Phase-Dependent Damping Rate of Josephson Plasmons in the Nonequilibrium Steady State

We give the expression of the current from which the damping rate is derived. This quantity is experimentally obtained from the impedance measurements, [7] and the expression of the impedance is given as follows. (We set $\hbar=c=1$ with $c$ the velocity of light.) By introducing the capacitance $C=\epsilon S/4\pi d$ ($d$ and $S$ are the thickness and the area of the junction, respectively), we obtain $Cd\omega^{2}A_{\omega}+J_{\omega}+J^{ext}_{\omega}=0$ from the Maxwell equation, [18] in which $J_{\omega}$ and $J^{ext}_{\omega}$ are the induced and external currents, respectively [19] ($A_{\omega}$ is the vector potential with its frequency $\omega$). With use of $J_{\omega}=-CdA_{\omega}(\omega_{\phi}^{2}-i\omega\gamma_{\omega})$, which is obtained below (Sect. 2.3), the impedance ($Z_{\omega}$) is written as $Z_{\omega}=-i\omega dA_{\omega}/J_{\omega}^{ext}=(i\omega/C)/(\omega^{2}-\omega_{\phi}^{2}+i\omega\gamma_{\omega})$. Here, $\omega_{\phi}^{2}=\omega_{J}^{2}{\rm cos}(\phi_{L}-\phi_{R})$ with $\omega_{J}=\sqrt{2eI_{0}/C}$ and $I_{0}=(\pi\Delta/2eR_{N}){\rm tanh}(\Delta/2T)$ being the plasma frequency [20] and the critical current [21] in the Josephson junction, respectively. $\Delta e^{i\phi_{L}}$ and $\Delta e^{i\phi_{R}}$ are the superconducting order parameters of two superconductors consisting the Josephson junction (the magnitude of the gap $\Delta$ is assumed to be equal for two superconductors), and $R_{N}$ is the tunnel resistance in the normal state. $\gamma_{\omega}$ is the damping rate and its expression is given in Sect. 2.3.

The expression of the current $J_{\omega}$ is obtained by using the quasiclassical Green functions. The following relation holds from the kinetic equations for quasiclassical Green functions: [22]

$$\begin{split}\int d\epsilon{\rm Tr}[\hat{\tau}_{3}\omega\hat{g}^{LK}_{\epsilon+\omega/2,\epsilon-\omega/2}]=&\iint d\epsilon d\epsilon_{1}{\rm Tr}[\hat{\tau}_{3}\hat{\Sigma}^{L+}_{\epsilon+\omega/2,\epsilon_{1}}\hat{g}^{LK}_{\epsilon_{1},\epsilon-\omega/2}+\hat{\tau}_{3}\hat{\Sigma}^{LK}_{\epsilon+\omega/2,\epsilon_{1}}\hat{g}^{L-}_{\epsilon_{1},\epsilon-\omega/2}\\ &-\hat{g}^{L+}_{\epsilon+\omega/2,\epsilon_{1}}\hat{\Sigma}^{LK}_{\epsilon_{1},\epsilon-\omega/2}\hat{\tau}_{3}-\hat{g}^{LK}_{\epsilon+\omega/2,\epsilon_{1}}\hat{\Sigma}^{L-}_{\epsilon_{1},\epsilon-\omega/2}\hat{\tau}_{3}].\end{split}$$

(1)

$\hat{g}^{L+(-)}$ and $\hat{g}^{LK}$ are the retarded (advanced) and Keldysh quasiclassical Green functions, respectively, and $\hat{\Sigma}^{L+(-)}$ and $\hat{\Sigma}^{LK}$ are the self-energies. The superscripts $L$ and $R$ are indices representing two superconductors consisting the Josephson junction. $\hat{\tau}_{3}=(\begin{smallmatrix}1&0\\ 0&-1\end{smallmatrix})$, $\hat{\cdot}$ means a matrix in the Nambu space, and ${\rm Tr}(\cdot)$ indicates taking the trace. This leads to the following relation between the current ($J_{t}$) and the temporal variation of the charge ($Q_{t}$):

$$\begin{split}&\frac{dQ_{t}^{L}}{dt}=-e\rho_{0}\int\frac{d\omega}{2\pi}\omega\int\frac{d\epsilon}{4}{\rm Tr}[\hat{\tau}_{3}\hat{g}^{LK}_{\epsilon+\omega/2,\epsilon-\omega/2}]e^{-i\omega t}\\ &=-e\rho_{0}\int\frac{d\omega}{2\pi}\iint\frac{d\epsilon d\epsilon_{1}}{4}{\rm Tr}[\hat{\tau}_{3}\hat{\Sigma}^{L+}_{\epsilon+\omega/2,\epsilon_{1}}\hat{g}^{LK}_{\epsilon_{1},\epsilon-\omega/2}+\hat{\tau}_{3}\hat{\Sigma}^{LK}_{\epsilon+\omega/2,\epsilon_{1}}\hat{g}^{L-}_{\epsilon_{1},\epsilon-\omega/2}\\ &-\hat{g}^{L+}_{\epsilon+\omega/2,\epsilon_{1}}\hat{\Sigma}^{LK}_{\epsilon_{1},\epsilon-\omega/2}\hat{\tau}_{3}-\hat{g}^{LK}_{\epsilon+\omega/2,\epsilon_{1}}\hat{\Sigma}^{L-}_{\epsilon_{1},\epsilon-\omega/2}\hat{\tau}_{3}]e^{-i\omega t}=-J_{t}.\end{split}$$

(2)

($\rho_{0}$ is the density of states which is assumed to be equal for two superconductors, and the volumes of superconductors are set to be 1 and not written explicitly below.)

We take account of the tunneling effect as the self-energy correction [23] as follows.

$$\hat{\Sigma}^{L\pm}_{\epsilon,\epsilon^{\prime}}=\pi\rho_{0}\int\frac{d\epsilon_{1}d\epsilon_{2}}{(2\pi)^{2}}\hat{t}^{LR}_{\epsilon-\epsilon_{1}}\hat{g}^{R\pm}_{\epsilon_{1},\epsilon_{2}}\hat{t}^{RL}_{\epsilon_{2}-\epsilon^{\prime}}.$$

(3)

Here,

$$\hat{t}^{LR}_{\omega}=\int dte^{i\omega t}\begin{pmatrix}t^{\prime}e^{-ied\tilde{A}_{t}}&0\\ 0&-t^{\prime}e^{ied\tilde{A}_{t}}\end{pmatrix}.$$

(4)

($\tilde{A}_{t}=\int\frac{d\omega}{2\pi}e^{-i\omega t}A_{\omega}$. The case of zero voltage is studied in this paper.) $e^{\pm ied\tilde{A}_{t}}$ is replaced by $e^{\mp ied\tilde{A}_{t}}$ in $\hat{t}^{RL}_{\omega}$. With use of the perturbative expansion by $edA_{\omega}$

$$\hat{\Sigma}^{L\pm}_{\epsilon,\epsilon^{\prime}}\simeq\pi\rho_{0}{t^{\prime}}^{2}\Bigl{[}\hat{\tau}_{3}\hat{g}^{R\pm}_{\epsilon,\epsilon^{\prime}}\hat{\tau}_{3}+\int\frac{d\epsilon_{1}}{2\pi}ied\left(\hat{\tau}_{3}\hat{g}^{R\pm}_{\epsilon,\epsilon_{1}}A_{\epsilon_{1}-\epsilon^{\prime}}-A_{\epsilon-\epsilon_{1}}\hat{g}^{R\pm}_{\epsilon_{1},\epsilon^{\prime}}\hat{\tau}_{3}\right)-\int\frac{d\epsilon_{1}d\epsilon_{1}^{\prime}}{(2\pi)^{2}}(ied)^{2}A_{\epsilon-\epsilon_{1}}\hat{g}^{R\pm}_{\epsilon_{1},\epsilon_{1}^{\prime}}A_{\epsilon_{1}^{\prime}-\epsilon^{\prime}}+\cdots\Bigr{]}.$$

(5)

We use the monochromatic field such as

$$A_{\omega}=2\pi\bar{A}[\delta(\omega-\omega_{0})+\delta(\omega+\omega_{0})],$$

(6)

and the oscillating terms ($\epsilon-\epsilon^{\prime}=n\omega_{0}$ with $n\neq 0$) will be neglected below:

$$\hat{g}^{L+}_{\epsilon,\epsilon^{\prime}}\simeq\hat{g}^{L+}_{\epsilon}2\pi\delta(\epsilon-\epsilon^{\prime}).$$

(7)

This approximation holds in the case that the strength of the external field is much smaller than $\omega_{0}$: [24] $\pi\rho_{0}(t^{\prime}ed\bar{A})^{2}\ll\omega_{0}$. From Eqs. (2) and (5) up to the linear response the current $J_{t}$ is written as follows.

$$\begin{split}&J_{t}=\frac{1}{4eR_{N}}\int\frac{d\omega}{2\pi}\int\frac{d\epsilon}{4}\bigl{\{}{\rm Tr}[\hat{g}^{R+}_{\epsilon}\hat{\tau}_{3}\hat{g}^{LK}_{\epsilon-\omega/2}+\hat{g}^{RK}_{\epsilon}\hat{\tau}_{3}\hat{g}^{L-}_{\epsilon-\omega/2}-\hat{g}^{L+}_{\epsilon+\omega/2}\hat{\tau}_{3}\hat{g}^{RK}_{\epsilon}-\hat{g}^{LK}_{\epsilon+\omega/2}\hat{\tau}_{3}\hat{g}^{R-}_{\epsilon}]2\pi\delta(\omega)\\ &-i{\rm Tr}[\hat{\tau}_{3}\hat{g}^{R+}_{\epsilon-\omega/2}\hat{\tau}_{3}\hat{g}^{LK}_{\epsilon-\omega/2}+\hat{\tau}_{3}\hat{g}^{RK}_{\epsilon-\omega/2}\hat{\tau}_{3}\hat{g}^{L-}_{\epsilon-\omega/2}+\hat{\tau}_{3}\hat{g}^{L+}_{\epsilon+\omega/2}\hat{\tau}_{3}\hat{g}^{RK}_{\epsilon+\omega/2}+\hat{\tau}_{3}\hat{g}^{LK}_{\epsilon+\omega/2}\hat{\tau}_{3}\hat{g}^{R-}_{\epsilon+\omega/2}\\ &-\hat{g}^{L+}_{\epsilon+\omega/2}\hat{g}^{RK}_{\epsilon-\omega/2}-\hat{g}^{LK}_{\epsilon+\omega/2}\hat{g}^{R-}_{\epsilon-\omega/2}-\hat{g}^{R+}_{\epsilon+\omega/2}\hat{g}^{LK}_{\epsilon-\omega/2}-\hat{g}^{RK}_{\epsilon+\omega/2}\hat{g}^{L-}_{\epsilon-\omega/2}]edA_{\omega}\bigr{\}}e^{-i\omega t}\end{split}$$

(8)

with $R_{N}=1/4\pi e^{2}\rho_{0}^{2}{t^{\prime}}^{2}$. The first line of Eq. (8) reduces to the supercurrent $I_{0}{\rm sin}(\phi_{L}-\phi_{R})$.

In this subsection we derive the one-particle Green functions which are used to calculate the current [Eq. (8)]. These functions are modified by the tunneling effect and the external field. The latter effect makes the system in the nonequilibrium state, and the kinetic equations for quasiclassical Green’s functions are written as follows. [25]

$$\hat{\tau}_{3}(\epsilon\hat{\tau}_{0}-\hat{\Delta}_{L})\hat{g}^{L\pm}_{\epsilon,\epsilon^{\prime}}-\hat{g}^{L\pm}_{\epsilon,\epsilon^{\prime}}(\epsilon^{\prime}\hat{\tau}_{0}-\hat{\Delta}_{L})\hat{\tau}_{3}-\int d\epsilon_{1}\left(\hat{\tau}_{3}\hat{\Sigma}^{L\pm}_{\epsilon,\epsilon_{1}}\hat{g}^{L\pm}_{\epsilon_{1},\epsilon^{\prime}}-\hat{g}^{L\pm}_{\epsilon,\epsilon_{1}}\hat{\Sigma}^{L\pm}_{\epsilon_{1}\epsilon^{\prime}}\hat{\tau}_{3}\right)=0$$

(9)

and

$$\begin{split}&\hat{\tau}_{3}(\epsilon\hat{\tau}_{0}-\hat{\Delta}_{L})\hat{g}^{L(a)}_{\epsilon,\epsilon^{\prime}}-\hat{g}^{L(a)}_{\epsilon,\epsilon^{\prime}}(\epsilon^{\prime}\hat{\tau}_{0}-\hat{\Delta}_{L})\hat{\tau}_{3}-\int d\epsilon_{1}\left(\hat{\tau}_{3}\hat{\Sigma}^{L+}_{\epsilon,\epsilon_{1}}\hat{g}^{L(a)}_{\epsilon_{1},\epsilon^{\prime}}+\hat{\tau}_{3}\hat{\Sigma}^{L(a)}_{\epsilon,\epsilon_{1}}\hat{g}^{L-}_{\epsilon_{1},\epsilon^{\prime}}-\hat{g}^{L+}_{\epsilon,\epsilon_{1}}\hat{\Sigma}^{L(a)}_{\epsilon_{1}\epsilon^{\prime}}\hat{\tau}_{3}-\hat{g}^{L(a)}_{\epsilon,\epsilon_{1}}\hat{\Sigma}^{L-}_{\epsilon_{1}\epsilon^{\prime}}\hat{\tau}_{3}\right)\\ &-\pi\rho_{0}\int d\epsilon_{1}d\epsilon_{2}d\epsilon_{2}^{\prime}\Bigl{\{}\hat{\tau}_{3}\hat{t}^{LR}_{\epsilon-\epsilon_{2}}\left[(T^{h}_{\epsilon_{2}^{\prime}}-T^{h}_{\epsilon_{1}})\hat{g}^{R+}_{\epsilon_{2},\epsilon_{2}^{\prime}}-(T^{h}_{\epsilon_{2}}-T^{h}_{\epsilon})\hat{g}^{R-}_{\epsilon_{2},\epsilon_{2}^{\prime}}\right]\hat{t}^{RL}_{\epsilon_{2}^{\prime}-\epsilon_{1}}\hat{g}^{L-}_{\epsilon_{1},\epsilon^{\prime}}\\ &-\hat{g}^{L+}_{\epsilon,\epsilon_{1}}\hat{t}^{LR}_{\epsilon_{1}-\epsilon_{2}}\left[(T^{h}_{\epsilon_{2}^{\prime}}-T^{h}_{\epsilon^{\prime}})\hat{g}^{R+}_{\epsilon_{2},\epsilon_{2}^{\prime}}-(T^{h}_{\epsilon_{2}}-T^{h}_{\epsilon_{1}})\hat{g}^{R-}_{\epsilon_{2},\epsilon_{2}^{\prime}}\right]\hat{t}^{RL}_{\epsilon_{2}^{\prime}-\epsilon^{\prime}}\hat{\tau}_{3}\Bigr{\}}=0\end{split}$$

(10)

with $\hat{\tau}_{0}=\left(\begin{smallmatrix}1&0\\ 0&1\end{smallmatrix}\right)$, $\hat{\Delta}_{L}=\left(\begin{smallmatrix}0&\Delta^{i\phi_{L}}\\ \Delta^{-i\phi_{L}}&0\end{smallmatrix}\right)$, and

$$T^{h}_{\epsilon}:={\rm tanh}\left(\frac{\epsilon}{2T}\right).$$

(11)

[The superscript (a) in $\hat{g}^{L(a)}$ means the anomalous term in Keldysh Green function.] By omitting the oscillating part as noted above in Eq. (7) and using the perturbative expansion Eq. (5) and

$$\hat{g}^{LK}_{\epsilon}=T^{h}_{\epsilon}(\hat{g}^{L+}_{\epsilon}-\hat{g}^{L-}_{\epsilon})+\hat{g}^{L(a)}_{\epsilon},$$

(12)

Eqs. (9) and (10) are rewritten as follows.

$$\hat{\tau}_{3}(\epsilon\hat{\tau}_{0}-\hat{\Delta}_{L}-\hat{\Sigma}^{L\pm}_{\epsilon})\hat{g}^{L\pm}_{\epsilon}-\hat{g}^{L\pm}_{\epsilon}(\epsilon\hat{\tau}_{0}-\hat{\Delta}_{L}-\hat{\Sigma}^{L\pm}_{\epsilon})\hat{\tau}_{3}=0$$

(13)

with

$$\hat{\Sigma}^{L\pm}_{\epsilon}=\hat{\Sigma^{\prime}}^{L\pm}_{\epsilon}+\pi\rho_{0}{t^{\prime}}^{2}\hat{\tau}_{3}\hat{g}^{R\pm}_{\epsilon}\hat{\tau}_{3}+\pi\rho_{0}(t^{\prime}ed\bar{A})^{2}\sum_{s=\pm 1}\hat{g}^{R\pm}_{\epsilon+s\omega_{0}},$$

(14)

and

$$\hat{\tau}_{3}(\epsilon\hat{\tau}_{0}-\hat{\Delta}_{L}-\hat{\Sigma}^{L+}_{\epsilon})\hat{g}^{L(a)}_{\epsilon}-\hat{g}^{L(a)}_{\epsilon}(\epsilon\hat{\tau}_{0}-\hat{\Delta}_{L}-\hat{\Sigma}^{L-}_{\epsilon})\hat{\tau}_{3}+\hat{\tau}_{3}\hat{z}^{L(a)}_{\epsilon}\hat{g}^{L-}_{\epsilon}-\hat{g}^{L+}_{\epsilon}\hat{z}^{L(a)}_{\epsilon}\hat{\tau}_{3}=0$$

(15)

with

$$\hat{z}^{L(a)}_{\epsilon}=-\hat{\Sigma}^{L(a)}_{\epsilon}-\pi\rho_{0}(t^{\prime}ed\bar{A})^{2}\sum_{\pm}(T^{h}_{\epsilon\pm\omega_{0}}-T^{h}_{\epsilon})(\hat{g}^{R+}_{\epsilon\pm\omega_{0}}-\hat{g}^{R-}_{\epsilon\pm\omega_{0}})$$

(16)

and

$$\hat{\Sigma}^{L(a)}_{\epsilon}=\hat{\Sigma^{\prime}}^{L(a)}_{\epsilon}+\pi\rho_{0}{t^{\prime}}^{2}\hat{\tau}_{3}\hat{g}^{R(a)}_{\epsilon}\hat{\tau}_{3}+\pi\rho_{0}(t^{\prime}ed\bar{A})^{2}\sum_{\pm}\hat{g}^{R(a)}_{\epsilon\pm\omega_{0}}.$$

(17)

Here, $\hat{\Sigma^{\prime}}_{\epsilon}^{L\pm}$ and $\hat{\Sigma^{\prime}}_{\epsilon}^{L(a)}$ in the above equations represent the inelastic scattering effect which is needed to dissipate absorbed energies and maintain the nonequilibrium steady state under the external fields, and the expression of this effect will be specified below. There also exist the equations with $L$ and $R$ interchanged in the above equations.

We introduce the following phase transformation to describe the current-carrying state due to the finite phase difference ($\phi:=\phi_{L}-\phi_{R}\neq 0$);

$$\hat{\varphi}=\begin{pmatrix}e^{i(\phi_{L}+\phi_{R})/4}&0\\ 0&e^{-i(\phi_{L}+\phi_{R})/4}\end{pmatrix}$$

(18)

and multiply the right (left) of Eq. (13) by $\hat{\varphi}$ (its complex conjugate $\hat{\varphi}^{*}$). This equation is rewritten as

$$\hat{\tau}_{3}(\tilde{\epsilon}^{\pm}\hat{\tau}_{0}-\tilde{\Delta}^{\pm}_{1,\epsilon}\hat{\tau}_{1}+i\tilde{\Delta}^{\pm}_{2,\epsilon}\hat{\tau}_{2})\hat{g}^{\pm}_{\epsilon}-\hat{g}^{\pm}_{\epsilon}(\tilde{\epsilon}^{\pm}\hat{\tau}_{0}-\tilde{\Delta}^{\pm}_{1,\epsilon}\hat{\tau}_{1}+i\tilde{\Delta}^{\pm}_{2,\epsilon}\hat{\tau}_{2})\hat{\tau}_{3}=0.$$

(19)

Here, we put

$$\hat{g}^{\pm}_{\epsilon}=\begin{pmatrix}g^{\pm}_{\epsilon}&f^{\pm}_{\epsilon}-h^{\pm}_{\epsilon}\\ f^{\pm}_{\epsilon}+h^{\pm}_{\epsilon}&g^{\pm}_{\epsilon}\end{pmatrix}:=\hat{\varphi}^{*}\hat{g}^{L\pm}_{\epsilon}\hat{\varphi},$$

(20)

$$\tilde{\epsilon}^{\pm}=\epsilon-{\Sigma^{\prime}}^{\pm}_{n,\epsilon}-\pi\rho_{0}{t^{\prime}}^{2}g^{\pm}_{\epsilon}-\pi\rho_{0}(t^{\prime}ed\bar{A})^{2}\sum_{s=\pm 1}g^{\pm}_{\epsilon+s\omega_{0}},$$

(21)

$$\tilde{\Delta}^{\pm}_{1,\epsilon}=\Delta_{1}+{\Sigma^{\prime}}^{\pm}_{a,\epsilon}-\pi\rho_{0}{t^{\prime}}^{2}f^{\pm}_{\epsilon}+\pi\rho_{0}(t^{\prime}ed\bar{A})^{2}\sum_{s=\pm 1}f^{\pm}_{\epsilon+s\omega_{0}},$$

(22)

$$\tilde{\Delta}^{\pm}_{2,\epsilon}=\Delta_{2}+{\Sigma^{\prime}}^{\pm}_{b,\epsilon}+\pi\rho_{0}{t^{\prime}}^{2}h^{\pm}_{\epsilon}-\pi\rho_{0}(t^{\prime}ed\bar{A})^{2}\sum_{s=\pm 1}h^{\pm}_{\epsilon+s\omega_{0}}$$

(23)

with $\Delta_{1}=\Delta{\rm cos}(\phi/2)$, $\Delta_{2}=-i\Delta{\rm sin}(\phi/2)$, $\hat{\tau}_{1}=\left(\begin{smallmatrix}0&1\\ 1&0\end{smallmatrix}\right)$, $\hat{\tau}_{2}=\left(\begin{smallmatrix}0&-i\\ i&0\end{smallmatrix}\right)$, and

$${\Sigma^{\prime}}^{\pm}_{n,\epsilon}\hat{\tau}_{0}+{\Sigma^{\prime}}^{\pm}_{a,\epsilon}\hat{\tau}_{1}-i{\Sigma^{\prime}}^{\pm}_{b,\epsilon}\hat{\tau}_{2}=\hat{\varphi}^{*}\hat{\Sigma^{\prime}}^{L\pm}_{\epsilon}\hat{\varphi}.$$

(24)

Then the solutions to Eq. (19) are written as

$$g^{\pm}_{\epsilon}=\frac{-\tilde{\epsilon}^{\pm}}{\tilde{\zeta}^{\pm}_{\epsilon}},\quad f^{\pm}_{\epsilon}=\frac{-\tilde{\Delta}^{\pm}_{1,\epsilon}}{\tilde{\zeta}^{\pm}_{\epsilon}},\quad\text{and}\quad h^{\pm}_{\epsilon}=\frac{-\tilde{\Delta}^{\pm}_{2,\epsilon}}{\tilde{\zeta}^{\pm}_{\epsilon}}$$

(25)

with $\tilde{\zeta}^{\pm}_{\epsilon}=\sqrt{(\tilde{\Delta}^{\pm}_{1,\epsilon})^{2}-(\tilde{\Delta}^{\pm}_{2,\epsilon})^{2}-(\tilde{\epsilon}^{\pm})^{2}}$. From $\hat{\varphi}^{*}\hat{g}^{R\pm}_{\epsilon}\hat{\varphi}$ and $\hat{\varphi}^{*}\hat{\Sigma^{\prime}}^{R\pm}_{\epsilon}\hat{\varphi}$, only the sign of $\hat{\tau}_{2}$ component is reversed, and other than that, the same results as the above equations are obtained.

Similarly Eq. (15) is rewritten as

$$\begin{split}&\hat{\tau}_{3}(\tilde{\epsilon}^{+}\hat{\tau}_{0}-\tilde{\Delta}^{+}_{1,\epsilon}\hat{\tau}_{1}+i\tilde{\Delta}^{+}_{2,\epsilon}\hat{\tau}_{2})\hat{g}^{(a)}_{\epsilon}-\hat{g}^{(a)}_{\epsilon}(\tilde{\epsilon}^{-}\hat{\tau}_{0}-\tilde{\Delta}^{-}_{1,\epsilon}\hat{\tau}_{1}+i\tilde{\Delta}^{-}_{2,\epsilon}\hat{\tau}_{1})\hat{\tau}_{3}\\ &+\hat{\tau}_{3}(z_{0,\epsilon}\hat{\tau}_{0}+z_{1,\epsilon}\hat{\tau}_{1}-iz_{2,\epsilon}\hat{\tau}_{2})\hat{g}^{-}_{\epsilon}-\hat{g}^{+}_{\epsilon}(z_{0,\epsilon}\hat{\tau}_{0}+z_{1,\epsilon}\hat{\tau}_{1}-iz_{2,\epsilon}\hat{\tau}_{2})\hat{\tau}_{3}=0.\end{split}$$

(26)

Here, $\hat{g}^{(a)}_{\epsilon}=\hat{\varphi}^{*}\hat{g}^{(a)L}_{\epsilon}\hat{\varphi}$,

$$z_{0,\epsilon}=-{\Sigma^{\prime}}^{(a)}_{n,\epsilon}-\pi\rho_{0}{t^{\prime}}^{2}g^{(a)}_{\epsilon}-\pi\rho_{0}(t^{\prime}ed\bar{A})^{2}\sum_{\pm}g^{(a)}_{\epsilon\pm\omega_{0}}-\pi\rho_{0}(t^{\prime}ed\bar{A})^{2}\sum_{\pm}(T^{h}_{\epsilon\pm\omega_{0}}-T^{h}_{\epsilon})(g^{+}_{\epsilon\pm\omega_{0}}-g^{-}_{\epsilon\pm\omega_{0}}),$$

(27)

$$z_{1,\epsilon}=-{\Sigma^{\prime}}^{(a)}_{a,\epsilon}+\pi\rho_{0}{t^{\prime}}^{2}f^{(a)}_{\epsilon}-\pi\rho_{0}(t^{\prime}ed\bar{A})^{2}\sum_{\pm}f^{(a)}_{\epsilon\pm\omega_{0}}-\pi\rho_{0}(t^{\prime}ed\bar{A})^{2}\sum_{\pm}(T^{h}_{\epsilon\pm\omega_{0}}-T^{h}_{\epsilon})(f^{+}_{\epsilon\pm\omega_{0}}-f^{-}_{\epsilon\pm\omega_{0}}),$$

(28)

and

$$z_{2,\epsilon}=-{\Sigma^{\prime}}^{(a)}_{b,\epsilon}-\pi\rho_{0}{t^{\prime}}^{2}h^{(a)}_{\epsilon}+\pi\rho_{0}(t^{\prime}ed\bar{A})^{2}\sum_{\pm}h^{(a)}_{\epsilon\pm\omega_{0}}+\pi\rho_{0}(t^{\prime}ed\bar{A})^{2}\sum_{\pm}(T^{h}_{\epsilon\pm\omega_{0}}-T^{h}_{\epsilon})(h^{+}_{\epsilon\pm\omega_{0}}-h^{-}_{\epsilon\pm\omega_{0}}).$$

(29)

[${\Sigma^{\prime}}^{(a)}_{n,\epsilon}$, ${\Sigma^{\prime}}^{(a)}_{a,\epsilon}$, and ${\Sigma^{\prime}}^{(a)}_{b,\epsilon}$ are the components of $\hat{\varphi}^{*}\hat{\Sigma^{\prime}}^{L(a)}_{\epsilon}\hat{\varphi}$ as in Eq. (24). The $\hat{\tau}_{3}$-component is omitted because it is confirmed to be small by numerical calculations and its contribution to the damping rate is negligible.] By introducing the quantity

$$x^{(a)}_{\epsilon}:=\frac{g^{(a)}_{\epsilon}}{g^{+}_{\epsilon}-g^{-}_{\epsilon}}=\frac{f^{(a)}_{\epsilon}}{f^{+}_{\epsilon}-f^{-}_{\epsilon}}=\frac{h^{(a)}_{\epsilon}}{h^{+}_{\epsilon}-h^{-}_{\epsilon}},$$

(30)

in which the equality of these three quantities follows from the normalization conditions for the quasiclassical Green functions, Eq. (26) is reduced to

$$\begin{split}&x^{(a)}_{\epsilon}[(-{\Sigma^{\prime}}^{+}_{n,\epsilon}+{\Sigma^{\prime}}^{-}_{n,\epsilon})(g^{+}_{\epsilon}-g^{-}_{\epsilon})-({\Sigma^{\prime}}^{+}_{a,\epsilon}-{\Sigma^{\prime}}^{-}_{a,\epsilon})(f^{+}_{\epsilon}-f^{-}_{\epsilon})+({\Sigma^{\prime}}^{+}_{b,\epsilon}-{\Sigma^{\prime}}^{-}_{b,\epsilon})(h^{+}_{\epsilon}-h^{-}_{\epsilon})]\\ &+{\Sigma^{\prime}}^{(a)}_{n,\epsilon}(g^{+}_{\epsilon}-g^{-}_{\epsilon})+{\Sigma^{\prime}}^{(a)}_{a,\epsilon}(f^{+}_{\epsilon}-f^{-}_{\epsilon})-{\Sigma^{\prime}}^{(a)}_{b,\epsilon}(h^{+}_{\epsilon}-h^{-}_{\epsilon})+\pi\rho_{0}(t^{\prime}ed\bar{A})^{2}\sum_{\pm}(\tilde{T}^{h}_{\epsilon\pm\omega_{0}}-\tilde{T}^{h}_{\epsilon})\\ &\times[(g^{+}_{\epsilon}-g^{-}_{\epsilon})(g^{+}_{\epsilon\pm\omega_{0}}-g^{-}_{\epsilon\pm\omega_{0}})+(f^{+}_{\epsilon}-f^{-}_{\epsilon})(f^{+}_{\epsilon\pm\omega_{0}}-f^{-}_{\epsilon\pm\omega_{0}})-(h^{+}_{\epsilon}-h^{-}_{\epsilon})(h^{+}_{\epsilon\pm\omega_{0}}-h^{-}_{\epsilon\pm\omega_{0}})]=0,\end{split}$$

(31)

here,

$$\tilde{T}^{h}_{\epsilon}:=T^{h}_{\epsilon}+x^{(a)}_{\epsilon}$$

(32)

and $x^{(a)}_{\epsilon}$ has a meaning of the nonequilibrium correction to the distribution function. $x^{(a)}_{\epsilon}$ is obtained by solving Eq. (31).

From Eq. (8), the rf current term is written as

$$\begin{split}&J_{\omega}=\frac{-i}{4eR_{N}}\int\frac{d\epsilon}{4}{\rm Tr}[\hat{\tau}_{3}\hat{g}^{R+}_{\epsilon-\omega/2}\hat{\tau}_{3}\hat{g}^{LK}_{\epsilon-\omega/2}+\hat{\tau}_{3}\hat{g}^{RK}_{\epsilon-\omega/2}\hat{\tau}_{3}\hat{g}^{L-}_{\epsilon-\omega/2}+\hat{\tau}_{3}\hat{g}^{L+}_{\epsilon+\omega/2}\hat{\tau}_{3}\hat{g}^{RK}_{\epsilon+\omega/2}+\hat{\tau}_{3}\hat{g}^{LK}_{\epsilon+\omega/2}\hat{\tau}_{3}\hat{g}^{R-}_{\epsilon+\omega/2}\\ &-\hat{g}^{L+}_{\epsilon+\omega/2}\hat{g}^{RK}_{\epsilon-\omega/2}-\hat{g}^{LK}_{\epsilon+\omega/2}\hat{g}^{R-}_{\epsilon-\omega/2}-\hat{g}^{R+}_{\epsilon+\omega/2}\hat{g}^{LK}_{\epsilon-\omega/2}-\hat{g}^{RK}_{\epsilon+\omega/2}\hat{g}^{L-}_{\epsilon-\omega/2}]edA_{\omega}.\end{split}$$

(33)

Using the quasiclassical Green functions in the previous subsection, this current is rewritten as $J_{\omega}=-CdA_{\omega}(\tilde{\omega}_{\phi}^{2}-i\omega\gamma_{\omega})$ with $\tilde{\omega}_{\phi}^{2}$ and $\gamma_{\omega}$ being given as follows.

$$\tilde{\omega}_{\phi}^{2}=\frac{1}{R_{N}C}\int d\epsilon\tilde{T}^{h}_{\epsilon}\left[-{\rm Re}(g^{+}_{\epsilon}-g^{+}_{\epsilon-\omega}){\rm Im}g^{+}_{\epsilon}+{\rm Re}(f^{+}_{\epsilon}-f^{+}_{\epsilon-\omega}){\rm Im}f^{+}_{\epsilon}+{\rm Im}(h^{+}_{\epsilon}-h^{+}_{\epsilon-\omega}){\rm Re}h^{+}_{\epsilon}\right]$$

(34)

and

$$\gamma_{\omega}=\frac{1}{R_{N}C}\int d\epsilon\frac{(\tilde{T}^{h}_{\epsilon+\omega/2}-\tilde{T}^{h}_{\epsilon-\omega/2})}{2\omega}({\rm Im}g^{+}_{\epsilon+\omega/2}{\rm Im}g^{+}_{\epsilon-\omega/2}+{\rm Im}f^{+}_{\epsilon+\omega/2}{\rm Im}f^{+}_{\epsilon-\omega/2}-{\rm Re}h^{+}_{\epsilon+\omega/2}{\rm Re}h^{+}_{\epsilon-\omega/2}).$$

(35)

We introduce the dimensionless damping rate $\Gamma_{\omega}$ such as

$$\gamma_{\omega}=\frac{1}{R_{N}C}\Gamma_{\omega}.$$

(36)

$\Gamma_{\omega}=\Gamma^{g}_{\omega}+\Gamma^{f}_{\omega}$ with

$$\Gamma^{g}_{\omega}:=\int d\epsilon\frac{(\tilde{T}^{h}_{\epsilon+\omega/2}-\tilde{T}^{h}_{\epsilon-\omega/2})}{2\omega}{\rm Im}g^{+}_{\epsilon+\omega/2}{\rm Im}g^{+}_{\epsilon-\omega/2}$$

(37)

and

$$\Gamma^{f}_{\omega}:=\int d\epsilon\frac{(\tilde{T}^{h}_{\epsilon+\omega/2}-\tilde{T}^{h}_{\epsilon-\omega/2})}{2\omega}({\rm Im}f^{+}_{\epsilon+\omega/2}{\rm Im}f^{+}_{\epsilon-\omega/2}-{\rm Re}h^{+}_{\epsilon+\omega/2}{\rm Re}h^{+}_{\epsilon-\omega/2}).$$

(38)

We call $\Gamma^{g}_{\omega}$ and $\Gamma^{f}_{\omega}$ the quasiparticle term and the interference term, [3] respectively. The finite values of these two terms are brought about by the thermal excitation of quasiparticles.

In this subsection we show results of the damping rate in which the corrections by the tunneling effect and the nonequilibrium distribution are not included. In this case quasiclassical Green functions, Eq. (25), are approximated as follows.

$$g^{\pm}_{\epsilon}=\frac{-\epsilon}{\zeta^{\pm}_{\epsilon}},\quad f^{\pm}_{\epsilon}=\frac{-\Delta{\rm cos}(\phi/2)}{\zeta^{\pm}_{\epsilon}},\quad\text{and}\quad h^{\pm}_{\epsilon}=\frac{i\Delta{\rm sin}(\phi/2)}{\zeta^{\pm}_{\epsilon}},$$

(39)

here $\zeta^{\pm}_{\epsilon}=\sqrt{\Delta^{2}-(\epsilon\pm i0)^{2}}$. When these quantities and $x^{(a)}_{\epsilon}=0$ are used, Eq. (35) results in

$$\gamma_{\omega}^{0}=\frac{1}{R_{N}C}(\Gamma^{g0}_{\omega}+\Gamma^{f0}_{\omega})$$

(40)

with

$$\Gamma^{g0}_{\omega}:=\int d\epsilon\frac{({T}^{h}_{\epsilon+\omega/2}-{T}^{h}_{\epsilon-\omega/2})|\epsilon+\omega/2||\epsilon-\omega/2|\theta(|\epsilon+\omega/2|-\Delta)\theta(|\epsilon-\omega/2|-\Delta)}{2\omega\sqrt{(\epsilon+\omega/2)^{2}-\Delta^{2}}\sqrt{(\epsilon-\omega/2)^{2}-\Delta^{2}}}$$

(41)

and

$$\Gamma^{f0}_{\omega}:={\rm cos}\phi\int d\epsilon\frac{({T}^{h}_{\epsilon+\omega/2}-{T}^{h}_{\epsilon-\omega/2})\Delta^{2}\theta(|\epsilon+\omega/2|-\Delta)\theta(|\epsilon-\omega/2|-\Delta)}{{\rm sgn}(\epsilon+\omega/2){\rm sgn}(\epsilon-\omega/2)2\omega\sqrt{(\epsilon+\omega/2)^{2}-\Delta^{2}}\sqrt{(\epsilon-\omega/2)^{2}-\Delta^{2}}},$$

(42)

here, $\theta(\cdot)$ means a step function. This result is similar to that obtained previously in Refs. 5 and 26.