# Use of Position Hilbert Space in Quantum Walk

To perform quantum walk first we need to define combine hilbert space for the position and coin, which is represented by: $$H=H_p\bigotimes H_c$$
Now, my question is what is the use of above representation, if all the operation are defined on initial state. For example here, $$| \psi(0)\rangle = |0\rangle\bigotimes\mid n=0\rangle$$
Here, first vector is coin with initial state $$0$$ and second vector is walker at initial node $$0$$.

• I'm not sure what your question is, could you elaborate? Are you asking why you have a tensor product structure? If so, it is because you are dealing with two different systems, one for the coin and one for the position. Combining them, the joint system is defined on the tensor product of the two Hilbert spaces. – Rammus Nov 7 at 21:14
• No, my question is what is point of defining hilbert space H, when all the operations related to quantum walk are going to be performed on initial state. – Binshumesh sachan Nov 8 at 14:33
• @Binshumeshsachan The point of a Hilbert space is to have all possible states of a quantum system. This is the set where not just the initial states but the time-evolved states reside as well, i.e., not just $|\psi(0)\rangle = |0\rangle_p \otimes |0\rangle_c$ but also $|\psi(t)\rangle$. – keisuke.akira Nov 9 at 13:36
• So, defining that is just for representational purpose. All the operations we define, will be actually on the initial state. – Binshumesh sachan Nov 9 at 14:32
• @Binshumeshsachan As I said, it is the set of all possible states; just like in classical mechanics, where we specify the entire phase space, even if a specific dynamical system only traces out a specific curve in the phase space. No, the operations are defined on an arbitrary initial state, just like a function, $f : \mathbb{R} \rightarrow \mathbb{R}$ and so it helps to define the set of all states, namely the Hilbert space. – keisuke.akira Nov 9 at 23:27