Why do discrete-time quantum walks need an extra coin space, but continuous-time quantum walks do not? As defined in the literature, discrete-time quantum walks use an extra coin space to decide in which direction to move.
To get an idea consider this classical example: Assume the walker on an integer line at 0 with a coin in his hands. The classical random walk proceeds as follows. He tosses the coin and depending on head or tail he decides to go left or right respectively. After a certain number of steps, he will reach to a random position.
On the other hand, in the quantum case, the coin can be in a superposition of head and tail. Now, in the discrete case, there is this extra coin space attached to the position space to go over the graph. In the continuous-time case, there is no coin. The process evolves similarly to continuous time classical random walk.
I am not sure why this happens. It will be great if somebody helps me understand it.
 A: They are different dynamical models.
In its simplest form, a Discrete-Time Quantum Walk (DTQW) is defined over a bipartite space of the form $\newcommand{\HC}{\mathcal H_{\mathcal C}}\newcommand{\HW}{\mathcal H_{\mathcal W}}\HC\otimes\HW$, where $\HC$ is the (often two-dimensional) coin space and $\HW$ a higher-dimensional walker space. A step of a DTQW is described by an operator $\mathcal W=\mathcal S\,\mathcal C$ that is obtained as the combined action of a controlled-shift operation $\mathcal S$ and a coin flipping operation $\mathcal C$.
Note that in this model the underlying Hamiltonian is not important. You are assuming to have access to the unitaries $\mathcal S,\mathcal C$, without worrying about how these operations are actually implemented physically.
This is a simple model that can describe interactions between low- and high-dimensional systems.
On the other hand, in a Continuous-Time Quantum Walk (CTQW) there is no partition of the underlying space, nor you have different degrees of freedom (thus in particular there is no "coin"). The dynamics is defined via the Hamiltonian and follows Schrodinger's equation: $$\lvert\Psi(t=0)\rangle\to\lvert\Psi(t)\rangle=e^{-i H t}\lvert \Psi(t=0)\rangle.$$
CTQWs and DTQWs are related in the sense that one can simulate one with the other, but this is not a straightforward result. See e.g. section $2.2$ in these notes by Mack Johnson (Link to pdf) for more details.
