# Confusion about position operator in QM

1. Nov 21, 2014

### ShayanJ

In quantum mechanics, the position operator(for a single particle moving in one dimension) is defined as $Q(\psi)(x)=x\psi(x)$, with the domain $D(Q)=\{\psi \epsilon L^2(\mathbb R) | Q\psi\epsilon L^2 (\mathbb R) \}$. But this means no square-integrable function in the domain becomes non-square-integrable after being acted by this operator which, in turn, means there exist no function in the domain for which you can't have a M that satisfies $||Q \psi|| \leq M ||\psi||$ and so this operator should be bounded. But people say its not bounded!
I'm really confused. What's the point here?
Thanks

2. Nov 22, 2014

### rubi

The point is that there needs to be an $M$ such that this inequality holds for all $\psi$. However, for the position operator, you can easily check that there cannot be such an $M$, for example by picking a sequence $\psi_n$ of normed, centered Gaussians with increasing standard deviation $\sigma_n = n$ (since $\lVert Q\psi_n\rVert = \sigma^2$, so you would need an $M$ such that $\forall n\in\mathbb N:M \geq n^2$, which doesn't exist).