On the propagation of Dirac fermions in graphene with strain-induced inhomogeneous Fermi velocity

First, let us briefly review a simple model where the nonvanishing component of the strain tensor and v₁₁(x) are piece-wise constant

$$\begin{equation}{\mathbf{u}}_{11}\left(x\right)=\begin{cases}0,\hfill & x{< }0,\hfill \\ {u}_{11}\ne 0,\hfill & x{ >}0\hfill \end{cases},\quad {\mathbf{v}}_{11}\left(x\right)=\begin{cases}{v}_{0},\hfill & x{< }0,\hfill \\ {v}_{11}{ >}0,\hfill & x{ >}0.\hfill \end{cases}\end{equation} \tag{ 8 }$$

Here we set v₀ = 1. The electrostatic potential V(x) is vanishing for x < 0 but it acquires non-zero constant value for x > 0. This model will help to understand the influence of the inhomogeneous Fermi velocity and the electrostatic potential on the scattering of the Dirac fermions, and it will also motivate our further analysis. This setting was discussed in detail in [38] for homogeneous Fermi velocity, see also [36, 37]. For inhomogeneous piece-wise constant Fermi velocity, it was elaborated in [39].

One can find explicit solutions F of the stationary equation separately for x < 0 and x > 0,

$$\begin{equation}F=\begin{cases}{F}_{i,r}^{{\pm}}=\frac{{\mathrm{e}}^{\mathrm{i}{k}_{x}^{i,r}x+\mathrm{i}{k}_{y}^{i,r}y}}{\sqrt{2}}{\left(1,{\pm}{\mathrm{e}}^{\mathrm{i}{\phi }_{i,r}}\right)}^{\mathrm{T}},\quad \hfill & x{< }0,\hfill \\ {F}_{t}^{{\pm}}=\frac{{\mathrm{e}}^{\mathrm{i}{k}_{x}^{t}x+\mathrm{i}{k}_{y}^{t}y}}{\sqrt{2}}{\left(1,{\pm}{\mathrm{e}}^{\mathrm{i}{\phi }_{t}}\right)}^{\mathrm{T}},\quad \hfill & x{ >}0,\hfill \end{cases}\end{equation} \tag{ 9 }$$

where the indices i, r, t denote incident, reflected and transmitted wave and $\mathrm{tan}\;{\phi }_{i,r,t}={k}_{y}^{i,r,t}/{v}_{11}\left(x\right){k}_{x}^{i,r,t}$. The index ± corresponds to the sign of the kinetic energy E − V and T denotes here the transposition. The normalization constant comes from integration over a unit volume. We fix the wave function as $F={F}_{i}^{+}+r\;{F}_{r}^{+}$ for x < 0 and $F=t\;{F}_{t}^{-}$ for x > 0 and E − V < 0 whereas $F=t\;{F}_{t}^{+}$ for x > 0 and E − V > 0. The coefficients r and t can be found when taking into account conservation of energy, the momentum k_y and conservation of probability current at x = 0. The calculation is straightforward and, for the sake of completeness, we present it in the appendix A. It gives us ϕ_r = π − ϕ_i both for E > V and E < V. In case of ϕ_t, we have to distinguish between the two cases

$$\begin{equation}{\phi }_{t}=\begin{cases}\pi +\mathrm{arcsin}\frac{E}{E-V}\mathrm{sin}\;{\phi }_{i},\quad \hfill & E{< }V,\hfill \\ \mathrm{arcsin}\frac{E}{E-V}\mathrm{sin}\;{\phi }_{i},\quad \hfill & E{ >}V.\hfill \end{cases}\end{equation} \tag{ 10 }$$

Taking into account that we have ϕ_i ∈ (−π/2, π/2), we can find the coefficients r and t as

$$\begin{equation}\begin{aligned}\hfill r& =\begin{cases}\frac{{\mathrm{e}}^{\mathrm{i}{\phi }_{i}}\left({\mathrm{e}}^{\mathrm{i}{\phi }_{i}}+{\mathrm{e}}^{\mathrm{i}{\phi }_{t}}\right)}{1-{\mathrm{e}}^{\mathrm{i}\left({\phi }_{i}+{\phi }_{t}\right)}},\quad E{< }V,\hfill \\ \frac{{\mathrm{e}}^{\mathrm{i}{\phi }_{i}}\left({\mathrm{e}}^{\mathrm{i}{\phi }_{i}}-{\mathrm{e}}^{\mathrm{i}{\phi }_{t}}\right)}{1+{\mathrm{e}}^{\mathrm{i}\left({\phi }_{i}+{\phi }_{t}\right)}},\quad E{ >}V,\quad \hfill \end{cases}\hfill \\ \hfill t& =\begin{cases}\frac{1+{\mathrm{e}}^{2\mathrm{i}{\phi }_{i}}}{\sqrt{{v}_{11}}\left(1-{\mathrm{e}}^{\mathrm{i}\left({\phi }_{i}+{\phi }_{t}\right)}\right)},\quad E{< }V,\hfill \\ \frac{1+{\mathrm{e}}^{2\mathrm{i}{\phi }_{i}}}{\sqrt{{v}_{11}}\left(1+{\mathrm{e}}^{\mathrm{i}\left({\phi }_{i}+{\phi }_{t}\right)}\right)},\quad E{ >}V.\quad \hfill \end{cases}\hfill \end{aligned}\end{equation} \tag{ 11 }$$

The transmission and reflection amplitudes are

$$\begin{equation}\begin{aligned}\hfill R& =\begin{cases}\frac{1+\mathrm{cos}\left({\phi }_{i}-{\phi }_{t}\right)}{1-\mathrm{cos}\left({\phi }_{i}+{\phi }_{t}\right)},\quad E{< }V,\hfill \\ \frac{1-\mathrm{cos}\left({\phi }_{i}-{\phi }_{t}\right)}{1+\mathrm{cos}\left({\phi }_{i}+{\phi }_{t}\right)},\quad E{ >}V,\quad \hfill \end{cases}\hfill \\ \hfill T& =\begin{cases}\frac{\mathrm{cos}\left({\phi }_{i}-{\phi }_{t}\right)+\mathrm{cos}\left({\phi }_{i}+{\phi }_{t}\right)}{-1+\mathrm{cos}\left({\phi }_{i}+{\phi }_{t}\right)},\quad E{< }V,\hfill \\ \frac{\mathrm{cos}\left({\phi }_{i}-{\phi }_{t}\right)+\mathrm{cos}\left({\phi }_{i}+{\phi }_{t}\right)}{1+\mathrm{cos}\left({\phi }_{i}+{\phi }_{t}\right)},\quad E{ >}V.\quad \hfill \end{cases}\hfill \end{aligned}\end{equation} \tag{ 12 }$$

The velocity operators v_x and v_y can be computed as $\mathbf{v}=\frac{1}{i}\left[\mathbf{x},H\right]$, giving explicitly v_x = v₁₁(x)σ₁ and v_y = σ₂. Their expectation values for the incident and the transmitted wave are

$$\begin{align}\hfill \begin{aligned}\hfill \quad \quad {v}_{x,{\pm}}^{i}& =\langle {F}_{i}^{{\pm}},{\sigma }_{1}{F}_{i}^{{\pm}}\rangle ={\pm}\mathrm{cos}\;{\phi }_{i},\hfill \\ \hfill {v}_{x,{\pm}}^{t}& =\langle {F}_{t}^{{\pm}},{v}_{11}{\sigma }_{1}{F}_{t}^{{\pm}}\rangle ={\pm}{v}_{11}\;\mathrm{cos}\;{\phi }_{t},\hfill \end{aligned}\end{align} \tag{ 13 }$$

$$\begin{equation}{v}_{y,{\pm}}^{i,t}=\langle {F}_{i,t}^{{\pm}},{\sigma }_{2}{F}_{i,t}^{{\pm}}\rangle ={\pm}\mathrm{sin}\;{\phi }_{i,t}.\end{equation} \tag{ 14 }$$

A few comments are in order. First, one can see that there occurs unusual transmission for E < 0. Suppose that ${v}_{x,+}^{i}{ >}0$ and ${v}_{y,+}^{i}{ >}0$ and the incidence angle ϕ_i ranges between 0 and π/2. Then, taking into account ϕ_t ∈ (π/2, 3π/2), see (10), the velocity of the transmitted wave is ${v}_{x,-}^{t}{ >}0$ and ${v}_{y,-}^{t}{< }0$. The fact that the ${v}_{y}^{t}$ has different sign then ${v}_{y}^{i}$ gives rise to the unusual transmission typical for metamaterials in optics and was utilized for design of Veselago lens in graphene [36]. Let us mention that classical particle with energy E > V and with the velocity $\mathbf{v}=\left({v}_{x,+}^{t},{v}_{y,+}^{t}\right)$ would move along the trajectory y = v₁₁ tan ϕ_tx + y₀.

Now, let us consider the situation where the electrostatic potential V is small. Taking the limit V → 0 in the formulas (10), (11) and (12), we get (assuming that E > 0)

$$\begin{equation}{\phi }_{t}={\phi }_{i},\quad r=0,\quad t={\sqrt{{v}_{11}}}^{-1},\quad R=0,\quad T=1.\end{equation} \tag{ 15 }$$

Hence, in the case of the vanishing potential term, the interface becomes completely transparent for particles of any energy and any incident angle. The total transmission T = 1 independent on the incident angle was discussed in [39] for specific values of energy. In the limit V → 0, the total transmission occurs for any angle and energy. Similarly to [39], it is marked by ϕ_i = ϕ_t. For x > 0, the trajectory of the particles changes to

$$\begin{equation}y={v}_{11}\;\mathrm{tan}\;{\phi }_{i}\;x+{y}_{0}.\end{equation} \tag{ 16 }$$

We can write the wave function as

$$\begin{align}F\hfill & ={\mathrm{e}}^{\mathrm{i}{\phi }_{i}}{\mathrm{e}}^{-\mathrm{i}{\sigma }_{3}{\phi }_{i}/2}{\mathbf{v}}_{11}{\left(x\right)}^{-1/2}{\mathrm{e}}^{\mathrm{i}E\left(\mathrm{cos}{\phi }_{i}x/{v}_{11}\left(x\right)+\mathrm{sin}{\phi }_{i}y\right)}{\left(1,1\right)}^{\mathrm{T}}\hfill \\ \hfill & ={\mathrm{e}}^{\mathrm{i}{\phi }_{i}}{\mathrm{e}}^{-\mathrm{i}{\sigma }_{3}{\phi }_{i}/2}{\mathbf{v}}_{11}{\left(x\right)}^{-1/2}{\sim}{F}\left(r\right),\end{align} \tag{ 17 }$$

where ${\sim}{F}\left(r\right)={\mathrm{e}}^{\mathrm{i}Er}{\left(1,1\right)}^{\mathrm{T}}$ and $r=\mathrm{sin}\;{\phi }_{i}\;y+\mathrm{cos}\;{\phi }_{i}\;\frac{x}{{\mathbf{v}}_{11}\left(x\right)}$. The wave function ${\sim}{F}\left(r\right)$ satisfies the stationary equation

$$\begin{equation}\left(-i{\sigma }_{1}{\partial }_{r}-i{\sigma }_{2}{\partial }_{s}-E\right){\sim}{F}\left(r\right)=0.\end{equation} \tag{ 18 }$$

Hence, in the regime V(x) = 0 we can transform the model into the one of free particle. It provides a natural explanation for the fact that there is no backscattering on the interface as T = 1. Let us elaborate the link the free particle system in the next subsection in the context of general profiles of strain and v₁₁(x).

We shall show that the Hamiltonian in (7) can be mapped to the free-particle energy operator for V(x) = 0. We suppose that the electrostatic potential induced by the deformation is compensated by the external electrostatic field. The Dirac equation for the free particle reads as

$$\begin{equation}{H}_{0}\left(r,s\right){\Phi}\left(r,s,t\right)={v}_{0}\left(-i{\sigma }_{1}{\partial }_{r}-i{\sigma }_{2}{\partial }_{s}\right){\Phi}\left(r,s,t\right)=i{\partial }_{t}{\Phi}\left(r,s,t\right).\end{equation} \tag{ 19 }$$

We define transformation of coordinates

$$\begin{align}\hfill \begin{aligned}\hfill r& =\mathrm{sin}\;\gamma \;y+\mathrm{cos}\;\gamma \;{\int }_{0}^{x}\frac{{v}_{0}}{{\mathbf{v}}_{11}\left(q\right)}\mathrm{d}q,\hfill \\ \hfill s& =\mathrm{cos}\;\gamma \;y-\mathrm{sin}\;\gamma \;{\int }_{0}^{x}\frac{{v}_{0}}{{\mathbf{v}}_{11}\left(q\right)}\mathrm{d}q,\quad \gamma \in \left(0,\pi /2\right)\hfill \end{aligned}\end{align} \tag{ 20 }$$

and the operator

$$\begin{equation}G\left(x\right)={\left(\frac{{\mathbf{v}}_{11}\left(x\right)}{{v}_{0}}\right)}^{1/2}\mathrm{exp}\left(\mathrm{i}{\sigma }_{3}\frac{\gamma }{2}+\frac{\mathrm{i}\;{v}_{0}\;\beta }{2a}{\int }_{0}^{x}\frac{{\mathbf{u}}_{11}\left(q\right)}{{\mathbf{v}}_{11}\left(q\right)}\mathrm{d}q\right).\end{equation} \tag{ 21 }$$

Then we have

$$\begin{equation}H\left(x,y\right)={G}^{-1}\left(x\right){H}_{0}\left(r\left(x,y\right),s\left(x,y\right)\right)G\left(x\right).\end{equation} \tag{ 22 }$$

The solutions of i∂_tΦ(r, s, t) = H₀Φ(r, s, t) transform as

$$\begin{equation}\begin{aligned}\hfill {\Psi}\left(x,y,t\right)& =G{\left(x\right)}^{-1}{\Phi}\left(r\left(x,y\right),s\left(x,y\right),t\right),\hfill \\ H\left(x,y\right){\Psi}\left(x,y,t\right)\hfill & =i{\partial }_{t}{\Psi}\left(x,y,t\right).\end{aligned}\end{equation} \tag{ 23 }$$

It can be verified by direct calculation that the transformation preserves the norm,

$$\begin{align}{\int }_{{\mathbb{R}}^{2}}\vert {\Psi}\left(x,y,t\right){\vert }^{2}\mathrm{d}x\mathrm{d}y={\int }_{{\mathbb{R}}^{2}}{\left\vert {\left(\frac{{\mathbf{v}}_{11}\left(x\right)}{{v}_{0}}\right)}^{-1/2}{\Phi}\left(r\left(x,y\right),s\left(x,y\right),t\right)\right\vert }^{2}\mathrm{d}x\mathrm{d}y\hfill \\ ={\int }_{{\mathbb{R}}^{2}}\vert {\Phi}\left(r,s,t\right){\vert }^{2}\mathrm{d}r\mathrm{d}s.\end{align} \tag{ 24 }$$

It is worth noticing that the transformation discussed in [17] coincides with the one discussed here for γ = 0. By varying γ, we can adjust the angle of the incident wave Ψ(x, y, t) while keeping Φ(r, s, t) independent of γ. It is also worth noting that a similar transformation can be used to provide alternative explanation for Klein tunneling on the electrostatic barriers [42]. In our current case, it demonstrates that elimination of V(x) in (7) implies the total transmission T = 1.

To get further insight into the influence of the strain on the propagation of the Dirac fermions, we find it illustrative to consider the localized wave packets. Let us suppose that we have a localized (normalizable) function Φ(r, s, t) that satisfies i∂_tΦ(r, s, t) = H₀(r, s)Φ(r, s, t). Then we can get Ψ(x, y, t) that satisfies the Dirac equation of the strained system (23) and it is localized by construction. Now, let us suppose that the Gaussian-like wave packet Φ(r, s, t) propagates along the r axis (defined by s = 0) and disperses symmetrically with respect to the s axis. Then the wave packet Ψ(r, s, t) does no longer move along a straight line, but it rather follows the curve s(x, y) = 0 corresponding to

$$\begin{equation}y\left(x\right)=\mathrm{tan}\;\gamma {\int }_{0}^{x}\frac{{v}_{0}}{{\mathbf{v}}_{11}\left(q\right)}\mathrm{d}q.\end{equation} \tag{ 25 }$$

The relation (25) generalizes the formula (16) for the smoothly changing velocity v₁₁(x).

The change of trajectory is one effect that the wave packet Ψ(x, y, t) undergoes when passing through the strained region. The other one is its squeezing or stretching along x-axis. To see this, let us suppose that the probability density of the wave packet Φ(r, s, t) is practically zero outside a domain Ω(r, s, t). The domain changes its position and shape in time as the wave packet moves and disperses. We want to know how is the domain ${\sim}{{\Omega}}\left(x,y,t\right)$ where the wave packet Ψ(x, y, t) is localized. We have

$$\begin{align}\hfill & {\int }_{{\Omega}\left(r,s,t\right)}\vert {\Phi}\left(r,s,t\right){\vert }^{2}\mathrm{d}r\mathrm{d}s\hfill \\ \hfill & \quad ={\int }_{{\sim}{{\Omega}}\left(x,y,t\right)}\vert {\mathbf{v}}_{11}^{-1/2}\left(x\right){\Phi}\left(r\left(x,y\right),s\left(x,y\right),t\right){\vert }^{2}\mathrm{d}x\mathrm{d}y\hfill \\ \hfill & \quad ={\int }_{{\sim}{{\Omega}}\left(x,y,t\right)}\vert {\Psi}\left(x,y,t\right){\vert }^{2}\mathrm{d}x\mathrm{d}y,\quad {\sim}{{\Omega}}\left(x,y,t\right)={\Omega}\left(r\left(x,y\right),s\left(x,y\right),t\right).\hfill \end{align} \tag{ 26 }$$

Therefore, the wave packet Ψ(x, y, t) propagating in the strained material is confined in the deformed domain ${\sim}{{\Omega}}$. The change of the domain where the Dirac particle is localized is compensated by the term ${\mathbf{v}}_{11}^{-1/2}\left(x\right)$ that alters the probability density. For illustration, we can suppose that Ω(r, s, t) is a disc of radius ρ(t) centered⁷ in the maximum (r₀(t), s₀(t)) of the probability density of the wave packet. Then the deformed domain ${\sim}{{\Omega}}\left(x,y,t\right)$ is composed from the points (x, y) such that

$$\begin{align}\rho {\left(t\right)}^{2}\hfill & {\geqslant}{\left(r\left(x,y\right)-{r}_{0}\left(x,y,t\right)\right)}^{2}+{\left(s\left(x,y\right)-{s}_{0}\left(x,y,t\right)\right)}^{2}\\ \hfill & ={\left({\int }_{{x}_{0}\left(t\right)}^{x}\frac{{v}_{0}}{{\mathbf{v}}_{11}\left(q\right)}\mathrm{d}q\right)}^{2}+{\left(y-{y}_{0}\left(t\right)\right)}^{2}.\end{align} \tag{ 27 }$$

If v₁₁(x) is constant on the domain where the wave packet is spread, the disk gets deformed into an ellipse,

$$\begin{equation}\rho {\left(t\right)}^{2}{\geqslant}{\left(r-{r}_{0}\right)}^{2}+{\left(r-{r}_{0}\right)}^{2}=\frac{{\left(x-{x}_{0}\left(t\right)\right)}^{2}}{{v}_{11}}+{\left(y-{y}_{0}\left(t\right)\right)}^{2}.\end{equation} \tag{ 28 }$$

When v₁₁ < v₀, we can see that the wave packet gets squeezed along the x axis. This way, the strain can focus the wave packets. In the figure 1, the focusing of the wave packet when passing through the region with v₁₁(x) < v₀ is depicted. As an approximation of Φ(r, s, t), we took the standard dispersing Gaussian wave packet.

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In the rest of the section, let us discuss briefly the trajectory of the wave packet in dependence on several configurations of the strain u₁₁(x) in more detail.

2.2.1. Asymptotically constant Fermi velocity

First, let us suppose that the Fermi velocity is asymptotically equal to v₀ and the strain induces only a localized fluctuation of v₁₁(x),

$$\begin{equation}{\mathbf{v}}_{11}\left(x\right)={v}_{0}\left(1+{\Delta}v\right),\quad {\mathrm{lim}}_{x\to {\pm}\infty }{\Delta}v=0.\end{equation} \tag{ 29 }$$

Deformation of this kind can be produced by folds of graphene sheet, see e.g. [12]. We suppose that the fluctuation Δv is small in the following sense,

$$\begin{align}\hfill \quad {\int }_{0}^{\infty }\frac{{\Delta}v\left(q\right)}{1+{\Delta}v\left(q\right)}\mathrm{d}x={X}_{+},\quad \hfill \\ \hfill -{\int }_{-\infty }^{0}\frac{{\Delta}v\left(q\right)}{1+{\Delta}v\left(q\right)}\mathrm{d}x={X}_{-},\quad \vert {X}_{{\pm}}\vert {< }\infty .\hfill \end{align} \tag{ 30 }$$

Then the trajectory defined in (25) tends asymptotically to two straight lines

$$\begin{equation}y\left(x\right)\to \begin{cases}{y}_{\mathrm{i}\mathrm{n}}=\mathrm{tan}\;\gamma \;\left(x-{X}_{-}\right),\quad \hfill & x\to -\infty \hfill \\ {y}_{\mathrm{o}\mathrm{u}\mathrm{t}}=\mathrm{tan}\;\gamma \;\left(x-{X}_{+}\right),\quad \hfill & x\to \infty .\hfill \end{cases}\end{equation} \tag{ 31 }$$

The two asymptotic trajectories y_in and y_out are parallel but mutually shifted. Therefore, the wave packet traveling along the trajectory (25) gets deflected by the mechanical deformation that gives rise to the inhomogeneous Fermi velocity (29). It is straightforward to compute explicitly the length of the normal vector n connecting the two lines,

$$\begin{equation}\vert n\vert =\mathrm{sin}\;\gamma \vert {X}_{+}-{X}_{-}\vert .\end{equation} \tag{ 32 }$$

see figure 2 for illustration.

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Now, we shall consider the situation where v₁₁(x) tends asymptotically to two, possibly different, constant values. It can be induced by the unidirectional strain which vanishes for x → −∞ but it converges to a constant positive value for x → ∞. In this case, the displacement of the atoms in the crystal have linear-like growth for large values of x. We suppose that such a deformation can be achieved by the corresponding strain of the substrate on which the graphene sheet is positioned.

Let us suppose that the deformation should be such that the induced Fermi velocity satisfies

$$\begin{equation}\frac{{v}_{0}}{{\mathbf{v}}_{11}\left(x\right)}=\begin{cases}{\omega }_{-}+O\left({x}^{-\alpha }\right),\quad \hfill & x\to -\infty ,\hfill \\ {\omega }_{+}+O\left({x}^{-\alpha }\right),\quad \hfill & x\to \infty ,\;\alpha { >}1.\hfill \end{cases}\end{equation} \tag{ 33 }$$

For these values of α, the O(x^−α) term gives convergent contribution to the trajectory (25) of the wave packet so that y(x) can be written as

$$\begin{equation}y\left(x\right)=\begin{cases}{\omega }_{-}\mathrm{tan}\;\gamma \;x+O\left(1\right),\quad \hfill & x\to -\infty ,\hfill \\ {\omega }_{+}\mathrm{tan}\;\gamma \;x+O\left(1\right),\quad \hfill & x\to \infty .\hfill \end{cases}\end{equation} \tag{ 34 }$$

This type of deformation is illustrated in figure 1 where we fixed v₁₁(x) = v₀(1 − μ(1 + tanh νx)) for constant μ and ν.

2.2.2. Periodic fluctuation of Fermi velocity

Let us focus on the situation where the Fermi velocity is constant for all x < 0 and periodic for x ⩾ 0,

$$\begin{equation}{\mathbf{v}}_{11}\left(x\right)=\begin{cases}{v}_{0},\quad \hfill & x{< }0,\hfill \\ {\mathbf{v}}_{11}\left(x+L\right)={\mathbf{v}}_{11}\left(x\right),\quad \hfill & x{\geqslant}0.\hfill \end{cases}\end{equation} \tag{ 35 }$$

We suppose that the corresponding periodic strain can be achieved by the flexural modes [17] or by interaction with the substrate that gives rise to the Moire patterns [5]. Then the trajectory (25) can be written as

$$\begin{equation}y\left(x\right)=\begin{cases}\mathrm{tan}\;\gamma \;x,\hfill & x{< }0,\hfill \\ \mathrm{tan}\;\gamma \;\left(\left[\frac{x}{L}\right]A+{\int }_{\left[\frac{x}{L}\right]L}^{x}\frac{{v}_{0}}{{\mathbf{v}}_{11}\left(q\right)}\mathrm{d}q\right),\quad A={\int }_{0}^{L}\frac{{v}_{0}}{{\mathbf{v}}_{11}\left(q\right)}\mathrm{d}q,\hfill & x{\geqslant}0,\hfill \end{cases}\end{equation} \tag{ 36 }$$

where [x] denotes integer part of x. In general, y(x) is no longer linear for x ⩾ 0. However, it can be confined between two parallel lines. Utilizing $x-1{\leqslant}\left[x\right]{\leqslant}x$ together with positivity of v₁₁(x), we get

$$\begin{equation}\mathrm{tan}\;\omega \;x-A\;\mathrm{tan}\;\gamma {\leqslant}y\left(x\right){\leqslant}\mathrm{tan}\;\omega \;x+A\;\mathrm{tan}\;\gamma ,\end{equation} \tag{ 37 }$$

where

$$\begin{equation}\mathrm{tan}\;\omega =\frac{A}{L}\mathrm{tan}\;\gamma =\frac{1}{L}\mathrm{tan}\;\gamma {\int }_{0}^{L}\frac{{v}_{0}}{{\mathbf{v}}_{11}\left(q\right)}\mathrm{d}q.\end{equation} \tag{ 38 }$$

Hence, the periodic strain deflects the wave packet by the angle ω − γ. The deflection angle ω is just function of the incidence angle γ, periodicity L and the integral $A={\int }_{0}^{L}\frac{{v}_{0}}{{\mathbf{v}}_{11}\left(q\right)}\mathrm{d}q$ so that it is the same for a family of Fermi velocities that share these quantities. It is worth noticing that ω also depends on the material constants β and a that stay hidden in definition of v₁₁(x). There is no difference in propagation of the wave packets formed in the K and K' valleys.

Before we proceed to the spectral analysis of the waveguides formed by the strains, let us make a brief comment on the influence of the strain on the wave packets. Likewise in the previous section, we can transform the Hamiltonian in order to eliminate the x-dependent potential and to simplify the kinetic term. Similarly to (20), we define transformation of coordinates

$$\begin{align}\hfill \begin{aligned}\hfill r& =\mathrm{cos}\;\gamma {\int }_{0}^{x}\frac{{v}_{0}}{{\mathbf{v}}_{11}\left(q\right)}\mathrm{d}q+\mathrm{sin}\;\gamma {\int }_{0}^{y}\frac{{v}_{0}}{{\mathbf{v}}_{22}\left(q\right)}\mathrm{d}q,\hfill \\ \hfill s& =-\mathrm{sin}\;\gamma {\int }_{0}^{x}\frac{{v}_{0}}{{\mathbf{v}}_{11}\left(q\right)}\mathrm{d}q+\mathrm{cos}\;\gamma {\int }_{0}^{y}\frac{{v}_{0}}{{\mathbf{v}}_{22}\left(q\right)}\mathrm{d}q\hfill \end{aligned}\end{align} \tag{ 41 }$$

and the operator

$$\begin{equation}G\left(x,y\right)=\left(\frac{\sqrt{{\mathbf{v}}_{11}\left(x\right){\mathbf{v}}_{22}\left(y\right)}}{{v}_{0}}\right)\mathrm{exp}\left(\mathrm{i}{\sigma }_{3}\frac{\gamma }{2}+\frac{\mathrm{i}\;{v}_{0}\beta }{2a}{\int }_{0}^{x}\frac{{\mathbf{u}}_{11}\left(q\right)}{{\mathbf{v}}_{11}\left(q\right)}\mathrm{d}q\right).\end{equation} \tag{ 42 }$$

Then we have

$$\begin{equation}H\left(x,y\right)=G{\left(x,y\right)}^{-1}{H}_{0}\left(r\left(x,y\right),s\left(x,y\right)\right)G\left(x,y\right).\end{equation} \tag{ 43 }$$

Here, the Hamiltonian with the flat Fermi velocity v_ij = v₀η_ij reads as

$$\begin{align}\hfill {H}_{0}\left(r,s\right)& =-{iv}_{0}{\sigma }_{1}{\partial }_{r}-i{v}_{0}{\sigma }_{2}{\partial }_{s}-\frac{{v}_{0}\beta }{2a}{\sigma }_{1}{\mathrm{e}}^{-\mathrm{i}{\sigma }_{3}\gamma }{\mathbf{u}}_{22}\left(y\left(r,s\right)\right)\hfill \\ \hfill & \quad +V\left(x\left(r,s\right),y\left(r,s\right)\right).\hfill \end{align} \tag{ 44 }$$

The solutions of i∂_tΦ(r, s, t) = H₀(r, s)Φ(r, s, t) transform as

$$\begin{equation}\begin{aligned}{\Psi}\left(x,y,t\right)\hfill & ={G}^{-1}\left(x,y\right){\Phi}\left(r\left(x,y\right),s\left(x,y\right),t\right),\\ H\left(x,y\right){\Psi}\left(x,y,t\right)& =i{\partial }_{t}{\Psi}\left(x,y,t\right).\end{aligned}\end{equation} \tag{ 45 }$$

Considering the transformation of coordinates (41), it can be rather complicated to get an explicit form of their inverse, x = x(r, s), y = y(r, s). Therefore, analytical calculation of Φ(r, s, t) together with the scattering characteristics can be impossible due to the complicated form of u₁₁(x(r, s)) and u₂₂(y(r, s)). However, we know how the wave packet Ψ(x, y, t) gets deformed in the regions where the Fermi velocity (39) fluctuates. The domain of its localization can be found in analogy with (26). For instance the circular domain ${\Omega}\left(r,s,t\right)=\left\{\left(r,s\right),{\left(r-{r}_{0}\left(t\right)\right)}^{2}+{\left(s-{s}_{0}\left(t\right)\right)}^{2}{\leqslant}\rho {\left(t\right)}^{2}\right\}$ transforms into ${\sim}{{\Omega}}\left(x,y,t\right)$ that is composed by $\left(x,y\right)\in {\mathbb{R}}^{2}$, $\rho {\left(t\right)}^{2}{\geqslant}{\left({\int }_{{x}_{0}\left(t\right)}^{x}\frac{{v}_{0}}{{\mathbf{v}}_{11}\left(q\right)}\mathrm{d}q\right)}^{2}+{\left({\int }_{{y}_{0}\left(t\right)}^{y}\frac{{v}_{0}}{{\mathbf{v}}_{22}\left(q\right)}\mathrm{d}q\right)}^{2}$. If v₁₁(x) ⩽ v₀ and v₂₂(y) ⩽ v₀, the wave packet gets focused (the density of probability gets denser) by the strain.

Let us focus on the ability of the strain to form a wave guide for Dirac fermions. We are no longer interested in the scattering but rather in the existence of localized (guided) modes in the strained region. We fix γ = 0 in (A.16) and we consider the electrostatic potential V(x, y) to be vanishing again. Then the stationary equation (44) gets the following form

$$\begin{align}\hfill {\sim}{H}\left(r,s\right)\psi \left(r,s\right)& ={v}_{0}\left(-i{\sigma }_{1}{\partial }_{r}-i{\sigma }_{2}{\partial }_{s}+{\sigma }_{1}F\left(s\right)\right)\psi \left(r,s\right)=E\psi \left(r,s\right),\\ F\left(s\right)& \equiv -\frac{\beta }{2a}{\mathbf{u}}_{22}\left(y\left(s\right)\right).\end{align} \tag{ 46 }$$

We require r and s to be mappings from $\mathbb{R}$ onto $\mathbb{R}$. They should be invertible, monotonic functions of x and y (there holds r'(x) ≠ 0 and s'(y) ≠ 0). The derivatives r'(x) as well as s'(y) should be bounded. For convenience, we fix

$$\begin{equation}{\mathrm{lim}}_{x\to {\pm}\infty }r\left(x\right)={\pm}\infty ,\quad {\mathrm{lim}}_{y\to {\pm}\infty }s\left(y\right)={\pm}\infty .\end{equation} \tag{ 47 }$$

It is granted that there is also an inverse function y = y(s) that satisfies y(s(y)) = y.

We can take advantage of the translational invariance of the system and focus on subspaces with a conserved value of the momentum −i∂_r. We make the partial Fourier transformation

$$\begin{equation}\left(\mathcal{F}\psi \right)\left(k,s\right)=\frac{1}{\sqrt{2\pi }}{\int }_{\mathbb{R}}{\mathrm{e}}^{-\mathrm{i}kr}\psi \left(r,s\right)\mathrm{d}r.\end{equation} \tag{ 48 }$$

The Hamitonian ${\sim}{H}\left(r,s\right)$ can be rewritten as direct integral

$$\begin{equation}{\sim}{H}\left(r,s\right)={\int }_{\mathbb{R}}^{\oplus }{{\sim}{H}}_{k}\left(s\right),\quad {{\sim}{H}}_{k}\left(s\right)={v}_{0}\left({\sigma }_{1}k-i{\sigma }_{2}{\partial }_{s}+{\sigma }_{1}F\left(s\right)\right).\end{equation} \tag{ 49 }$$

The direct integral can be understood as a generalization of the partial wave decomposition for the case where the conserved quantum number is not discretized but acquires values from a real interval. As the potential term F(s) is bounded and continuous, the Hamiltonian H_k(s) is self-adjoint on the space of functions that are square integrable together with their first derivative.

We are interested in the configurations of mechanical strain where the Hamiltonian ${{\sim}{H}}_{k}\left(s\right)$ possesses discrete energies. The reason is that these discrete energies give rise to discrete energy bands in the spectrum of ${\sim}{H}\left(r,s\right)$ and H(x, y) that can be associated with existence of (partially) dispersionless wave packets [43]. Indeed, let us suppose that we can find the solution of

$$\begin{equation}{{\sim}{H}}_{k}\left(s\right){\psi }_{k,n}\left(s\right)={E}_{n}\left(k\right){\psi }_{k,n}\left(s\right),\quad \forall \;k\in {\mathcal{I}}_{n}\subset \mathbb{R},\end{equation} \tag{ 50 }$$

where E_n(k) is a discrete energy of ${{\sim}{H}}_{k}\left(s\right)$ labeled⁸ by n, i.e. ψ_k,n(s) is square integrable together with its first derivative. The intervals ${\mathcal{I}}_{n}$ can be finite but also (semi-)infinite.

We can get eigenstates of H(x, y) from those of ${{\sim}{H}}_{k}\left(s\right)$,

$$\begin{equation}\begin{aligned}\hfill H\left(x,y\right){{\Psi}}_{k,n}\left(x,y\right)& ={E}_{n}\left(k\right){{\Psi}}_{k,n}\left(x,y\right),\\ {{\Psi}}_{k,n}\left(x,y\right)& =G{\left(x,y\right)}^{-1}{\mathrm{e}}^{\mathrm{i}kr\left(x\right)}{\psi }_{k,n}\left(s\left(y\right)\right).\end{aligned}\end{equation} \tag{ 51 }$$

The discrete energies E_n(k) form discrete energy bands in the spectrum of the two-dimensional Hamiltonian, see figure 3 for illustration. We can use them to construct the following wave packets associated with the energy bands E_n(k),

$$\begin{equation}{{\Psi}}_{n}\left(x,y\right)=\frac{1}{\sqrt{2\pi }}{\int }_{{\mathcal{I}}_{n}}{\rho }_{n}\left(k\right){{\Psi}}_{k,n}\left(x,y\right)\mathrm{d}k.\end{equation} \tag{ 52 }$$

This wave packet is normalizable provided that ρ(k) is normalizable (see appendix B for details),

$$\begin{equation}{\int }_{{\mathbb{R}}^{2}}{\left\vert {{\Psi}}_{n}\left(x,y\right)\right\vert }^{2}\mathrm{d}x\mathrm{d}y={\int }_{\mathbb{R}}{\left\vert {\rho }_{n}\left(k\right)\right\vert }^{2}\mathrm{d}k.\end{equation} \tag{ 53 }$$

The wave packet (52) has a remarkable property—it does not disperse along y axis. Indeed, there holds (again, see appendix for details)

$$\begin{equation}{\int }_{a}^{b}\mathrm{d}y{\int }_{\mathbb{R}}\mathrm{d}x{\left\vert {\mathrm{e}}^{-\mathrm{i}tH\left(x,y\right)}{{\Psi}}_{n}\left(x,y\right)\right\vert }^{2}={\int }_{a}^{b}\mathrm{d}y{\int }_{\mathbb{R}}\mathrm{d}x{\left\vert {{\Psi}}_{n}\left(x,y\right)\right\vert }^{2},\end{equation} \tag{ 54 }$$

where a and b are arbitrary real numbers, i.e. the probability density in the y direction is conserved during the time evolution.

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The speed of the wave packets can be approximated by the averaged group velocity

$$\begin{equation}{v}_{\mathrm{g}}={v}_{0}\frac{{\int }_{{I}_{n}}{\partial }_{k}{E}_{n}\left(k\right)\mathrm{d}k}{\vert {\mathcal{I}}_{n}\vert }.\end{equation} \tag{ 55 }$$

The spectrum of H(x, y) is symmetric with respect to 0. The transport of the wave packets (52) is bidirectional when ∂_kE_n(k) of the positive energy bands⁹ can be both positive and negative for $k\in {\mathcal{I}}_{n}$, see figure 4(a). When ∂_kE_n(k) of the positive energy bands is positive (negative) for all $k\in {\mathcal{I}}_{n}$, the corresponding wave packets defined in (52) can move in one direction only, they are unidirectional, see figure 4(c). When the derivative of the positive energy bands has negative (positive) sign on a finite subinterval of ${\mathcal{I}}_{n}$ and positive (negative, respectively) sign for all other $k\in {\mathcal{I}}_{n}$, we say that the transport is essentially unidirectional, see figure 4(b).

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It is worth noticing that the construction of the partially dispersionless wave packets is not limited to our model but can be applied to broad class of systems with translational symmetry, see [43] for more details. Examples can be found in the literature where explicit models were solved numerically. See e.g. [11–13, 44] for the discrete energy bands corresponding to the unidirectional, essentially unidirectional and bidirectional wave packets.

Explicit solutions of (51) are needed for construction of the wave packets (52). There are well known models where the stationary equation (51) is exactly solvable, let us mention the model with F(s) = tanh s. However, for reconstruction of the initial system described by H(x, y), we would need the explicit form of u₂₂(y) (or v₂₂(y)). It would be rather difficult to extract it from $F\left(s\left(y\right)\right)=\mathrm{tanh}\left({\int }_{0}^{y}\frac{{v}_{0}}{{\mathbf{v}}_{22}\left(q\right)}\mathrm{d}q\right)=-\frac{\beta }{2a}{\mathbf{u}}_{22}\left(y\right)$. Nevertheless, there is one scenario where we can get a partial solution immediately. When

$$\begin{equation}{\mathrm{lim}}_{s\to {\pm}\infty }\;F\left(s\right)={F}_{{\pm}},\quad \vert {F}_{+}\vert \ne \vert {F}_{-}\vert ,\end{equation} \tag{ 56 }$$

i.e. the strain along y axis is asymptotically constant. Then we can obtain the following zero modes for any fixed k

$$\begin{equation}{\psi }_{+}={\mathrm{e}}^{\mathrm{i}kr}{\left({\mathrm{e}}^{\int \left(k+F\left(s\right)\right)\mathrm{d}s},0\right)}^{\mathrm{T}},\quad {\psi }_{-}={\mathrm{e}}^{\mathrm{i}kr}{\left(0,{\mathrm{e}}^{-\int \left(k+F\left(s\right)\right)\mathrm{d}s}\right)}^{\mathrm{T}},\end{equation} \tag{ 57 }$$

$$\begin{equation}{\sim}{H}\left(r,s\right){\psi }_{{\pm}}\left(r,s\right)=0,\end{equation} \tag{ 58 }$$

where one of them is vanishing exponentially for |x| → ∞ provided that (F₋ + k)(F₊ + k) < 0. This way, we get the zero energy E₀(k) = 0 of ${\sim}{H}\left(r,s\right)$. The wave packets (52) associated with this energy band do not move as its average group velocity would be identically zero.

3.2.1. Criteria for existence of the discrete energy bands

Instead of looking for exactly solvable configurations of (50), we focus on the qualitative spectral analysis of the system described by (46). It provides us with useful information on the energy bands E_n(k) without the need to solve the stationary equation. We will find the sufficient conditions for the strain such that it gives rise to waveguides supporting unidirectional or bidirectional transport.

Let us suppose that the second component of the strain tensor u₂₂(y) is asymptotically constant, lim_|y|→∞ u₂₂(y) = 0. It implies that

$$\begin{equation}{\mathrm{lim}}_{\vert y\vert \to \infty }{\mathbf{v}}_{22}\left(y\right)={v}_{0},\quad {\mathrm{lim}}_{\vert s\vert \to \infty }F\left(s\right)=0.\end{equation} \tag{ 59 }$$

We can utilize directly the results presented in [40, 41]. They are based on the fact that the square of ${{\sim}{H}}_{k}\left(s\right)$ is Pauli-type diagonal Hamiltonian¹⁰, ${{\sim}{H}}_{k}^{2}\left(s\right)=-{\partial }_{s}^{2}+{\left(k+F\left(s\right)\right)}^{2}+{\sigma }_{3}{F}^{\prime }\left(s\right)$, with the spectrum bounded from below. It is possible to use the variational principle to find criteria for existence of its discrete energies. Existence of discrete energies of ${{\sim}{H}}_{k}^{2}\left(s\right)$ then implies existence of discrete energies in the spectrum of ${{\sim}{H}}_{k}\left(s\right)$ and also of the associated discrete energy bands in the spectrum of H(x, y). The corresponding eigenstates can be used in the construction of the dispersionless wave packets (51) and (52).

Let us summarize some of the relevant criteria below:

Let us suppose that F(s) is continuous together with its first derivative and ${\int }_{\mathbb{R}}\vert F\left(s\right)\left(F\left(s\right)+2k\right)\vert \mathrm{d}s{< }\infty$ or there exists s₀ such that F(s)(F(s) + 2k) ⩽ 0 for all s < s₀. Then we can make the following conclusions [41]:

• (a)  
  If F(s) ⩾ 0 (or F(s) ⩽ 0) for all $s\in \mathbb{R}$, then there are no discrete energies in the spectrum of ${{\sim}{H}}_{k}\left(s\right)$ for all k ⩾ 0 (or for all k ⩽ 0, respectively).
• (b)  
  If, for some $a\in \mathbb{R}$, ${\int }_{-\infty }^{a}F\left(s\right)\mathrm{d}s{< }0$ (or if ${\int }_{-\infty }^{a}F\left(s\right)\mathrm{d}s{ >}0$), then there exists K > 0 such that for all k > K (or for all k < −K, respectively) there are discrete energies in the spectrum of ${{\sim}{H}}_{k}\left(s\right)$.
• (c)  
  If, for some $a\in \mathbb{R}$, ${\int }_{a}^{\infty }F\left(s\right)\mathrm{d}s{< }0$ (or if ${\int }_{a}^{\infty }F\left(s\right)\mathrm{d}s{ >}0$), then there exists K > 0 such that for all k > K (or for all k < −K, respectively) there are discrete energies in the spectrum of ${{\sim}{H}}_{k}\left(s\right)$.
• (d)  
  If F(s) ⩾ 0 and F(s) ≠ 0 then ${{\sim}{H}}_{k}\left(s\right)$ has discrete energies for all $k{< }-\frac{1}{2}{\mathrm{max}}_{s\in \mathbb{R}}F\left(s\right)$.
• (e)  
  If F(s) ⩽ 0 and F(s) ≠ 0 then ${{\sim}{H}}_{k}\left(s\right)$ has discrete energies for all $k{ >}\frac{1}{2}\vert {\mathrm{min}}_{s\in \mathbb{R}}F\left(s\right)\vert$.

Suppose that F(s) is integrable.

• (f)  
  If ${\int }_{\mathbb{R}}F\left(s\right)\mathrm{d}s{ >}0$ then there are discrete energy values in the spectrum of ${{\sim}{H}}_{k}\left(s\right)$ for all $k{< }-\frac{{\int }_{\mathbb{R}}F{\left(s\right)}^{2}\mathrm{d}s}{2{\int }_{\mathbb{R}}F\left(s\right)\mathrm{d}s}$.
• (g)  
  If ${\int }_{\mathbb{R}}F\left(s\right)\mathrm{d}s{< }0$ then there are discrete energy values in the spectrum of ${{\sim}{H}}_{k}\left(s\right)$ for all $k{ >}\frac{{\int }_{\mathbb{R}}F{\left(s\right)}^{2}\mathrm{d}s}{2}\left\vert {\int }_{\mathbb{R}}F\left(s\right)\mathrm{d}s\right\vert$.

A remark is in order. The criteria (b)–(g) represent sufficient conditions for existence of the discrete energies of ${{\sim}{H}}_{k}\left(s\right)$. It is possible that the discrete energy levels exist also outside of the specified interval for k. However, we do not have any tool how to guarantee their existence in that case. When F(s) ⩾ 0 (F(s) ⩽ 0), then (d) and (f) [or (e) and (g)] provide two different values for the threshold values of k. It is not possible to decide which one provides better estimate without evaluating them for an explicit F(s).

3.2.2. Waveguides for essentially unidirectional wave packets

We can use these criteria to show that the deformation with

$$\begin{equation}{\mathbf{u}}_{22}\left(y\right){\geqslant}0\end{equation} \tag{ 60 }$$

induces a waveguide for the essentially unidirectional wave packets (52). Let us suppose that $F\left(s\left(y\right)\right)=-\frac{\beta }{2a}{\mathbf{u}}_{22}\left(y\right)$ meets the assumptions of the criteria above. As u₂₂(y) is positive, there holds F(s) ⩽ 0. Then it follows from (a) that there are no discrete energies in the spectrum of ${{\sim}{H}}_{k}\left(s\right)$ for any k ⩽ 0. We also know from (e) and (g) that there is a threshold

$$\begin{equation*}{k}_{0}=\mathrm{min}\left\{\frac{{\int }_{\mathbb{R}}F{\left(s\right)}^{2}\mathrm{d}s}{2\left\vert {\int }_{\mathbb{R}}F\left(s\right)\mathrm{d}s\right\vert },\frac{1}{2}\vert {\mathrm{min}}_{s\in \mathbb{R}}F\left(s\right)\vert \right\}\end{equation*}$$

such that the effective Hamiltonian ${{\sim}{H}}_{k}\left(s\right)$ has discrete energies for all k > k₀. The value of k₀ can be expressed in terms of u₂₂(y) and the vertical displacement h = h(y) with the use of (41) in the following manner

$$\begin{equation}{k}_{0}=\mathrm{min}\left\{\vert \frac{\beta }{4a}\vert \frac{{\int }_{\mathbb{R}}\frac{{\mathbf{u}}_{22}{\left(y\right)}^{2}}{1+\left(1-\beta \right){\mathbf{u}}_{22}\left(y\right)-\frac{1}{2}{\left({\partial }_{y}h\right)}^{2}}\mathrm{d}y}{\vert {\int }_{\mathbb{R}}\frac{{\mathbf{u}}_{22}\left(y\right)}{1+\left(1-\beta \right){\mathbf{u}}_{22}\left(y\right)-\frac{1}{2}{\left({\partial }_{y}h\right)}^{2}}\mathrm{d}y\vert },\frac{\beta }{4a}{\mathrm{max}}_{y\in \mathbb{R}}\vert {\mathbf{u}}_{22}\left(y\right)\vert \right\},\end{equation} \tag{ 61 }$$

We can conclude that whenever there holds u₂₂(y) ⩾ 0 (but not identically u₂₂ = 0), the strain induces a waveguide for essentially unidirectional transport of the dispersionless wave packets (52).

An inhomogeneous unidirectional strain is a good example of the deformation that gives rise to (60). The associated deformation vector is

$$\begin{equation}u=\left(-\mu {\epsilon}f\left(x\right),{\epsilon}f\left(y\right)\right),\quad {f}^{\prime }\left(y\right){\geqslant}0,\end{equation} \tag{ 62 }$$

where is the strain and ν is the Poisson ratio¹¹ gives rise to the following strain tensor and Fermi velocity

$$\begin{equation}\begin{aligned}\hfill \mathbf{u}& =\left(\begin{pmatrix}{cc}\hfill -\mu {f}^{\prime }\left(x\right)\hfill & 0\hfill \\ 0\hfill & \hfill {\epsilon}{f}^{\prime }\left(y\right)\hfill \end{pmatrix}\right),\hfill \\ \mathbf{v}& ={v}_{0}\left(\begin{pmatrix}{cc}\hfill 1-\mu \left(1-\beta \right){\epsilon}{f}^{\prime }\left(x\right)\hfill & \hfill 0\hfill \\ 0\hfill & \hfill 1+\left(1-\beta \right){\epsilon}{f}^{\prime }\left(y\right)\hfill \end{pmatrix}\right).\end{aligned}\end{equation} \tag{ 63 }$$

3.2.3. Waveguides for bidirectional wave packets

3.2.4. Waveguides for valleytronics

Let us suppose that strained graphene with Fermi velocity (39) is in presence of an external magnetic field perpendicular to the crystal, B = (0, 0, ∂_yA_x(y)). Contrary to the pseudo-magnetic gauge field induced by the mechanical deformations, the magnetic field breaks the time-reversal symmetry and comes with opposite sign when we consider Dirac fermions in the vicinity of the second Dirac point K' ≡ −K. The Dirac Hamiltonians at the Dirac points K and −K can be written in the following form

$$\begin{align}\hfill {H}_{{\pm}\mathbf{K}}=& \mp i{\sigma }_{1}\sqrt{{\mathbf{v}}_{11}\left(x\right)}{\partial }_{x}\sqrt{{\mathbf{v}}_{11}\left(x\right)}-i{\sigma }_{2}\sqrt{{\mathbf{v}}_{22}\left(y\right)}{\partial }_{y}\sqrt{{\mathbf{v}}_{22}\left(y\right)}\hfill \\ \hfill & {\pm}{\sigma }_{1}\frac{{v}_{0}\beta }{2a}\left({\mathbf{u}}_{11}\left(x\right)-{\mathbf{u}}_{22}\left(y\right)\right)+\frac{{v}_{0}\beta }{2a}{\sigma }_{1}{A}_{x}\left(y\right).\hfill \end{align} \tag{ 64 }$$

Let us set the strain and the magnetic field such that

$$\begin{equation}{A}_{x}\left(y\right)={\mathbf{u}}_{22}\left(y\right).\end{equation} \tag{ 65 }$$

Then we follow the steps (46)–(49) for both H_±K and get the effective one dimensional operators

$$\begin{equation}{{\sim}{H}}_{k}^{\mathbf{K}}\left(s\right)=-i{\sigma }_{2}{\partial }_{s}+{\sigma }_{1}k,\end{equation} \tag{ 66 }$$

$$\begin{equation}{{\sim}{H}}_{k}^{{\mathbf{K}}^{\prime }}\left(s\right)=-i{\sigma }_{2}{\partial }_{s}+{\sigma }_{1}\left(-k+\frac{\beta }{a}{\mathbf{u}}_{22}\left(y\left(s\right)\right)\right).\end{equation} \tag{ 67 }$$

Therefore, the Dirac fermions at the vicinity of the Dirac point K are effectively governed by free-particle Hamiltonian which has no discrete energies in its spectrum. On the other hand, the Dirac fermions at the Dirac point K' are subject to the vector potential $\frac{{v}_{0}\beta }{a}{\mathbf{u}}_{22}\left(y\right)$ that can induce discrete energies. In particular, when u₂₂ ⩾ 0, the strain forms waveguide for essentially unidirectional wave packets in the K'-valley and the spectrum of H_−K looks like figures 4(b) or (c). The combination of the strain and the magnetic field does not confine Dirac fermions in the K-valley. When the condition (65) for the perfect valley polarization is not met, there can be discrete energy bands both in the spectrum of H_K and H_−K. However, their number goes to zero as the effective vector potential gets small in the corresponding valley. The combination of the external magnetic field with the strain in order to control propagation of the electrons in K and K' valleys appeared e.g. in construction of valley filters [10, 33, 46]. Valley currents in the nanoribbons induced by unidirectional strain were analyzed in [47].

In the end of the section, let us make few comments. The qualitative approach to the spectral analysis was based on the results obtained via standard variational principle. It was applied on the square on the effective Hamiltonian (49) that coincides with Pauli Hamiltonian [40]. If the electrostatic field was included, this approach would be no longer feasible as the square of the effective Hamiltonian acquires more complicated form. Nevertheless, it should be possible to study the influence of the small V(x) on the discrete energy bands of H(x, y) perturbatively, with the use of the test functions that provide estimates of the ground state energy of ${{\sim}{H}}_{k}^{2}\left(s\right)$, see [40] for more information.