# Proving shift operator on infinite dimensional Hilbert space is not unitary

Consider an infinite-dimensional Hilbert space with an orthonormal basis: $$\{\lvert{n}\rangle\}, n = 0,1,2,...$$ Define a shift operator such that $$S\lvert{n}\rangle = \lvert{n+1}\rangle$$

Then find a representation of $$S$$ in terms of the above basis and prove that it is not unitary namely: $$S^\dagger S = 1 \ne S S^{\dagger}$$

I tried to consider the representation $$S = \sum_{n} \lvert n+1 \rangle\langle n \lvert$$ $$S^{\dagger} = \sum_{n} \lvert n \rangle\langle n+ 1\lvert$$

But I'm not sure why $$S S^{\dagger} \ne 1.$$

• You should be able to compute $SS^\dagger$ explicitly with little effort. What do you get when you try? Oct 9, 2021 at 18:39

Look at what happens with $$|0\rangle$$ when you try to apply $$SS^{\dagger}$$ to it. The summation in the last two expressions run from $$n=0$$. As a result, the combination $$SS^{\dagger}$$ does not contain a term that operates on $$|0\rangle$$.