Rashba effect in cylindrical quantum well wire with finite confining potential

Abstract

The effect of the Rashba spin-orbit interaction has been investigated in an infinite cylindrical quantum well wire (QWW) with finite confining potential. The effect has been considered in the absence and presence of an axial magnetic field. The solution of the radial Schrödinger equation has been obtained in the wire and in the barrier, in the absence of the Rashba spin-orbit interactions. The general solution in the presence of these interactions has then been expanded in terms of the obtained two solutions and in terms of the step function. The study should, however, be performed in two dimensions. For this reason, the wave vector along the axis of the wire has been chosen to be zero. The orthogonality of the elements of the resulting basis has been proved. The dependence of the Rashba coupling coefficient on the electron effective mass has also been taken into consideration. The results have been applied to the case of $GaAs - Ga_{1 - x} Al_{x} As$ cylindrical quantum wire.

Graphical Abstract

The energy levels (E_o01, E_o02) and (E₀₁, E₀₂) in the absence and presence of the Rashba effect against the radius (R) of the QWW. The magnetic field B = 10 T, the confining potential V_o = 25 R^* and two different values for the Rashba coupling coefficient have been used in the wire and barrier.

Appendix 1 Orthogonality and some important relations

In this appendix, we give and prove some important relations which have been used throughout the text.

• (1) It can be shown by using the relations between the Bessel functions (Lebedev [44]) and the analogous relations between the modified Bessel functions that

  $$\mathop \int \limits_{0}^{R} J_{\nu } \left( {\alpha \rho } \right)J_{\nu } \left( {\beta \rho } \right) \rho d\rho = \frac{R}{{\left( {\alpha^{2} - \beta^{2} } \right)}}\left[ {\beta J_{\nu } \left( {\alpha R} \right)J_{\nu }^{^{\prime}} \left( {\beta R} \right) - \alpha J_{\nu } \left( {\beta R} \right)J_{\nu }^{^{\prime}} \left( {\alpha R} \right)} \right],$$

  (73)
  $$\mathop \int \limits_{0}^{R} J_{\nu }^{2} \left( {\alpha \rho } \right) \rho d\rho = \frac{{R^{2} }}{2}\left[ { - J_{\nu + 1} \left( {\alpha R} \right) J_{\nu - 1} \left( {\alpha R} \right) + J_{\nu }^{2} \left( {\alpha R} \right)} \right],$$

  (74)
  $$\mathop \int \limits_{R}^{\infty } K_{\nu } \left( {\alpha \rho } \right)K_{\nu } \left( {\beta \rho } \right)\rho d\rho = \frac{R}{{\left( {\alpha^{2} - \beta^{2} } \right)}}\left[ {\beta K_{\nu } \left( {\alpha R} \right)K_{\nu }^{^{\prime}} \left( {\beta R} \right) - \alpha K_{\nu } \left( {\beta R} \right)K_{\nu }^{^{\prime}} \left( {\alpha R} \right)} \right]$$

  (75)

  and

  $$\mathop \int \limits_{R}^{\infty } K_{\nu }^{2} \left( {\alpha \rho } \right) \rho d\rho = \frac{{R^{2} }}{2}\left[ {K_{\nu + 1} \left( {\alpha R} \right) K_{\nu - 1} \left( {\alpha R} \right) - K_{\nu }^{2} \left( {\alpha R} \right)} \right],$$

  (76)

  The dash in Eqs. (73), (75) refers to the differentiation of $J_{\nu }$ and $K_{\nu }$ with respect to their arguments.

• (2) The above relations can be used to prove the orthogonality of the elements of the basis $\left\{ {{\mathcal{T}}_{n,m}^{\left( 1 \right)} \left( \rho \right)} \right\}$ in the absence of a magnetic field. For this purpose, we restrict $\alpha , \beta$ in Eqs. (73), (74) to be two values of the parameter $a_{nm}$ being defined by Eq. (6). Also, $\alpha , \beta$ in Eqs. (75), (76) are taken to be two values of the parameter $b_{nm}$ which is defined by Eq. (10). Thus, the four parameters involved are related to each other and to the energy by the relation

  $$\frac{1}{{m_{2} }} \left( {b_{n\prime m\prime }^{2} - b_{{{\text{nm}}}}^{2} } \right) = \frac{1}{{m_{1} }} \left( {a_{{{\text{nm}}}}^{2} - a_{n\prime m\prime }^{2} } \right) = \frac{2}{{\hbar^{2} }} \left( {E_{{{\text{onmk}}}} - E_{{{\text{on}}\prime m\prime k\prime }} } \right).$$

  (77)

  They are also related by the boundary conditions (11), (12). The orthogonality can consequently be proved by defining ${\mathcal{T}}_{n,m}^{\left( 1 \right)} \left( \rho \right)$ from the second part of Eq. (18) and using Eqs. (73), (75), (77). Accordingly, for fixed $m$ and $n \ne n^{\prime}$:

  $$\mathop \int \limits_{0}^{\infty } {\mathcal{T}}_{n,m}^{\left( 1 \right)} \left( \rho \right){ \mathcal{T}}_{n\prime ,m}^{\left( 1 \right)} \left( \rho \right) \rho d\rho = \frac{{Rm_{1} }}{{\left( {a_{{{\text{nm}}}}^{2} - a_{n\prime m}^{2} } \right)}}\left[ {J_{m} \left( {a_{{{\text{nm}}}} R} \right)B_{n\prime } - J_{m} \left( {a_{n\prime m} R} \right)B_{n} } \right],$$

  (78)

  where

  $$B_{n} = \frac{{a_{{{\text{nm}}}} }}{{m_{1} }}J_{m}^{^{\prime}} \left( {a_{{{\text{nm}}}} R} \right) - \frac{{b_{{{\text{nm}}}} }}{{m_{2} }} \frac{{J_{m} \left( {a_{{{\text{nm}}}} R} \right)}}{{K_{m} \left( {b_{{{\text{nm}}}} R} \right)}} K_{m}^{^{\prime}} \left( {b_{{{\text{nm}}}} R} \right).$$

  (79)

  But from the boundary condition (12)

  $$B_{n} = B_{{n^{\prime}}} = 0.$$

  (80)

  It then follows from (78) that

  $$\mathop \int \limits_{0}^{\infty } {\mathcal{T}}_{n,m}^{\left( 1 \right)} \left( \rho \right){ \mathcal{T}}_{{n^{\prime},m}}^{\left( 1 \right)} \left( \rho \right) \rho d\rho = 0 \quad for \quad n \ne n^{\prime}$$

  (81)

  On the other hand, for $n = n^{\prime}$ the normalization condition follows from Eqs. (74), (76). According to these equations

  $$\begin{gathered} \mathop \int \limits_{0}^{\infty } \left( {{\mathcal{T}}_{n,m}^{\left( 1 \right)} \left( \rho \right)} \right)^{2} \rho d\rho \hfill \\ = \frac{{R^{2} }}{2}\left[ { - J_{m + 1} \left( {a_{{{\text{nm}}}} R} \right) J_{m - 1} \left( {a_{{{\text{nm}}}} R} \right) + \frac{{J_{m}^{2} \left( {a_{{{\text{nm}}}} R} \right)}}{{K_{m}^{2} \left( {b_{{{\text{nm}}}} R} \right)}} K_{m + 1} \left( {b_{{{\text{nm}}}} R} \right) K_{m - 1} \left( {b_{{{\text{nm}}}} R} \right)} \right], \hfill \\ \end{gathered}$$

  (82)

  The normalization condition (82) was given in Brown and Spector [38]. However, the present calculations have the advantage that they are derived in the general case of different masses and that the orthogonal relation (81) has been proved. In fact, Brown and Spector [38] have dealt only with one wave function and thus the orthonormal relation has not been discussed. Also, they have considered the special case of equal masses.

• (3) In the presence of a magnetic field, the elements of $\left\{ {{\mathcal{T}}_{n^{\prime},m}^{\left( 1 \right)} \left( \rho \right)} \right\}$ and $F_{{{\text{nm}}}} \left( \rho \right)$ are defined by the first part of Eq. (18) and by Eq. (21) where $F_{{{\text{in}}}} \left( \rho \right), F_{{{\text{out}}}} \left( \rho \right)$ are now defined by Eqs. (53), (54). Thus, the orthogonality of the elements of $\left\{ {{\mathcal{T}}_{n,m}^{\left( 1 \right)} \left( \rho \right)} \right\}$ depends on the orthogonality of the hypergeometric functions ${}_{1}^{{}} F_{1}^{{}} , U$ which was given in textbooks only over an infinite interval, for which $\rho$ varies from $0$ to $\infty$. In the present work, the proof of the orthogonality relation necessitates the consideration of integrals over intervals that represent $\rho$ from $0$ to $R$ or from $R$ to $\infty$. In order to deal with these integrals, we consider Eq. (48) for two different energy eigenvalues $E_{{{\text{on}}_{1} {\text{mk}}}}$ and $E_{{{\text{on}}_{2} {\text{mk}}}}$. Here, the quantum numbers $m, k$ are fixed. The difference between the corresponding values $\tilde{E}_{1} , \tilde{E}_{2}$ of $\tilde{E}$ is accordingly given from Eq. (49) by

  $$\tilde{E}_{1} - \tilde{E}_{2} = \frac{{m^{*} \left( \rho \right)}}{eB\hbar }\left( {E_{o1} - E_{o2} } \right).$$

  (83)

  In Eq. (83) $E_{{on_{i} mk}}$ has been denoted by $E_{oi}$, $i = 1,2$, for simplicity. It then follows from Eq. (48) that

  $$4\zeta \frac{{d^{2} F_{i} }}{{d\zeta^{2} }} + 4\frac{{dF_{i} }}{d\zeta } - \zeta F_{i} - \frac{{m^{2} F_{i} }}{\zeta } + 4\tilde{E}_{i} F_{i} = 0, i = 1,2$$

  (84)

  In Eq. (84), $F_{i}$ stands either for $\left( {F_{{{\text{in}}}} } \right)_{i}$ in the region $\rho < R$ or for $\left( {F_{{{\text{out}}}} } \right)_{i}$ in the region $\rho > R$. We now multiply Eq. (84) for $i = 1$ by $F_{2}$ and for $i = 2$ by $F_{1}$. We then subtract the resulting two equations and use Eq. (83) to find

  $$\frac{d}{d\zeta }\left[ {\frac{\zeta }{{m^{*} \left( \rho \right)}}\left( {F_{2} \frac{{dF_{1} }}{d\zeta } - F_{1} \frac{{dF_{2} }}{d\zeta }} \right)} \right] + \frac{1}{eB\hbar }\left( {E_{o1} - E_{o2} } \right)F_{1} F_{2} = 0.$$

  (85)

  For the region $\rho < R$ we integrate the above equation with respect to $\zeta$ from $0$ to $\zeta_{R}$ while for the region $\rho > R$ we integrate from $\zeta_{R}$ to $\infty$. It can consequently be shown after adding the resulting two equations that

  $$\left[ {\frac{\zeta }{{m_{1} }}\left( {F_{2} \frac{{dF_{1} }}{d\zeta } - F_{1} \frac{{dF_{2} }}{d\zeta }} \right)_{{{\text{in}}}} } \right]_{0}^{{\zeta_{R} }} + \left[ {\frac{\zeta }{{m_{2} }}\left( {F_{2} \frac{{dF_{1} }}{d\zeta } - F_{1} \frac{{dF_{2} }}{d\zeta }} \right)_{{{\text{out}}}} } \right]_{{\zeta_{R} }}^{\infty } = \frac{1}{eB\hbar }\left( {E_{o2} - E_{o1} } \right)\mathop \int \limits_{0}^{\infty } F_{{n_{1} m}} F_{{n_{2} m}} d\zeta .$$

  (86)

  The L.H.S of (86) vanishes due to the boundary conditions (11), (12), and due to the fact that.

  $\left( {F_{i} } \right)_{in} \to 0$ for $\zeta \to 0$ and $\left( {F_{i} } \right)_{{{\text{out}}}} \to 0$ for $\zeta \to \infty$.

  Accordingly, for $E_{o2} \ne E_{o1}$ we find

  $$\mathop \int \limits_{0}^{\infty } F_{{n_{1} m}} F_{{n_{2} m}} d\zeta = \mathop \int \limits_{0}^{\infty } {\mathcal{T}}_{{n_{1} ,m}}^{\left( 1 \right)} \left( \zeta \right) {\mathcal{T}}_{{n_{2} ,m}}^{\left( 1 \right)} \left( \zeta \right) d\zeta = 0, for n_{1} \ne n_{2} ,$$

  (87)

  which also implies that

  $$\mathop \int \limits_{0}^{\infty } {\mathcal{T}}_{{n_{1} ,m}}^{\left( 1 \right)} \left( \rho \right) {\mathcal{T}}_{{n_{2} ,m}}^{\left( 1 \right)} \left( \rho \right) \rho d\rho = 0, for n_{1} \ne n_{2} .$$

  (88)

  The last equation represents the orthogonality of the elements of the set $\left\{ {{\mathcal{T}}_{n,m}^{\left( 1 \right)} \left( \rho \right)} \right\}$. It can be alternatively expressed as

  $$\mathop \int \limits_{0}^{R} \left( {F_{in} } \right)_{1} \left( {F_{in} } \right)_{2} \rho d\rho + \mathop \int \limits_{R}^{\infty } \left( {F_{out} } \right)_{1} \left( {F_{out} } \right)_{2} \rho d\rho = 0, for n_{1} \ne n_{2}$$

  (89)

  where $F_{in} , F_{out}$ are defined by Eqs. (53), (54) receptively.

• (4) In this part we proceed to prove that

  $${\text{In}}\left( {I_{2} } \right)_{nn}^{m + 1} = \left( {I_{3} } \right)_{nn}^{m} ,$$

  (90)

  which has been used in the absence (Sec. 3, Eq. (29)) and presence (Sec. 5) of a magnetic field. We start by the relations

  $$\left( {I_{3} } \right)_{nn}^{m} = \mathop \int \limits_{0}^{\infty } {\mathcal{T}}_{n,m + 1}^{\left( 3 \right)} \left( \rho \right) {\mathcal{T}}_{n,m}^{\left( 1 \right)} \left( \rho \right) \rho d\rho$$

  (91)

  and

  $$\left( {I_{2} } \right)_{{{\text{nn}}}}^{m + 1} = \mathop \int \limits_{0}^{\infty } {\mathcal{T}}_{n,m}^{\left( 2 \right)} \left( \rho \right) {\mathcal{T}}_{n,m + 1}^{\left( 1 \right)} \left( \rho \right) \rho d\rho$$

  (92)

  which follows directly from Eqs. (26), (27). Substitution of the first part of Eq. (18) and of Eqs. (58), (59) in Eqs. (91), (92) consequently leads to

  $$\left( {I_{3} } \right)_{nn}^{m} = \mathop \int \limits_{0}^{\infty } \rho F_{nm} \left[ {\frac{d}{d\rho } + \frac{m + 1}{\rho } + \frac{eB}{{2\hbar }}\rho } \right]F_{n,m + 1} d\rho$$

  (93)

  and

  $$\left( {I_{2} } \right)_{nn}^{m + 1} = \mathop \int \limits_{0}^{\infty } \rho F_{n,m + 1} \left[ { - \frac{d}{d\rho } + \frac{m}{\rho } + \frac{eB}{{2\hbar }}\rho } \right]F_{{{\text{nm}}}} d\rho$$

  (94)

  In the absence of a magnetic $B = 0$ and accordingly, the third term in the above two equations does not exist. In all cases, this term is identical in both the expressions for $\left( {I_{3} } \right)_{nn}^{m}$, $\left( {I_{2} } \right)_{nn}^{m + 1}$.

  On the other hand, the first term in Eq. (93) gives rise to the integral

  $$J = \mathop \int \limits_{0}^{\infty } \left( {\frac{d}{d\rho }\left( {F_{n,m + 1} } \right)} \right)F_{n,m} \rho d\rho ,$$

  (95)

  which can be divided into two parts

  $$J = \mathop \int \limits_{0}^{R} \left( {\frac{d}{d\rho }\left( {F_{n,m + 1} } \right)_{in} } \right)\left( {F_{n,m} } \right)_{in} \rho d\rho + \mathop \int \limits_{R}^{\infty } \left( {\frac{d}{d\rho }\left( {F_{n,m + 1} } \right)_{{{\text{out}}}} } \right)\left( {F_{n,m} } \right)_{{{\text{out}}}} \rho d\rho .$$

  (96)

  The integration of the above two integrals by parts yields

  $$J = - \mathop \int \limits_{0}^{R} \left( {F_{n,m + 1} } \right)_{in} \frac{d}{d\rho }\left[ {\left( {F_{nm} } \right)_{in} \rho } \right]d\rho - \mathop \int \limits_{R}^{\infty } \left( {F_{n,m + 1} } \right)_{out} \frac{d}{d\rho }\left[ {\left( {F_{nm} } \right)_{out} \rho } \right]d\rho$$

  (97)
  $$= - \mathop \int \limits_{0}^{\infty } \rho F_{n,m + 1} \left[ {\frac{d}{d\rho }\left( {F_{{{\text{nm}}}} } \right) + \frac{{F_{{{\text{nm}}}} }}{\rho }} \right]d\rho .$$

  (98)

  The integrated two terms in Eq. (97) have been canceled due to the boundary condition (11) and due to the choice that $F_{nm}$ should vanish at $\rho = 0$ and $\rho \to \infty$. On substituting from Eq. (98) in (93) the required relation (90) can be obtained.