Can anyone see how the following is obtained:

In a section on Perturbation by an oscillating electric field. In book "Atomic Physics" by Foot. The following is stated: Consider the Shrodinger equation $$i \hbar \frac{\partial \Psi}{\partial t} = H \Psi.$$ The Hamiltonian has two parts $$H = H_0 + H_{I}(t).$$

The perturbation describing the Hamiltonian is: $$H_1(t) = e\mathbf{r} \cdot \mathbf{E}_0 \cos(\omega t)$$ The interaction mixes the two states: $$\Psi(\mathbf{r},t) = c_1(t)\psi_1(\mathbf{r}) e^{-\frac{iE_1 t}{\hbar}} + c_2(t)\psi_2(\mathbf{r}) e^{-\frac{i E_2 t}{\hbar}}$$ which can be written as $$\Psi(\mathbf{r},t) = c_1(t)|1 \rangle e^{-i \omega_1 t} + c_2|2\rangle e^{-i \omega_2 t}.$$

Question: Can anyone see how it follows that subsitution into the Shrodinger equation leads to $$\dot{1}\dot{c_1} = \Omega \cos(\omega t)e^{-i \omega_0 t}c_2,~~\dot{1}\dot{c_2} = \Omega^* \cos(\omega t)e^{-i \omega_0 t}c_1?$$ where $\omega_0 = \frac{E_2 - E_1}{\hbar}$ and the Rabi frequency $\Omega$ is defined by $\Omega = \frac{\langle 1| e \mathbf{r} \cdot \mathbf{E_0} |2 \rangle}{\hbar}$. Also what is the importance of $\dot{1}\dot{c_1}$ and $\dot{1}\dot{c_2}$ in describing the interaction of the two level system with radiation?

Thanks for any assistance.

  • $\begingroup$ what is $\dot 1$? $\endgroup$ Apr 3, 2017 at 17:32
  • $\begingroup$ @ZeroTheHero I don't know, this is part of what I'm asking, I assume that it's just a label for what he is defining it to be on the right of the equations, why he chose that I'm not sure... $\endgroup$
    – user101311
    Apr 3, 2017 at 19:50
  • $\begingroup$ actually: if this refers to Eqs.(7.9) and (7.10) it it not $\dot 1$ but rather "i", i.e. the imaginary unit. $\endgroup$ Apr 3, 2017 at 19:56
  • $\begingroup$ @ZeroTheHero Oh okay, yeah that is the equations I am referring to. Is his $\dot{c}$ the rate of change of the probability constants maybe? $\endgroup$
    – user101311
    Apr 3, 2017 at 20:31
  • $\begingroup$ The $c$'s are not constant otherwise they would have $0$ derivatives. They are just functions to solve for which give you probabilities as a function of time. $\endgroup$ Apr 3, 2017 at 22:13

1 Answer 1


As you wrote, we can generally express the state as $|\psi \rangle = c_1(t) e^{-i \omega_1 t} |1\rangle + c_2(t) e^{-i \omega_2 t} |2\rangle$. The time dependent Schrödinger equation is $i \hbar \frac{\partial |\psi\rangle}{\partial t} = H |\psi\rangle$. To calculate the coefficients $c_1$ and $c_2$ we can take the inner product of both sides with just one basis vector, such as $|1\rangle$: $$ i \hbar \langle 1 | \frac{\partial |\psi \rangle}{\partial t} = \langle 1 | H | \psi \rangle. $$

Since $\langle 1 | 1\rangle = 1$ and $\langle 1 | 2 \rangle = 0$, we obtain for the left hand side the time derivative only of the coefficient of $|1 \rangle$: $$ \langle 1 | \frac{\partial |\psi \rangle}{\partial t} =(\dot{c_1}(t) - i \omega_1 c_1(t))e^{-i \omega_1 t} $$ Meanwhile, for the righthand side, you know $H_0 |\psi\rangle = E_1 c_1(t) e^{-i \omega_1 t} |1\rangle + E_2 c_2(t) e^{-i \omega_2 t} |2\rangle$, so $\langle 1 | H_0 | \psi \rangle = E_1 c_1(t) e^{-i \omega_1 t}$.

The last piece we need is $\langle 1 | H_I(t) | \psi \rangle = e \cos(\omega t) \langle 1| r \cdot E_0 | \psi \rangle$. Now, $\langle 1 | r \cdot E_0 | 1 \rangle = 0$ (if you are unsure why this is true, then think explicitly about this inner product as an integral and consider the symmetry of the integrand). We are therefore left with $$ \langle 1 | H_I(t) | \psi \rangle = e \cos (\omega t) c_2(t) e^{-i \omega_2 t} \langle 1 | r \cdot E_0 | 2\rangle $$

Combining all of these calculations: $$ i \hbar (\dot{c_1}(t) - i \omega_1 c_1(t)) e^{-i \omega_1 t} = E_1 c_1(t)e^{-i \omega_1 t} + e \cos(\omega t) c_2(t) e^{-i \omega_2 t} \langle 1 | r \cdot E_0 | 2\rangle $$

Recalling that $E_1 = \hbar \omega_1$, the second term on the left hand side cancels with the first term on the right hand side. We are left only with $$ i \hbar \dot{c_1}(t) = \langle 1 | e r \cdot E_0 | 2 \rangle \cos(\omega t) e^{i (\omega_1 - \omega_2) t} c_2(t) $$

This is exactly the expression you wrote earlier up to substitution of the definitions of $\Omega$ and $\omega_0$. By an identical procedure you can construct the corresponding equation for $\dot{c_2}(t)$ (or just swap '1's with '2's).


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