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Rabi Oscillations Presentation.pdf
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We have a single qubit in state $\ket 0$, which alongside $\ket 1$ is a basis state of a two-dimensional vector space. The outcome of a measurement on this qubit is determined by a specific Hermitian matrix, the Hamiltonian, the eigenstates of which are $\ket {E_+}$ and $\ket {E_-}$, which are not the basis vectors.
$$ H\ket {E_\pm}=E_\pm\ket {E_\pm} $$
The question is then, if we let the qubit evolve in time, what is the probability that at any given time $t$ we measure the qubit in state $\ket 0$ or state $\ket 1$.
$$ P_0(t)=? \\P_1(t)=? $$
In A First Introduction to Quantum Mechanics we learned that all physical states evolve under the unitary matrix
$$ U=e^{iH(t-t_0)/\hbar} $$
$$ e^M \stackrel {def}{=}\sum_{n=0}^{\infty} {\frac{M^n}{n!}} $$
Our single qubit state will evolve under this matrix as well, as it is not equal to either of the eigenstates of the Hamiltonian.
Eigenstates do not change spontaneously in time because when acted upon by the unitary matrix, they only get multiplied by an overall phase. This is physically equivalent to the original state.
$$ \begin{aligned}U\ket{E_+}&=\ket{E_+}+\frac{i}{\hbar}tH\ket{E_+}+\frac{1}{2}\left(\frac{i}{\hbar}t\right)^2H^2\ket{E_+}+\dots\\ &=\ket{E_+}+\frac{i}{\hbar}tE_+\ket{E_+}+\frac{1}{2}\left(\frac{i}{\hbar}t\right)^2E_+^2\ket{E_+}+\dots\\ &=\ket{E_+}\left[1+\frac{i}{\hbar}tE_++\frac{1}{2}\left(\frac{i}{\hbar}t\right)^2E_+^2+\dots+\frac{1}{n!}\left(\frac{i}{\hbar }tE_+\right)^n\right]\\ &=e^{itE_+/\hbar}\ket{E_+}\\&=\text{phase} *\ket{E_+} \end{aligned} $$
Remember that $t_0 =0$. The equivalent goes for $\ket{E_-}$.
From the set-up of the problem, we know that at $t=0$ the qubit is in the state $\ket {\Psi(0)}=\ket 0$. We now have to derive how the unitary matrix $U$ acts on this state in time, assuming a general Hamiltonian.
A general Hamiltonian is simply a general Hermitian matrix as described in Matrices and Determinants.