# [Quantum Optics] Scully and Zubairy Section 6.2

• theuselessone
In summary, the conversation discusses the calculation of atomic inversion for the Jaynes-Cummings model Hamiltonian in quantum optics. The atomic inversion is derived from the definition of the atomic inversion given by equation (6.2.20) to equation (6.2.21), which involves computing |c_{a,n}(t)|^2 and |c_{b,n}(t)|^2 using equations (6.2.16) and (6.2.17), respectively. The conversation also mentions that the second sum in the resulting equation should start from n=0 instead of n=-1.
theuselessone

## Homework Statement

I've been reading through Quantum Optics by Scully and Zubairy and have been stuck in section 6.2 getting from the definition of the atomic inversion given by equation (6.2.20, pg 199)

$W(t)=\sum_{n}\left[|c_{a,n}(t)|^2-|c_{b,n}(t)|^2\right]$

to the atomic inversion for the Jaynes-Cummings model Hamiltonian between a single-mode quantized field and a single two-level atom given by (6.2.21, pg 200)

$W(t)=\sum_{n=0}^{\infty}\rho_{nn}(0)\left[\frac{\Delta^2}{\Omega_n^2}+\frac{4g^2(n+1)}{\Omega_n^2}\cos(\Omega_n t)\right]$

where $|c_n(0)|^2=\rho_{nn}(0)$ is the initial probability of n-photons. The coefficient $|c_{a,n}(t)|^2$ is the probability of being in atomic state $\left|a\right\rangle$ and photonic state $\left|n\right\rangle$ (similarly for $|c_{b,n}(t)|^2$). In the book, they solve for the case of the atom initially in the excited state $\left|a\right\rangle$ such that $c_{a,n}(0)=c_n(0)$ and $c_{b,n+1}(0)=0$.

## Homework Equations

Computing $|c_{a,n}(t)|^2$ and $|c_{b,n}(t)|^2$ using equations (6.2.16) and (6.2.17), respectively (or reading ahead to 6.2.18), I get

$|c_{a,n}(t)|^2=\rho_{nn}(0)\left[\cos^2(\frac{\Omega_n t}{2})+\frac{\Delta^2}{\Omega_n^2}\sin^2(\frac{\Omega_n t}{2})\right]$

and

$|c_{b,n}(t)|^2=\rho_{n-1,n-1}(0)(\frac{4g^2n}{\Omega_{n-1}^2})\sin^2(\frac{\Omega_{n-1}t}{2})$

## The Attempt at a Solution

Plugging these into (6.2.20), I find

\begin{align} W(t)&=\sum_{n=0}^\infty\rho_{nn}(0)\left[\cos^2(\frac{\Omega_n t}{2})+\frac{\Delta^2}{\Omega_n^2}\sin^2(\frac{\Omega_n t}{2})\right]-\sum_{n=0}^\infty\rho_{n-1,n-1}(0)(\frac{4g^2n}{\Omega_{n-1}^2})\sin^2(\frac{\Omega_{n-1}t}{2}) \\ &=\sum_{n=0}^\infty\rho_{nn}(0)\left[\cos^2(\frac{\Omega_n t}{2})+\frac{\Delta^2}{\Omega_n^2}\sin^2(\frac{ \Omega_n t}{2})\right]-\sum_{n=-1}^\infty\rho_{nn}(0)(\frac{4g^2(n+1)}{\Omega_{n}^2})\sin^2(\frac{\Omega_{n}t}{2}) \end{align}

And that's where I get stuck... So far, the only justification I've come up with is that $\rho_{nn}(0)=0$ for $n<0$ since there shouldn't be less than zero photons. I'm able to get the desired equation (6.2.21) if the second sum runs from $n=0$ to $n=\infty$ instead of starting from $n=-1$.

Thanks in advance for any help.

Note: It's showing \Omeg$a_n t$ instead of $\Omega_n t$ for me in the last equation above.

Last edited by a moderator:
It's probably because your lines of LaTeX are so long. I tried to fix it, but was unsuccessful. The align bits made it so that I couldn't break up the long TeX expression. Once the expressions get up around 70 characters, they don't render right.

## 1. What is the main focus of Section 6.2 in Scully and Zubairy's "Quantum Optics"?

The main focus of Section 6.2 is on the quantum theory of spontaneous emission and its relation to the classical theory of radiation.

## 2. What is the difference between the quantum and classical theories of spontaneous emission?

The quantum theory of spontaneous emission takes into account the discrete energy levels of an atom and the probabilistic nature of quantum mechanics. In contrast, the classical theory of radiation assumes a continuous radiation field and does not consider the internal structure of an atom.

## 3. How does the quantum theory of spontaneous emission explain the phenomenon of photon emission?

The quantum theory states that when an atom transitions from a higher energy level to a lower energy level, it emits a photon with a specific energy and frequency. This process is known as spontaneous emission.

## 4. What is the role of the density matrix in the quantum theory of spontaneous emission?

The density matrix is used to describe the state of a quantum system and its evolution over time. In the quantum theory of spontaneous emission, the density matrix is used to calculate the transition rates between energy levels and the probability of photon emission.

## 5. How does the quantum theory of spontaneous emission impact the field of quantum optics?

The quantum theory of spontaneous emission is a fundamental concept in quantum optics and is used to understand and manipulate the behavior of light and matter at the quantum level. It has also led to advances in technologies such as lasers and quantum computing.

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