## 2.1. Energy spectrum and wave function

The potential of the quantum system under consideration is defined as

\(U\left( {\rho ,z} \right)=\left\{ {\begin{array}{*{20}{c}} {0,\frac{{{\rho ^2}}}{{R_{0}^{2}}}+\frac{{{z^2}}}{{L_{0}^{2}}} \leqslant 1} \\ {\infty ,\frac{{{\rho ^2}}}{{R_{0}^{2}}}+\frac{{{z^2}}}{{L_{0}^{2}}}>1} \end{array},{R_0} \gg {L_0}} \right\}\) . (1)

The potential we define has ellipsoid geometry, and it is accepted that the value of the potential is zero inside the ellipsoid, and the potential is infinite outside the ellipsoid. The electron, which we limit with its potential known as the hard wall potential, is confined in this quantum system. In the potential, the \({R_0}\) parameter is defined in the \(XOY\) plane and determines the radius of the ellipsoid and is defined as \(\rho =\sqrt {{x^2}+{y^2}}\); The \({L_0}\) parameter is defined in the \(OZ\) direction and determines the thickness of the ellipsoid. The thickness of the ellipsoid reaches its maximum value at \(\rho =0\), that is, at the center of the ellipsoid, and as \(\rho \to {R_0}\) approaches, the thickness of the ellipsoid gradually approaches zero [20].

In Eq. (1), we assume that \({R_0} \gg {L_0}\). This condition tells us that the diameter of the ellipsoid is much larger than the height of the ellipsoid. In such a system, the movement of the electron in the \(XOY\) plane can be considered as an adiabatic distortion in the \(OZ\) direction. That is, it is assumed that the state change of the electron in the \(OZ\) direction changes very quickly compared to the state change in the \(XOY\) plane. Taking this into account, we can divide the Hamilton of our system under consideration into two subsystems. Let's define the change of the electron in the \(OZ\) direction as the "fast" subsystem, and the change in the \(XOY\) plane as the "slow" subsystem. The wave function of the system we have defined can be written as

$$\psi \left( {\rho ,\phi ,z} \right)={e^{im\phi }}R\left( \rho \right){\chi _\rho }\left( z \right)$$

2

,

depending on the variables \(\rho\) and . Here, is the magnetic quantum number and takes \(0, \pm 1, \pm 2,..\) values, \(R\left( \rho \right)\) determines the state of the electron in the \(XOY\) plane, and \({\chi _\rho }\left( z \right)\) determines the state of the electron in the \(OZ\) direction.

Considering what we wrote above, let's write the Hamiltonian expression of the quantum system. If we look at Eq. (1), the potential value is zero inside the ellipsoid. Therefore, the motion of the electron is assumed to be free within the ellipsoid. In addition, we assume that the free electron in the ellipsoid is influenced by the Rashba interaction and the DMS-induced exchange interaction. We also assume that a constant magnetic field is applied to the ellipsoid in the \(OZ\) direction. Accordingly, the Hamilton expression of the system is written as

$$H=\frac{1}{{2\mu }}{\left( {{\mathbf{p}}+e{\mathbf{A}}} \right)^2}+U\left( {\rho ,z} \right)+{H_R}+{H_Z}+{H_{DMS}}$$

3

.

Here, \({\mathbf{p}}= - i\hbar \nabla\) is the momentum operator; \({\mathbf{A}}\) is a vector potential, \({\mathbf{A}}=\left( {0,\frac{{B\rho }}{2},0} \right)\); \(\mu\) is the effective mass of the electron. In the Hamiltonian expression, the term \({H_R}\) is the Rashba term and is written as

Here, \({\mathbf{\sigma }}\) is Puali matrices and \({\alpha _R}\) Rashba parameter. The expression is unit vector and is taken parallel to the growth direction of the structure. Let's assume that the growth direction of the system under consideration is in the direction. Thus, as a result of vector multiplication in the Rashba term, only the terms related to the direction are different from zero. We can write this as

$${H_R}=\frac{{i{\alpha _R}}}{\hbar }\left( { - {\sigma _x}\sin \phi +{\sigma _y}\cos \phi } \right)\frac{\partial }{{\partial \rho }} - \frac{{{\alpha _R}}}{{\hbar \rho }}\left( {{\sigma _x}\cos \phi +{\sigma _y}\sin \phi } \right)\left( {i\frac{\partial }{{\partial \phi }}+\frac{\Phi }{{{\Phi _0}}}} \right)$$

5

.

Here, \(\Phi =\pi {\rho ^2}B\) is the magnetic flux and \({\Phi _0}=\frac{{2\pi \hbar }}{e}\) is the magnetic flux constant. The fourth term in the Hamilton expression is the Zeeman term, written as

$${H_Z}=\frac{1}{2}g{\mu _B}{B_z}{\sigma _z}$$

6

.

Here, is the Lande factor, \({\mu _B}\) is the Bohr magneton, \({\sigma _z}\) is the -direction component of the Pauli matrix. The last term in Hamilton is the term expressing the exchange interaction resulting from the DMS feature of the structure and is written as [9]

$${H_{DMS}}=\frac{1}{2}x{N_0}\tau \left\langle {{S_z}} \right\rangle {\sigma _z}$$

7

.

Also here,

$$\tau =\left\langle {X\left| J \right|X} \right\rangle =\left\langle {Y\left| J \right|Y} \right\rangle =\left\langle {Z\left| J \right|Z} \right\rangle$$

8

Equation (7) defines the \(sp - d\) exchange interaction that occurs between the \(3{d^5}\) electron of the \(M{n^{2+}}\) ion and the \(sp\)-band electron in DMS structures formed under the applied constant magnetic effect. Here, \({J_{s - d}}\) is the exchange interaction coefficient; The integral \(\tau\) express the exchange interaction integrals between the \(s - d\) and \(p - d\) bands, respectively; \({N_0}\) density parameter per cell; The \(\left\langle {{S_z}} \right\rangle\) parameter is the thermodynamic average of the component of the spin of the \(Mn\) atom and is written as [9]

$$\left\langle {{S_z}} \right\rangle = - S{B_S}\left( {\frac{{S{g_{Mn}}{\mu _B}B}}{{{k_B}T}}} \right)$$

9

.

Here, \({B_{5/2}}\left( \xi \right)\) is the Brillouin function, \({g_{Mn}}=2\) is the Lande factor of the atom, \({k_B}\) is the Boltzmann constant; temperature; \(S=\frac{5}{2}\) is the total spin value of the \(M{n^{2+}}\) ion. Let's define the wave function corresponding to Hamilton that we defined in Eq. (3) as

$$\Psi =\left( {\begin{array}{*{20}{c}} {{\psi _1}} \\ {{\psi _2}} \end{array}} \right)$$

10

.

After this, we write the Schrödinger Equation in the following form

$$\left( {\begin{array}{*{20}{c}} {{H_0}+\frac{1}{2}{\mu _B}B{g^*} - E}&{{\alpha _R}{e^{ - i\phi }}\left( {\frac{\partial }{{\partial \rho }} - \frac{1}{\rho }\left( {i\frac{\partial }{{\partial \phi }}+\frac{\Phi }{{{\Phi _0}}}} \right)} \right)} \\ { - {\alpha _R}{e^{i\phi }}\left( {\frac{\partial }{{\partial \rho }}+\frac{1}{\rho }\left( {i\frac{\partial }{{\partial \phi }}+\frac{\Phi }{{{\Phi _0}}}} \right)} \right)}&{{H_0} - \frac{1}{2}{\mu _B}B{g^*} - E} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {{C_1}{\psi _1}} \\ {{C_2}{\psi _2}} \end{array}} \right)=0$$

11

.

Here,

$${H_0}= - \frac{{{\hbar ^2}}}{{2\mu }}\left( {\frac{{{\partial ^2}}}{{\partial {\rho ^2}}}+\frac{1}{\rho }\frac{\partial }{{\partial \rho }}+\frac{1}{{{\rho ^2}}}\frac{{{\partial ^2}}}{{\partial {\phi ^2}}}+\frac{{{\partial ^2}}}{{\partial {z^2}}}} \right) - i\frac{{\hbar {\omega _c}}}{2}\frac{\partial }{{\partial \phi }}+\frac{{\mu \omega _{c}^{2}{\rho ^2}}}{8}$$

12

is the Hamiltonian expression that does not include perturbation terms. Let's define the wave functions \({\psi _1}\) and \({\psi _2}\) in Eq. (11) as

\({\psi _m}={e^{im\phi }}{R_m}\left( \rho \right){\chi _\rho }\left( z \right)\) , \({\psi _{m+1}}={e^{i\left( {m+1} \right)\phi }}{R_{m+1}}\left( \rho \right){\chi _\rho }\left( z \right)\) (13)

respectively. Then, taking into account the wave functions we defined, let's write the Schrödinger equation as

\({H_0}{\psi _m}={E_m}{\psi _m}\) , \({H_0}{\psi _{m+1}}={E_{m+1}}{\psi _{m+1}}\) (14)

for Hamilton \({H_0}\) in Eq. (12). As can be seen from these two equations, the energy spectra in the equations differ from each other only according to the quantum number . If we find the energy spectrum of one of these equations, the energy spectrum of the other can be found by substituting instead of \(m+1\) or, conversely, substituting \(m+1\) instead of . Thus, it is enough to obtain the solution of one of these equations.

Let's try to get the energy spectrum of the Schrödinger equation for in Eq. (14). To do this, we write \({H_0}{\psi _m}={E_m}{\psi _m}\) as

$$- \frac{{{\hbar ^2}}}{{2\mu }}\left( {\frac{{{\partial ^2}}}{{\partial {\rho ^2}}}+\frac{1}{\rho }\frac{\partial }{{\partial \rho }}+\frac{1}{{{\rho ^2}}}\frac{{{\partial ^2}}}{{\partial {\phi ^2}}}+\frac{{{\partial ^2}}}{{\partial {z^2}}}} \right){\psi _m} - i\frac{{\hbar {\omega _c}}}{2}\frac{\partial }{{\partial \phi }}{\psi _m}+\frac{{\mu \omega _{c}^{2}{\rho ^2}}}{8}{\psi _m}={E_m}{\psi _m}$$

15

.

It has been explained in the studies of [20], [21] that in the quantum system with ellipsoidal geometry, the movement of the electron in the \(OZ\) direction acts as an effective potential in the \(XOY\) plane. Using these studies, let's write the Schrödinger equation as

$$- \frac{{{\hbar ^2}}}{{2\mu }}\frac{{{\partial ^2}}}{{\partial {z^2}}}{\chi _\rho }\left( z \right)={E_\chi }{\chi _\rho }\left( z \right)$$

16

for the "fast" subsystem of the electron in the ellipsoid quantum dot. Since the potential is zero in the ellipsoid according to Eq. (1) (\(U\left( {\rho ,z} \right)=0\)), it can be considered as a quantum system with an infinite well potential with a base of length \(L\left( \rho \right)\). Since the thickness of the ellipsoid varies according to \(\rho\) (as \(\rho\) increases, the thickness of the ellipsoid decreases) we can write \(L\left( \rho \right)\) as

$$L\left( \rho \right)={L_0}\sqrt {1 - \frac{{{\rho ^2}}}{{{R_0}^{2}}}}$$

17

.

Assuming that the probability of the electron being outside the ellipsoid is zero and accepting the boundary condition \({\chi _{_{\rho }}}\left( {z=L\left( \rho \right)} \right)=0\), we find the solution of Eq. (16) as

$${\chi _\rho }\left( z \right)=\sqrt {\frac{2}{{L\left( \rho \right)}}} \sin \left( {\frac{{n\pi z}}{{L\left( \rho \right)}}+n\pi } \right)$$

18

.

The \({E_\chi }\) energy spectrum of the electron is equal to

$${E_\chi }=\frac{{{\hbar ^2}{\pi ^2}{n^2}}}{{2\mu L{{\left( \rho \right)}^2}}}$$

19

.

Assuming that the electron is located in the \(\left| \rho \right|<<{R_0}\) region, let's expand the \(L\left( \rho \right)\) function in Eq. (19) to its second term using the Taylor method. Then let's substitute it into Eq. (19). Thus, we obtain

$${E_\chi }=\frac{{{\hbar ^2}}}{{2\mu }}\frac{{{n^2}{\pi ^2}}}{{L_{0}^{2}}}+\frac{{{\hbar ^2}}}{{2\mu }}\frac{{{n^2}{\pi ^2}{\rho ^2}}}{{R_{0}^{2}L_{0}^{2}}}=\frac{{{\hbar ^2}}}{{2\mu }}\left( {\gamma +{\beta ^2}{\rho ^2}} \right)$$

20

energy spectrum for the "fast" subsystem. Here, parameters

\(\gamma =\frac{{{\pi ^2}{n^2}}}{{{L_0}^{2}}}\) , \(\beta =\frac{{\pi n}}{{{R_0}{L_0}}}\) (21)

are defined. Here, \(n=0,1,2,...,\). The expression in Eq. (20) is added as an effective potential to the Schrödinger equation with variable \(\rho\) [20]. The Schrödinger equation depending on the variable \(\rho\) is written as

\(- \frac{{{\hbar ^2}}}{{2\mu }}\left( {\frac{{{\partial ^2}}}{{\partial {\rho ^2}}}+\frac{1}{\rho }\frac{\partial }{{\partial \rho }} - \frac{{{m^2}}}{{{\rho ^2}}}} \right){R_m}\left( \rho \right) - \frac{{2\mu }}{{{\hbar ^2}}}\frac{{\hbar {\omega _c}m}}{2}{R_m}\left( \rho \right)\)

$$- \frac{{\mu \omega _{c}^{2}{\rho ^2}}}{8}{R_m}\left( \rho \right) - \frac{{{\hbar ^2}}}{{2\mu }}\left( {\gamma +{\beta ^2}{\rho ^2}} \right){R_m}\left( \rho \right)+{E_m}{R_m}\left( \rho \right)=0$$

22

.

As a result of solving this equation, we obtain

$${R_m}\left( \rho \right)=\frac{1}{{\lambda \sqrt {2\pi } }}{\left( {\frac{{{n_\rho }!}}{{\Gamma \left( {{n_\rho }+m+1} \right)}}} \right)^{\frac{1}{2}}}{e^{ - \frac{{{\rho ^2}}}{{4{\lambda ^2}}}}}{\left( {\frac{1}{2}\frac{{{\rho ^2}}}{{{\lambda ^2}}}} \right)^{\frac{m}{2}}}L_{{{n_\rho }}}^{m}\left( {\frac{1}{2}\frac{{{\rho ^2}}}{{{\lambda ^2}}}} \right)$$

23

wave function and

$${E_m}=\frac{{{\hbar ^2}}}{{2\mu }}{\gamma ^2} - \frac{{\hbar {\omega _c}m}}{2}+\hbar \Omega \left( {{n_\rho }+\frac{{m+1}}{2}} \right)$$

24

energy spectrum. We obtain a similar solution for equation

$${H_0}{\psi _{m+1}}={E_{m+1}}{\psi _{m+1}}$$

25

,

that is, the expressions

$${R_{m+1}}\left( \rho \right)=\frac{1}{{\lambda \sqrt {2\pi } }}{\left( {\frac{{{n_\rho }!}}{{\Gamma \left( {{n_\rho }+\left( {m+1} \right)+1} \right)}}} \right)^{\frac{1}{2}}}{e^{ - \frac{{{\rho ^2}}}{{4{\lambda ^2}}}}}{\left( {\frac{1}{2}\frac{{{\rho ^2}}}{{{\lambda ^2}}}} \right)^{\frac{{m+1}}{2}}}L_{{{n_\rho }}}^{{m+1}}\left( {\frac{1}{2}\frac{{{\rho ^2}}}{{{\lambda ^2}}}} \right)$$

26

and

$${E_{m+1}}=\frac{{{\hbar ^2}}}{{2\mu }}{\gamma ^2} - \frac{{\hbar {\omega _c}\left( {m+1} \right)}}{2}+\hbar \Omega \left( {{n_\rho }+\frac{{m+2}}{2}} \right)$$

27

.

Here

$$\Omega =\sqrt {{\omega ^2}+\frac{{4{\hbar ^2}}}{{{\mu ^2}}}{\beta ^2}}$$

28

.

Let's write the expressions we obtained in Equations (23)–(24) and Equations (26)–(27) in Eq. (11). Afterwards, we obtain the expression

$$\left( {{E_m}+\frac{1}{2}{g^*}{\mu _B}B - E} \right){C_1}{\psi _m}+{\alpha _R}{e^{ - i\phi }}\left( {\frac{\partial }{{\partial \rho }} - \frac{i}{\rho }\frac{\partial }{{\partial \phi }}+\frac{{eB\rho }}{{2\hbar }}} \right){C_2}{\psi _{m+1}}=0$$

29

,

$$\left( {{E_{m+1}} - \frac{1}{2}{g^*}{\mu _B}B - E} \right){C_2}{\psi _{m+1}} - {\alpha _R}{e^{i\phi }}\left( {\frac{\partial }{{\partial \rho }}+\frac{i}{\rho }\frac{\partial }{{\partial \phi }} - \frac{{eB\rho }}{{2\hbar }}} \right){C_1}{\psi _m}=0$$

30

.

Then, we multiply Eq. (29) from the left to \(\psi _{m}^{*}\) and likewise multiply Eq. (30) from the left to \(\psi _{{m+1}}^{*}\). As a result of this process, we obtain the

$$\left( {{E_m}+\frac{1}{2}{g^*}{\mu _B}B - E} \right){C_1}\int\limits_{0}^{\infty } {R_{m}^{*}{R_m}\rho d\rho } +{\alpha _R}{C_2}\int\limits_{0}^{\infty } {R_{m}^{*}\left( {\frac{\partial }{{\partial \rho }}+\frac{{m+1}}{\rho }+\frac{{eB\rho }}{{2\hbar }}} \right){R_{m+1}}\rho d\rho } =0$$

31

,

$$\left( {{E_{m+1}} - \frac{1}{2}{g^*}{\mu _B}B - E} \right){C_2}\int\limits_{0}^{\infty } {R_{{m+1}}^{*}{R_{m+1}}\rho d\rho } +{\alpha _R}{C_1}\int\limits_{0}^{\infty } {R_{{m+1}}^{*}\left( { - \frac{\partial }{{\partial \rho }}+\frac{m}{\rho }+\frac{{eB\rho }}{{2\hbar }}} \right){R_m}\rho d\rho } =0$$

32

.

equations. Thus, as we see from Eq. (31) and Eq. (32), we obtain integral terms consisting of the products of the \({R_m}\) and \({R_{m+1}}\) functions and the products of the conjugates of these functions. Taking into account the expression [22]

\(\int\limits_{0}^{\infty } {{x^{\alpha - 1}}{e^{ - cx}}L_{m}^{\gamma }\left( {cx} \right)L_{n}^{\lambda }\left( {cx} \right)dx=\frac{{{{\left( {1+\gamma } \right)}_m}{{\left( {1 - \alpha +\lambda } \right)}_n}\Gamma \left( \alpha \right)}}{{m!n!{c^\alpha }}}}\)

$${}_{3}{F_2}\left( { - m,\alpha ,\alpha - \lambda ;1+\gamma ,\alpha - \lambda - n;1} \right)$$

33

,

we find the solution to the integrals in Eq. (31) and Eq. (32). After writing the solution to the integrals in Eq. (31) and Eq. (32), we obtain the following expression

$$\left( {\begin{array}{*{20}{c}} {\left( {{E_m}+\frac{1}{2}{g^*}{\mu _B}B - E} \right){\tau _1}}&{{\alpha _R}\left( {{\upsilon _1}+\left( {m+1} \right)\kappa +\frac{{eB}}{{2\hbar }}\eta } \right)} \\ {{\alpha _R}\left( { - {\upsilon _2}+m\kappa +\frac{{eB}}{{2\hbar }}\eta } \right)}&{\left( {{E_{m+1}} - \frac{1}{2}{g^*}{\mu _B}B - E} \right){\tau _2}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {{C_1}} \\ {{C_2}} \end{array}} \right)=0$$

34

.

Here,

\({\tau _1}={\lambda ^2}\frac{{\Gamma \left( {m+{n_\rho }+1} \right)}}{{{n_\rho }!}}\) , \({\tau _2}={\lambda ^2}\frac{{\Gamma \left( {m+{n_\rho }+2} \right)}}{{{n_\rho }!}}\) \(\kappa =\frac{\lambda }{{\sqrt 2 }}\frac{{{{\left( {1+m} \right)}_{{n_\rho }}}}}{{{n_\rho }!}}\Gamma \left( {m+1} \right)\), (35)

$$\eta ={\lambda ^3}\sqrt 2 \frac{{{{\left( {m+1} \right)}_{{n_\rho }}}{{\left( 0 \right)}_{{n_\rho }}}\Gamma \left( {m+2} \right)}}{{{{\left( {{n_\rho }!} \right)}^2}}}{}_{3}{F_2}\left( { - {n_\rho },m+2,1;m+1,1 - {n_\rho };1} \right)$$

36

\({\upsilon _1}=\frac{{\lambda \left( {m+1} \right)}}{{\sqrt 2 }}\frac{{{{\left( {m+1} \right)}_{{n_\rho }}}}}{{{n_\rho }!}}\Gamma \left( {m+1} \right)+ - \frac{{2\lambda }}{{\sqrt 2 }}\frac{{{{\left( {m+1} \right)}_{{n_\rho }}}}}{{{n_\rho }!}}\Gamma \left( {m+2} \right)\)

$$\frac{\lambda }{{\sqrt 2 }}\frac{{{{\left( {m+2} \right)}_{{n_\rho }}}{{\left( { - 1} \right)}_{{n_\rho }}}\Gamma \left( {m+2} \right)}}{{{{\left( {{n_\rho }!} \right)}^2}}}{}_{3}{F_2}\left( { - {n_\rho },m+2,2;m+2,2 - {n_\rho };1} \right)$$

37

\({\upsilon _2}=\frac{\lambda }{{\sqrt 2 }}\frac{{{{\left( {m+2} \right)}_{{n_\rho }}}{{\left( { - 1} \right)}_{{n_\rho }}}\Gamma \left( {m+2} \right)}}{{{{\left( {{n_\rho }!} \right)}^2}}}{}_{3}{F_2}\left( { - {n_\rho },m+2,2;m+2,2 - {n_\rho };1} \right)+\)

$$\frac{{\lambda m}}{{\sqrt 2 }}\frac{{{{\left( {m+1} \right)}_{{n_\rho }}}\Gamma \left( {m+1} \right)}}{{{n_\rho }!}} - \frac{{2\lambda }}{{\sqrt 2 }}\frac{{\Gamma \left( {m+{n_\rho }+2} \right)}}{{{n_\rho }!}}$$

38

,

From here, assuming that \({C_1}\) and \({C_2}\) are not equal to zero, we can write the matrix \(2 \times 2\) in Eq. (34) as

$$\left| {\begin{array}{*{20}{c}} {\left( {{E_m}+\frac{1}{2}{g^*}{\mu _B}B - E} \right){\tau _1}}&{{\alpha _R}\left( {{\upsilon _1}+\left( {m+1} \right)\kappa +\frac{{eB}}{{2\hbar }}\eta } \right)} \\ {{\alpha _R}\left( { - {\upsilon _2}+m\kappa +\frac{{eB}}{{2\hbar }}\eta } \right)}&{\left( {{E_{m+1}} - \frac{1}{2}{g^*}{\mu _B}B - E} \right){\tau _2}} \end{array}} \right|=0$$

39

.

Thus, by solving the above equation according to , we find the energy spectrum of the ellipsoidal quantum dot as

$${E_\sigma }=\frac{{\left( {{\varepsilon _m}+{\varepsilon _{m+1}}} \right)+\sigma \sqrt C }}{2}$$

40

.

Here \(\sigma = \pm 1\) and the parameters

\({\varepsilon _{m+1}}={E_{m+1}} - \frac{1}{2}{g^*}{\mu _B}B\) , \({\varepsilon _m}={E_m}+\frac{1}{2}{g^*}{\mu _B}B\) (41)

and

$$C={\left( {{\varepsilon _m} - {\varepsilon _{m+1}}} \right)^2}+4\frac{{{\alpha _R}^{2}}}{{{\tau _1}{\tau _2}}}\left( {m\kappa - {\upsilon _2}+\frac{{eB}}{{2\hbar }}\eta } \right)\left( {\left( {m+1} \right)\kappa +{\upsilon _1}+\frac{{eB}}{{2\hbar }}\eta } \right)$$

42

are defined. Finally, taking into account the operations we have done above, the wave function in Eq. (10) can be written as

$$\Psi =\left( {\begin{array}{*{20}{c}} {\cos {\theta _\sigma }{\psi _m}} \\ {\sin {\theta _\sigma }{\psi _{m+1}}} \end{array}} \right)$$

43

.

Here,

\(\cos {\theta _\sigma }=\sqrt {\frac{{{a^2}}}{{{a^2}+{t^2}}}}\) , \(\sin {\theta _\sigma }= - \sqrt {\frac{{{t^2}}}{{{a^2}+{t^2}}}}\) (44)

and

\(t={\alpha _R}\left( {{\upsilon _1}+\left( {m+1} \right)\kappa +\frac{{eB}}{{2\hbar }}\eta } \right)\) , \(a=\left( {{E_m}+\frac{1}{2}{g^*}{\mu _B}B - {E_\sigma }} \right){\tau _1}\) (45)

## 2.2. Absorption coefficient

In this part of our study, let's try to obtain the absorption coefficient expression for the intra-band optical transitions of the quantum system we are considering, taking into account the expressions we obtained in Eq. (40) and Eq. (43). According to Anselm, the absorption coefficient is defined as [23]

$$\alpha \left( \omega \right)=\frac{{4{\pi ^2}{e^2}}}{{{\varepsilon _0}c{n_0}{\mu _0}^{2}\omega V}}\sum\limits_{{if}} {{{\left| {{\mathbf{e}}{{\mathbf{p}}_{if}}} \right|}^2}} \left( {f\left( {{E_i}} \right) - f\left( {{E_f}} \right)} \right)\delta \left( {{E_f} - {E_i} - \hbar \omega } \right)$$

46

.

Here, \(\hbar \omega\) is the photon energy, \({\mathbf{e}}\) is the unit polarization vector, \({n_0}\) is the refractive index, \({\mu _0}\) is the effective mass of the electron, \({\varepsilon _0}\) is the electric permittivity constant, is the charge of the electron, is the speed of light and is the volume of the ellipsoid. The summation takes place over the initial and final states of the quantum system. Also \(f\left( E \right)\) is the Fermi-Dirac function and we assume \(f\left( {{E_i}} \right)=1\) and \(f\left( {{E_f}} \right)=0\). In Eq. (46), \({{\mathbf{p}}_{if}}\) is the momentum matrix element and for intraband optical transitions we can write it as [23]

$${{\mathbf{p}}_{if}}=\int\limits_{V} {\Psi _{f}^{*}{\mathbf{p}}{\Psi _i}d{\mathbf{r}}}$$

47

.

Here \({\Psi _i}\) and \({\Psi _f}\) are the wave functions before and after the optical transition, respectively. We assume that the polarization vector is directed along the axis. As a result of the scalar multiplication \({{\mathbf{p}}_{if}}\) with the polarized vector \({\mathbf{e}}\), we obtain

$${\mathbf{e}}{{\mathbf{p}}_{if}}= - i\hbar {e_z}\int\limits_{V} {\Psi _{f}^{*}\frac{d}{{dz}}{\Psi _i}\rho d\rho d\phi dz}$$

48

.

As a result of some mathematical calculations, the integral solution in Eq. (48) is found as

$${\mathbf{e}}{{\mathbf{p}}_{if}}= - i\hbar \left( {\Theta \Lambda Y} \right)\left( {\frac{M}{K}+\frac{G}{W}} \right)$$

49

.

Here,

$$Y=\frac{1}{{2\pi \lambda \lambda '}}{\left( {\frac{{{n_\rho }'!{n_\rho }!}}{{\Gamma \left( {{n_\rho }'+m'+1} \right)\Gamma \left( {{n_\rho }+m+1} \right)}}} \right)^{\frac{1}{2}}}$$

50

,

\(K=\frac{1}{{\cos {\theta ^{\sigma f}}\cos {\theta ^{\sigma i}}}}\) , \(W=\frac{1}{{\sin {\theta ^{\sigma f}}\sin {\theta ^{\sigma i}}}}\) (51)

\(\lambda =\sqrt {\frac{\hbar }{{\mu \sqrt {\omega _{c}^{2}+\frac{{4{\hbar ^2}{\pi ^2}{n^2}}}{{{L_0}^{2}{R_0}^{2}{\mu ^2}}}} }}}\) , \(\lambda '=\sqrt {\frac{\hbar }{{\mu \sqrt {\omega _{c}^{2}+\frac{{4{\hbar ^2}{\pi ^2}n{'^2}}}{{{L_0}^{2}{R_0}^{2}{\mu ^2}}}} }}}\), (52)

$$\Theta =\int\limits_{0}^{{2\pi }} {{e^{i\left( { - m'+m} \right)\phi }}d\phi } =2\pi {\delta _{mm'}}$$

53

,

$$\Lambda =\frac{2}{{{L_0}}}\int\limits_{0}^{{{L_0}}} {\sin \left( {\frac{{\pi zn'}}{{{L_0}}}+\pi n'} \right)\frac{d}{{dz}}\sin \left( {\frac{{\pi zn}}{{{L_0}}}+\pi n} \right)dz}$$

54

.

In these expressions \({n_\rho },m,n\) and \({n_\rho }',m',n'\) are quantum numbers before and after the optical transition, respectively.

Equtaion, (53) and (54) establish the rule for selecting optical transitions of our quantum system. As can be seen from these expressions, optical transitions are possible only in cases where \(m=m'\) and \(\left| {n - n'} \right|=odd\).

\(M=\frac{1}{2}{\left( {d'd} \right)^{\frac{m}{2}}}\frac{{{{\left( {b - d'} \right)}^{{n_\rho }'}}{{\left( {b - d} \right)}^{{n_\rho }}}}}{{{b^{{n_\rho }'+{n_\rho }+m+1}}}}\frac{{\Gamma \left( {{n_\rho }+{n_\rho }'+m+1} \right)}}{{{n_\rho }!{n_\rho }'!}} \times\)

$$F\left( { - {n_\rho }, - {n_\rho }'; - {n_\rho } - {n_\rho }' - m;\frac{{b\left( {b - d' - d} \right)}}{{\left( {b - d'} \right)\left( {b - d} \right)}}} \right)$$

55

,

\(G=\frac{1}{2}{\left( {d'd} \right)^{\frac{{m+1}}{2}}}\frac{{{{\left( {b - d'} \right)}^{{n_\rho }'}}{{\left( {b - d} \right)}^{{n_\rho }}}}}{{{b^{{n_\rho }'+{n_\rho }+m+2}}}}\frac{{\Gamma \left( {{n_\rho }+{n_\rho }'+m+2} \right)}}{{{n_\rho }!{n_\rho }'!}} \times\)

$$F\left( { - {n_\rho }, - {n_\rho }'; - {n_\rho } - {n_\rho }' - m - 1;\frac{{b\left( {b - d' - d} \right)}}{{\left( {b - d'} \right)\left( {b - d} \right)}}} \right)$$

56

,

Here,

\(b=\frac{1}{2}\left( {d'+d} \right)\) , \(d'=\frac{1}{{2\lambda {'^2}}}\), \(d=\frac{1}{{2{\lambda ^2}}}\). (57)

Equations (55) and (56) were obtained based on the fact that \(m=m'\) and taking into account the integral expression [22]

\(\int\limits_{0}^{\infty } {{e^{bx}}{x^\alpha }L_{n}^{\alpha }} \left( {\lambda x} \right)L_{m}^{\alpha }\left( {\mu x} \right)=\frac{{\Gamma \left( {m+n+\alpha +1} \right)}}{{m!n!}}\frac{{{{\left( {b - \lambda } \right)}^n}{{\left( {b - \mu } \right)}^m}}}{{{b^{m+n+\alpha +1}}}} \times\)

$$F\left( { - m, - n; - m - n - \alpha ;\frac{{b\left( {b - \lambda - \mu } \right)}}{{\left( {b - \lambda } \right)\left( {b - \mu } \right)}}} \right)$$

58

.

In Eq. (49), we defined the momentum matrix element. As can be seen from equations (53) and (54), this expression determines the allowed and forbidden optical transitions between the initial and final quantum states of an ellipsoidal quantum dot.Next we can write the Dirac delta function \(\delta \left( {{E_f} - {E_i} - \hbar \omega } \right)\) in Eq. (46) as [12]

$$\delta \left( {{E_\sigma }} \right) \approx \frac{1}{\pi }\frac{\Gamma }{{E_{\sigma }^{2}+{\Gamma ^2}}}$$

59

.

Here, \(\Gamma\) is the relaxation rate constant in units of \(eV\), which expresses the time required for the system to reach thermodynamic equilibrium again after the optical transition. Also,

$${E_\sigma }=E_{\sigma }^{f} - E_{\sigma }^{i} - \hbar \omega$$

60

.

Here \(E_{\sigma }^{f}\) and \(E_{\sigma }^{i}\) are the energy spectrum of the electron before and after the optical transition, respectively. Finally, considering Eq. (49) and Eq. (59), we can write the absorption coefficient in Eq. (46) as

$$\alpha \left( {\hbar \omega } \right)=\frac{{4\pi {e^2}{\hbar ^3}}}{{{\varepsilon _0}c{n_0}{\mu _0}^{2}\hbar \omega V}}{\left( {\Theta \Lambda Y} \right)^2}{\left( {\frac{M}{K}+\frac{G}{W}} \right)^2}\frac{\Gamma }{{\Delta E_{\sigma }^{2}+{\Gamma ^2}}}$$

61

for the case of intra-band optical transitions.

## 2.1. Energy spectrum and wave function

The potential of the quantum system under consideration is defined as

\(U\left( {\rho ,z} \right)=\left\{ {\begin{array}{*{20}{c}} {0,\frac{{{\rho ^2}}}{{R_{0}^{2}}}+\frac{{{z^2}}}{{L_{0}^{2}}} \leqslant 1} \\ {\infty ,\frac{{{\rho ^2}}}{{R_{0}^{2}}}+\frac{{{z^2}}}{{L_{0}^{2}}}>1} \end{array},{R_0} \gg {L_0}} \right\}\) . (1)

The potential we define has ellipsoid geometry, and it is accepted that the value of the potential is zero inside the ellipsoid, and the potential is infinite outside the ellipsoid. The electron, which we limit with its potential known as the hard wall potential, is confined in this quantum system. In the potential, the \({R_0}\) parameter is defined in the \(XOY\) plane and determines the radius of the ellipsoid and is defined as \(\rho =\sqrt {{x^2}+{y^2}}\); The \({L_0}\) parameter is defined in the \(OZ\) direction and determines the thickness of the ellipsoid. The thickness of the ellipsoid reaches its maximum value at \(\rho =0\), that is, at the center of the ellipsoid, and as \(\rho \to {R_0}\) approaches, the thickness of the ellipsoid gradually approaches zero [20].

In Eq. (1), we assume that \({R_0} \gg {L_0}\). This condition tells us that the diameter of the ellipsoid is much larger than the height of the ellipsoid. In such a system, the movement of the electron in the \(XOY\) plane can be considered as an adiabatic distortion in the \(OZ\) direction. That is, it is assumed that the state change of the electron in the \(OZ\) direction changes very quickly compared to the state change in the \(XOY\) plane. Taking this into account, we can divide the Hamilton of our system under consideration into two subsystems. Let's define the change of the electron in the \(OZ\) direction as the "fast" subsystem, and the change in the \(XOY\) plane as the "slow" subsystem. The wave function of the system we have defined can be written as

$$\psi \left( {\rho ,\phi ,z} \right)={e^{im\phi }}R\left( \rho \right){\chi _\rho }\left( z \right)$$

2

,

depending on the variables \(\rho\) and . Here, is the magnetic quantum number and takes \(0, \pm 1, \pm 2,..\) values, \(R\left( \rho \right)\) determines the state of the electron in the \(XOY\) plane, and \({\chi _\rho }\left( z \right)\) determines the state of the electron in the \(OZ\) direction.

Considering what we wrote above, let's write the Hamiltonian expression of the quantum system. If we look at Eq. (1), the potential value is zero inside the ellipsoid. Therefore, the motion of the electron is assumed to be free within the ellipsoid. In addition, we assume that the free electron in the ellipsoid is influenced by the Rashba interaction and the DMS-induced exchange interaction. We also assume that a constant magnetic field is applied to the ellipsoid in the \(OZ\) direction. Accordingly, the Hamilton expression of the system is written as

$$H=\frac{1}{{2\mu }}{\left( {{\mathbf{p}}+e{\mathbf{A}}} \right)^2}+U\left( {\rho ,z} \right)+{H_R}+{H_Z}+{H_{DMS}}$$

3

.

Here, \({\mathbf{p}}= - i\hbar \nabla\) is the momentum operator; \({\mathbf{A}}\) is a vector potential, \({\mathbf{A}}=\left( {0,\frac{{B\rho }}{2},0} \right)\); \(\mu\) is the effective mass of the electron. In the Hamiltonian expression, the term \({H_R}\) is the Rashba term and is written as

Here, \({\mathbf{\sigma }}\) is Puali matrices and \({\alpha _R}\) Rashba parameter. The expression is unit vector and is taken parallel to the growth direction of the structure. Let's assume that the growth direction of the system under consideration is in the direction. Thus, as a result of vector multiplication in the Rashba term, only the terms related to the direction are different from zero. We can write this as

$${H_R}=\frac{{i{\alpha _R}}}{\hbar }\left( { - {\sigma _x}\sin \phi +{\sigma _y}\cos \phi } \right)\frac{\partial }{{\partial \rho }} - \frac{{{\alpha _R}}}{{\hbar \rho }}\left( {{\sigma _x}\cos \phi +{\sigma _y}\sin \phi } \right)\left( {i\frac{\partial }{{\partial \phi }}+\frac{\Phi }{{{\Phi _0}}}} \right)$$

5

.

Here, \(\Phi =\pi {\rho ^2}B\) is the magnetic flux and \({\Phi _0}=\frac{{2\pi \hbar }}{e}\) is the magnetic flux constant. The fourth term in the Hamilton expression is the Zeeman term, written as

$${H_Z}=\frac{1}{2}g{\mu _B}{B_z}{\sigma _z}$$

6

.

Here, is the Lande factor, \({\mu _B}\) is the Bohr magneton, \({\sigma _z}\) is the -direction component of the Pauli matrix. The last term in Hamilton is the term expressing the exchange interaction resulting from the DMS feature of the structure and is written as [9]

$${H_{DMS}}=\frac{1}{2}x{N_0}\tau \left\langle {{S_z}} \right\rangle {\sigma _z}$$

7

.

Also here,

$$\tau =\left\langle {X\left| J \right|X} \right\rangle =\left\langle {Y\left| J \right|Y} \right\rangle =\left\langle {Z\left| J \right|Z} \right\rangle$$

8

Equation (7) defines the \(sp - d\) exchange interaction that occurs between the \(3{d^5}\) electron of the \(M{n^{2+}}\) ion and the \(sp\)-band electron in DMS structures formed under the applied constant magnetic effect. Here, \({J_{s - d}}\) is the exchange interaction coefficient; The integral \(\tau\) express the exchange interaction integrals between the \(s - d\) and \(p - d\) bands, respectively; \({N_0}\) density parameter per cell; The \(\left\langle {{S_z}} \right\rangle\) parameter is the thermodynamic average of the component of the spin of the \(Mn\) atom and is written as [9]

$$\left\langle {{S_z}} \right\rangle = - S{B_S}\left( {\frac{{S{g_{Mn}}{\mu _B}B}}{{{k_B}T}}} \right)$$

9

.

Here, \({B_{5/2}}\left( \xi \right)\) is the Brillouin function, \({g_{Mn}}=2\) is the Lande factor of the atom, \({k_B}\) is the Boltzmann constant; temperature; \(S=\frac{5}{2}\) is the total spin value of the \(M{n^{2+}}\) ion. Let's define the wave function corresponding to Hamilton that we defined in Eq. (3) as

$$\Psi =\left( {\begin{array}{*{20}{c}} {{\psi _1}} \\ {{\psi _2}} \end{array}} \right)$$

10

.

After this, we write the Schrödinger Equation in the following form

$$\left( {\begin{array}{*{20}{c}} {{H_0}+\frac{1}{2}{\mu _B}B{g^*} - E}&{{\alpha _R}{e^{ - i\phi }}\left( {\frac{\partial }{{\partial \rho }} - \frac{1}{\rho }\left( {i\frac{\partial }{{\partial \phi }}+\frac{\Phi }{{{\Phi _0}}}} \right)} \right)} \\ { - {\alpha _R}{e^{i\phi }}\left( {\frac{\partial }{{\partial \rho }}+\frac{1}{\rho }\left( {i\frac{\partial }{{\partial \phi }}+\frac{\Phi }{{{\Phi _0}}}} \right)} \right)}&{{H_0} - \frac{1}{2}{\mu _B}B{g^*} - E} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {{C_1}{\psi _1}} \\ {{C_2}{\psi _2}} \end{array}} \right)=0$$

11

.

Here,

$${H_0}= - \frac{{{\hbar ^2}}}{{2\mu }}\left( {\frac{{{\partial ^2}}}{{\partial {\rho ^2}}}+\frac{1}{\rho }\frac{\partial }{{\partial \rho }}+\frac{1}{{{\rho ^2}}}\frac{{{\partial ^2}}}{{\partial {\phi ^2}}}+\frac{{{\partial ^2}}}{{\partial {z^2}}}} \right) - i\frac{{\hbar {\omega _c}}}{2}\frac{\partial }{{\partial \phi }}+\frac{{\mu \omega _{c}^{2}{\rho ^2}}}{8}$$

12

is the Hamiltonian expression that does not include perturbation terms. Let's define the wave functions \({\psi _1}\) and \({\psi _2}\) in Eq. (11) as

\({\psi _m}={e^{im\phi }}{R_m}\left( \rho \right){\chi _\rho }\left( z \right)\) , \({\psi _{m+1}}={e^{i\left( {m+1} \right)\phi }}{R_{m+1}}\left( \rho \right){\chi _\rho }\left( z \right)\) (13)

respectively. Then, taking into account the wave functions we defined, let's write the Schrödinger equation as

\({H_0}{\psi _m}={E_m}{\psi _m}\) , \({H_0}{\psi _{m+1}}={E_{m+1}}{\psi _{m+1}}\) (14)

for Hamilton \({H_0}\) in Eq. (12). As can be seen from these two equations, the energy spectra in the equations differ from each other only according to the quantum number . If we find the energy spectrum of one of these equations, the energy spectrum of the other can be found by substituting instead of \(m+1\) or, conversely, substituting \(m+1\) instead of . Thus, it is enough to obtain the solution of one of these equations.

Let's try to get the energy spectrum of the Schrödinger equation for in Eq. (14). To do this, we write \({H_0}{\psi _m}={E_m}{\psi _m}\) as

$$- \frac{{{\hbar ^2}}}{{2\mu }}\left( {\frac{{{\partial ^2}}}{{\partial {\rho ^2}}}+\frac{1}{\rho }\frac{\partial }{{\partial \rho }}+\frac{1}{{{\rho ^2}}}\frac{{{\partial ^2}}}{{\partial {\phi ^2}}}+\frac{{{\partial ^2}}}{{\partial {z^2}}}} \right){\psi _m} - i\frac{{\hbar {\omega _c}}}{2}\frac{\partial }{{\partial \phi }}{\psi _m}+\frac{{\mu \omega _{c}^{2}{\rho ^2}}}{8}{\psi _m}={E_m}{\psi _m}$$

15

.

It has been explained in the studies of [20], [21] that in the quantum system with ellipsoidal geometry, the movement of the electron in the \(OZ\) direction acts as an effective potential in the \(XOY\) plane. Using these studies, let's write the Schrödinger equation as

$$- \frac{{{\hbar ^2}}}{{2\mu }}\frac{{{\partial ^2}}}{{\partial {z^2}}}{\chi _\rho }\left( z \right)={E_\chi }{\chi _\rho }\left( z \right)$$

16

for the "fast" subsystem of the electron in the ellipsoid quantum dot. Since the potential is zero in the ellipsoid according to Eq. (1) (\(U\left( {\rho ,z} \right)=0\)), it can be considered as a quantum system with an infinite well potential with a base of length \(L\left( \rho \right)\). Since the thickness of the ellipsoid varies according to \(\rho\) (as \(\rho\) increases, the thickness of the ellipsoid decreases) we can write \(L\left( \rho \right)\) as

$$L\left( \rho \right)={L_0}\sqrt {1 - \frac{{{\rho ^2}}}{{{R_0}^{2}}}}$$

17

.

Assuming that the probability of the electron being outside the ellipsoid is zero and accepting the boundary condition \({\chi _{_{\rho }}}\left( {z=L\left( \rho \right)} \right)=0\), we find the solution of Eq. (16) as

$${\chi _\rho }\left( z \right)=\sqrt {\frac{2}{{L\left( \rho \right)}}} \sin \left( {\frac{{n\pi z}}{{L\left( \rho \right)}}+n\pi } \right)$$

18

.

The \({E_\chi }\) energy spectrum of the electron is equal to

$${E_\chi }=\frac{{{\hbar ^2}{\pi ^2}{n^2}}}{{2\mu L{{\left( \rho \right)}^2}}}$$

19

.

Assuming that the electron is located in the \(\left| \rho \right|<<{R_0}\) region, let's expand the \(L\left( \rho \right)\) function in Eq. (19) to its second term using the Taylor method. Then let's substitute it into Eq. (19). Thus, we obtain

$${E_\chi }=\frac{{{\hbar ^2}}}{{2\mu }}\frac{{{n^2}{\pi ^2}}}{{L_{0}^{2}}}+\frac{{{\hbar ^2}}}{{2\mu }}\frac{{{n^2}{\pi ^2}{\rho ^2}}}{{R_{0}^{2}L_{0}^{2}}}=\frac{{{\hbar ^2}}}{{2\mu }}\left( {\gamma +{\beta ^2}{\rho ^2}} \right)$$

20

energy spectrum for the "fast" subsystem. Here, parameters

\(\gamma =\frac{{{\pi ^2}{n^2}}}{{{L_0}^{2}}}\) , \(\beta =\frac{{\pi n}}{{{R_0}{L_0}}}\) (21)

are defined. Here, \(n=0,1,2,...,\). The expression in Eq. (20) is added as an effective potential to the Schrödinger equation with variable \(\rho\) [20]. The Schrödinger equation depending on the variable \(\rho\) is written as

\(- \frac{{{\hbar ^2}}}{{2\mu }}\left( {\frac{{{\partial ^2}}}{{\partial {\rho ^2}}}+\frac{1}{\rho }\frac{\partial }{{\partial \rho }} - \frac{{{m^2}}}{{{\rho ^2}}}} \right){R_m}\left( \rho \right) - \frac{{2\mu }}{{{\hbar ^2}}}\frac{{\hbar {\omega _c}m}}{2}{R_m}\left( \rho \right)\)

$$- \frac{{\mu \omega _{c}^{2}{\rho ^2}}}{8}{R_m}\left( \rho \right) - \frac{{{\hbar ^2}}}{{2\mu }}\left( {\gamma +{\beta ^2}{\rho ^2}} \right){R_m}\left( \rho \right)+{E_m}{R_m}\left( \rho \right)=0$$

22

.

As a result of solving this equation, we obtain

$${R_m}\left( \rho \right)=\frac{1}{{\lambda \sqrt {2\pi } }}{\left( {\frac{{{n_\rho }!}}{{\Gamma \left( {{n_\rho }+m+1} \right)}}} \right)^{\frac{1}{2}}}{e^{ - \frac{{{\rho ^2}}}{{4{\lambda ^2}}}}}{\left( {\frac{1}{2}\frac{{{\rho ^2}}}{{{\lambda ^2}}}} \right)^{\frac{m}{2}}}L_{{{n_\rho }}}^{m}\left( {\frac{1}{2}\frac{{{\rho ^2}}}{{{\lambda ^2}}}} \right)$$

23

wave function and

$${E_m}=\frac{{{\hbar ^2}}}{{2\mu }}{\gamma ^2} - \frac{{\hbar {\omega _c}m}}{2}+\hbar \Omega \left( {{n_\rho }+\frac{{m+1}}{2}} \right)$$

24

energy spectrum. We obtain a similar solution for equation

$${H_0}{\psi _{m+1}}={E_{m+1}}{\psi _{m+1}}$$

25

,

that is, the expressions

$${R_{m+1}}\left( \rho \right)=\frac{1}{{\lambda \sqrt {2\pi } }}{\left( {\frac{{{n_\rho }!}}{{\Gamma \left( {{n_\rho }+\left( {m+1} \right)+1} \right)}}} \right)^{\frac{1}{2}}}{e^{ - \frac{{{\rho ^2}}}{{4{\lambda ^2}}}}}{\left( {\frac{1}{2}\frac{{{\rho ^2}}}{{{\lambda ^2}}}} \right)^{\frac{{m+1}}{2}}}L_{{{n_\rho }}}^{{m+1}}\left( {\frac{1}{2}\frac{{{\rho ^2}}}{{{\lambda ^2}}}} \right)$$

26

and

$${E_{m+1}}=\frac{{{\hbar ^2}}}{{2\mu }}{\gamma ^2} - \frac{{\hbar {\omega _c}\left( {m+1} \right)}}{2}+\hbar \Omega \left( {{n_\rho }+\frac{{m+2}}{2}} \right)$$

27

.

Here

$$\Omega =\sqrt {{\omega ^2}+\frac{{4{\hbar ^2}}}{{{\mu ^2}}}{\beta ^2}}$$

28

.

Let's write the expressions we obtained in Equations (23)–(24) and Equations (26)–(27) in Eq. (11). Afterwards, we obtain the expression

$$\left( {{E_m}+\frac{1}{2}{g^*}{\mu _B}B - E} \right){C_1}{\psi _m}+{\alpha _R}{e^{ - i\phi }}\left( {\frac{\partial }{{\partial \rho }} - \frac{i}{\rho }\frac{\partial }{{\partial \phi }}+\frac{{eB\rho }}{{2\hbar }}} \right){C_2}{\psi _{m+1}}=0$$

29

,

$$\left( {{E_{m+1}} - \frac{1}{2}{g^*}{\mu _B}B - E} \right){C_2}{\psi _{m+1}} - {\alpha _R}{e^{i\phi }}\left( {\frac{\partial }{{\partial \rho }}+\frac{i}{\rho }\frac{\partial }{{\partial \phi }} - \frac{{eB\rho }}{{2\hbar }}} \right){C_1}{\psi _m}=0$$

30

.

Then, we multiply Eq. (29) from the left to \(\psi _{m}^{*}\) and likewise multiply Eq. (30) from the left to \(\psi _{{m+1}}^{*}\). As a result of this process, we obtain the

$$\left( {{E_m}+\frac{1}{2}{g^*}{\mu _B}B - E} \right){C_1}\int\limits_{0}^{\infty } {R_{m}^{*}{R_m}\rho d\rho } +{\alpha _R}{C_2}\int\limits_{0}^{\infty } {R_{m}^{*}\left( {\frac{\partial }{{\partial \rho }}+\frac{{m+1}}{\rho }+\frac{{eB\rho }}{{2\hbar }}} \right){R_{m+1}}\rho d\rho } =0$$

31

,

$$\left( {{E_{m+1}} - \frac{1}{2}{g^*}{\mu _B}B - E} \right){C_2}\int\limits_{0}^{\infty } {R_{{m+1}}^{*}{R_{m+1}}\rho d\rho } +{\alpha _R}{C_1}\int\limits_{0}^{\infty } {R_{{m+1}}^{*}\left( { - \frac{\partial }{{\partial \rho }}+\frac{m}{\rho }+\frac{{eB\rho }}{{2\hbar }}} \right){R_m}\rho d\rho } =0$$

32

.

equations. Thus, as we see from Eq. (31) and Eq. (32), we obtain integral terms consisting of the products of the \({R_m}\) and \({R_{m+1}}\) functions and the products of the conjugates of these functions. Taking into account the expression [22]

\(\int\limits_{0}^{\infty } {{x^{\alpha - 1}}{e^{ - cx}}L_{m}^{\gamma }\left( {cx} \right)L_{n}^{\lambda }\left( {cx} \right)dx=\frac{{{{\left( {1+\gamma } \right)}_m}{{\left( {1 - \alpha +\lambda } \right)}_n}\Gamma \left( \alpha \right)}}{{m!n!{c^\alpha }}}}\)

$${}_{3}{F_2}\left( { - m,\alpha ,\alpha - \lambda ;1+\gamma ,\alpha - \lambda - n;1} \right)$$

33

,

we find the solution to the integrals in Eq. (31) and Eq. (32). After writing the solution to the integrals in Eq. (31) and Eq. (32), we obtain the following expression

$$\left( {\begin{array}{*{20}{c}} {\left( {{E_m}+\frac{1}{2}{g^*}{\mu _B}B - E} \right){\tau _1}}&{{\alpha _R}\left( {{\upsilon _1}+\left( {m+1} \right)\kappa +\frac{{eB}}{{2\hbar }}\eta } \right)} \\ {{\alpha _R}\left( { - {\upsilon _2}+m\kappa +\frac{{eB}}{{2\hbar }}\eta } \right)}&{\left( {{E_{m+1}} - \frac{1}{2}{g^*}{\mu _B}B - E} \right){\tau _2}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {{C_1}} \\ {{C_2}} \end{array}} \right)=0$$

34

.

Here,

\({\tau _1}={\lambda ^2}\frac{{\Gamma \left( {m+{n_\rho }+1} \right)}}{{{n_\rho }!}}\) , \({\tau _2}={\lambda ^2}\frac{{\Gamma \left( {m+{n_\rho }+2} \right)}}{{{n_\rho }!}}\) \(\kappa =\frac{\lambda }{{\sqrt 2 }}\frac{{{{\left( {1+m} \right)}_{{n_\rho }}}}}{{{n_\rho }!}}\Gamma \left( {m+1} \right)\), (35)

$$\eta ={\lambda ^3}\sqrt 2 \frac{{{{\left( {m+1} \right)}_{{n_\rho }}}{{\left( 0 \right)}_{{n_\rho }}}\Gamma \left( {m+2} \right)}}{{{{\left( {{n_\rho }!} \right)}^2}}}{}_{3}{F_2}\left( { - {n_\rho },m+2,1;m+1,1 - {n_\rho };1} \right)$$

36

\({\upsilon _1}=\frac{{\lambda \left( {m+1} \right)}}{{\sqrt 2 }}\frac{{{{\left( {m+1} \right)}_{{n_\rho }}}}}{{{n_\rho }!}}\Gamma \left( {m+1} \right)+ - \frac{{2\lambda }}{{\sqrt 2 }}\frac{{{{\left( {m+1} \right)}_{{n_\rho }}}}}{{{n_\rho }!}}\Gamma \left( {m+2} \right)\)

$$\frac{\lambda }{{\sqrt 2 }}\frac{{{{\left( {m+2} \right)}_{{n_\rho }}}{{\left( { - 1} \right)}_{{n_\rho }}}\Gamma \left( {m+2} \right)}}{{{{\left( {{n_\rho }!} \right)}^2}}}{}_{3}{F_2}\left( { - {n_\rho },m+2,2;m+2,2 - {n_\rho };1} \right)$$

37

\({\upsilon _2}=\frac{\lambda }{{\sqrt 2 }}\frac{{{{\left( {m+2} \right)}_{{n_\rho }}}{{\left( { - 1} \right)}_{{n_\rho }}}\Gamma \left( {m+2} \right)}}{{{{\left( {{n_\rho }!} \right)}^2}}}{}_{3}{F_2}\left( { - {n_\rho },m+2,2;m+2,2 - {n_\rho };1} \right)+\)

$$\frac{{\lambda m}}{{\sqrt 2 }}\frac{{{{\left( {m+1} \right)}_{{n_\rho }}}\Gamma \left( {m+1} \right)}}{{{n_\rho }!}} - \frac{{2\lambda }}{{\sqrt 2 }}\frac{{\Gamma \left( {m+{n_\rho }+2} \right)}}{{{n_\rho }!}}$$

38

,

From here, assuming that \({C_1}\) and \({C_2}\) are not equal to zero, we can write the matrix \(2 \times 2\) in Eq. (34) as

$$\left| {\begin{array}{*{20}{c}} {\left( {{E_m}+\frac{1}{2}{g^*}{\mu _B}B - E} \right){\tau _1}}&{{\alpha _R}\left( {{\upsilon _1}+\left( {m+1} \right)\kappa +\frac{{eB}}{{2\hbar }}\eta } \right)} \\ {{\alpha _R}\left( { - {\upsilon _2}+m\kappa +\frac{{eB}}{{2\hbar }}\eta } \right)}&{\left( {{E_{m+1}} - \frac{1}{2}{g^*}{\mu _B}B - E} \right){\tau _2}} \end{array}} \right|=0$$

39

.

Thus, by solving the above equation according to , we find the energy spectrum of the ellipsoidal quantum dot as

$${E_\sigma }=\frac{{\left( {{\varepsilon _m}+{\varepsilon _{m+1}}} \right)+\sigma \sqrt C }}{2}$$

40

.

Here \(\sigma = \pm 1\) and the parameters

\({\varepsilon _{m+1}}={E_{m+1}} - \frac{1}{2}{g^*}{\mu _B}B\) , \({\varepsilon _m}={E_m}+\frac{1}{2}{g^*}{\mu _B}B\) (41)

and

$$C={\left( {{\varepsilon _m} - {\varepsilon _{m+1}}} \right)^2}+4\frac{{{\alpha _R}^{2}}}{{{\tau _1}{\tau _2}}}\left( {m\kappa - {\upsilon _2}+\frac{{eB}}{{2\hbar }}\eta } \right)\left( {\left( {m+1} \right)\kappa +{\upsilon _1}+\frac{{eB}}{{2\hbar }}\eta } \right)$$

42

are defined. Finally, taking into account the operations we have done above, the wave function in Eq. (10) can be written as

$$\Psi =\left( {\begin{array}{*{20}{c}} {\cos {\theta _\sigma }{\psi _m}} \\ {\sin {\theta _\sigma }{\psi _{m+1}}} \end{array}} \right)$$

43

.

Here,

\(\cos {\theta _\sigma }=\sqrt {\frac{{{a^2}}}{{{a^2}+{t^2}}}}\) , \(\sin {\theta _\sigma }= - \sqrt {\frac{{{t^2}}}{{{a^2}+{t^2}}}}\) (44)

and

\(t={\alpha _R}\left( {{\upsilon _1}+\left( {m+1} \right)\kappa +\frac{{eB}}{{2\hbar }}\eta } \right)\) , \(a=\left( {{E_m}+\frac{1}{2}{g^*}{\mu _B}B - {E_\sigma }} \right){\tau _1}\) (45)

## 2.2. Absorption coefficient

In this part of our study, let's try to obtain the absorption coefficient expression for the intra-band optical transitions of the quantum system we are considering, taking into account the expressions we obtained in Eq. (40) and Eq. (43). According to Anselm, the absorption coefficient is defined as [23]

$$\alpha \left( \omega \right)=\frac{{4{\pi ^2}{e^2}}}{{{\varepsilon _0}c{n_0}{\mu _0}^{2}\omega V}}\sum\limits_{{if}} {{{\left| {{\mathbf{e}}{{\mathbf{p}}_{if}}} \right|}^2}} \left( {f\left( {{E_i}} \right) - f\left( {{E_f}} \right)} \right)\delta \left( {{E_f} - {E_i} - \hbar \omega } \right)$$

46

.

Here, \(\hbar \omega\) is the photon energy, \({\mathbf{e}}\) is the unit polarization vector, \({n_0}\) is the refractive index, \({\mu _0}\) is the effective mass of the electron, \({\varepsilon _0}\) is the electric permittivity constant, is the charge of the electron, is the speed of light and is the volume of the ellipsoid. The summation takes place over the initial and final states of the quantum system. Also \(f\left( E \right)\) is the Fermi-Dirac function and we assume \(f\left( {{E_i}} \right)=1\) and \(f\left( {{E_f}} \right)=0\). In Eq. (46), \({{\mathbf{p}}_{if}}\) is the momentum matrix element and for intraband optical transitions we can write it as [23]

$${{\mathbf{p}}_{if}}=\int\limits_{V} {\Psi _{f}^{*}{\mathbf{p}}{\Psi _i}d{\mathbf{r}}}$$

47

.

Here \({\Psi _i}\) and \({\Psi _f}\) are the wave functions before and after the optical transition, respectively. We assume that the polarization vector is directed along the axis. As a result of the scalar multiplication \({{\mathbf{p}}_{if}}\) with the polarized vector \({\mathbf{e}}\), we obtain

$${\mathbf{e}}{{\mathbf{p}}_{if}}= - i\hbar {e_z}\int\limits_{V} {\Psi _{f}^{*}\frac{d}{{dz}}{\Psi _i}\rho d\rho d\phi dz}$$

48

.

As a result of some mathematical calculations, the integral solution in Eq. (48) is found as

$${\mathbf{e}}{{\mathbf{p}}_{if}}= - i\hbar \left( {\Theta \Lambda Y} \right)\left( {\frac{M}{K}+\frac{G}{W}} \right)$$

49

.

Here,

$$Y=\frac{1}{{2\pi \lambda \lambda '}}{\left( {\frac{{{n_\rho }'!{n_\rho }!}}{{\Gamma \left( {{n_\rho }'+m'+1} \right)\Gamma \left( {{n_\rho }+m+1} \right)}}} \right)^{\frac{1}{2}}}$$

50

,

\(K=\frac{1}{{\cos {\theta ^{\sigma f}}\cos {\theta ^{\sigma i}}}}\) , \(W=\frac{1}{{\sin {\theta ^{\sigma f}}\sin {\theta ^{\sigma i}}}}\) (51)

\(\lambda =\sqrt {\frac{\hbar }{{\mu \sqrt {\omega _{c}^{2}+\frac{{4{\hbar ^2}{\pi ^2}{n^2}}}{{{L_0}^{2}{R_0}^{2}{\mu ^2}}}} }}}\) , \(\lambda '=\sqrt {\frac{\hbar }{{\mu \sqrt {\omega _{c}^{2}+\frac{{4{\hbar ^2}{\pi ^2}n{'^2}}}{{{L_0}^{2}{R_0}^{2}{\mu ^2}}}} }}}\), (52)

$$\Theta =\int\limits_{0}^{{2\pi }} {{e^{i\left( { - m'+m} \right)\phi }}d\phi } =2\pi {\delta _{mm'}}$$

53

,

$$\Lambda =\frac{2}{{{L_0}}}\int\limits_{0}^{{{L_0}}} {\sin \left( {\frac{{\pi zn'}}{{{L_0}}}+\pi n'} \right)\frac{d}{{dz}}\sin \left( {\frac{{\pi zn}}{{{L_0}}}+\pi n} \right)dz}$$

54

.

In these expressions \({n_\rho },m,n\) and \({n_\rho }',m',n'\) are quantum numbers before and after the optical transition, respectively.

Equtaion, (53) and (54) establish the rule for selecting optical transitions of our quantum system. As can be seen from these expressions, optical transitions are possible only in cases where \(m=m'\) and \(\left| {n - n'} \right|=odd\).

\(M=\frac{1}{2}{\left( {d'd} \right)^{\frac{m}{2}}}\frac{{{{\left( {b - d'} \right)}^{{n_\rho }'}}{{\left( {b - d} \right)}^{{n_\rho }}}}}{{{b^{{n_\rho }'+{n_\rho }+m+1}}}}\frac{{\Gamma \left( {{n_\rho }+{n_\rho }'+m+1} \right)}}{{{n_\rho }!{n_\rho }'!}} \times\)

$$F\left( { - {n_\rho }, - {n_\rho }'; - {n_\rho } - {n_\rho }' - m;\frac{{b\left( {b - d' - d} \right)}}{{\left( {b - d'} \right)\left( {b - d} \right)}}} \right)$$

55

,

\(G=\frac{1}{2}{\left( {d'd} \right)^{\frac{{m+1}}{2}}}\frac{{{{\left( {b - d'} \right)}^{{n_\rho }'}}{{\left( {b - d} \right)}^{{n_\rho }}}}}{{{b^{{n_\rho }'+{n_\rho }+m+2}}}}\frac{{\Gamma \left( {{n_\rho }+{n_\rho }'+m+2} \right)}}{{{n_\rho }!{n_\rho }'!}} \times\)

$$F\left( { - {n_\rho }, - {n_\rho }'; - {n_\rho } - {n_\rho }' - m - 1;\frac{{b\left( {b - d' - d} \right)}}{{\left( {b - d'} \right)\left( {b - d} \right)}}} \right)$$

56

,

Here,

\(b=\frac{1}{2}\left( {d'+d} \right)\) , \(d'=\frac{1}{{2\lambda {'^2}}}\), \(d=\frac{1}{{2{\lambda ^2}}}\). (57)

Equations (55) and (56) were obtained based on the fact that \(m=m'\) and taking into account the integral expression [22]

\(\int\limits_{0}^{\infty } {{e^{bx}}{x^\alpha }L_{n}^{\alpha }} \left( {\lambda x} \right)L_{m}^{\alpha }\left( {\mu x} \right)=\frac{{\Gamma \left( {m+n+\alpha +1} \right)}}{{m!n!}}\frac{{{{\left( {b - \lambda } \right)}^n}{{\left( {b - \mu } \right)}^m}}}{{{b^{m+n+\alpha +1}}}} \times\)

$$F\left( { - m, - n; - m - n - \alpha ;\frac{{b\left( {b - \lambda - \mu } \right)}}{{\left( {b - \lambda } \right)\left( {b - \mu } \right)}}} \right)$$

58

.

In Eq. (49), we defined the momentum matrix element. As can be seen from equations (53) and (54), this expression determines the allowed and forbidden optical transitions between the initial and final quantum states of an ellipsoidal quantum dot.Next we can write the Dirac delta function \(\delta \left( {{E_f} - {E_i} - \hbar \omega } \right)\) in Eq. (46) as [12]

$$\delta \left( {{E_\sigma }} \right) \approx \frac{1}{\pi }\frac{\Gamma }{{E_{\sigma }^{2}+{\Gamma ^2}}}$$

59

.

Here, \(\Gamma\) is the relaxation rate constant in units of \(eV\), which expresses the time required for the system to reach thermodynamic equilibrium again after the optical transition. Also,

$${E_\sigma }=E_{\sigma }^{f} - E_{\sigma }^{i} - \hbar \omega$$

60

.

Here \(E_{\sigma }^{f}\) and \(E_{\sigma }^{i}\) are the energy spectrum of the electron before and after the optical transition, respectively. Finally, considering Eq. (49) and Eq. (59), we can write the absorption coefficient in Eq. (46) as

$$\alpha \left( {\hbar \omega } \right)=\frac{{4\pi {e^2}{\hbar ^3}}}{{{\varepsilon _0}c{n_0}{\mu _0}^{2}\hbar \omega V}}{\left( {\Theta \Lambda Y} \right)^2}{\left( {\frac{M}{K}+\frac{G}{W}} \right)^2}\frac{\Gamma }{{\Delta E_{\sigma }^{2}+{\Gamma ^2}}}$$

61

for the case of intra-band optical transitions.