Skip to content

Lesson01

Single qubit

\[ \begin{align*} \ket\psi &= \cos\dfrac\theta2 \ket0 + e^{i\phi}\sin\dfrac\theta2 \ket1 \\ \theta&\in[0, \pi] \\ \phi&\in [0,2\pi] \end{align*} \]

, or in matrix representation: \(\ket\psi = \begin{bmatrix} \cos\frac\theta2 \\ e^{i\phi}\sin\frac\theta2 \end{bmatrix}\).

State evolution

Equation of motion:

\[ i\hbar \frac{\partial}{\partial t}\ket\psi = \hat H \ket\psi \]

Solution to this E.O.M.:

\[ \begin{align*} \ket{\psi(t)} &= \mathcal U(t)\ket{\psi(0)} \\ &= e^{-i\hat Ht/\hbar} \ket{\psi(0)} \\ &= e^{-i\hat Ht} \ket{\psi(0)} \end{align*} \]
  • the Evolution Operator: \(\mathcal U(t) = e^{-i\hat Ht/\hbar} = e^{-i\hat Ht}\), and people often let \(\hbar = 1\) for convenience.

Pauli Matrices

The Hamiltonian in the above E.O.M. can always be decomposed as

\[ \hat H = \frac12 \sum_{j=0,1,2,3} \omega_j\sigma_j = \frac12(\omega_0I + \vec \omega\cdot\vec\sigma) \]
  • \(\omega_0I\) term : only gives rise to global phase (shift energy equally for both levels)
  • \(\vec \omega\cdot\vec\sigma\) term : is relevant to dynamics

Properties

\[ \sigma_j^2 = I \newline \{\sigma_j, \sigma_k\} = 2I\delta_{jk} \newline [\sigma_j, \sigma_k] = 2i\sigma_l\epsilon_{jkl} \]

Thus, we can rewrite the evolution operator as:

\[ \begin{align*} \mathcal U(t) &= e^{-iHt} = \exp(-i\cdot\frac12\vec \omega\cdot\vec\sigma\cdot t) \\ &= \exp(-i\cdot\frac{1}{2}\vert{\vec \omega}\vert \cdot \hat\sigma_n\cdot t) \\ &= \left(\cos\frac{\vert{\vec \omega}\vert t}2\right) \hat I - i \left(\sin\frac{\vert{\vec \omega}\vert t}{2}\right) \hat\sigma_n \end{align*} \]

where we’ve used the fact of the unit vector \(\hat\sigma_n\) (in Pauli basis) that

\[ (\hat\sigma_n)^2 = \sum_{i,j}n_i\sigma_i n_j\sigma_j = \frac12\sum_{i,j}n_in_j \{\sigma_i, \sigma_j\} \newline =\frac12\sum_{i,j}n_in_j\cdot2I\delta_{ij} = \sum_i n_i^2I = I. \]

Precession

If \(\hat\sigma_n = \hat z\), then

\[ \mathcal U(t) = \begin{bmatrix} e^{-i\omega t/2} & 0 \\ 0 & e^{i\omega t/2} \end{bmatrix} \]

, which makes state evolutes \(\ket{\psi(0)}\mapsto\ket{\psi(t)}\) as \(\begin{bmatrix} \cos\frac\theta2 \\ e^{i\phi}\sin\frac\theta2 \end{bmatrix} \mapsto e^{-i\omega t/2} \begin{bmatrix} \cos\frac\theta2 \\ e^{i(\phi+\omega t)}\sin\frac\theta2 \end{bmatrix}\).

So in the Bloch vector picture, \(\theta\mapsto\theta\) and \(\phi\mapsto\phi+\omega t\).

For general \(\hat\sigma_n\), the Bloch vector undergoes a precession about axis \(\hat\sigma_n\), with angular velocity \(\vec \omega\)​. (the detailed proof is in the next lesson)

Rotating Wave Approximation

framework

  • \(H(t)\) is still an arbitrary Hamiltonian, but now it could be time-dependent.
    (It's called the Hamiltonian in the interaction picture.)

  • Introduce a rotating frame-related quantity \(T(t) = e^{iAt}\), here \(A\) is time-independent and Hermitian.

Now we want to transform from the resting frame into the “rotating frame” (which is characterized by \(A\)), and the logic is:

\[ \begin{align*} i\frac{∂}{∂t} \ket\psi &= H(t) \ket\psi \\ i\frac{∂}{∂t} (T\ket\psi) &= H_I(t)(T\ket\psi), \\ \ket\psi &\mapsto T\ket\psi \\ H(t) &\mapsto H_I(t). \end{align*} \]

How do we express \(H_I(t)\) in terms of the original \(H(t)\) and \(T\)?
Notice that:

\[ \begin{align*} (THT^\dagger)(T\ket\psi) &= TH(T^\dagger T\ket\psi) \\ &= T i\frac{∂}{∂t}(T^\dagger T\ket\psi) \\ &= T i\left[ \frac{∂T^\dagger}{∂t}\cdot (T\ket\psi) + T^\dagger\cdot\frac{∂}{∂t}(T\ket\psi) \right] \\ &= iT\frac{∂T^\dagger}{∂t} \cdot (T\ket\psi) + i \frac{∂}{∂t}(T\ket\psi) \\ \implies H_I(t) &= (THT^\dagger) - iT\frac{∂T^\dagger}{∂t} \end{align*} \]

application to RWA

Following the previous definitions for \(H\), \(A\) and \(T\), one specific case that's considered widely is

\[ \begin{cases} &H = \boxed{\alpha H_0} + H_1 = \boxed{\frac{\omega_0}{2}\sigma_z} + \Omega\cos(\omega t + \phi)\sigma_x \\ &A = \boxed{\beta H_0} = \boxed{\frac{\omega}{2}\sigma_z} \end{cases} \]

often we consider the case where \(\omega\gg\Omega\) and \(\omega\gg|\omega-\omega_0|\).

Then, according to the changing rule for \(H(t) \mapsto H_I(t)\) just derived above,

\[ \begin{align*} H_I(t) &= (THT^\dagger) - iT\frac{∂T^\dagger}{∂t} \\ &= e^{iAt} (\alpha H_0 + H_1) e^{-iAt} - ie^{iAt}\frac{∂e^{-iAt}}{∂t} \\ &= \boxed{e^{iAt} (\alpha H_0) e^{-iAt}} + e^{iAt} (H_1) e^{-iAt} - A \\ &= \boxed{(\alpha H_0)} + e^{iAt} (\Omega\cos(\omega t + \phi)\sigma_x) e^{-iAt} - A \\ &= \frac{\omega_0}{2}\sigma_z + \Omega\cos(\omega t + \phi) e^{i\omega\sigma_zt/2} \sigma_x e^{-i\omega\sigma_zt/2} - \frac{\omega}{2}\sigma_z \\ &= \frac{\omega_0-\omega}{2}\sigma_z + \Omega\cos(\omega t + \phi) \begin{bmatrix} e^{i\omega t/2} & 0 \\ 0 & e^{-i\omega t/2} \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} e^{-i\omega t/2} & 0 \\ 0 & e^{i\omega t/2} \end{bmatrix} \\ &= \frac{\omega_0-\omega}{2}\sigma_z + \Omega\cos(\omega t + \phi) \begin{bmatrix} 0 & e^{i\omega t} \\ e^{-i\omega t} & 0 \end{bmatrix} \\ &= \frac{\omega_0-\omega}{2}\sigma_z + \frac\Omega 2 \begin{bmatrix} 0 & \textcolor{red}{e^{2i\omega t+i\phi}} + e^{-i\phi} \\ e^{i\phi} + \textcolor{red}{e^{-2i\omega t-i\phi}} & 0 \end{bmatrix} \end{align*} \]

(note that the boxed \(\boxed{e^{iAt} (\alpha H_0) e^{-iAt}}\) term can be reduced to \(\boxed{(\alpha H_0)}\) because all terms only involve \(\sigma_z\)​, hence commute with each other.)

We can ignore the term \(\textcolor{red}{\cancel{e^{2i\omega t+i\phi}}}\) and \(\textcolor{red}{\cancel{e^{-2i\omega t-i\phi}}}\) if we are only interested in the system's long-time behavior, and this leads to:

\[ H_I(t) \approx \frac{\omega_0-\omega}{2}\sigma_z + \frac\Omega 2 (\sigma_x\cos\phi + \sigma_y\sin\phi), \]

which is the effective Hamiltonian in the rotating frame \((\omega)\).