Luttinger–Kohn model

All bands can be subdivided into two classes:

• Class A: six valence bands (heavy hole, light hole, split off band and their spin counterparts) and two conduction bands.
• Class B: all other bands.

The method concentrates on the bands in Class A, and takes into account Class B bands perturbatively.

We can write the perturbed solution ${\displaystyle \phi _{}^{}}$ as a linear combination of the unperturbed eigenstates ${\displaystyle \phi _{i}^{(0)}}$:

${\displaystyle \phi =\sum _{n}^{A,B}a_{n}\phi _{n}^{(0)}}$

Assuming the unperturbed eigenstates are orthonormalized, the eigenequation are:

${\displaystyle (E-H_{mm})a_{m}=\sum _{n\neq m}^{A}H_{mn}a_{n}+\sum _{\alpha \neq m}^{B}H_{m\alpha }a_{\alpha }}$,

where

${\displaystyle H_{mn}=\int \phi _{m}^{(0)\dagger }H\phi _{n}^{(0)}d^{3}\mathbf {r} =E_{n}^{(0)}\delta _{mn}+H_{mn}^{'}}$.

From this expression we can write:

${\displaystyle a_{m}=\sum _{n\neq m}^{A}{\frac {H_{mn}}{E-H_{mm}}}a_{n}+\sum _{\alpha \neq m}^{B}{\frac {H_{m\alpha }}{E-H_{mm}}}a_{\alpha }}$,

where the first sum on the right-hand side is over the states in class A only, while the second sum is over the states on class B. Since we are interested in the coefficients ${\displaystyle a_{m}}$ for m in class A, we may eliminate those in class B by an iteration procedure to obtain:

${\displaystyle a_{m}=\sum _{n}^{A}{\frac {U_{mn}^{A}-\delta _{mn}H_{mn}}{E-H_{mm}}}a_{n}}$,

${\displaystyle U_{mn}^{A}=H_{mn}+\sum _{\alpha \neq m}^{B}{\frac {H_{m\alpha }H_{\alpha n}}{E-H_{\alpha \alpha }}}+\sum _{\alpha ,\beta \neq m,n;\alpha \neq \beta }{\frac {H_{m\alpha }H_{\alpha \beta }H_{\beta n}}{(E-H_{\alpha \alpha })(E-H_{\beta \beta })}}+\ldots }$

Equivalently, for ${\displaystyle a_{n}}$ (${\displaystyle n\in A}$):

${\displaystyle a_{n}=\sum _{n}^{A}(U_{mn}^{A}-E\delta _{mn})a_{n}=0,m\in A}$

and

${\displaystyle a_{\gamma }=\sum _{n}^{A}{\frac {U_{\gamma n}^{A}-H_{\gamma n}\delta _{\gamma n}}{E-H_{\gamma \gamma }}}a_{n}=0,\gamma \in B}$.

When the coefficients ${\displaystyle a_{n}}$ belonging to Class A are determined so are ${\displaystyle a_{\gamma }}$.

The Hamiltonian including the spin-orbit interaction can be written as:

${\displaystyle H=H_{0}+{\frac {\hbar }{4m_{0}^{2}c^{2}}}{\bar {\sigma }}\cdot \nabla V\times \mathbf {p} }$,

where ${\displaystyle {\bar {\sigma }}}$ is the Pauli spin matrix vector. Substituting into the Schrödinger equation we obtain

${\displaystyle Hu_{n\mathbf {k} }(\mathbf {r} )=\left(H_{0}+{\frac {\hbar }{m_{0}}}\mathbf {k} \cdot \mathbf {\Pi } +{\frac {\hbar ^{2}k^{2}}{4m_{0}^{2}c^{2}}}\nabla V\times \mathbf {p} \cdot {\bar {\sigma }}\right)u_{n\mathbf {k} }(\mathbf {r} )=E_{n}(\mathbf {k} )u_{n\mathbf {k} }(\mathbf {r} )}$,

where

${\displaystyle \mathbf {\Pi } =\mathbf {p} +{\frac {\hbar }{4m_{0}^{2}c^{2}}}{\bar {\sigma }}\times \nabla V}$

and the perturbation Hamiltonian can be defined as

${\displaystyle H'={\frac {\hbar }{m_{0}}}\mathbf {k} \cdot \mathbf {\Pi } .}$

The unperturbed Hamiltonian refers to the band-edge spin-orbit system (for k=0). At the band edge, conduction band Bloch waves exhibit s-like symmetry, while valence band states are p-like (3-fold degenerate without spin). Let us denote these states as ${\displaystyle |S\rangle }$, and ${\displaystyle |X\rangle }$, ${\displaystyle |Y\rangle }$ and ${\displaystyle |Z\rangle }$ respectively. These Bloch functions can be pictured as periodic repetition of atomic orbitals, repeated at intervals corresponding to the lattice spacing. The Bloch function can be expanded in the following manner:

${\displaystyle u_{n\mathbf {k} }(\mathbf {r} )=\sum _{j'}^{A}a_{j'}(\mathbf {k} )u_{j'0}(\mathbf {r} )+\sum _{\gamma }^{B}a_{\gamma }(\mathbf {k} )u_{\gamma 0}(\mathbf {r} )}$,

where j' is in Class A and ${\displaystyle \gamma }$ is in Class B. The basis functions can be chosen to be

${\displaystyle u_{10}(\mathbf {r} )=u_{el}(\mathbf {r} )=\left|S{\frac {1}{2}},{\frac {1}{2}}\right\rangle =\left|S\uparrow \right\rangle }$

${\displaystyle u_{20}(\mathbf {r} )=u_{SO}(\mathbf {r} )=\left|{\frac {1}{2}},{\frac {1}{2}}\right\rangle ={\frac {1}{\sqrt {3}}}|(X+iY)\downarrow \rangle +{\frac {1}{\sqrt {3}}}|Z\uparrow \rangle }$

${\displaystyle u_{30}(\mathbf {r} )=u_{lh}(\mathbf {r} )=\left|{\frac {3}{2}},{\frac {1}{2}}\right\rangle =-{\frac {1}{\sqrt {6}}}|(X+iY)\downarrow \rangle +{\sqrt {\frac {2}{3}}}|Z\uparrow \rangle }$

${\displaystyle u_{40}(\mathbf {r} )=u_{hh}(\mathbf {r} )=\left|{\frac {3}{2}},{\frac {3}{2}}\right\rangle =-{\frac {1}{\sqrt {2}}}|(X+iY)\uparrow \rangle }$

${\displaystyle u_{50}(\mathbf {r} )={\bar {u}}_{el}(\mathbf {r} )=\left|S{\frac {1}{2}},-{\frac {1}{2}}\right\rangle =-|S\downarrow \rangle }$

${\displaystyle u_{60}(\mathbf {r} )={\bar {u}}_{SO}(\mathbf {r} )=\left|{\frac {1}{2}},-{\frac {1}{2}}\right\rangle ={\frac {1}{\sqrt {3}}}|(X-iY)\uparrow \rangle -{\frac {1}{\sqrt {3}}}|Z\downarrow \rangle }$

${\displaystyle u_{70}(\mathbf {r} )={\bar {u}}_{lh}(\mathbf {r} )=\left|{\frac {3}{2}},-{\frac {1}{2}}\right\rangle ={\frac {1}{\sqrt {6}}}|(X-iY)\uparrow \rangle +{\sqrt {\frac {2}{3}}}|Z\downarrow \rangle }$

${\displaystyle u_{80}(\mathbf {r} )={\bar {u}}_{hh}(\mathbf {r} )=\left|{\frac {3}{2}},-{\frac {3}{2}}\right\rangle =-{\frac {1}{\sqrt {2}}}|(X-iY)\downarrow \rangle }$.

Using Löwdin's method, only the following eigenvalue problem needs to be solved

${\displaystyle \sum _{j'}^{A}(U_{jj'}^{A}-E\delta _{jj'})a_{j'}(\mathbf {k} )=0,}$

where

${\displaystyle U_{jj'}^{A}=H_{jj'}+\sum _{\gamma \neq j,j'}^{B}{\frac {H_{j\gamma }H_{\gamma j'}}{E_{0}-E_{\gamma }}}=H_{jj'}+\sum _{\gamma \neq j,j'}^{B}{\frac {H_{j\gamma }^{'}H_{\gamma j'}^{'}}{E_{0}-E_{\gamma }}}}$,

${\displaystyle H_{j\gamma }^{'}=\left\langle u_{j0}\right|{\frac {\hbar }{m_{0}}}\mathbf {k} \cdot \left(\mathbf {p} +{\frac {\hbar }{4m_{0}c^{2}}}{\bar {\sigma }}\times \nabla V\right)\left|u_{\gamma 0}\right\rangle \approx \sum _{\alpha }{\frac {\hbar k_{\alpha }}{m_{0}}}p_{j\gamma }^{\alpha }.}$

The second term of ${\displaystyle \Pi }$ can be neglected compared to the similar term with p instead of k. Similarly to the single band case, we can write for ${\displaystyle U_{jj'}^{A}}$

${\displaystyle D_{jj'}\equiv U_{jj'}^{A}=E_{j}(0)\delta _{jj'}+\sum _{\alpha \beta }D_{jj'}^{\alpha \beta }k_{\alpha }k_{\beta },}$

${\displaystyle D_{jj'}^{\alpha \beta }={\frac {\hbar ^{2}}{2m_{0}}}\left[\delta _{jj'}\delta _{\alpha \beta }+\sum _{\gamma }^{B}{\frac {p_{j\gamma }^{\alpha }p_{\gamma j'}^{\beta }+p_{j\gamma }^{\beta }p_{\gamma j'}^{\alpha }}{m_{0}(E_{0}-E_{\gamma })}}\right].}$

We now define the following parameters

${\displaystyle A_{0}={\frac {\hbar ^{2}}{2m_{0}}}+{\frac {\hbar ^{2}}{m_{0}^{2}}}\sum _{\gamma }^{B}{\frac {p_{x\gamma }^{x}p_{\gamma x}^{x}}{E_{0}-E_{\gamma }}},}$

${\displaystyle B_{0}={\frac {\hbar ^{2}}{2m_{0}}}+{\frac {\hbar ^{2}}{m_{0}^{2}}}\sum _{\gamma }^{B}{\frac {p_{x\gamma }^{y}p_{\gamma x}^{y}}{E_{0}-E_{\gamma }}},}$

${\displaystyle C_{0}={\frac {\hbar ^{2}}{m_{0}^{2}}}\sum _{\gamma }^{B}{\frac {p_{x\gamma }^{x}p_{\gamma y}^{y}+p_{x\gamma }^{y}p_{\gamma y}^{x}}{E_{0}-E_{\gamma }}},}$

and the band structure parameters (or the Luttinger parameters) can be defined to be

${\displaystyle \gamma _{1}=-{\frac {1}{3}}{\frac {2m_{0}}{\hbar ^{2}}}(A_{0}+2B_{0}),}$

${\displaystyle \gamma _{2}=-{\frac {1}{6}}{\frac {2m_{0}}{\hbar ^{2}}}(A_{0}-B_{0}),}$

${\displaystyle \gamma _{3}=-{\frac {1}{6}}{\frac {2m_{0}}{\hbar ^{2}}}C_{0},}$

These parameters are very closely related to the effective masses of the holes in various valence bands. ${\displaystyle \gamma _{1}}$ and ${\displaystyle \gamma _{2}}$ describe the coupling of the ${\displaystyle |X\rangle }$, ${\displaystyle |Y\rangle }$ and ${\displaystyle |Z\rangle }$ states to the other states. The third parameter ${\displaystyle \gamma _{3}}$ relates to the anisotropy of the energy band structure around the ${\displaystyle \Gamma }$ point when ${\displaystyle \gamma _{2}\neq \gamma _{3}}$.

The Luttinger-Kohn Hamiltonian ${\displaystyle \mathbf {D_{jj'}} }$ can be written explicitly as a 8X8 matrix (taking into account 8 bands - 2 conduction, 2 heavy-holes, 2 light-holes and 2 split-off)

${\displaystyle \mathbf {H} =\left({\begin{array}{cccccccc}E_{el}&P_{z}&{\sqrt {2}}P_{z}&-{\sqrt {3}}P_{+}&0&{\sqrt {2}}P_{-}&P_{-}&0\\P_{z}^{\dagger }&P+\Delta &{\sqrt {2}}Q^{\dagger }&-S^{\dagger }/{\sqrt {2}}&-{\sqrt {2}}P_{+}^{\dagger }&0&-{\sqrt {3/2}}S&-{\sqrt {2}}R\\E_{el}&P_{z}&{\sqrt {2}}P_{z}&-{\sqrt {3}}P_{+}&0&{\sqrt {2}}P_{-}&P_{-}&0\\E_{el}&P_{z}&{\sqrt {2}}P_{z}&-{\sqrt {3}}P_{+}&0&{\sqrt {2}}P_{-}&P_{-}&0\\E_{el}&P_{z}&{\sqrt {2}}P_{z}&-{\sqrt {3}}P_{+}&0&{\sqrt {2}}P_{-}&P_{-}&0\\E_{el}&P_{z}&{\sqrt {2}}P_{z}&-{\sqrt {3}}P_{+}&0&{\sqrt {2}}P_{-}&P_{-}&0\\E_{el}&P_{z}&{\sqrt {2}}P_{z}&-{\sqrt {3}}P_{+}&0&{\sqrt {2}}P_{-}&P_{-}&0\\E_{el}&P_{z}&{\sqrt {2}}P_{z}&-{\sqrt {3}}P_{+}&0&{\sqrt {2}}P_{-}&P_{-}&0\\\end{array}}\right)}$