How is Electromagnetically-Induced Transparency a result of "destructive quantum interference between two pathways"?

Question

It has been described that Electromagnetically-Induced Transparency (EIT) is a result of "destructive quantum interference between two pathways." To quote from this source:

In these simple cases, the characteristic EIT interference effects arise because the control beam induces a splitting of the excited state into a doublet of “dressed” states, providing two interfering excitation pathways for the probe beam. The null in the probe absorption at the normal atomic resonance frequency is a result of destructive quantum interference between these two pathways.

Another resource says:

Ordinarily, when one applies an electromagnetic field of frequency ω31, atoms in the ground state can absorb energy and transition to state |3>. But, when an electromagnetic field resonant with the |2> → |3> transition is also applied, there are now two transition pathways by which an atom can get from the ground state to the excited state: it can either transition in the same way as before, |1> → |3>, or it can transition along |1> → |3> → |2> → |3>. These two allowed transition paths can destructively interfere, and under the appropriate conditions this results in zero probe absorption at resonance.

In the same resource it is derived that the term that produces the radiation in this case can be expressed as:

$$\tilde{\rho}_{13}=-\frac{2\left(\gamma_{12}-i\left(\Delta_{c}-\Delta_{p}\right)\right) \Omega_{p}}{4\left(\gamma_{13}+i \Delta_{p}\right)\left(-i \gamma_{12}-\Delta_{c}+\Delta_{p}\right)-i \Omega_{c}^{2}} $$

I completely understand the math behind the derivation of this quantity, and I see that it goes to zero when $\Delta_c = \Delta_p$, but how does one show that this is because of an interference due to different transition pathways?

How can I show that this effect is due to interference of the |1$\rangle$ → |3$\rangle$ and |1$\rangle$ → |3$\rangle$ → |2$\rangle$ → |3$\rangle$ pathways?

Answer 1

To my understanding you have done the proof already.

• Step 1. Without the coupling laser the probe laser is absorbed due to the coupling of the bare atomic states $\langle3| \vec d\cdot \vec E_{13}|1\rangle \ne 0$.
• Step 2a. By coupling the bare atomic states $\{|2\rangle,|3\rangle\}$ using the "coupling lasers", we obtain the dressed states $|\pm\rangle$.
• Step 2b. The dressed states are exactly the superpositions of the bare atomic states you were looking for: $|3\rangle$, $|3\rangle \to |2\rangle \to |3\rangle$, $|3\rangle \to |2\rangle \to |3\rangle\to |2\rangle \to |3\rangle$, and all other higher orders.
• Step 2c. A "weak" probe laser field is no longer absorbed. Hence, $\langle\pm| \vec d\cdot \vec E_{13}|1\rangle = 0$.

I understand that this argument is not fully satisfactory, because you are probably looking for a calculation where we calculate the transition coeffs. In order to do so take the "fully mixed states", $|\pm\rangle = \frac{|2\rangle \pm |3\rangle}{\sqrt{2}}$ and calc $$ \langle -|\vec d\cdot \vec E_{13}|1\rangle \propto -\langle 3|\vec d\cdot \vec E_{13}|1\rangle $$ and $$ \langle +|\vec d\cdot \vec E_{13}|1\rangle \propto +\langle 3|\vec d\cdot \vec E_{13}|1\rangle $$ Again the coherent superposition of these two states vanish.

The fully mixed states are the result of an adiabatic change of the two bare atomic states, which takes place if we "slowly" increase the power of the coupling laser. E.g. the state $|3\rangle$ acquires an admixture of the state $|2\rangle$ and vice versa. This can be written as \begin{align} |2^*\rangle &= \cos(\alpha) |2\rangle + \sin(\alpha) e^{i\varphi_{2}}|3\rangle \\ |3^*\rangle &= \cos(\alpha) |3\rangle + \sin(\alpha) e^{i\varphi_{3}}|2\rangle \end{align}

Answer 2

I think I now understand why this is sometimes said (and why it is inaccurate).

Our polarization in its full form is written as:

$$\frac{2 \text{$\Omega $}_p (\text{$\gamma $}_{12}-i (\text{$\Delta $}_c-\text{$\Delta $p}))}{4 (\text{$\gamma $}_{13}+i \text{$\Delta $p}) (-i \text{$\gamma $}_{12}-\text{$\Delta $c}+\text{$\Delta $p})-i \text{$\Omega $c}^2}$$

In EIT the control field is typically very strong. But we can consider the situation when it's weak and taylor expand $\Omega_c$:

$$\frac{\text{$\Omega $p}}{2 (\text{$\Delta $p}-i \text{$\gamma $13})}+\frac{i \text{$\Omega $c}^2 \text{$\Omega $p}}{8 (\text{$\Delta $p}-i \text{$\gamma $13})^2 (\text{$\gamma $12}-i \text{$\Delta $c}+i \text{$\Delta $p})}+O\left(\text{$\Omega $c}^3\right)$$

(I've added some simple mathematica code that does this if anyone wants it at the end.) So now we can see that the polarizability can be broken down into a linear sum of different terms. The first term is:

$$ \frac{\text{$\Omega $p}}{2 (\text{$\Delta $p}-i \text{$\gamma $13})} $$

Here we can see that this term can only represent transitions between $|1\rangle$ and $|3\rangle$ (since no $\Omega_c$ is involved). (the atom is in a Lambda configuration where the probe $E_p$ causes transitions from $|1\rangle$ and $|3\rangle$ and the control causes transitions between $|2\rangle$ and $|3\rangle$. $|1\rangle$ and $|2\rangle$ are ground states and $|3\rangle$ is the excited state.)

So when the control field is off, we can think of the strength of the polarization as being due to the transitions from $|1\rangle$ and $|3\rangle$.

The second term has the form:

$$\frac{i \text{$\Omega $c}^2 \text{$\Omega $p}}{8 (\text{$\Delta $p}-i \text{$\gamma $13})^2 (\text{$\gamma $12}-i \text{$\Delta $c}+i \text{$\Delta $p})}$$

Now if we think of each unit of $\Omega_c$ or $\Omega_p$ as being a single photon that causes the atom to transition to another state, we can think of $\text{$\Omega $}_p\text{$\Omega $}_c^2 = \text{$\Omega $}_p\text{$\Omega $}_c \text{$\Omega $}_c$ as the situation where the atom goes from $|1\rangle$ to $|3\rangle$ via $\Omega_p$ then $|3\rangle$ to $|2\rangle$ via $\Omega_c$ and finally $|2\rangle$ back to $|3\rangle$ via the second $\Omega_c$.

Now intuitively, our atom has two pathways to transition to the state $|3\rangle$, one via $|1\rangle \rightarrow |3\rangle$ and one via $|1\rangle \rightarrow |3\rangle \rightarrow |2\rangle \rightarrow |3\rangle$. Now we have a quantum system that can travel two different pathways, and therefore there can be some quantum interference between the two possible paths, like the double-slit experiment or a quantum Mach-Zhender interferometer.

Now from our expansion we know the probability amplitudes. The first path will have an amplitude $\propto \frac{1}{2 (\text{$\Delta $p}-i \text{$\gamma $13})}$ while the second path will have an amplitude $\propto \frac{i }{8 (\text{$\Delta $p}-i \text{$\gamma $13})^2 (\text{$\gamma $12}-i \text{$\Delta $c}+i \text{$\Delta $p})}$. I admit though that it gets a bit confusing at this point, because we are referring to $\rho_{13}$,the quantum superposition strength of $|1\rangle$ and $|3\rangle$, and how that value changes depending on quantum interference.

In any case, this is what I believe to be what is meant when people referring to EIT being due to quantum interference between different transitions. Obviously, EIT with its strong control field cannot be Taylor expanded like this, so this particular path interference only works in this very weak limit. But, intuitively we can still technically think of interference by taking the infinite limit of the sum of all of the contributing terms.

Mathematica code for anyone who wants it:

Series[(2*(\[Gamma]12 - 
     I (\[CapitalDelta]c - \[CapitalDelta]p)) \[CapitalOmega]p)/(
 4*(\[Gamma]13 + 
     I \[CapitalDelta]p) (-I \[Gamma]12 - \[CapitalDelta]c + \
\[CapitalDelta]p) - I \[CapitalOmega]c^2), {\[CapitalOmega]c, 0, 2}]