On the Transition Probability for the Near or Exact Resonance with the RWA

Abstract

Rotating wave approximation (RWA) has been used to evaluate the transition probability and solve the Schrödinger equation approximately in quantum optics. Examples include the invalidity of the traditional adiabatic condition for the adiabaticity invoking a two-level coupled system near resonance. Here, using a two-state system driven by an oscillatory force, we derive the exact transition probability by solving the Schrödinger equation analytically for a general wave function. Our results reveal that the application of the RWA may lead to false conclusions on the transition probability for the near resonance with weak coupling, especially when the coupling strength is about a half of the transition frequency. We also investigate conditions for which RWA may work or fail.

Appendices

Appendix A: the Wave Function of the System Obtained with the RWA

Fujii considered a two-state system in [20],

$$\begin{array}{@{}rcl@{}} H(t) &=&-\frac{\Delta }{2}\sigma_{z}+ 2g\cos (\omega t+\phi )\sigma_{x} \end{array} $$

(A1)

$$\begin{array}{@{}rcl@{}} &=&\left( \begin{array}{cc} -\frac{\Delta }{2} & 2g\cos (\omega t+\phi ) \\ 2g\cos (\omega t+\phi ) & \frac{\Delta }{2} \end{array} \right). \end{array} $$

(A2)

With the RWA, H in (A2) is reduced to (see Eq. (9) in [20])

$$ H^{\prime }(t)=\left( \begin{array}{cc} -\frac{\Delta }{2} & ge^{i(\omega t+\phi )} \\ ge^{-i(\omega t+\phi )} & \frac{\Delta }{2} \end{array} \right) . $$

(A3)

We consider the Schrödinger equation

$$ i\frac{\partial \psi (t)}{\partial t}=H^{\prime }(t)\psi (t). $$

(A4)

Let

$$ \psi (t)=\left( \begin{array}{cc} e^{i(\omega t+\phi )/2} & 0 \\ 0 & e^{-i(\omega t+\phi )/2} \end{array} \right) {\Phi} (t). $$

(A5)

Substituting (A5) into (A4), we obtain the following Schrödinger equation (see Eq. (12) in [20])

$$ i\frac{\partial {\Phi} (t)}{\partial t}=\left( \begin{array}{cc} -\frac{\Delta -\omega }{2} & g \\ g & \frac{\Delta -\omega }{2} \end{array} \right) {\Phi} (t). $$

(A6)

Let $\displaystyle r=\frac {\Delta -\omega }{2}$. The exact solution of the Schrödinger equation in (A6) for r = 0 (i.e. Δ = ω) was discussed in [20]. We next solve the Schrö dinger equation in (A6) for r≠ 0 (i.e. Δ≠ω).

Let $\displaystyle \mathcal {A}=g\sigma _{x}-r\sigma _{z}$. We rewrite (A6) as follows

$$ \frac{\partial {\Phi} (t)}{\partial t}=-i\mathcal{A}{\Phi} (t). $$

(A7)

From (A7) we obtain

$$ {\Phi} (t)=e^{-i\mathcal{A}t}{\Phi} (0). $$

(A8)

A calculation yields

$$\begin{array}{@{}rcl@{}} &&e^{-i\mathcal{A}t} \\ &=&\left( \begin{array}{cc} \cos ({\Pi} t)+\frac{ir}{\Pi }\sin ({\Pi} t) & -\frac{ig}{\Pi }\sin ({\Pi} t) \\ -\frac{ig}{\Pi }\sin ({\Pi} t) & \cos ({\Pi} t)-\frac{ir}{\Pi }\sin ({\Pi} t) \end{array} \right) , \\ && \end{array} $$

(A9)

where $\displaystyle {\Pi } =\sqrt {g^{2}+r^{2}}$. When r = 0, then π = g. The detail for the derivation of $e^{-i\mathcal {A}t}$ is given in B. Note that Φ(0) = ψ(0) = |0〉. From (A8, A9), we obtain

$$ {\Phi} (t)=\left( \begin{array}{c} \cos ({\Pi} t)+\frac{ir}{\Pi }\sin ({\Pi} t) \\ -\frac{ig}{\Pi }\sin ({\Pi} t) \end{array} \right) . $$

(A10)

From (A10, A5), after omitting the phase factor e^{i(ωt + ϕ)/2}, we obtain

$$ \psi (t)=\left( \begin{array}{c} \cos ({\Pi} t)+\frac{ir}{\Pi }\sin ({\Pi} t) \\ -\frac{ig}{\Pi }\sin ({\Pi} t)e^{-i(\omega t+\phi )} \end{array} \right) . $$

(A11)

Appendix B: Calculation of $e^{\protect \lambda \mathcal {A}}$

Note that $\displaystyle e^{\lambda \mathcal {A}}={\sum }_{k = 0}^{\infty }\frac { \lambda ^{k}}{k!}\mathcal {A}^{k}$, where $\displaystyle \mathcal {A}^{0}=I_{2}$ .

Taking $\displaystyle {\Pi } =\sqrt {g^{2}+r^{2}}$ leads to $\displaystyle \mathcal {A}^{2}={\Pi }^{2}I_{2}$. A calculation yields

$$ \mathcal{A}^{2k}=(g^{2}+r^{2})^{k}I_{2}={\Pi}^{2k}I_{2}, $$

(B1)

and

$$ \mathcal{A}^{2k + 1}=(g^{2}+r^{2})^{k}\mathcal{A}={\Pi}^{2k}\mathcal{A}. $$

(B2)

Then,

$$\begin{array}{@{}rcl@{}} &&e^{-it\mathcal{A}} \\ &=&{\sum}_{m = 0}^{\infty }\frac{(-it)^{2m}}{(2m)!}\mathcal{A} ^{2m}+{\sum}_{m = 0}^{\infty }\frac{(-it)^{2m + 1}}{(2m + 1)!}\mathcal{A}^{2m + 1} \\ &=&\cos ({\Pi} t)I_{2}-\frac{i}{\Pi }\sin ({\Pi} t)\mathcal{A}. \end{array} $$

(B3)

Appendix C: the Numerical Solution

Let a(t) = u(t) + iv(t) and b(t) = x(t) + iy(t), where u(0) = 1 and v(0) = x(0) = y(0) = 0. From (15) we obtain the first-order real linear system

$$ \left( \begin{array}{c} \dot{x} \\ \dot{v} \end{array} \right) =A\left( \begin{array}{c} u \\ y \end{array} \right) ,\left( \begin{array}{c} \dot{u} \\ \dot{y} \end{array} \right) =-A\left( \begin{array}{c} x \\ v \end{array} \right) , $$

(C1)

where $\dot {x}$ denotes $\frac {dx}{dt}$ and $\displaystyle A=\left (\begin {array}{cc} -\beta (t) & \frac {\Omega (t)}{2} \\ \frac {\Omega (t)}{2} & \beta (t) \end {array} \right )$. Let $\displaystyle X=\left (\begin {array}{c} x \\ v \end {array} \right )$ and $\displaystyle Y=\left (\begin {array}{c} u \\ y \end {array} \right )$, then the first-order real linear system becomes a simpler form $\dot {X}=AY,\dot {Y}=-AX$. We then apply a forth-order Runge-Kutta scheme to the linear system

$$\begin{array}{@{}rcl@{}} X_{1} &=&A_{n}Y_{n},\text{ } \\ Y_{1} &=&-A_{n}X_{n}, \\ X_{2} &=&A_{n + 1/2}(Y_{n}+\frac{\Delta t}{2}Y_{1}),\text{ } \\ Y_{2} &=&-A_{n + 1/2}(X_{n}+\frac{\Delta t}{2}X_{1}), \\ X_{3} &=&A_{n + 1/2}(Y_{n}+\frac{\Delta t}{2}Y_{2}),\text{ } \\ Y_{3} &=&-A_{n + 1/2}(X_{n}+\frac{\Delta t}{2}X_{2}), \\ X_{4} &=&A_{n + 1}(Y_{n}+{\Delta} tY_{3}),\text{ } \\ Y_{4} &=&-A_{n + 1}(X_{n}+{\Delta} tX_{3}), \\ X_{n + 1} &=&X_{n}+\frac{\Delta t}{6}(X_{1}+ 2X_{2}+ 2X_{3}+X_{4}), \\ Y_{n + 1} &=&Y_{n}+\frac{\Delta t}{6}(Y_{1}+ 2Y_{2}+ 2Y_{3}+Y_{4}), \end{array} $$

where A _n is the matrix A evaluated at t _n = nΔt, A_n+ 1/2 is the matrix A evaluated at $\displaystyle t_{n+\frac {1}{2} }=(n + 1/2){\Delta } t$, and A_n+ 1 is the matrix A evaluated at t_n+ 1 = (n + 1)Δt.