Evolution of a Qubit
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This section is to illustrate on a simple example the evolution of a quantum system.
Fundamental relation
The Schrödinger equation rules the evolution of any isolated quantum system:
\[i \hbar \frac{\partial \ket{\psi(t)} }{\partial t} = \hat{H} \ket{\psi(t)}\]
We assume that the Hamiltonian \(\hat{H}\) does not depend on time,
and for the sake of simplicity,
that all its eigenvalues \(\{ E_1, \dots, E_n \}\) are nondegenerated.
All eigenvectors \(\{ \ket{\psi_1}, \dots, \ket{\psi_n} \}\) are orthogonal.
Any state vector can be expressed in this basis of eigenvectors,
thus we define the coefficients:
\[\begin{split}& \ket{\psi(0)} = \sum_\alpha c_\alpha \ket{\psi_\alpha} \\
& \ket{\psi(t)} = \sum_\alpha \lambda_\alpha(t) \ket{\psi_\alpha}\end{split}\]
The Schrödinger equation writes:
\[\begin{split}i \hbar \frac{\partial}{\partial t}
\begin{bmatrix}
\lambda_1(t) \\ \dots \\ \lambda_n(t)
\end{bmatrix}
= \hat{H}
\begin{bmatrix}
\lambda_1(t) \\ \dots \\ \lambda_n(t)
\end{bmatrix}
=
\begin{bmatrix}
E_1 \lambda_1(t) \\ \dots \\ E_2 \lambda_n(t)
\end{bmatrix}\end{split}\]
i.e. for each of the (orthogonal) eigenvectors:
\[i \hbar \frac{\partial}{\partial t} \lambda_\alpha(t) = E_\alpha \lambda_\alpha(t)\]
and we end up with:
\[\ket{\psi(t)} = \sum_\alpha c_\alpha e^{-i E_\alpha t / \hbar} \ket{\psi_\alpha}\]
This means:
If \(\ket{\psi(0)}\) is one of the eigenvectors,
than the state will always remain in this eigenstate!
In case of an initial superposition of eigenstates, the coefficients will evolve,
such that the amplitude of each component will stay the same, but not the phases.
Illustration
Let us first imagine that after some time we measure in the eigenbasis
of the Hamiltonian: we will always get the outcomes with the same probabilities
given by \(| c_\alpha |^2\).
But what happens if we measure an other observable, i.e. in an other basis?
Imagine we consider a two-level system with the energy levels \(E_0, \, E_1\),
with \(\hbar \Omega = E_0 - E_1\),
and the initial state
\(\ket{\psi(0)} = \ket{+} = \frac{1}{\sqrt{2}} \left( \ket{0} + \ket{1} \right)\).
This state’s evolution is given as (the absolute phase can be left out):
\[\ket{\psi(t)} = \frac{1}{\sqrt{2}} \left(
e^{-i \Omega t / 2} \ket{0} +
e^{+i \Omega t / 2} \ket{1}
\right)\]
By rearranging the terms, we can infer what will happen if we measure in the basis
\({\ket{+}, \ket{-}}\) i.e. we measure an observable that has these eigenstates:
\[\begin{split}\ket{\psi(t)}
& = \frac{1}{\sqrt{2}} \left(
e^{-i \Omega t / 2} \frac{\ket{+} + \ket{-}}{2} +
e^{+i \Omega t / 2} \frac{\ket{+} - \ket{-}}{2}
\right) \\
& = \frac{1}{\sqrt{2}} \left(
\cos (\frac{\Omega}{2} t) \ket{+} +
\sin (\frac{\Omega}{2} t) \ket{-}
\right)\end{split}\]
Thus the probabilities to measure either \(\ket{+}\) or \(\ket{-}\)
will oscillate in time with the frequency given by the energy gap between the two levels!
References: