Let us take the initial state with the particle located at the origin $|n=0\rangle$ and the coin state with spin up $|0\rangle$. So, $$ |\psi(0)\rangle = |0\rangle|n=0\rangle, $$ where $|\psi(0)\rangle$ denotes the at the initial time and $|\psi(t)\rangle$ denotes the state of the quantum walk at time $t$.

Please, help me understand that, if on RHS we have a matrix product of two column vectors?


1 Answer 1


This is common shorthand for the tensor product. That is, you should read it as $|0 \rangle \otimes | n=0\rangle$.

  • $\begingroup$ |a⟩⊗|b⟩=|a,b⟩ |a⟩⊗|b⟩=|ab⟩ , Are these two notations also true? $\endgroup$ Nov 5, 2020 at 16:06
  • $\begingroup$ They're also notations I've seen elsewhere, yeah. $\endgroup$
    – Rammus
    Nov 5, 2020 at 16:07
  • $\begingroup$ I would doubt it. What kind of matrix product are you thinking of? If written $|a \rangle \langle b|$ then it denotes the outer product usually but this would not make sense given the LHS of your equation which is a vector $|\psi(0)\rangle$. The only other possibility I could imagine (but which I very very much doubt is the case) is that it denotes an elementwise product but this only makes sense if $|\psi(0)\rangle$, $|0\rangle$ and $|n=0\rangle$ are all vectors from the same dimensional vector space. If this notation is not specified explicitly then I am confident it is the tensor product. $\endgroup$
    – Rammus
    Nov 5, 2020 at 16:14
  • $\begingroup$ I just checked the reference of the book which I am reading. It denotes the tensor product. Thanks for your help. These different notations made me confuse. $\endgroup$ Nov 5, 2020 at 16:17

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