab frame.

At resonance, \(\omega=\omega_0\) and with \(U(t)=e^{-i\omega t \hat S_z}\) we have

\[ \ket{\psi(t)} = U(t)\ket{\psi_{\rm rot}(t)} = e^{-i\omega_0 t \hat S_z}\ket{\psi_{\rm rot}(t)}. \]

Since \(\hat S_z\ket{m}=m\ket{m}\), the unitary contributes phases \(e^{-im\omega_0 t}\) to each \(\ket{m}\) component:

\[ e^{-i\omega_0 t \hat S_z}\ket{+1}=e^{-i\omega_0 t}\ket{+1}, \qquad e^{-i\omega_0 t \hat S_z}\ket{0}=\ket{0}, \qquad e^{-i\omega_0 t \hat S_z}\ket{-1}=e^{+i\omega_0 t}\ket{-1}. \]

workout

Therefore the lab-frame state is

\[ \boxed{ \ket{\psi(t)} = -\sin^2\!\left(\frac{\Omega_1 t}{2}\right)e^{-i\omega_0 t}\ket{+1} -\frac{i}{\sqrt{2}}\sin(\Omega_1 t)\ket{0} +\cos^2\!\left(\frac{\Omega_1 t}{2}\right)e^{+i\omega_0 t}\ket{-1}. } \]

Step 5: probabilities \(P_m(t)=|\braket{m}{\psi(t)}|^2\).

The lab-frame phases \(e^{\mp i\omega_0 t}\) do not affect probabilities, so \(P_m(t)=|c_m(t)|^2\):

workout

\[ \boxed{ P_{+1}(t)=\sin^4\!\left(\frac{\Omega_1 t}{2}\right), \qquad P_{0}(t)=\frac{1}{2}\sin^2(\Omega_1 t), \qquad P_{-1}(t)=\cos^4\!\left(\frac{\Omega_1 t}{2}\right). } \]

Normalization check:

\[ P_{+1}(t)+P_0(t)+P_{-1}(t) = \sin^4\theta + 2\sin^2\theta\cos^2\theta + \cos^4\theta = (\sin^2\theta+\cos^2\theta)^2 = 1, \quad \theta=\frac{\Omega_1 t}{2}. \]

Plot: example Python/Matplotlib snippet.

import numpy as np
import matplotlib.pyplot as plt

Omega1 = 1.0  # set scale; you can change this
t = np.linspace(0, 2*np.pi/Omega1, 2000)

P_p1 = np.sin(Omega1*t/2)**4
P_0  = 0.5*np.sin(Omega1*t)**2
P_m1 = np.cos(Omega1*t/2)**4

plt.figure()
plt.plot(t, P_p1, label=r"$P_{+1}(t)$")
plt.plot(t, P_0,  label=r"$P_{0}(t)$")
plt.plot(t, P_m1, label=r"$P_{-1}(t)$")
plt.xlabel(r"$t$")
plt.ylabel(r"$P_m(t)$")
plt.ylim(-0.05, 1.05)
plt.legend()
plt.tight_layout()
plt.show()

For plotting, you can use whatever tool like Python/Matplotlib, Mathematica, Desmos, etc. The plot will show oscillations between the three levels with the given probabilities.

(g)¶

We stay at resonance \(\delta=0\), so the rotating-frame Hamiltonian is the same as in part (f), with eigenpairs

\[ E_+=+\Omega_1,\quad \ket{E_+} = \frac{1}{2}\ket{+1} +\frac{1}{\sqrt{2}}\ket{0} +\frac{1}{2}\ket{-1}, \]

\[ E_0=0,\quad \ket{E_0} = \frac{1}{\sqrt{2}}\ket{+1} -\frac{1}{\sqrt{2}}\ket{-1}, \]

\[ E_-=-\Omega_1,\quad \ket{E_-} = \frac{1}{2}\ket{+1} -\frac{1}{\sqrt{2}}\ket{0} +\frac{1}{2}\ket{-1}. \]

workout

Step 1: decompose the initial state \(\ket{\psi(0)}=\ket{0}\) in this eigenbasis.

Compute overlaps:

\(\braket{E_0}{0}=0\) since \(\ket{E_0}\) has no \(\ket{0}\) component.
\(\braket{E_+}{0}=1/\sqrt{2}\).
\(\braket{E_-}{0}=-1/\sqrt{2}\).

Hence

\[ \ket{0} = \frac{1}{\sqrt{2}}\ket{E_+} -\frac{1}{\sqrt{2}}\ket{E_-}. \]

Step 2: time evolution in the rotating frame.

\[ \ket{\psi_{\rm rot}(t)} = \frac{1}{\sqrt{2}}e^{-i\Omega_1 t}\ket{E_+} -\frac{1}{\sqrt{2}}e^{+i\Omega_1 t}\ket{E_-}. \]

Expand in the \(\{\ket{+1},\ket{0},\ket{-1}\}\) basis. Write

\[ \ket{\psi_{\rm rot}(t)} = c_{+1}(t)\ket{+1} + c_0(t)\ket{0} + c_{-1}(t)\ket{-1}. \]

For \(\ket{0}\):

\[ c_0(t) = \frac{1}{\sqrt{2}}e^{-i\Omega_1 t}\cdot\frac{1}{\sqrt{2}} -\frac{1}{\sqrt{2}}e^{+i\Omega_1 t}\cdot\left(-\frac{1}{\sqrt{2}}\right) = \frac{1}{2}\left(e^{-i\Omega_1 t}+e^{+i\Omega_1 t}\right) = \cos(\Omega_1 t). \]

For \(\ket{+1}\):

\[ c_{+1}(t) = \frac{1}{\sqrt{2}}e^{-i\Omega_1 t}\cdot\frac{1}{2} -\frac{1}{\sqrt{2}}e^{+i\Omega_1 t}\cdot\frac{1}{2} = \frac{1}{2\sqrt{2}}\left(e^{-i\Omega_1 t}-e^{+i\Omega_1 t}\right) = -\frac{i}{\sqrt{2}}\sin(\Omega_1 t). \]

For \(\ket{-1}\):

\[ c_{-1}(t) = \frac{1}{\sqrt{2}}e^{-i\Omega_1 t}\cdot\frac{1}{2} -\frac{1}{\sqrt{2}}e^{+i\Omega_1 t}\cdot\frac{1}{2} = -\frac{i}{\sqrt{2}}\sin(\Omega_1 t). \]

So

\[ \ket{\psi_{\rm rot}(t)} = -\frac{i}{\sqrt{2}}\sin(\Omega_1 t)\ket{+1} +\cos(\Omega_1 t)\ket{0} -\frac{i}{\sqrt{2}}\sin(\Omega_1 t)\ket{-1}. \]

Step 3: transform back to the lab frame (still at resonance \(\omega=\omega_0\)).

As in part (f), \(U(t)=e^{-i\omega_0 t \hat S_z}\) multiplies \(\ket{m}\) by \(e^{-im\omega_0 t}\). Therefore

\[ \boxed{ \ket{\psi(t)} = -\frac{i}{\sqrt{2}}\sin(\Omega_1 t)e^{-i\omega_0 t}\ket{+1} +\cos(\Omega_1 t)\ket{0} -\frac{i}{\sqrt{2}}\sin(\Omega_1 t)e^{+i\omega_0 t}\ket{-1}. } \]

Step 4: probabilities.

Again, the lab-frame phases do not affect probabilities, so

\[ \boxed{ P_{+1}(t)=\frac{1}{2}\sin^2(\Omega_1 t), \qquad P_{0}(t)=\cos^2(\Omega_1 t), \qquad P_{-1}(t)=\frac{1}{2}\sin^2(\Omega_1 t). } \]

Normalization check:

\[ P_{+1}(t)+P_0(t)+P_{-1}(t)=\sin^2(\Omega_1 t)+\cos^2(\Omega_1 t)=1. \]

Plot: example Python/Matplotlib snippet.

import numpy as np
import matplotlib.pyplot as plt

Omega1 = 1.0
t = np.linspace(0, 2*np.pi/Omega1, 2000)

P_p1 = 0.5*np.sin(Omega1*t)**2
P_0  = np.cos(Omega1*t)**2
P_m1 = 0.5*np.sin(Omega1*t)**2

plt.figure()
plt.plot(t, P_p1, label=r"$P_{+1}(t)$")
plt.plot(t, P_0,  label=r"$P_{0}(t)$")
plt.plot(t, P_m1, label=r"$P_{-1}(t)$")
plt.xlabel(r"$t$")
plt.ylabel(r"$P_m(t)$")
plt.ylim(-0.05, 1.05)
plt.legend()
plt.tight_layout()
plt.show()

Problem 5-2: Sudden perturbation to a harmonic oscillator¶

(a)¶

Write the Hamiltonian as \(\hat H(t)=\hat H_0+\hat V(t)\) with

\[ \hat V(t)=\theta(t)\,\frac{\Delta k}{2}\,\hat x^2. \]

Define the interaction-picture state (with respect to \(\hat H_0\)) by

\[ \ket{\psi_I(t)} \equiv e^{i \hat H_0 t}\ket{\psi_S(t)}. \]

Differentiate and use the Schrödinger-picture equation \(i\,\partial_t\ket{\psi_S(t)}=\hat H(t)\ket{\psi_S(t)}\):

\[ \begin{aligned} i\,\frac{d}{dt}\ket{\psi_I(t)} &= i\left(\frac{d}{dt}e^{i\hat H_0 t}\right)\ket{\psi_S(t)} + i e^{i\hat H_0 t}\frac{d}{dt}\ket{\psi_S(t)} \\ &= -\hat H_0 e^{i\hat H_0 t}\ket{\psi_S(t)} + e^{i\hat H_0 t}\hat H(t)\ket{\psi_S(t)} \\ &= e^{i\hat H_0 t}\bigl(\hat H(t)-\hat H_0\bigr)\ket{\psi_S(t)} \\ &= e^{i\hat H_0 t}\hat V(t)e^{-i\hat H_0 t}\ket{\psi_I(t)}. \end{aligned} \]

Thus the interaction-picture Schrödinger equation is (you can also find this on page 12 of Aash's Lecture Note #10)

\[ i\,\frac{d}{dt}\ket{\psi_I(t)}=\hat V_I(t)\ket{\psi_I(t)}, \qquad \hat V_I(t)\equiv e^{i\hat H_0 t}\hat V(t)e^{-i\hat H_0 t}. \]

workout

Now express \(\hat V_I(t)\) using ladder operators of the unperturbed oscillator. Let

\[ \omega_0 \equiv \sqrt{\frac{k}{m}}, \qquad \hat H_0=\omega_0\left(\hat a^\dagger \hat a+\frac{1}{2}\right), \qquad \hat x=\sqrt{\frac{1}{2m\omega_0}}\left(\hat a+\hat a^\dagger\right). \]

The interaction-picture ladder operators evolve under \(\hat H_0\) as

\[ \hat a_I(t)=e^{i\hat H_0 t}\hat a e^{-i\hat H_0 t}=\hat a\,e^{-i\omega_0 t}, \qquad \hat a_I^\dagger(t)=e^{i\hat H_0 t}\hat a^\dagger e^{-i\hat H_0 t}=\hat a^\dagger e^{+i\omega_0 t}. \]

Therefore

\[ \hat x_I(t)\equiv e^{i\hat H_0 t}\hat x e^{-i\hat H_0 t} =\sqrt{\frac{1}{2m\omega_0}}\left(\hat a\,e^{-i\omega_0 t}+\hat a^\dagger e^{+i\omega_0 t}\right). \]

Square this (and use \(\hat a\hat a^\dagger=\hat a^\dagger\hat a+1\)):

\[ \begin{aligned} \hat x_I^2(t) &=\frac{1}{2m\omega_0}\left(\hat a\,e^{-i\omega_0 t}+\hat a^\dagger e^{+i\omega_0 t}\right)^2 \\ &=\frac{1}{2m\omega_0}\left(\hat a^2 e^{-2i\omega_0 t}+(\hat a^\dagger)^2 e^{+2i\omega_0 t}+\hat a\hat a^\dagger+\hat a^\dagger\hat a\right) \\ &=\frac{1}{2m\omega_0}\left(\hat a^2 e^{-2i\omega_0 t}+(\hat a^\dagger)^2 e^{+2i\omega_0 t}+2\hat a^\dagger\hat a+1\right). \end{aligned} \]

Finally, since \(\theta(t)\) is a \(c\)-number,

\[ \boxed{ \hat V_I(t) = \theta(t)\,\frac{\Delta k}{2}\,\hat x_I^2(t) = \theta(t)\,\frac{\Delta k}{4m\omega_0} \left( \hat a^2 e^{-2i\omega_0 t} + (\hat a^\dagger)^2 e^{+2i\omega_0 t} + 2\hat a^\dagger\hat a + 1 \right). } \]

(b)¶

For \(t>0\), the interaction-picture state is expanded in the \(\hat H_0\) number basis as (see Lecture Note #11b)

\[ \ket{\psi_I(t)}=\sum_{n=0}^\infty c_n(t)\ket{n},\qquad c_n(0)=\delta_{n0}. \]

First-order time-dependent perturbation theory gives, for each \(n\),

\[ c_n^{(1)}(t)= -i\int_0^t dt'\,\matrixel{n}{\hat V_I(t')}{0}. \]

workout

From part (a), with \(\omega_0=\sqrt{k/m}\) and \(\hat x=\sqrt{\frac{1}{2m\omega_0}}(\hat a+\hat a^\dagger)\), we had (for \(t>0\))

\[ \hat V_I(t)=\frac{\Delta k}{4m\omega_0}\Big(\hat a^2 e^{-2i\omega_0 t}+\hat a^{\dagger 2} e^{+2i\omega_0 t}+2\hat a^\dagger \hat a+1\Big). \]

Selection rule from \(\hat x^2\): it changes \(n\) by \(0,\pm2\). Starting from \(\ket{0}\), the only off-diagonal state reachable at first order is \(\ket{2}\) (since \(\hat a^2\ket{0}=0\) and \(\hat a^\dagger \hat a\ket{0}=0\)).

Amplitude to reach \(\ket{2}\)

Only the \(\hat a^{\dagger 2}e^{+2i\omega_0 t'}\) term contributes:

\[ \matrixel{2}{\hat V_I(t')}{0} =\frac{\Delta k}{4m\omega_0}\,e^{+2i\omega_0 t'}\matrixel{2}{\hat a^{\dagger 2}}{0} =\frac{\Delta k}{4m\omega_0}\,e^{+2i\omega_0 t'}\sqrt{2}. \]

Therefore,

\[ c_2^{(1)}(t) =-i\frac{\Delta k\sqrt{2}}{4m\omega_0}\int_0^t dt'\,e^{+2i\omega_0 t'} =-i\frac{\Delta k\sqrt{2}}{4m\omega_0}\cdot \frac{e^{2i\omega_0 t}-1}{2i\omega_0} =\frac{\Delta k\sqrt{2}}{8m\omega_0^2}\Big(1-e^{2i\omega_0 t}\Big). \]

workout

Hence the first-order transition probability to \(n=2\) (which is already order \((\Delta k)^2\)) is

\[ p_2(t)=|c_2^{(1)}(t)|^2 =\frac{\Delta k^2}{32m^2\omega_0^4}\,\big|1-e^{2i\omega_0 t}\big|^2 =\frac{\Delta k^2}{32m^2\omega_0^4}\cdot 4\sin^2(\omega_0 t) =\boxed{\frac{\Delta k^2}{8m^2\omega_0^4}\sin^2(\omega_0 t)}. \]

Using \(m\omega_0^2=k\), this can be written more compactly as

\[ \boxed{p_2(t)=\frac{\Delta k^2}{8k^2}\sin^2(\omega_0 t)}. \]

All other \(n\ge 0\) at first order

workout

For \(n\neq 0,2\), one has \(\matrixel{n}{\hat V_I(t')}{0}=0\), hence

\[ c_n^{(1)}(t)=0\quad (n\neq 0,2) \qquad\Rightarrow\qquad p_n(t)=0\quad (n\neq 0,2) \]

at this order.

additional reading

Survival probability \(p_0(t)\) (important bookkeeping note)

The diagonal matrix element produces only a phase at first order:

\[ \matrixel{0}{\hat V_I(t')}{0}=\frac{\Delta k}{4m\omega_0} \quad\Rightarrow\quad c_0(t)=1-i\frac{\Delta k}{4m\omega_0}\,t+\cdots, \]

so \(|c_0(t)|^2=1+O((\Delta k)^2)\). To see the depletion \(p_0(t)=1-p_2(t)+\cdots\) explicitly requires going to second order for \(c_0(t)\). If one enforces normalization only to leading nontrivial order in probabilities, you can summarize the results as

\[ p_n(t)\approx \begin{cases} 1-\dfrac{\Delta k^2}{8k^2}\sin^2(\omega_0 t), & n=0,\\[8pt] \dfrac{\Delta k^2}{8k^2}\sin^2(\omega_0 t), & n=2,\\[8pt] 0, & \text{otherwise}, \end{cases} \]

understanding that the correction to \(p_0\) is effectively a second-order effect.

(c)¶

For \(t>0\), \(\theta(t)=1\), so the Hamiltonian becomes time-independent:

\[ \hat H=\hat H_0+\frac{\Delta k}{2}\hat x^2 =\frac{\hat p^2}{2m}+\frac{1}{2}(k+\Delta k)\hat x^2. \]

This is again a harmonic oscillator, but with an updated spring constant

\[ k' \equiv k+\Delta k, \]

so the new angular frequency is

\[ \omega' \equiv \sqrt{\frac{k'}{m}}=\sqrt{\frac{k+\Delta k}{m}}. \]

Define the new ladder operators (built from the same \(\hat x,\hat p\) but using \(\omega'\)):

\[ \hat b \equiv \sqrt{\frac{m\omega'}{2}}\,\hat x+\frac{i}{\sqrt{2m\omega'}}\,\hat p, \qquad \hat b^\dagger \equiv \sqrt{\frac{m\omega'}{2}}\,\hat x-\frac{i}{\sqrt{2m\omega'}}\,\hat p. \]

workout

They satisfy \([\hat b,\hat b^\dagger]=1\) (this follows directly from \([\hat x,\hat p]=i\)). Inverting these relations gives

\[ \hat x=\frac{1}{\sqrt{2m\omega'}}(\hat b+\hat b^\dagger), \qquad \hat p=-i\sqrt{\frac{m\omega'}{2}}(\hat b-\hat b^\dagger). \]

Now rewrite \(\hat H\) in terms of \(\hat b,\hat b^\dagger\). Substitute \(\hat x,\hat p\) into

\[ \hat H=\frac{\hat p^2}{2m}+\frac{1}{2}m\omega'^2\hat x^2. \]

Using the above expressions, one finds (standard HO algebra)

\[ \hat H=\omega'\left(\hat b^\dagger \hat b+\frac{1}{2}\right). \]

Therefore, \(\hat H\) is diagonal in the number basis of \(\hat b^\dagger\hat b\). Its energy eigenstates are \(\ket{n'}\) satisfying \(\hat b^\dagger\hat b\ket{n'}=n\ket{n'}\), and the eigenvalues are

\[ \boxed{E_n=\omega'\left(n+\frac{1}{2}\right),\qquad n=0,1,2,\dots} \]

with \(\omega'=\sqrt{(k+\Delta k)/m}\).

(d)¶

For \(t>0\) the Hamiltonian is time-independent and has a discrete spectrum

\[ \hat H=\omega'\left(\hat b^\dagger \hat b+\frac{1}{2}\right), \qquad E_n=\omega'\left(n+\frac{1}{2}\right). \]

Let \(\{\ket{n'}\}_{n\ge 0}\) be the corresponding energy eigenbasis, so that

\[ \hat H\ket{n'}=E_n\ket{n'}. \]

The initial state at \(t=0\) is the old ground state \(\ket{0}\) of \(\hat H_0\), which is not (in general) an eigenstate of \(\hat H\). Expand it in the new eigenbasis:

\[ \ket{0}=\sum_{n=0}^\infty c_n \ket{n'}, \qquad c_n\equiv\braket{n'}{0}. \]

Exact time evolution for \(t>0\) is then

\[ \ket{\psi(t)}=e^{-i\hat H t}\ket{0} =\sum_{n=0}^\infty c_n e^{-iE_n t}\ket{n'}. \]

workout

We are asked about \(p_n(t)\), the probability to find (upon measuring \(\hat H_0\) quanta) the original number state \(\ket{n}\):

\[ p_n(t)=|\braket{n}{\psi(t)}|^2. \]

Compute the amplitude:

\[ A_n(t)\equiv \braket{n}{\psi(t)} =\sum_{m=0}^\infty c_m e^{-iE_m t}\braket{n}{m'}. \]

Insert \(E_m=\omega'(m+1/2)\) and factor out the global phase \(e^{-i\omega' t/2}\) (which drops out of probabilities):

\[ A_n(t)=e^{-i\omega' t/2}\sum_{m=0}^\infty c_m \braket{n}{m'}\,e^{-im\omega' t}. \]

Thus \(A_n(t)\) is a Fourier series in harmonics \(e^{-im\omega' t}\), so it is periodic with period

\[ T=\frac{2\pi}{\omega'}. \]

Since \(p_n(t)=|A_n(t)|^2\), it follows that \(p_n(t)\) is also periodic with the same period:

\[ \boxed{p_n(t+T)=p_n(t)\quad\text{for all }t>0,\qquad T=\frac{2\pi}{\omega'}.} \]

(Equivalently: all energy gaps are integer multiples of \(\omega'\), since \(E_m-E_{m'}=\omega'(m-m')\), so any observable built from superpositions of energy eigenstates can only contain frequencies \(|m-m'|\omega'\), hence strict periodicity.)

Now you can compare this with part (b).

additional reading

In part (b) (first-order time-dependent perturbation theory in the interaction picture with respect to \(\hat H_0\)), one finds transition amplitudes proportional to

\[ \int_0^t dt'\,e^{i\omega_{n0}t'} \quad\text{with}\quad \omega_{n0}=E_n^{(0)}-E_0^{(0)}=n\omega_0, \]

leading to probabilities that (for short times) behave like powers of \(t\) (e.g. \(p_2(t)\propto t^2\) for sufficiently small \(t\)), i.e. a secular growth that is not periodic.

There is no contradiction because first-order perturbation theory is only valid for short times and small perturbation strength. More precisely:

The exact dynamics are periodic because the post-quench Hamiltonian has a perfectly discrete, equally spaced spectrum.
The perturbative result expands the exact periodic functions in a Taylor series around \(t=0\). For example, a generic oscillatory factor satisfies

\[ \sin^2(\Omega t)=\Omega^2 t^2+O(t^4), \qquad 1-\cos(\Omega t)=\frac{1}{2}\Omega^2 t^2+O(t^4). \]

So at early times the exact periodic probabilities look like polynomials in \(t\) and can appear to “grow” without showing the eventual turn-around.

At longer times \(t\sim 1/\Omega\) (here the relevant scale is set by \(\omega'\) and also by the perturbation strength through overlaps), higher-order terms in \(\Delta k\) become important and resum into bounded periodic functions. The apparent non-periodicity/secular behavior is an artifact of truncating the perturbation expansion.

So the consistent statement is:

Exact \(p_n(t)\) is periodic with \(T=2\pi/\omega'\), while first-order PT reproduces only the short-time expansion and misses recurrences.