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[tex] explain why stuff does not work

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commit ab2083aefccdcafd3c35746ca2d8ea4c5c50aca3 1 parent 8658c9f
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11 tex/intro.tex
@@ -67,8 +67,17 @@ \chapter{Introduction}
We present both analytical calculations and numerical simulations based on
tight-binding approximation. The analytical calculations show an analogy to
light optics: an interface between two media with different optical densities
-can result in polarized light, and polarization dependent reflection. In
+can result in polarized light, and polarization-dependent reflection. In
analogy, an interface between two regions of different spin-orbit coupling
strength splits up an electron beam into two beams of defined chirality.
+The fact that the beams are of different chirality -- as opposed to
+linear polarization, as in the case of light optics -- makes it much hard to
+build an efficient spin polarizer.
+The numerical results show that a certain degree of spin separation can be
+achieved for appropriate interface angles and spin-orbit strengths, even in
+the new case where there is non-zero spin-orbit interaction on both sides of
+the interface.
% vim: spell
23 tex/khodas.tex
@@ -220,6 +220,19 @@ \section{Interface Between Normal and Spin-Orbit Coupling Regions}
Since the wave with $-$ chirality is transmitted at smaller angles
$\theta^- < \phi$, no critical phenomena arise.
+The analogy to classical wave optics is obvious, even if some of the
+details are different: a light beam comes from an optical dense
+region to a medium with lower refractive index. The transmitted beam
+is deflected away from the surface perpendicular, and if the notional
+angle of transmission
+exceeds $90^\circ$, the wave can't propagate and is reflected instead.
+The reflected beam is linearly polarized at the Brewster angle.
+In our case a beam hitting a boundary between normal and spin-orbit
+interacting region is also deflected away from the boundary
+perpendicular, and for large deflection angles one of the chiralities
\section{Generalization to two Spin-Orbit Regions}
The system can be generalized to two regions with non-zero spin-orbit
@@ -276,17 +289,21 @@ \section{Generalization to two Spin-Orbit Regions}
coupling strength in the right regime:
- \phi^+_{B,c} = -\sin^{-1}(\ta_B - \sqrt{1+\ta_B^2})
+ \phi^+_{c1} = -\sin^{-1}(\ta_B - \sqrt{1+\ta_B^2})
This is the first gray vertical line in figure
\ref{fig:plots-nonzero}, and in the region $\phi < \phi+_{B,c}$ the various transmission and reflection
coefficients look very similar to the case with $\ta_A = 0$.
-The second gray line is critical angle associated with $\ta_A$. For
+The second gray line is critical angle $\phi^+_{c2}$ associated with
+$\ta_A$. For
$\phi > \phi^+_{A,c}$ the momentum in $x$ direction $p_{x,A}$ is again
imaginary, just like if we had another interface to a normal region
-left of the $A$ region.
+left of the $A$ region.
+It follows the same $\ta$ dependency, namely $\phi^+_{c2} =
+-\sin^{-1}(\ta_A - \sqrt{1+\ta_A^2})$.
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95 tex/numerics.tex
@@ -305,14 +305,15 @@ \subsection{Comparison to Analytical Results}
and that the part of wave function in the sample which travels from left to
right is a smooth continuation of the wave function in the lead:
\mathbf{\Psi^\uparrow}(x=x_1) &=
a\cdot e^{i p_x x_1}e^{i p_z z} \chi_N^+
- + b\cdot e^{i p_x x_1}e^{i p_z z} \chi_N^-\\
+ + b\cdot e^{i p_x x_1}e^{i p_z z} \chi_N^-\nonumber\\
\mathbf{\Psi^\downarrow}(x=x_1) &=
c\cdot e^{i p_x x_1}e^{i p_z z} \chi_N^+
+ d\cdot e^{i p_x x_1}e^{i p_z z} \chi_N^-
+ \label{eq:a-n-left}
where $x_1$ is the location where the left lead is connected to the sample, the
interface is at $x = 0$ and the right lead is connected at $x = x_2$.
@@ -326,16 +327,17 @@ \subsection{Comparison to Analytical Results}
T_{2\uparrow,1\uparrow} = \left| \left(
a \mathbf{\Psi^+}(x=x_2) + b \mathbf{\Psi^-}(x=x_2)
- \right)^\dagger \cdot \mathbf{\Psi}^\uparrow(x=x_2) \right|^2\\
+ \right)^\dagger \cdot \mathbf{\Psi}^\uparrow(x=x_2) \right|^2\nonumber\\
T_{2\downarrow,1\uparrow} = \left| \left(
a \mathbf{\Psi^+}(x=x_2) + b \mathbf{\Psi^-}(x=x_2)
- \right)^\dagger \cdot \mathbf{\Psi}^\downarrow(x=x_2) \right|^2\\
+ \right)^\dagger \cdot \mathbf{\Psi}^\downarrow(x=x_2) \right|^2\nonumber\\
T_{2\uparrow,1\downarrow} = \left| \left(
c \mathbf{\Psi^+}(x=x_2) + d \mathbf{\Psi^-}(x=x_2)
- \right)^\dagger \cdot \mathbf{\Psi}^\uparrow(x=x_2) \right|^2\\
+ \right)^\dagger \cdot \mathbf{\Psi}^\uparrow(x=x_2) \right|^2\nonumber\\
T_{2\downarrow,1\downarrow} = \left| \left(
c \mathbf{\Psi^+}(x=x_2) + d \mathbf{\Psi^-}(x=x_2)
\right)^\dagger \cdot \mathbf{\Psi}^\downarrow(x=x_2) \right|^2
+% \label{eq:an-right}
where $\mathbf{\Psi^\pm}$ are the wave functions introduced in
@@ -387,6 +389,18 @@ \subsection{Comparison to Analytical Results}
+ \begin{center}
+ \includegraphics[width=0.8\textwidth]{relative-polarization-n-so.pdf}
+ \end{center}
+ \caption{$T_S^{rel} = \frac{T_{2\uparrow,1\uparrow} - T_{2\downarrow,1\downarrow}}
+ {T_{2\uparrow,1\uparrow} + T_{2\downarrow,1\downarrow}}$ as
+ a function of the interface angle $\phi$ (in degrees), and otherwise
+ identical parameters as in fig. \ref{fig:a-n-matching-phi}.
+ We observe a maximal spin polarization of about $20\%$.
+ }
+ \label{fig:n-so-rel}
Figure \ref{fig:a-n-matching-alpha} shows the signal for two different
angles of the interface. For weak spin-orbit coupling, the theory and the
@@ -406,6 +420,75 @@ \subsection{Comparison to Analytical Results}
cosine form that the tight binding model implies. This explains why there is a
shift between the two curves in figure \ref{fig:a-n-matching-phi}.
+For a rather small spin-orbit coupling strength of $\frac{\tso}{2 t a} = 0.02$
+we already get a quite respectable relative spin-polarization of $20\%$.
+To understand better why we don't get a very clear picture of the critical
+angle the numerical results, we look at \ref{eq:a-n-left} and for a moment
+ignore the global phases that the $\exp$ functions provide, and obtain a
+simplified expression for $a$ and $b$.
+ a &= \frac{1-\sin \phi}{1 + \sin\phi} \nonumber\\
+ b &= \sqrt{1-a^2}
+Since $t_{-+}$ and $t_{+-}$ are rather small, we also neglect them, as well as
+the $exp$ functions in $\mathbf{\Psi^\pm}$, which only contribute phases for
+$\phi < \phi_c$. Thus we obtain approximate,
+simplified expression for our matrix elements:
+ T_{2\uparrow,1\uparrow} &\approx \left|a \chi_{SO}^{+U} t_{++}
+ + b \chi_{SO}^{-U} t_{--} \right|^2\\
+ T_{2\downarrow,1\downarrow} &\approx \left|c \chi_{SO}^{-D} t_{--}
+ + d \chi_{SO}^{+D} t_{--} \right|^2
+where the superscript index $U$ means \emph{upper component of}, and $D$ means
+\emph{lower component of}.
+When $\phi$ is varied in the range of 0 to $\pi/2$, the coefficients $a, b, c,
+d$ and the absolute values of the spinor components cover the range of 0 to 1.
+The critical phenomena and the variation of $t_{++}$ and $t_{--}$ drown in
+eight parameters which oscillate roughly with the same frequency and
+magnitude, washing out a clear signature from the chiral waves.
+Or speaking more in terms of physical quantities, the interface efficiently
+selects waves of $-$ chirality over waves of $+$ chirality (for large angles),
+but injecting and measuring the waves in the $\uparrow,\downarrow$ bases
+hides a significant part of this effect.
+\section{Interface Between Two Spin-Orbit Coupling Regions}
+ \begin{center}
+ \includegraphics[width=0.8\textwidth]{relative-polarization-so-so.pdf}
+ \end{center}
+ \caption{$T_S^{rel} = \frac{T_{2\uparrow,1\uparrow} - T_{2\downarrow,1\downarrow}}
+ {T_{2\uparrow,1\uparrow} + T_{2\downarrow,1\downarrow}}$ as
+ a function of the interface angle $\phi$ (in degrees), for an
+ interface of two non-zero spin-orbit coupling regions. $t_{SO,B} =
+ \frac{0.1}{2 a t}$, $t_{SO,A} = 0.5 t_{SO,B}$. As in the case of
+ a N and SO region, we can observe a spin polarization of about $20\%$,
+ with additional peaks going up to about $40\%$.
+ ($150 \times 150$ sample of $100~nm \times 100~nm$ at $E_F = 2 t$)
+ }
+ \label{fig:n-so-rel}
+When a sample contains a 2-dimensional electron gas with Rashba spin-orbit
+coupling, it is very hard to create a region without any spin-orbit coupling.
+While it is hard to switch it off entirely, it is quite possible to tune the
+strength of an individual region by using a gate electrode on top of the
+sample to apply an electric field.
+Figure \ref{fig:n-so-rel} shows the relative spin polarization as a function
+of the interface angle $\phi$, for $\ta_B = 2 \ta_A$. Again a spin
+polarization of $20\%$ can be observed, with some few spikes going up as high
+as $40\%$ (but with stronger SO interaction on the right-hand side than in the
+case of the N-SO interface).
%For $\phi > \phi_c$, the wave $\exp{i p_x^+ x}\exp{i p_z z} t_{++}\chi_{SO}^+$
%does not propagate, because $p_x^+$ is imaginary. That means that the relative
%spin polarization caused by the $\mathbf{\Psi^+}$ wave
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2  tex/thesis.tex
@@ -389,12 +389,12 @@ \section{Rashba Spin-Orbit Coupling}
\section{Tight Binding Hamiltonian}
In order to numerically evaluate the Green's function, we
discretize the sample into sites, and assume that electrons only travel from
one site to an immediate neighbor. This allows us to represent the
Hamiltonian and other differential operators by a finite matrix.
%% TODO: remove bullshit
%% TODO: references
%The nearest neighbor approximation is quite common in transport calculations,
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