Case Study
Evolution of two-states system
(1) Stationary(time-independent)
Assume that \( \left|\psi_1 \right\rangle \) and \( \left|\psi_2 \right\rangle \) represent state 1 and state 2 , respectively, and \( E_1 \) and \( E_2 \) represent the energy of state 1 and state 2, respectively.
If two electronic states are independent(without the coupling),then:
\[
\hat{H}_0\left|\psi_1 \right \rangle = E_1\left|\psi_1 \right \rangle
\]
\[
\hat{H}_0\left|\psi_2 \right \rangle = E_2\left|\psi_2 \right \rangle
\]
where $ H_0 $ is the Hamitonian of system, which without coupling.
If two electronic states are coupling,then the hamitonian changes into:
\[
\hat{H} = \hat{H}_0 + \hat{W}
\]
In this condition, the time-independent Schrödinger equation can be written as:
$$
\hat{H} \left|\psi \right \rangle=E\left|\psi \right \rangle
$$
We can linearly combine \( \left|\psi_1 \right\rangle \) and \( \left|\psi_2 \right\rangle \) to construct \( \left|\psi \right\rangle \) :
\[
\left|\psi \right\rangle = c_1 \left|\psi_1 \right\rangle + c_2 \left|\psi_2 \right\rangle
\]
where \( c_1 \) and \( c_2 \) are complex coefficients.
\[
\left|\psi \right \rangle = c_1 \left|\psi_1 \right \rangle + c_2\left|\psi_2 \right \rangle
\]
Multiply the both sides of time-independent Schrödinger equation left by\(\left\langle\psi_1\right|\) and \(\left\langle\psi_2\right|\) respectively, then we obtain:
\[
\left\langle\psi_1\right|\hat{H} [c_1 \left|\psi_1 \right \rangle + c_2\left|\psi_2 \right \rangle]=\left\langle\psi_1\right|E[c_1 \left|\psi_1 \right \rangle + c_2\left|\psi_2 \right \rangle]
\]
\[
\left\langle\psi_2\right|\hat{H} [c_1 \left|\psi_1 \right \rangle + c_2\left|\psi_2 \right \rangle]=\left\langle\psi_2\right|E[c_1 \left|\psi_1 \right \rangle + c_2\left|\psi_2 \right \rangle]
\]
or simplely:
\[
c_1 \hat{H}_{11} + c_2 \hat{H}_{12} = c_1 E
\]
\[
c_1 \hat{H}_{21} + c_2 \hat{H}_{22} = c_2 E
\]
where \(H_{ij} = \left\langle\psi_i\right|\hat{H}\left|\psi_j \right \rangle\) , and \(\left\langle\psi_i | \psi_j \right \rangle = \delta_{ij}\) is used.
In matrix representation, the equation above change into the form as follow:
\[\begin{align}
\begin{bmatrix} H_{11} & H_{12} \\
H_{21} & H_{22} \end{bmatrix} \begin{bmatrix} c_1 \\
c_2 \end{bmatrix} = \begin{bmatrix} E_{11} & 0 \\
0 & E_{22} \end{bmatrix} \begin{bmatrix} c_1 \\
c_2 \end{bmatrix}
\end{align}
\]
Now we introduce a common method for solving the equation above.
At first, we divide the total hamitonian matrix H into two parts:
\[
\begin{align}
H &=
\begin{bmatrix}
\dfrac{1}{2}(H_{11} + H_{12}) & 0 \\ 0 & \dfrac{1}{2}(H_{11} + H_{12})
\end{bmatrix} \\
&\quad +
\begin{bmatrix}
\dfrac{1}{2}(H_{11} - H_{12}) & H_{12} \\ H_{21} & -\dfrac{1}{2}(H_{11} - H_{12})
\end{bmatrix}
\end{align}
\]
Then the the total hamitonian matrix H can be written as:
\[
H = \dfrac{1}{2}(H_{11} + H_{12}) I + \dfrac{1}{2}(H_{11} - H_{12}) K
\]
where I is identity matirx, K is a matrix that:
\[
K = \begin{bmatrix} 1 & \dfrac{2H_{12}}{H_{11} - H_{12}} \\ \dfrac{2H_{21}}{H_{11} - H_{12}} & -1 \end{bmatrix}
\]
We set:
\[
tan \theta = \dfrac{2|H_{12}|}{H_{11} - H_{12}}
\]
\[
|H_{12}| = |H_{21}|
\]
\[
H_{12} = |H_{12}| e^{-i\varphi }
\]
\[
H_{21} = |H_{21}| e^{i\varphi }
\]
Then matrix K becomes:
\[
K = \begin{bmatrix} 1 & tan \theta e^{-i\varphi} \\ tan \theta e^{i\varphi} & -1 \end{bmatrix}
\]
Because matirx H satisfies \(\hat{H} \left|\psi \right \rangle=E\left|\psi \right \rangle\) , then matirx K also satisfies:
\[
\hat{K} \left|\psi \right \rangle=k\left|\psi \right \rangle
\]
or in matrix representation:
\[
\begin{aligned}
\begin{bmatrix}
1 - k & \tan \theta \, e^{-i\varphi} \\
\tan \theta \, e^{i\varphi} & -1 - k
\end{bmatrix}
\begin{bmatrix}
c_1 \\ c_2
\end{bmatrix}
&= 0
\end{aligned}
\]
If there has a non-trivial solution of the equation above, it must satisfies:
\[
\begin{aligned}
\begin{bmatrix}
1 - k & \tan \theta \, e^{-i\varphi} \\
\tan \theta \, e^{i\varphi} & -1 - k
\end{bmatrix}
\begin{bmatrix}
c_1 \\ c_2
\end{bmatrix}
&= 0
\end{aligned}
\]
Solve for:
\[
k_+ = \dfrac{1}{cos \theta}
\]
\[
k_- = -\dfrac{1}{cos \theta}
\]
So we obtain the expression of E:
\[
E_+ = \dfrac{1}{2}(H_{11} + H_{12}) + \sqrt{\dfrac{1}{4} * (H_{11} - H_{22})^2 + |H_{12}|^2}
\]
\[
E_- = \dfrac{1}{2}(H_{11} + H_{12}) - \sqrt{\dfrac{1}{4} * (H_{11} - H_{22})^2 + |H_{12}|^2}
\]
Finally, we can solve the expression of constant \(c_1\) and \(c_2\) . With the example of \(k_+\) , we obtain:
\[
(1-k)c_1 + tan \theta e^{-i\varphi}c_2 = 0
\\tan \theta e^{i\varphi}c_1 -(1+k)c_2 = 0
\]
Solve for:
\[
c_1 = cos\frac{\theta}{2} * e^{-i \dfrac{\varphi}{2}}\\
c_2 = sin\frac{\theta}{2} * e^{i \dfrac{\varphi}{2}}\\
\]
That is the constant of \(\left|\psi_+ \right \rangle\) , whose eigenvalue is \(E_+\) .
Similarly, we can know the constant of \(\left|\psi_- \right \rangle\) , whose eigenvalue is \(E_-\) :
\[
c_1 = -sin\frac{\theta}{2} * e^{-i \dfrac{\varphi}{2}}\\
c_2 = cos\frac{\theta}{2} * e^{i \dfrac{\varphi}{2}}\\
\]
where normalization condition \(|c_1|^2 + |c_2|^2 = 1\) is used.
(2) Evolution of Two-States(time-dependent)
The evolution of two-states system can be descibed by time-dependent Schrödinger equation:
\[
i\hbar \dfrac{\partial }{\partial t}\left|\Psi(t) \right \rangle = \hat{H}\left|\Psi(t) \right \rangle
\]
\(\left|\Psi(t) \right \rangle\) can be written as:
\[
\left|\Psi(t) \right \rangle = a_1(t)\left|\psi_1 \right \rangle + a_2(t)\left|\psi_2 \right \rangle
\]
Similarly, we multiply the both sides of time-dependent Schrödinger equation left by\(\left\langle\psi_1\right|\) and \(\left\langle\psi_2\right|\) respectively, then we obtain:
\[
\left\langle\psi_1\right|i\hbar \dfrac{\partial }{\partial t}[a_1(t)\left|\psi_1 \right \rangle + a_2(t)\left|\psi_2 \right \rangle] = \left\langle\psi_1\right|\hat{H}[a_1(t)\left|\psi_1 \right \rangle + a_2(t)\left|\psi_2 \right \rangle]
\]
\[
\left\langle\psi_2\right|i\hbar \dfrac{\partial }{\partial t}[a_1(t)\left|\psi_1 \right \rangle + a_2(t)\left|\psi_2 \right \rangle] = \left\langle\psi_2\right|\hat{H}[a_1(t)\left|\psi_1 \right \rangle + a_2(t)\left|\psi_2 \right \rangle]
\]
Then:
\[
i\hbar \dfrac{\partial }{\partial t}a_1(t) = a_1(t)H_{11} + a_2(t)H_{12}
\]
\[
i\hbar \dfrac{\partial }{\partial t}a_2(t) = a_1(t)H_{21} + a_2(t)H_{22}
\]
We can choose:
\[
a_i(t) = exp(\frac{-iE_it}{\hbar})
\]
Then the total wavefunction becomes:
\[
\left|\Psi(t) \right \rangle = exp(\frac{-iE_+t}{\hbar})\left|\psi_+ \right \rangle + exp(\frac{-iE_-t}{\hbar})\left|\psi_- \right \rangle
\]
We choose \(\left|\psi_1 \right \rangle\) as initial state, then:
\[
\left|\Psi(0) \right \rangle = \left|\psi_1 \right \rangle
\]
and \(\left|\psi_1 \right \rangle\) is given by:
\[
\left|\psi_1 \right \rangle = [cos\dfrac{\theta}{2} \left|\psi_+ \right \rangle - sin\dfrac{\theta}{2} \left|\psi_- \right \rangle] * exp(i \dfrac{\varphi}{2})
\]
The Probability of state 2 is defined as:
\[
P_2 = |\left\langle\psi_2|\Psi(t)\right \rangle|^2
\]
and \(a_2(t)\) :
\[
\left\langle\psi_2|\Psi(t)\right \rangle = exp(i \varphi)*sin\dfrac{\theta}{2}cos\dfrac{\theta}{2}*[exp(\frac{-iE_+t}{\hbar}) - exp(\frac{-iE_-t}{\hbar})]
\]
Thus:
\[
P_2 = \dfrac{1}{4}sin\theta * [2-exp(-i\frac{E_+-E_-}{\hbar}t) - exp(-i\frac{E_--E_+}{\hbar}t)]
\]
by Euler's formula:
\[
e^{i\theta} = isin\theta + cos\theta
\]
\(P_2\) changes into:
\[
P_2 = sin^2\theta * sin^2\frac{E_+-E_-}{2\hbar}t
\]
Finally, \(P_2\) becomes:
\[
P_2 = \frac{4|H_{12}|^2}{4|H_{12}|^2 + (H_{11} - H_{22})^2} * sin^2\sqrt{\frac{(H_{11} - H_{22})^2 + 4|H_{12}|^2}{2\hbar}t}
\]
That is so_called Rabi equaition.
(3) Rabi frequency
According to:
\[
P_2 = \frac{4|H_{12}|^2}{4|H_{12}|^2 + (H_{11} - H_{22})^2} * sin^2\sqrt{\frac{(H_{11} - H_{22})^2 + 4|H_{12}|^2}{2\hbar}t}
\]
we obtain the f, or so-called Rabi frequency:
\[
f = \sqrt{\frac{(H_{11} - H_{22})^2 + 4|H_{12}|^2}{2\hbar}}
\]
The period of \(P_2\) satisfies:
\[
\frac{(H_{11} - H_{22})^2 + 4|H_{12}|^2}{2\hbar}t = (2k+1)\pi
\]
\[
t = \frac{2(2k+1)\pi \hbar}{(H_{11} - H_{22})^2 + 4|H_{12}|^2}
\]