Zero modes $a_j\sim e^{-\kappa j}$ in a semi-infinite quantum Ising chain? As a way of analyzing the performance of quantum annealing, I've been studying quantum diffusion in fermionizable lattice models with zero modes. 
In particular, the 1+1D quantum Ising model, semi-infinite in the spatial direction, is the simplest possible example of a fermionizable lattice model with a zero mode just outside of its paramagnetic phase. However, I've been having trouble producing these modes in ab initio calculations. I have that the semi-infinite quantum Ising model is dual to the non-interacting fermion model
$$H=i\sum_{j\geq 1}\left(B\gamma_{2j-1}\gamma_{2j}+J\gamma_{2j}\gamma_{2j+1}\right)\in \mathfrak{spin}_\infty$$
Utilizing the identification $\mathfrak{spin}_\infty\simeq \mathfrak{so}_\infty$ induced by the universal covering map, one gets a representation of the model as an antisymmetric matrix $\tilde{H}$, of which the zero modes should be spatially decaying eigenvectors. Perhaps the reason I can't find these modes is that I'm extracting the eigenvectors of $\tilde H$ via a bulk ansatz, where we note that the infinite Ising chain
$$H'=i\sum_{j\in \mathbb{Z}}\left(B\gamma_{2j-1}\gamma_{2j}+J\gamma_{2j}\gamma_{2j+1}\right)~~~~~~~~~~~$$
can be solved exactly using translational symmetry, and admits a formal translational eigenspace decomposition. The zero modes should then correspond to those formal eigenvectors with imaginary wave vector $k=i\kappa$ that also happen to satisfy the boundary condition of the semi-infinite chain. Furthermore, normalization constraints guarantee a maximum amplitude at the boundary of the lattice and an exponentially decaying amplitude into the interior, i.e. a pure waveform. However, when I try and fit the mode to the semi-infinite chain, the boundary condition reduces to
$$B^2\,\frac{Je^{\kappa}-B}{B-Je^{-\kappa}}=B^2+J^2-2JB\cosh \kappa\,$$
Which reduces to $B=B-Je^{-\kappa}$, an equation which admits only the completely localized mode $\kappa =\infty$. I wonder then, about how to obtain gently decaying solutions, and, furthermore, conditions on the transverse field that guarantee their existence. Some clarity in precisely where my logic/method fails would be great!
 A: The upshot is that you forgot to allow a complex component into the wave vector. The fact that your dispersion includes a trigonometric function like $\cosh$ or $\cos$, which are only defined on the real and imaginary axes, shows that you were hoping to find a purely real or imaginary wave vector. The true wave vector of the low energy mode lives on neither axis, so it is no surprise that you could not find it.
How do we find its explicit form? Better than an "bulk ansatz" is a rigorous mathematical approach. In fact, the Toeplitz extension from K-theory tells us that this low-energy mode in fact has zero energy, because the topological index of your model is nonzero and equal to one, and because of the fact that essential spectra are invariant under perturbations by a compact operator. This helps us immensely. 
Also, you should have used one more important symmetry of your Hamiltonian: it does not mix odd-index basis vectors or even-index basis vectors among themselves. This implies a block-decomposition
$$H=\begin{pmatrix}0&JS_L+B\\ -JS_L^\dagger-B&0\end{pmatrix}$$
Where $S_L$ is the left-shift operator. Always, always use symmetries in the problem to your advantage. Furthermore, from the formula for the analytical index (because all quadratic fermion Hamiltonians are Dirac-type operators, they share the same index formula):
$$\text{a-ind}(H)=\dim\ker (JS_L+B)-\dim\ker (-JS_L^\dagger-B)$$
The first term is invertible, and so we have the formula
$$\text{a-ind}(H)=\dim\ker (-JS_L^\dagger-B)=\text{t-ind}(H)=1$$
(Recall that your hamiltonian is a smooth deformation of the Kitaev model at zero chemical potential, so it shares its topological index, which is one.) Therefore, our zero mode is the unique solution to this equation guaranteed by the topological index computation:
$$S_L=-B/J$$
This implies that the zero-mode is an eigenvector of left-shift, with negative eigenvalue $-B/J$. This has a solution with wave vector
$$\boxed{\kappa = \log B/J +i\pi/2}$$
Which does not lie on the real or imaginary axes, and is instead in the interior of the complex plane.
