0
$\begingroup$

After making the electric-dipole approx., I can express the interaction of a monochromatic field with angular frequency $\omega$ and a dipole moment ${\bf \mu(x)}$ as

$V({\bf x},t) = - {\bf \mu(x)} \cdot E_0\, sin(\omega t + \phi)$.

I read that in many circumstances, truncation to only a finite number of quantum states is adequate. Apparently the two state approximation results in this expression:

$H_{TLS}(t) = -\frac{1}{2} (E_2 - E_1) \sigma_z - \mu E_0\, sin(\omega t + \phi)\, \sigma_x$.

How does one arrive at this two-level approximation?

$\endgroup$
1
$\begingroup$

You are essentially asking for the secular approximation. The time evolution of the system is governed by the hamiltonian $$ \hat{H} = \hat{H}_0 + V(x,t),$$ where $$V(x,t) = -exE_0\frac{1}{2i}\left(e^{i\omega t}-e^{-i\omega t}\right).$$ Suppose we know the solutions of the time-dependent Schrödinger equation governed by $\hat{H}_0$ (unperturbed problem) to be $$ \psi_n(t) = \left|\left. n\right>\right. e^{-i\omega_n t},$$ where the states $\left|\left. n\right>\right.$ are eigenstates of $\hat{H}_0$, and $\hbar\omega_n$ are the corresponding eigenvalues. Then we can expand the solutions of the time-dependent problem in terms of the eigenstates of the unperturbed problem according to $$ \psi(t) = \sum_n a_n(t)\left|\left. n\right>\right. e^{-i\omega_nt}.$$ Inserting this Ansatz into the time-dependent Schrödinger equation governed by $\hat{H}$ gives a system of equations for the coefficients $a_n(t)$, namely, $$ i\hbar\partial_t a_m(t) = -\frac{eE}{2i}\sum_n a_n(t)x_{mn}\left(e^{i(\omega_{mn}+\omega)t}-e^{i(\omega_{mn}-\omega)t}\right)$$ with $\omega_{mn}=\omega_m-\omega_n$.

Now we assume that the excitation frequency $\omega$ is very close to the transition frequency $\omega_{fi}$ between an initial state $i$ and a final state $f$, i.e., $$ \omega = \omega_{fi}+\epsilon.$$ We further assume that at $t<0$ the system is in the initial state $\left|\left. i\right>\right.$, and that the excitation $V(x,t)$ is switched on abruptly at $t=0$.

The secular approximation takes in such a case only the coefficients $a_i(t)$ and $a_f(t)$ into account. This approximation leads to the two coupled equations $$ i\hbar\partial_t a_i(t) = -\frac{eE}{2i}a_f(t)x_{if}e^{+i\epsilon t}$$ $$ i\hbar\partial_t a_f(t) = +\frac{eE}{2i}a_i(t)x_{if}^\star e^{-i\epsilon t}.$$ This system of equations can be solved exactly. To this end, we define new coefficients $b_i(t)$ and $b_f(t)$ according to $$ a_i(t) = e^{+i\epsilon t/2}b_i(t)$$ $$ a_f(t) = e^{-i\epsilon t/2}b_f(t)$$ and insert them into the pair of equations above. This leads to $$ i\hbar\partial_t b_i(t) = \frac{\hbar\epsilon}{2}b_i(t) - \frac{eE}{2i}x_{if}b_f(t)$$ $$ i\hbar\partial_t b_f(t) = \frac{eE}{2i}x_{if}^\star b_i(t) - \frac{\hbar\epsilon}{2}b_f(t). $$ This pair of equations reduces to the harmonic oscillator equations $$ \partial_t^2 b_{i/f}(t) + \Omega^2b_{i/f}(t) = 0,$$ where $$ \Omega = \sqrt{\left(\frac{eE}{2\hbar}\right)^2|x_{if}|^2+\left(\frac{\epsilon}{2}\right)^2},$$ which is called the Rabi frequency.

A more elaborate discussion of the secular approximation is found, e.g, in Cohen-Tannoudji, Quantum Mechanics, Vol. II, chapter XIII (see in particular the complement C$_{\mathrm{XIII}}$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.