Show that if $[Q,P]=it\Bbb{I}$ then the operators are unbounded In the Hilbert space $\mathcal{H} = L^2(\mathbb{R},dx)$, let 2 symmetrical operators $P$ and $Q$ be given, with the following properties: 


*

*$D(P) = D(Q) = \mathcal{S}(\mathbb{R})$

*$P\mathcal{S}(\mathbb{R}) \subset \mathcal{S}(\mathbb{R})$

*$Q\mathcal{S}(\mathbb{R}) \subset \mathcal{S}(\mathbb{R})$

*$[Q,P]=it\Bbb{I}$ on $\mathcal{S}(\mathbb{R})$


Where $\mathcal{S}(\mathbb{R})$ is Schwartz space, $t$ is a real constant different from $0$ and $[Q,P] = QP - PQ$. 
Show that operators cannot be bounded.
I've tried to define an operator $C=[Q,P]$ and prove that it's unbounded using sequence $\Phi_n(x)=char_{[n,n+1)}x$,where $char$ is characteristic function. And using the definition for a bounded operator I have the following:
$$||C\Phi_n(x)||=(\int_n^{n+1}|it|^2dx)^{1/2}=t$$
Which means $C$ in bounded. And from this moment I'm not sure in which direction I should move. Or maybe it was wrong from the beginning.
 A: The proof has nothing to do with the Schwartz space per se; nor with $i$ or $t$, or that $P$ and $Q$ are symmetric. If $P,Q$ are operators on a Hilbert space $H$ with domain $D$ and such that $PD\subset D$, $QD\subset D$, and $QP-PQ=\mathbb I$, then at least one of $P$ and $Q$ is unbounded. 
This applies to the case in the question because if we have $QP-PQ=it\, I$, we can replace $P$ by $P/i$ and $Q$ by $Q/t$. 
Actually, what one proves is that the equality $QP-PQ=I$ is impossible in a Banach algebra; below are three different arguments. In the three cases we show that at least one of $P,Q$ is unbounded. We cannot improve on that: for instance, on $L^2([0,1],dx)$ we can take $Q$ to be differentiation, and $P$ multiplication by $x$; then $P$ is bounded and $QP-PQ=I$ (and, as all separable Hilbert spaces are isomorphic, we can translate this example to $L^2(\mathbb R,dx)$).

First Proof: via the spectrum. If both $P,Q\in B(H)$, then $PQ$ is bounded, and we know that $\sigma(PQ)\cup\{0\}=\sigma(QP)\cup\{0\}$. Also, from $QP=\, I+PQ$, we have that
$$
\sigma(QP)=1+\sigma(PQ).
$$
So, if $\lambda\in\sigma(PQ)$, then $1+\lambda\in\sigma(QP)$, and then $1+\lambda\in\sigma(PQ)$. It follows that $n+\lambda\in\sigma(PQ)$ for all $n\in\mathbb N$, and thus $PQ$ is not bounded, a contradiction. This shows that at least one of $P,Q$ has to be unbounded. 

Second Proof: algebra and norm. Assume both $P,Q$ are bounded. From $QP-PQ=I$, multiplying on the right and on the left respectively, 
$$
QP^2-PQP=P,\ \ \ \ \ \ \ PQP-P^2Q=P.
$$
Adding, $QP^2-P^2Q=2P$. Repeating the trick inductively, we get
$$QP^n-P^nQ=2^{n-1}P^{n-1}.$$ Then
$$
2^{n-1}\|P^{n-1}\|=\|QP^n-P^nQ\|\leq\|QP^n\|+\|P^nQ\|\leq2\|Q\|\,\|P\|\,\|P\|^{n-1}.
$$
Thus, for every $n$, 
$$
2^{n-2}\leq\|P\|\,\|Q\|,
$$
a contradiction. 

Third Proof: deeper theorem. The result is a particular case of the Kleinecke-Shirokov theorem: if $P,Q$ are elements of a Banach algebra such that $[P,Q]$ commutes with $P$, then $[P,Q]$ is quasinilpotent (spectrum equal to $\{0\}$). As the identity $I$ is obviously not quasinilpotent, the relation $[P,Q]=I$ is impossible between bounded operators. 
