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Investigating Hot-Carrier Effects using the Backward Monte Carlo Method

5.5 Implementation for Parabolic and Isotropic Bands

The next model we want to discuss assumes a parabolic dispersion for both the sample and the partner electron, which is the standard model often used in literature. Some integrals can be solved analytically if a parabolic and isotropic dispersion relation is assumed for the sample electrons:

$(5.78) \{begin}{align} \Epsilon (\kv _1) = \frac {\hbar ^2 \kv _1^2}{2m}\,. \{end}{align}$

5.5.1 The Total Scattering Rate for Boltzmann Statistics

The total scattering rate is defined as

$(5.79) \{begin}{align} \Gamma _1(\kv _1) = \int S(\kv _1,\kv _1+\qv )\dint ^3\,q\,. \label {eq:scatt-rate-general} \{end}{align}$

We use formulation (5.54) and introduce spherical polar coordinates with $\kv _1$ defining the polar axis.

$(5.80) \{begin}{align} \Gamma _1(\kv _1) = \frac {C}{2\pi }\int \limits _0^\infty \int \limits _0^\pi \frac {\beta _s^2}{q \left (q^2 + \beta _s^2\right )^2}\,\exp \left [-\left ( \frac {q + k_1\cos \vartheta _1}{\tau } \right )^2\right ]\,2\pi \,\sin \vartheta _1\,\dint \vartheta _1\,q^2\dint q \{end}{align}$

Here, we make use of the relation

$(5.81–5.80) \{begin}{align*} \kappa (\kv _1,\qv ) = q - k_1\cos \vartheta _1\,. \{end}{align*}$

Defining $\chi = \cos \vartheta _1$ and the scaled variables

$(5.81) \{begin}{align} p=\frac {k_1}{\tau }\„\qquad s=\frac {q}{\tau }\„\qquad \gamma =\frac {\beta _s}{\tau } \{end}{align}$

we obtain

$(5.82) \{begin}{align} \Gamma _1(\kv _1)= C \int \limits _0^\infty \int \limits _{-1}^1 \frac {\gamma ^2\,s}{(s^2+\gamma ^2)^2}\e ^{-(s+p\chi )^2}\dint \chi \dint s\,. \{end}{align}$

The double integral defines a function $F(p,\gamma )$ :

$(5.83) \{begin}{align} F(p,\gamma )=\int \limits _0^\infty \int \limits _{-1}^1 \frac {\gamma ^2\,s}{(s^2+\gamma ^2)^2}\e ^{-(s+p\chi )^2}\dint \chi \dint s\,. \{end}{align}$

Evaluating the $\chi$ -integral will result in an error function. However, by applying integration by parts in $s$ the error function can be avoided. We use

$(5.84) \{begin}{align} \int \limits _0^\infty \frac {\partial u}{\partial s} v \dint s = u\,v \bigg \rvert ^\infty _{s=0} - \int \limits _0^\infty u \frac {\partial v}{\partial s}\dint s \label {eq:int-by-parts} \{end}{align}$

where $\frac {\partial u}{\partial s}$ and $v$ are defined as:

$(5.85–5.84) \{begin}{align*} \frac {\partial u(s,\gamma )}{\partial s} &= \frac {\gamma ^2\,s}{(s^2+\gamma ^2)^2}\\ v(p,s,\chi ) &= \e ^{-(s+p\chi )^2} \{end}{align*}$

For the antiderivative $u$ and the partial derivative of $v$ we find:

$(5.85–5.84) \{begin}{align*} u(s,\gamma ) &= \int \limits _0^s \frac {\gamma ^2\,\mean {s}}{(\mean {s}^2+\gamma ^2)^2}\dint \mean {s} = \frac {s^2}{2(s^2+\gamma ^2)}\\ \frac {\partial v}{\partial s} &= -2(s + p\chi )\,\e ^{-(s+p\chi )^2}=\frac {1}{p}\,\frac {\partial v}{\partial \chi } \{end}{align*}$

Inserting these expression in (5.84) gives

$(5.85–5.84) \{begin}{align*} \int \limits _0^\infty \frac {\partial u}{\partial s}\,v\dint s = -\frac {1}{p} \int \limits _0^\infty u \frac {\partial v}{\partial \chi }\dint s \{end}{align*}$

Now integration over $\chi$ can be carried out.

$(5.85–5.84) \{begin}{align*} F(p,\gamma ) = - \frac {1}{p} \int \limits _0^\infty \dint s \int \limits _{-1}^1 u\frac {\partial v}{\partial \chi }\dint \chi = \frac {1}{p} \int \limits _0^\infty u(s,\gamma )[v(p,s,-1) - v(p,s,1)] \dint s \{end}{align*}$

Back substitution of the functions $u$ and $v$ gives

$(5.85) \{begin}{align} F(p,\gamma ) = \frac {1}{2p}\int \limits _0^\infty \frac {s^2}{s^2+\gamma ^2}\,\left ( \e ^{-(s-p)^2} - \e ^{-(s+p)^2} \right ) \dint s \label {eq:sp-F} \{end}{align}$

The asymptotic behavior of $F(p,\gamma )$ is discussed in Appendix A.5.2

$(5.86–5.85) \{begin}{align*} F(0,0) = 1\,. \{end}{align*}$

With back substitution of $p$ and $\tau$ , the single-particle scattering can be reformulated to

$(5.86) \{begin}{align} \Gamma _1(\Epsilon ) = C\, F\left ( \sqrt {\Epsilon /k_BT}, \beta _s/\tau \right )\,. \label {eq:Gamma1-integrated} \{end}{align}$

The pre-factor C (5.56) with (5.49) reads

$(5.87) \{begin}{align} C = \frac {n e^4 }{(2\pi )^{3/2}\hbar ^2\epsilon _s^2\beta _s^2}\, \sqrt {\frac {m}{k_B T}} \{end}{align}$

Inserting the definition of $\beta _s^2$ (5.2), the pre-factor $C$ becomes independent of the electron density $n$ .

$(5.88) \{begin}{align} C=\frac {n e^4 }{(2\pi )^{3/2}\hbar ^2\epsilon _s^2}\,\frac {\epsilon _s\,k_B T}{e^2\,n}\, \sqrt {\frac {m}{k_B T}}=\frac {e^2 \sqrt {m k_BT}}{(2\pi )^{3/2}\hbar ^2\epsilon _s} \{end}{align}$

The electron-electron scattering rate only depends on the electron density through the screening parameter $\gamma =\beta _s/\tau$ in the function $F$ . Since $\gamma$ shows up in the denominator of the integral in $F$ , the electron-electron scattering decreases at higher electron concentrations.

5.5.2 Two-particle Scattering Rate for Boltzmann Statistics

In the previous section the total scattering rate was calculated by integrating the scattering rate $S(\kv _1,\kv _1’)$ . The latter is the result of another integration. Thus the total scattering rate is obtained by two consecutive integrations. Inserting (5.26) in (5.79) and assuming Boltzmann statistics gives

$(5.89) \{begin}{align} \Gamma _1(\kv _1) = A \int \frac {\delta \bigl [\Epsilon (\kv _1 + \qv ) + \Epsilon (\kv _2 - \qv ) - \Epsilon (\kv _1) -\Epsilon (\kv _2)\bigr ]} {\left (q^2 + \beta _s^2\right )^2} p_0(\kv _2) \dint ^3k_2 \dint ^3q \label {eq:1p-scat-rate} \{end}{align}$

In this section we reverse the order of integration and perform the $\qv$ -integration first. The result of the first integration is the two-particle scattering rate $\Gamma _2(\kv _1,\kv _2)$ .

$(5.90) \{begin}{align} \Gamma _2(\kv _1,\kv _2) = A \int \frac {\delta \bigl [\Epsilon (\kv _1 + \qv ) + \Epsilon (\kv _2 - \qv ) - \Epsilon (\kv _1) -\Epsilon (\kv _2)\bigr ]} {\left (q^2 + \beta _s^2\right )^2} p_0(\kv _2) \dint ^3q \label {eq:2p-scat-rate} \{end}{align}$

For a parabolic band the argument of the $\delta$ -function in (5.90) becomes

$(5.91) \{begin}{multline} \Epsilon (\kv _1 + \qv ) - \Epsilon (\kv _1) + \Epsilon (\kv _2 - \qv ) -\Epsilon (\kv _2) = \frac {\hbar ^2}{2 m} \left ((\kv _1 + \qv )^2 + (\kv _2 - \qv )^2 - \kv _1^2 -\kv _2^2\right ) \\ = \frac {\hbar ^2}{m} \left (q^2 + (\kv _1 - \kv _2)\cdot \qv \right ) = \frac {\hbar ^2}{m} \left (q^2 - K q \cos \vartheta \right )\,. \label {eq:energy-argument-parabolic} \{end}{multline}$

Here we have introduced the difference vector $\Kv$ defined as

$(5.92) \{begin}{align} \Kv = \kv _2 - \kv _1\„ \label {eq:K-parabolic} \{end}{align}$

and $\vartheta$ , the angle between $\Kv$ and $\qv$ . In a spherical polar coordinate system, with $\Kv$ as the polar axis and $\vartheta$ as the polar angle, the energy conservation requires a positive $\cos \vartheta$

$(5.93) \{begin}{align} q^2 - q K \cos \vartheta = 0 \quad \Rightarrow \quad \cos \vartheta = \frac {q}{K} \ge 0\,. \label {eq:ees-polar-angle} \{end}{align}$

Since $q = \abs {\qv }$ and $K = \abs {\Kv }$ are positive by definition the allowed range of $\vartheta$ is

$(5.94–5.93) \{begin}{align*} \vartheta \in \left [0,\frac {\pi }{2}\right ]\,. \{end}{align*}$

With (5.91) the integral (5.90) becomes

$(5.94–5.93) \{begin}{align*} \Gamma _2(\kv _1,\kv _2)= A\int \limits _0^\infty \int \limits _0^{\pi /2} \frac {\delta [\frac {\hbar ^2}{m}(q^2-K\,q\,\cos \vartheta )]}{(q^2 +\beta _s^2)^2}\,2\pi \, \sin \vartheta \dint \vartheta \,q^2\dint q \{end}{align*}$

We carry out the $\vartheta$ -integration first. Defining $\xi = \cos \vartheta$ we obtain

$(5.94) \{begin}{multline} H(q)=\int \limits _0^1 \delta \left ( \frac {\hbar ^2}{m}q(q-K\xi ) \right )\dint \xi \\ = \frac {m}{\hbar ^2\,q}\int \limits _0^1 \delta (K\xi - q)\dint \xi = \frac {m}{\hbar ^2\,K\,q}\left [ \Theta (K-q) - \underbrace {\Theta (-q)}_0 \right ] \{end}{multline}$

Note that $H(q)$ restricts the domain of integration to $q \in [0,K]$ .

$(5.95) \{begin}{align} \Gamma _2(\kv _1,\kv _2) = 2\pi \,A \int \limits _0^\infty \frac {q^2}{(q^2-\beta _s^2)^2} H(q)\dint q = 2\pi \,A\,\frac {m}{\hbar ^2\,K} \int \limits _0^K \frac {q^2}{(q^2-\beta _s^2)^2} \dint q \label {eq:lambda-q-integral} \{end}{align}$

With the pre-factor $B$ (5.49), the scattering rate can be evaluated as [100]:

$(5.96) \{begin}{align} \Gamma _2(\kv _1,\kv _2) = \frac {B}{2} \frac {K}{K^2 + \beta _s^2} \,. \{end}{align}$

5.5.3 Single-particle Scattering Rate for Boltzmann Statistics

Having evaluated the $\qv$ -integral in (5.89), we now evaluate the $\kv _2$ -integral. The single-particle scattering rate $\Gamma _1$ can be calculated by integrating the two-particle scattering rate $\Gamma _2$ against the normalized probability density $p_0(\kv _2)$ , see [100].

$(5.97) \{begin}{align} \Gamma _1(\kv _1) =\int \Gamma _2(\kv _1,\kv _2)\,p_0(\kv _2) \;d^3k_2 = \frac {B}{2} \int \frac {\abs {\kv _2-\kv _1}}{\abs {\kv _2-\kv _1}^2 + \beta _s^2} \frac {f_0(\kv _2)}{C_\mathrm {MB}}\;d^3k_2 \{end}{align}$

The variable substitution

$(5.98–5.97) \{begin}{align*} \Kv =\kv _2-\kv _1, \qquad d^3k_2=d^3K,\qquad \kv _2=\kv _1+\Kv \{end}{align*}$

leads to the following expression for the single-particle scattering rate

$(5.98) \{begin}{align} \Gamma _1(\kv _1) = \frac {B}{2\, C_{MB}} \int \frac {K}{K^2 + \beta _s^2} \,f_0(\kv _1+\Kv )\;d^3K\,. \label {eq:sp-scatt-1} \{end}{align}$

Introducing scaled variables

$(5.99–5.98) \{begin}{align*} p=\frac {k_1}{\tau }, \qquad s=\frac {K}{\tau },\qquad \gamma =\frac {\beta _s}{\tau }\„ \{end}{align*}$

the Boltzmann term can be reformulated as

$(5.99) \{begin}{align} f_0(\kv _1+\Kv )=\e ^{\eta -(k_1^2+2k_1 K \cos \vartheta +K^2)/\tau ^2} = \e ^{\eta -p^2-s^2-2\,p\,s\,\cos \vartheta } \{end}{align}$

the $\Kv$ -integral in (5.98) can be evaluated analytically:

$(5.100) \{begin}{align} \int \frac {K}{K^2 + \beta _s^2} \,f_0(\kv _1+\Kv )\;d^3K &= \tau ^2 \,\e ^\eta \int \frac {s}{s^2+\gamma ^2}\e ^{-p^2-s^2-2\,p\,s\,\cos \vartheta } \,d^3s\nonumber \\ &=2\pi \tau ^2\,\e ^{\eta }\int \limits _0^\infty \frac {s}{s^2+\gamma ^2}\e ^{-s^2-p^2}s^2\,ds\int \limits _{-1}^{1}\e ^{-2ps\chi }\,d\chi \nonumber \\ &= 2\pi \tau \,\e ^{\eta } \int \limits _0^\infty \frac {s^2}{s^2+\gamma ^2} \frac {\e ^{-(s-p)^2}-\e ^{-(s+p)^2}}{2p}\,ds \nonumber \\ &= 2\pi \,\tau ^2\,\e ^\eta \,F(p,\gamma ) \{end}{align}$

The function $F(p,\gamma )$ is defined in (5.85), which leads finally to the single-particle scattering rate (5.86).

5.5.4 Random Selection of the Momentum Transfer q

In the first step the wave vector $\kv _2$ of the partner electron is selected randomly from an equilibrium Maxwellian distribution. With the current wave vector $\kv _1$ of the sample electron the difference vetor is computed

$(5.101–5.100) \{begin}{align*} \Kv = \kv _2 - \kv _1\„\qquad K=\abs {\Kv }\,. \{end}{align*}$

The two-particle scattering rate can be used to calculate the momentum transfer $\qv$ . For this purpose the integrand in (5.95) can be used as a probability density for $q$ . We define $F(q)$ as:

$(5.101) \{begin}{align} F(q) = \int \limits _0^q \frac {\mean {q} \dint \mean {q}}{\left ( \mean {q}^2 + \beta _s^2\right )^2} = \frac {q^2}{2\beta _s^2\left (q^2 + \beta _s^2\right )} \{end}{align}$

The normalized cumulative probability distribution is

$(5.102) \{begin}{align} P(q) = \frac {F(q)}{F(K)}, \qquad q\in [0, K]\,. \{end}{align}$

The random number $q$ is generated using the inversion method, see Section 3.3.2.

$(5.103–5.102) \{begin}{align*} P(q_r) = r_1\quad \Rightarrow \quad q_r^2 = \frac {r_1 K^2 \beta _s^2}{K^2(1 - r_1) + \beta _s^2} \{end}{align*}$

The polar angle $\vartheta _r$ is given by (5.93):

$(5.103) \{begin}{align} \cos \vartheta _r = \frac {q_r}{K}\,. \{end}{align}$

With a second random number $r_2$ the azimuthal angle $\varphi$ is generated from a uniform distribution in $[0,2\pi ]$ .

$(5.104) \{begin}{align} \varphi _r = 2\pi r_2 \{end}{align}$

From $(q_r, \cos \vartheta _r, \varphi _r)$ the 3D vector $\qv _r$ can be constructed. Note that the polar axis is given by the difference vector $\Kv = \kv _2 - \kv _1$ . In the last step the final state is computed.

$(5.105) \{begin}{align} \kv _1’ &= \kv _1 + \qv _r \{end}{align}$

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