Josephson effect of superconductors with $J=3/2$ electrons

Let us consider a superconductor in which two $J=3/2$ electrons form a Cooper pair belonging to even-parity symmetry. Its Hamiltonian of the normal state is given by

	$\displaystyle\mathcal{H}_{N}$	$\displaystyle=\sum_{\bm{k}}\Psi_{\bm{k}}^{\dagger}H_{N}(\bm{k})\Psi_{\bm{k}},$		(2)
	$\displaystyle\Psi_{\bm{k}}$	$\displaystyle=[c_{\bm{k},3/2},c_{\bm{k},1/2},c_{\bm{k},-1/2},c_{\bm{k},-3/2}]^{T},$		(3)

where $c_{\bm{k},j_{z}}$ is the annihilation operator of an electron with momentum $\bm{k}$ and the $z$-component of angular momentum being $j_{z}$. The Hamiltonian with spin-orbit coupling is described as

$\displaystyle H_{N}(\bm{k})$

$\displaystyle=a_{1}\bm{k}^{2}+2a_{2}\sum_{i}k_{i}^{2}J_{i}^{2}+a_{3}\sum_{i\neq j}k_{i}k_{j}J_{i}J_{j}-\mu,$

(4)

with $i=1-3$.Luttinger (1956) The three matrices describing angular momentum $J=3/2$ are given by

	$\displaystyle J_{1}$	$\displaystyle=\frac{1}{2}\begin{pmatrix}0&\sqrt{3}&0&0\\ \sqrt{3}&0&2&0\\ 0&2&0&\sqrt{3}\\ 0&0&\sqrt{3}&0\end{pmatrix},$
	$\displaystyle J_{2}$	$\displaystyle=\frac{i}{2}\begin{pmatrix}0&-\sqrt{3}&0&0\\ \sqrt{3}&0&-2&0\\ 0&2&0&-\sqrt{3}\\ 0&0&\sqrt{3}&0\end{pmatrix},$
	$\displaystyle J_{3}$	$\displaystyle=\frac{1}{2}\begin{pmatrix}3&0&0&0\\ 0&1&0&0\\ 0&0&-1&0\\ 0&0&0&-3\end{pmatrix}.$		(5)

The pseudospin part of a pairing state is described by $\gamma$ matrices,Brydon et al. (2018)

	$\displaystyle\gamma^{0}$	$\displaystyle=1_{4\times 4},\quad\gamma^{1}=\left(J_{1}\,J_{2}+J_{2}\,J_{1}\right)/\sqrt{3},$
	$\displaystyle\gamma^{2}$	$\displaystyle=\left(J_{2}\,J_{3}+J_{3}\,J_{2}\right)/\sqrt{3},\quad\gamma^{3}=\left(J_{1}\,J_{3}+J_{3}\,J_{1}\right)/\sqrt{3},$
	$\displaystyle\gamma^{4}$	$\displaystyle=\left(J_{1}^{2}-J_{2}^{2}\right)/\sqrt{3},\quad\gamma^{5}=\left(2J_{3}^{2}-J_{1}^{2}-J_{2}^{2}\right)/3.$		(6)

The five matrices $\gamma^{\nu}$ for $\nu=1-5$ anticommute to one another as $\gamma^{\mu}\gamma^{\nu}+\gamma^{\nu}\gamma^{\mu}=2\gamma^{0}\delta_{\mu\nu}$. The pseudospin matrices can be expressed in terms of $\gamma^{j}$,

	$\displaystyle J_{1}=$	$\displaystyle\frac{-i}{2}\left(\sqrt{3}\gamma^{2}\,\gamma^{5}+\gamma^{1}\,\gamma^{3}+\gamma^{2}\,\gamma^{4}\right),$		(7)
	$\displaystyle J_{2}=$	$\displaystyle\frac{i}{2}\left(\sqrt{3}\gamma^{3}\,\gamma^{5}+\gamma^{1}\,\gamma^{2}-\gamma^{3}\,\gamma^{4}\right),$		(8)
	$\displaystyle J_{3}=$	$\displaystyle\frac{i}{2}\left(\gamma^{2}\,\gamma^{3}+2\,\gamma^{1}\,\gamma^{4}\right).$		(9)

The Hamiltonian in Eq. (4) can be transformed into

$\displaystyle H_{N}(\bm{k})$

$\displaystyle=\varepsilon_{0}(\bm{k})+\sum_{\nu=1}^{5}\varepsilon_{\nu}(\bm{k})\gamma^{\nu}-\mu=\mathbb{E}_{\bm{k}},$

(10)

with coefficients,

	$\displaystyle\varepsilon_{0}(\bm{k})$	$\displaystyle=\frac{|\bm{k}|^{2}}{2m_{0}},$
	$\displaystyle\varepsilon_{j=1-3}(\bm{k})$	$\displaystyle=\frac{|\bm{k}|^{2}}{2m_{1}}e_{j=1-3}(\hat{\bm{k}}),$
	$\displaystyle\varepsilon_{j=4,5}(\bm{k})$	$\displaystyle=\frac{|\bm{k}|^{2}}{2m_{2}}e_{j=4,5}(\hat{\bm{k}}),$
	$\displaystyle m_{0}$	$\displaystyle=\frac{1}{2}\left(a_{1}+\frac{5}{4}a_{2}\right)^{-1},$
	$\displaystyle m_{1}$	$\displaystyle=\frac{1}{a_{3}},\quad m_{2}=\frac{1}{2a_{2}}.$		(11)

The functions are defined as

	$\displaystyle e_{1}(\hat{\bm{k}})$	$\displaystyle=\sqrt{3}\hat{k}_{x}\hat{k}_{y},\quad e_{2}(\hat{\bm{k}})=\sqrt{3}\hat{k}_{y}\hat{k}_{z},$
	$\displaystyle e_{3}(\hat{\bm{k}})$	$\displaystyle=\sqrt{3}\hat{k}_{z}\hat{k}_{x},\quad e_{4}(\hat{\bm{k}})=\frac{\sqrt{3}}{2}(\hat{k}_{x}^{2}-\hat{k}_{y}^{2}),$
	$\displaystyle e_{5}(\hat{\bm{k}})$	$\displaystyle=\frac{1}{2}(2\hat{k}_{z}^{2}-\hat{k}_{x}^{2}-\hat{k}_{y}^{2}),$		(12)

where $\hat{k}_{i}=k_{i}/|\bm{k}|$ and $e_{\nu}(\hat{\bm{k}})$ is normalized, i.e. $\sum_{\nu=1}^{5}e_{\nu}^{2}(\hat{\bm{k}})=1$. Time-reversal symmetry (TRS) of $H_{N}$ is represented by

	$\displaystyle\mathcal{T}H_{N}(-\bm{k})\mathcal{T}^{-1}$	$\displaystyle=H_{N}(\bm{k}),$		(13)
	$\displaystyle\mathcal{T}=U_{T}\mathcal{K},\quad U_{T}$	$\displaystyle=\gamma^{1}\gamma^{2},$		(14)

where $\mathcal{K}$ means the complex conjugation. Under the time-reversal operation, $\gamma^{\nu}$ and $J_{i}$ show the relation,

	$\displaystyle\mathcal{T}\,\gamma^{\nu}\,\mathcal{T}^{-1}=-U_{T}\,(\gamma^{\nu})^{\ast}\,U_{T}=\gamma^{\nu},\quad\nu=0-5,$		(15)
	$\displaystyle\mathcal{T}\,J_{i}\,\mathcal{T}^{-1}=-U_{T}\,(J_{i})^{\ast}\,U_{T}=-J_{i},\quad i=1-3,$		(16)

therefore $\gamma^{\nu}$ remains unchanged, whereas $J_{i}$ changes its sign.

A pair potential in even-parity symmetry is classified into two symmetry class in terms of its pseudospin configuration: pseudospin-singlet and pseudospin-quintet. Also, the symmetry class is divided to corresponding representation of crystal symmetry of the superconductor. We list the pair potentials in a cubic symmetric superconductor with on-site (s-wave) symmetry in Table 1.

In this paper, we focus on a superconductor preserving TRS and inversion symmetry. The pair potential of such a superconductor has the formBrydon et al. (2018)

$\displaystyle\Delta(\bm{k})=\eta_{\bm{k},0}U_{T}+\bm{\eta}_{\bm{k}}\cdot\bm{\gamma}U_{T}.$

(17)

The first term refers to the pair potential of pseudospin-singlet state and the second term refers to that of pseudospin-quintet states. The amplitudes $\eta_{\bm{k},0}$ and $\bm{\eta}_{\bm{k}}$ are even in momentum and real in the presence of TRS. It is easy to confirm that Eq. (17) preserves time-reversal symmetry,

$\displaystyle\mathcal{T}\,\Delta(\bm{k})\,\mathcal{T}^{-1}=\Delta(\bm{k}),$

(18)

due to Eq. (15). Because electrons obey the Fermi-Dirac statistics, the pair potential must be antisymmetric under the permutation of two electrons forming a Cooper pair,

$\displaystyle\Delta^{\mathrm{T}}(-\bm{k})=-\Delta(\bm{k}),$

(19)

where the permutation of pseudospin is carried out by taking the transpose of the matrix ($\mathrm{T}$). The pair potentials under the consideration are even-parity, [i.e., $\Delta(\bm{k})=\Delta(-\bm{k})]$, and odd-pseudospin symmetry. Therefore, the relation

$\displaystyle\Delta^{\mathrm{T}}(\bm{k})=-\Delta(\bm{k}),$

(20)

is satisfied because of $(\gamma^{\nu})^{\mathrm{T}}=(\gamma^{\nu})^{\ast}$ and Eq. (15). We will use an expression $\mathbb{D}$ defined by $\Delta(\bm{k})=\mathbb{D}_{\bm{k}}U_{T}$, for simplicity.

The BdG Hamiltonian is expressed in terms of the normal Hamiltonian $H_{N}$ and the pair potential $\Delta(\bm{k})$ as

$\displaystyle\check{H}_{\mathrm{BdG}}(\bm{k})=\begin{bmatrix}H_{N}(\bm{k})&\Delta(\bm{k})\\ -\Delta^{*}(-\bm{k})&-H_{N}^{*}(-\bm{k})\end{bmatrix},$

(21)

which preserves particle-hole symmetry,

$\displaystyle\mathcal{C}\check{H}_{\mathrm{BdG}}(-\bm{k})\mathcal{C}^{-1}=-\check{H}_{\mathrm{BdG}}(\bm{k}),\quad\mathcal{C}=\tau_{1}\mathcal{K},$

(22)

where $\tau_{i}$ for $i=1-3$ is the Pauli matrix in the particle-hole space. The Gor’kov equation becomes

$\displaystyle\begin{bmatrix}i{\omega_{n}}-\check{H}_{\mathrm{BdG}}(\bm{k})\end{bmatrix}\begin{bmatrix}{G}(\bm{k},i{\omega_{n}})&{F}(\bm{k},i{\omega_{n}})\\ \underline{{F}}(\bm{k},i{\omega_{n}})&\underline{{G}}(\bm{k},i{\omega_{n}})\end{bmatrix}=\check{1}_{8\times 8},$

(23)

where the Matsubara Green’s functions are defined by

	$\displaystyle G(\bm{k},i\omega_{n})=$	$\displaystyle-\int_{0}^{1/T}d\tau\,\left\langle T_{\tau}c_{\bm{k},s}(\tau)c^{\dagger}_{\bm{k},s^{\prime}}\right\rangle\,e^{i\omega_{n}\tau},$		(24)
	$\displaystyle F(\bm{k},i\omega_{n})=$	$\displaystyle-\int_{0}^{1/T}d\tau\,\left\langle T_{\tau}c_{\bm{k},s}(\tau)c_{\bm{k},s^{\prime}}\right\rangle\,e^{i\omega_{n}\tau}.$		(25)

Each Green’s function is a $4\times 4$ matrix. The relations

	$\displaystyle\underline{G}(\bm{k},i\omega_{n})=$	$\displaystyle-G^{\ast}(-\bm{k},i\omega_{n}),$		(26)
	$\displaystyle\underline{F}(\bm{k},i\omega_{n})=$	$\displaystyle-F^{\ast}(-\bm{k},i\omega_{n}),$		(27)

can be derived due to the particle-hole symmetry of the BdG Hamiltonian. The solution of the BdG equation is given by

	$\displaystyle\left[\begin{array}[]{cc}G_{(}\bm{k},i\omega_{n})&F(\bm{k},i\omega_{n})\\ \underline{F}(\bm{k},i\omega_{n})&\underline{G}(\bm{k},i\omega_{n})\end{array}\right]=\left[\begin{array}[]{cc}1&0\\ 0&-U_{T}\end{array}\right]$		(32)
	$\displaystyle\times\left[\begin{array}[]{cc}i\omega_{n}-\mathbb{E}_{\bm{k}}&-\mathbb{D}_{\bm{k}}\\ -\mathbb{D}_{\bm{k}}&i\omega_{n}+\mathbb{E}_{\bm{k}}\end{array}\right]^{-1}\left[\begin{array}[]{cc}1&0\\ 0&U_{T}\end{array}\right].$		(37)

The anomalous Green’s functions are calculated as

	$\displaystyle F(\bm{k},i\omega_{n})=$	$\displaystyle\mathbb{F}(\bm{k},i\omega_{n})\,U_{T},$		(38)
	$\displaystyle\underline{F}(\bm{k},i\omega_{n})=$	$\displaystyle-U_{T}\mathbb{F}(\bm{k},-i\omega_{n}),$		(39)
	$\displaystyle\mathbb{F}(\bm{k},i\omega_{n})=$	$\displaystyle\left\{-\mathbb{D}_{\bm{k}}+(i\omega_{n}+\mathbb{E}_{\bm{k}})\mathbb{D}_{\bm{k}}^{-1}(i\omega_{n}-\mathbb{E}_{\bm{k}})\right\}^{-1}.$		(40)

Especially, when the dispersion relation is independent of the pseudospin, i.e. $\mathbb{E}_{\bm{k}}=\varepsilon_{0}(\bm{k})\gamma^{0}$, Eq. (40) is simplified to

	$\displaystyle\mathbb{F}(\bm{k},i\omega_{n})$	$\displaystyle=-\frac{\mathbb{D}_{\bm{k}}}{\sum_{\nu=0-5}|\eta_{\nu}|^{2}+\omega_{n}^{2}+\xi_{\bm{k}}^{2}},$		(41)
	$\displaystyle\xi_{\bm{k}}$	$\displaystyle=\varepsilon_{0}(\bm{k})-\mu,$		(42)

for both pseudospin-singlet or pseudospin-quintet states.

We discuss the Josephson effect in a superconductor/insulator/superconductor(SIS) junction consisting of two $J=3/2$ superconductors. The Josephson current flows in the $x$ direction and the interface is located at $x=0$ as shown in Fig. 1.

The Hamiltonian of the SIS junction consists of three components,

	$\displaystyle H$	$\displaystyle=H_{L}+H_{R}+H_{T},$		(43)
	$\displaystyle H_{T}$	$\displaystyle=\sum_{\beta,\gamma}(t_{\gamma\beta}c_{\gamma}^{\dagger}c_{\beta}+t_{\beta\gamma}c_{\beta}^{\dagger}c_{\gamma}),$		(44)

where $H_{L}(H_{R})$ is the BdG Hamiltonian in Eq. (21) of the left (right) superconductor. $H_{T}$ is a tunneling Hamiltonian which refers to the interaction between two superconductors, where $\beta=\{\bm{p},s\}(\gamma=\{\bm{k},s^{\prime}\})$ is a combined index for an electron in the left(right) superconductor. To satisfy the Hermicity of the Hamiltonian, $t_{\beta\gamma}^{*}=t_{\gamma\beta}$ is required.

The electric Josephson current through the junction is derived within the linear response theory with respect to $H_{T}$.Mahan (1990) The result is expressed as

	$\displaystyle I$	$\displaystyle=2e\,{\mathrm{Im}}\,T\sum_{\omega_{n}}\sum_{\bm{k},\bm{p}}{}^{{}^{\prime}}$
		$\displaystyle\times{\text{Tr}}\left[t_{-\bm{p},-\bm{k}}^{\ast}\,\underline{F}^{R}(\bm{k},i\omega_{n})\,t_{\bm{k},\bm{p}}\,F^{L}(\bm{p},i\omega_{n})\right],$		(45)

where $\mathrm{Tr}$ means the summation over the pseudospin space, $F^{L(R)}$ is the anomalous Green’s function of the left(right) superconductor and $t_{\bm{k},\bm{p}}$ represents the tunnel element, $\{t_{\bm{k},\bm{p}}\}_{s,s^{\prime}}=t_{\beta\gamma}$. The wave number $\bm{k}$ is decomposed into a component in the $x$ direction $k_{x}$ and those parallel to the interface $\bm{k}_{\parallel}$. The summation $\sum_{\bm{k}}^{\prime}$ is carried out over $\bm{k}$ for $k_{x}>0$. We assume that the tunnel elements are independent of the wave numbers as

$\displaystyle t_{\bm{k},\bm{p}}=\mathbb{T}_{+}\,\delta_{\hat{\bm{k}},\hat{\bm{p}}},\quad\mathbb{T}_{\pm}=t_{0}\pm\sum_{i=1}^{3}t_{i}J_{i},$

(46)

where $\delta_{\hat{\bm{k}},\hat{\bm{p}}}$ implies the presence of the translation symmetry in the direction parallel to the interface. The tunnel event happens between two states on the Fermi surface : one is a state at $k_{F}\hat{\bm{p}}$ in the left superconductor and the other is a state at $k_{F}\hat{\bm{k}}$ in the right superconductor. The coefficients $t_{i}$ are real numbers for $i=0-3$. The element $t_{0}$ denotes the tunnel amplitude preserving pseudospin, whereas $t_{i}$ for $i=1-3$ refer to a Zeeman field which couples angular momentum of an electron. The complex conjugation of the hopping element is represented by

$\displaystyle t_{-\bm{k},-\bm{p}}^{\ast}=-U_{T}\mathbb{T}_{+}\,U_{T}\,\delta_{\hat{\bm{k}},\hat{\bm{p}}}=\mathbb{T}_{-}\,\delta_{\hat{\bm{k}},\hat{\bm{p}}},$

(47)

because of Eqs. (15) and (16). When the superconducting phase is $\varphi_{L}(\varphi_{R})$ on the left(right) superconductor, the Josephson current in Eq. (45) is calculated as

	$\displaystyle I=$	$\displaystyle 2e\,\mathrm{Im}{\sum_{\bm{k},\bm{p}}}^{\prime}\,e^{i\varphi}\;T\sum_{\omega_{n}}$
		$\displaystyle\times\mathrm{Tr}\left[\mathbb{T}^{\prime}\,\mathbb{F}^{R}(\bm{k},-i\omega_{n})\,\mathbb{T}_{+}\,\mathbb{F}^{L}(\bm{p},i\omega_{n})\right]\delta_{\hat{\bm{k}},\hat{\bm{p}}},$		(48)
	$\displaystyle\mathbb{T}^{\prime}$	$\displaystyle=U_{T}\mathbb{T}_{-}U_{T}^{-1}=t_{0}+t_{1}J_{1}-t_{2}J_{2}+t_{3}J_{3}.$		(49)

We will calculate the lowest order term of Josephson current between two $J=3/2$ superconductors by using Eq. (48) in Sec. III.

When both superconductors of a Josephson junction are in pseudospin-singlet state, the Josephson current should be represented by the Ambegaokar-Baratoff formula Ambegaokar and Baratoff (1963). We will confirm this feature at the beginning of this section. The s-wave pair potentials are given by

$\displaystyle\mathbb{D}^{L}_{\bm{p}}=\Delta^{L},\quad\mathbb{D}^{R}_{\bm{k}}=\Delta^{R},$

(50)

for the left and right superconductor, respectively. The Josephson current in Eq. (48) becomes

$\displaystyle I_{\mathrm{SS}}=$

$\displaystyle 8e\,t_{0}^{2}\,\sin\varphi\,T\sum_{\omega_{n}}\,K(i\omega_{n}),$

(51)

while the summation over momentum is carried out as follows,

	$\displaystyle K$	$\displaystyle(i\omega_{n})={\sum_{\bm{p}}}^{\prime}\frac{\Delta^{L}}{(\Delta^{L})^{2}+\omega_{n}^{2}+\xi_{\bm{p}}^{2}}$
		$\displaystyle\times\,{\sum_{\bm{k}}}^{\prime}\frac{\Delta^{R}}{(\Delta^{R})^{2}+\omega_{n}^{2}+\xi_{\bm{k}}^{2}}\;\delta_{\hat{\bm{k}},\hat{\bm{p}}},$		(52)
		$\displaystyle=\left\langle\int\frac{d\xi_{k}\,N_{0}}{(\Delta^{L})^{2}+\omega_{n}^{2}+\xi_{k}^{2}}\int\frac{d\xi_{k}\,N_{0}}{(\Delta^{R})^{2}+\omega_{n}^{2}+\xi_{k}^{2}}\right.$
		$\displaystyle\times\left.\Delta^{L}\Delta^{R}\right\rangle_{\hat{\bm{k}}},$		(53)
		$\displaystyle=\frac{\pi N_{0}}{\sqrt{\omega_{n}^{2}+(\Delta^{L})^{2}}}\frac{\pi N_{0}}{\sqrt{\omega_{n}^{2}+(\Delta^{R})^{2}}}\times\Delta^{L}\Delta^{R}.$		(54)

The summation over the momentum is decomposed into two parts. One part is the summation over its amplitude $|\bm{k}|$, which can be replaced by the integration over $\xi_{\bm{k}}$. The density of states at the Fermi level is denoted by $N_{0}$. The details of the integral is summarized in Appendix. A. The other part is summation over the direction of $\bm{k}$ on the Fermi surface for $k_{x}>0$. It is calculated as the average over the Fermi surface,

$\displaystyle\langle X\rangle_{\hat{\bm{k}}}=\frac{1}{\pi}\int_{0}^{\pi}d\theta\sin^{2}\theta\int_{-\pi/2}^{\pi/2}d\phi\cos\phi\;X(\hat{\bm{k}}),$

(55)

where $k_{x}=k_{F}\sin\theta\cos\phi>0$, $k_{y}=k_{F}\sin\theta\sin\phi$ and $k_{z}=k_{F}\cos\theta$. The Fermi momentum is obtained from $\varepsilon_{0}(k_{F})-\mu=0$. Finally the Josephson current is calculated as

	$\displaystyle I_{\mathrm{SS}}=$	$\displaystyle I_{0}\sin\varphi,$		(56)
	$\displaystyle I_{0}=$	$\displaystyle\frac{\pi}{eR_{N}}\,T\sum_{\omega_{n}}\left(\frac{\Delta^{L}}{\sqrt{\omega_{n}^{2}+(\Delta^{L})^{2}}}\right)\left(\frac{\Delta^{R}}{\sqrt{\omega_{n}^{2}+(\Delta^{R})^{2}}}\right),$		(57)
	$\displaystyle R_{N}^{-1}=$	$\displaystyle 4\times 2\pi e^{2}\left(N_{0}t_{0}\right)^{2},$		(58)

where $R_{N}$ is the normal resistance of the junction. As the most simple case, we assume that the amplitudes of the two pair potentials are identical to each other, $\Delta^{L}=\Delta^{R}=\Delta$. Then the Josephson current results in

$\displaystyle I_{\mathrm{SS}}=$

$\displaystyle\frac{\pi\Delta}{2eR_{N}}\,\tanh\left(\frac{\Delta}{2T}\right)\,\sin\varphi.$

(59)

The expression of the current is identical to the Ambegaokar-Baratoff’s formula Ambegaokar and Baratoff (1963).

Secondly, we discuss a junction where a pseudospin-singlet $s$-wave superconductor is on the left and a pseudospin-quintet superconductor is on the right side. The pair potentials are given by

$\displaystyle\mathbb{D}^{L}_{\bm{p}}=\Delta^{L},\quad\mathbb{D}^{R}_{\bm{k}}=\Delta^{R}\bm{q}\cdot\bm{\gamma}.$

(60)

The vector $\bm{q}$ is normalized vectors in five-dimensional pseudospin space, i.e. $\sum_{\nu=1-5}|q_{\nu}|^{2}=1$. When the tunneling elements are independent of pseudospin $\mathbb{T}_{\pm}=t_{0}\gamma^{0}$, the lowest order of the Josephson current vanishes by the selection rule due to the pseudospin configuration of Cooper pairs. Pseudospin of a Cooper pair is $J=0$ in a singlet superconductor, whereas it is $J=2$ in a quintet superconductor. To make the Josephson current finite, an electron must change its angular momentum while tunneling the insulating barrier because of the difference in the angular momentum, $\delta J=2$.

We switch on a weak Zeeman field to make the interface magnetically active. Then the tunnel elements become pseudospin dependent and can have finite $t_{i}(\ll t_{0})$ for $i=1-3$ in Eq. (46). The Josephson current is calculated as

	$\displaystyle I_{\mathrm{SQ}}$	$\displaystyle=\frac{2\sqrt{3}t_{1}t_{3}q_{3}+\sqrt{3}(t_{1}^{2}+t_{2}^{2})q_{4}-(t_{1}^{2}-t_{2}^{2}-2t_{3}^{2})q_{5}}{2t_{0}^{2}}$
		$\displaystyle\quad\times I_{0}\sin\varphi.$		(61)

Although the lowest order current can have finite value because of the Zeeman field, the contributions of $\nu=1,2$ terms vanish. As far as we study, the current-phase relationship is sinusoidal even if a Zeeman field breaks time-reversal symmetry. The results in Eq. (61) imply a possibility of $\pi$-junction caused by a magnetically active interface. Also, we find the second order terms in the coefficient $t_{i}$ of the Josephson current as a feature of $J=3/2$ superconductors. The second order implies that a magnetic field affects a Cooper pair twice during the tunneling. The reason can be explained by considering two facts that the difference in angular momentum of Cooper pairs is $\delta J=2$ and that $J_{i}$ matrices in the tunneling Hamiltonian can change the angular momentum only by $\delta J=1$ for each operation. Therefore, a Zeeman field affects a Cooper pair twice during the tunneling to compensate the difference in the angular momenta of Cooper pairs. In contrast, the Josephson current between $S=1/2$ spin-singlet and triplet superconductor is calculated as $I\propto\bm{d}\cdot\bm{U}_{M}$, where $\bm{U}_{M}$ is the magnetic scattering potential (Appendix. B). Thus the current is written in the first order terms in the coefficients of the spin-dependent tunneling Hamiltonian. This implies that a single operation of a Zeeman field can couple the Cooper pairs with difference of the angular momentum, $\delta S=1$.

In this subsection, we discuss the Josephson effect in a junction consisting of two superconductors in pseudospin-quintet states. First, we consider a junction of two $s$-wave pseudospin-quintet superconductors described by

$\displaystyle\mathbb{D}^{L}_{\bm{p}}=\Delta^{L}\bm{q}^{L}\cdot\bm{\gamma},\quad\mathbb{D}^{R}_{\bm{k}}=\Delta^{R}\bm{q}^{R}\cdot\bm{\gamma}.$

(62)

The anomalous Green’s function can be calculated as Eq. (41). By substituting the anomalous Green’s function in Eq. (41) to Eq. (48), the Josephson current is calculated as

$\displaystyle I_{\mathrm{QQ}}$

$\displaystyle=I_{0}\sin\varphi\left(\bm{q}^{R}\cdot\bm{q}^{L}\right).$

(63)

The result indicates that the Josephson current is given by the scalar product of two vectors, $I_{\mathrm{QQ}}\propto\bm{q}^{L}\cdot\bm{q}^{R}$, which describes the selection rule due to the pseudospin configuration of Cooper pairs. It is easy to confirm that Josephson current cannot flow between two superconductors with Cooper pairs corresponding to different representations in Table 1. This selection rule is similar to that in two spin-triplet superconductors of $S=1/2$ electrons,

	$\displaystyle I_{\mathrm{TT}}=$	$\displaystyle I_{1}\frac{\Big{\langle}\bm{d}^{L}\cdot\bm{d}^{R}\Big{\rangle}_{\hat{\bm{k}}}}{d_{l}\,d_{r}}\,\sin\varphi,$		(64)
	$\displaystyle d_{l(r)}=$	$\displaystyle\sqrt{\left\langle\left\{\bm{d}^{L(R)}(\hat{\bm{k}})\right\}^{2}\right\rangle_{\hat{\bm{k}}}}$		(65)

where $\bm{d}$ is the $\bm{d}$-vector in three-dimensional spin space of the spin-triplet Cooper pair with pair potential $\Delta(\hat{\bm{k}})=i\bm{d}(\hat{\bm{k}})\cdot\bm{\sigma}\sigma_{2}$. By comparing Eq. (63) and (64), we conclude that the selection rule for quintet states is severer than that for triplet states. This is because two vectors can be orthogonal to each other more easily in higher dimension.

Additionally, we study a case with two $J=3/2$ superconductors in which pair potentials are anisotropic in the momentum space. With the general even-parity pair potential

$\displaystyle\mathbb{D}^{L}_{\bm{p}}=\bm{\eta}_{\bm{p}}^{L}\cdot\bm{\gamma},\quad\mathbb{D}^{R}_{\bm{k}}=\bm{\eta}_{\bm{k}}^{R}\cdot\bm{\gamma},$

(66)

the Josephson current is calculated as

	$\displaystyle I_{\mathrm{QQ}}$	$\displaystyle=I_{1}\frac{\left\langle\bm{\eta}_{\bm{k}}^{R}\cdot\bm{\eta}_{\bm{k}}^{L}\right\rangle_{\hat{\bm{k}}}}{\eta^{R}\eta^{L}}\sin\varphi,$		(67)
	$\displaystyle I_{1}$	$\displaystyle=\frac{\pi}{eR_{N}}T\sum_{\omega_{n}}\frac{\eta^{R}}{\sqrt{\omega_{n}^{2}+(\eta^{R})^{2}}}\frac{\eta^{R}}{\sqrt{\omega_{n}^{2}+(\eta^{R})^{2}}},$		(68)
	$\displaystyle\eta^{L(R)}$	$\displaystyle=\sqrt{\left\langle\left(\bm{\eta}^{L(R)}_{\bm{k}}\right)^{2}\right\rangle_{\hat{\bm{k}}}}.$		(69)

The average of $\left\langle\bm{\eta}_{\bm{k}}^{R}\cdot\bm{\eta}_{\bm{k}}^{L}\right\rangle_{\hat{\bm{k}}}$ on the Fermi surface leads to additional selection rule for the pair potentials depending on momenta. For example in a case of $\bm{\eta}_{\bm{k}}^{L}\propto e_{1}(\hat{\bm{k}})$ and $\bm{\eta}_{\bm{k}}^{R}\propto e_{2}(\hat{\bm{k}})$, the Josephson current vanishes irrespective of relative pseudospin configurations of Cooper pairs in the two superconductors because of the selection rule in momentum space $\left\langle e_{1}(\hat{\bm{k}})e_{2}(\hat{\bm{k}})\right\rangle_{\hat{\bm{k}}}=0$.

In this section, we have assumed that electronic structures are independent of pseudospin, i.e., $\mathbb{E}_{\bm{k}}=\varepsilon_{0}(\bm{k})-\mu$. The effects of a pseudospin-dependent electronic structures on the selection rule will be clarified in the next section.

We first consider a junction that consists of two identical superconductors described by Eq. (70) and a barrier with tunnel elements $t_{\bm{k},\bm{p}}=t_{0}\gamma^{0}\delta_{\hat{\bm{k}},\hat{\bm{p}}}$. By substituting the Green’s function in Eq. (71) into Eq. (48), the Josephson current is calculated as

	$\displaystyle I=$	$\displaystyle 2e\,t_{0}^{2}\,\sin\varphi\,T\sum_{\omega_{n}}{\sum_{\bm{k},\bm{p}}}^{\prime}\,\mathrm{Tr}\left[\mathbb{F}^{R(0)}_{\nu}(\bm{k},-i\omega_{n})\,\mathbb{F}^{L(0)}_{\nu}(\bm{p},i\omega_{n})\right.$
		$\displaystyle+\left.\delta\mathbb{F}_{\nu\lambda}^{R}(\bm{k},-i\omega_{n})\,\delta\mathbb{F}_{\nu\lambda}^{L}(\bm{p},i\omega_{n})\right],$		(75)
	$\displaystyle=$	$\displaystyle I_{0}\sin\varphi\left(1-\frac{\langle\varepsilon_{\lambda}(\bm{k})\rangle_{\hat{\bm{k}}}}{8\eta^{2}}\right).$		(76)

The details of the derivation are shown in Appendix C. The first term of Eqs. (75)-(76) has been already explained in Sec. III C and it dominates the Josephson current. An induced odd-frequency pair carries the Josephson current at the second term in Eqs. (75)-(76). The contribution of an odd-frequency pair is negative, which is derived from the order of pseudospin matrices in the second term as $\mathrm{Tr}[\gamma^{\nu}\,\gamma^{\lambda}\,\gamma^{\nu}\,\gamma^{\lambda}]=-4$. Namely, the coupling between the induced odd-frequency pairing correlations stabilizes the $\pi$-state rather than the 0-state because an odd-frequency pair favors phase difference across the junction due to its paramagnetic property.Sasaki et al. (2020) When the amplitude of odd-frequency correlation in Eq. (73) exceeds that of even-frequency component in Eq. (72), the junction may undergo the transition to the $\pi$-state. However, such a superconducting state is impossible because an odd-frequency Cooper pair is thermodynamically unstable.

As shown in Eq. (76), the junction we consider should be always the 0-state because the even-frequency pairing correlations dominate the Josephson current. Simultaneously, it proposes a possibility of a $\pi$-junction in the system consisting of time-reversal symmetric components only, if the dominant term could be deleted. In such a context, we consider a Josephson junction where the severer pseudospin selection rule is applied. We choose its left superconductor described by Eq. (70) and the right superconductor described by

$\displaystyle\mathbb{E}_{\bm{k}}=\varepsilon_{0}+\varepsilon_{\nu}(\bm{k})\gamma^{\nu}-\mu,\quad\mathbb{D}_{\bm{k}}=\eta_{\bm{k}}^{R}\,\gamma^{\lambda}.$

(77)

The anomalous Green’s function of the right superconductor is described by

$\displaystyle\mathbb{F}_{\lambda\nu}(\bm{k},i\omega_{n})=\mathbb{F}_{\lambda}^{(0)}(\bm{k},i\omega_{n})+\delta\mathbb{F}_{\lambda\nu}(\bm{k},i\omega_{n}).$

(78)

The Josephson current in this junction is calculated as

	$\displaystyle I=$	$\displaystyle 2e\,t_{0}^{2}\,\sin\varphi\,T\sum_{\omega_{n}}{\sum_{\bm{k},\bm{p}}}^{\prime}\,\mathrm{Tr}\left[\mathbb{F}^{L(0)}_{\nu}(\bm{p},i\omega_{n})\,\mathbb{F}^{R(0)}_{\lambda}(\bm{k},-i\omega_{n})\right.$
		$\displaystyle+\left.\delta\mathbb{F}_{\nu\lambda}^{L}(\bm{p},i\omega_{n})\,\delta\mathbb{F}_{\lambda\nu}^{R}(\bm{k},-i\omega_{n})\right],$		(79)
	$\displaystyle=$	$\displaystyle\frac{\pi}{eR_{N}}\langle\varepsilon_{\lambda}(\bm{k})\,\varepsilon_{\nu}(\bm{k})\eta_{\bm{k}}^{R}\eta_{\bm{k}}^{L}\rangle_{\hat{\bm{k}}}\,\sin\varphi$
		$\displaystyle\times T\sum_{\omega_{n}}\frac{\omega_{n}^{2}}{\{\omega_{n}^{2}+(\eta^{R})^{2}\}^{3/2}\{\omega_{n}^{2}+(\eta^{L})^{2}\}^{3/2}}.$		(80)

The positive sign of the trace is derived from the order of pseudospin matrices as $\mathrm{Tr}[\gamma^{\nu}\,\gamma^{\lambda}\,\gamma^{\lambda}\,\gamma^{\nu}]=4$. The first term in Eq. (79) vanishes due to the pseudospin selection rule as expected. The second term is carried by induced odd-frequency and its sign is determined by the momentum dependence of the pair potentials. However, it is calculated as positive or zero if the momentum dependence is $s$-wave or $d$-wave symmetric. Unfortunately, we cannot draw any physical picture of why an odd-frequency pair stabilizes the 0-state. Even so, the result of the 0-junction is reasonable in the phenomenological point of view. Because all the parts of the Josephson junction preserve time-reversal symmetry as we have discussed at the beginning of this subsection. The superconducting phase should be uniform at the ground state of such a junction.