I am recording here some notes on a nice problem that Sorin Popa shared with me recently. To motivate the question, we begin with the basic observation that the differentiation operator {Df(x) := \frac{d}{dx} f(x)} and the position operator {Xf(x) := xf(x)} in one dimension formally obey the commutator equation

\displaystyle  [D,X] = 1 \ \ \ \ \ (1)

where {1} is the identity operator and {[D,X] := DX-XD} is the commutator. Among other things, this equation is fundamental in quantum mechanics, leading for instance to the Heisenberg uncertainty principle.

The operators {D,X} are unbounded on spaces such as {L^2({\bf R})}. One can ask whether the commutator equation (1) can be solved using bounded operators {D,X \in B(H)} on a Hilbert space {H} rather than unbounded ones. In the finite dimensional case when {D, X} are just {n \times n} matrices for some {n \geq 1}, the answer is clearly negative, since the left-hand side of (1) has trace zero and the right-hand side does not. What about in infinite dimensions, when the trace is not available? As it turns out, the answer is still negative, as was first worked out by Wintner and Wielandt. A short proof can be given as follows. Suppose for contradiction that we can find bounded operators {D, X} obeying (1). From (1) and an easy induction argument, we obtain the identity

\displaystyle  [D,X^n] = n X^{n-1} \ \ \ \ \ (2)

for all natural numbers {n}. From the triangle inequality, this implies that

\displaystyle  n \| X^{n-1} \|_{op} \leq 2 \|D\|_{op} \| X^n \|_{op}.

Iterating this, we conclude that

\displaystyle  \| X \|_{op} \leq \frac{(2 \|D\|_{op})^{n-1}}{n!} \|X^n \|_{op}

for any {n}. Bounding {\|X^n\|_{op} \leq \|X\|_{op}^n} and then sending {n \rightarrow \infty}, we conclude that {\|X\|_{op}=0}, which clearly contradicts (1). (Note the argument can be generalised without difficulty to the case when {D,X} lie in a Banach algebra, rather than be bounded operators on a Hilbert space.)

It was observed by Popa that there is a quantitative version of this result:

Theorem 1 Let {D, X \in B(H)} such that

\displaystyle  \| [D,X] - I \|_{op} \leq \varepsilon

for some {\varepsilon > 0}. Then we have

\displaystyle  \| X \|_{op} \|D \|_{op} \geq \frac{1}{2} \log \frac{1}{\varepsilon}. \ \ \ \ \ (3)

Proof: By multiplying {D} by a suitable constant and dividing {X} by the same constant, we may normalise {\|D\|_{op}=1/2}. Write {DX - XD = 1 + E} with {\|E\|_{op} \leq \varepsilon}. Then the same induction that established (2) now shows that

\displaystyle  [D,X^n]= n X^{n-1} + X^{n-1} E + X^{n-2} E X + \dots + E X^{n-1}

and hence by the triangle inequality

\displaystyle  n \| X^{n-1} \|_{op} \leq \| X^n \|_{op} + n \varepsilon \|X\|_{op}^{n-1}.

We divide by {n!} and sum to conclude that

\displaystyle  \sum_{n=0}^\infty \frac{\|X^n\|_{op}}{n!} \leq \sum_{n=1}^\infty \frac{\|X^n\|_{op}}{n!} + \varepsilon \exp( \|X\|_{op} )

giving the claim.
\Box

Again, the argument generalises easily to any Banach algebra. Popa then posed the question of whether the quantity {\frac{1}{2} \log \frac{1}{\varepsilon}} can be replaced by any substantially larger function of {\varepsilon}, such as a polynomial in {\frac{1}{\varepsilon}}. As far as I know, the above simple bound has not been substantially improved.

In the opposite direction, one can ask for constructions of operators {X,D} that are not too large in operator norm, such that {[D,X]} is close to the identity. Again, one cannot do this in finite dimensions: {[D,X]} has trace zero, so at least one of its eigenvalues must outside the disk {\{ z: |z-1| < 1\}}, and therefore {\|[D,X]-1\|_{op} \geq 1} for any finite-dimensional {n \times n} matrices {X,D}.

However, it was shown in 1965 by Brown and Pearcy that in infinite dimensions, one can construct operators {D,X} with {[D,X]} arbitrarily close to {1} in operator norm (in fact one can prescribe any operator for {[D,X]} as long as it is not equal to a non-zero multiple of the identity plus a compact operator). In the above paper of Popa, a quantitative version of the argument (based in part on some earlier work of Apostol and Zsido) was given as follows. The first step is to observe the following Hilbert space version of Hilbert’s hotel: in an infinite dimensional Hilbert space {H}, one can locate isometries {u, v \in B(H)} obeying the equation

\displaystyle  uu^* + vv^* = 1, \ \ \ \ \ (4)

where {u^*} denotes the adjoint of {u}. For instance, if {H} has a countable orthonormal basis {e_1, e_2, \dots}, one could set

\displaystyle  u := \sum_{n=1}^\infty e_{2n-1} e_n^*

and

\displaystyle  v := \sum_{n=1}^\infty e_{2n} e_n^*,

where {e_n^*} denotes the linear functional {x \mapsto \langle x, e_n \rangle} on {H}. Observe that (4) is again impossible to satisfy in finite dimension {n}, as the left-hand side must have trace {2n} while the right-hand side has trace {n}.

As {u,v} are isometries, we have

\displaystyle  v^* v = u^* u = 1; \ \ \ \ \ (5)

Multiplying (4) on the left by {v^*} and right by {u}, or on the left by {u^*} and right by {v}, then gives

\displaystyle  v^* u = u^* v = 0. \ \ \ \ \ (6)

From (4), (5) we see in particular that, while we cannot express {1} as a commutator of bounded operators, we can at least express it as the sum of two commutators:

\displaystyle  [u^*, u] + [v^*, v] =1.

We can rewrite this somewhat strangely as

\displaystyle  [\frac{1}{2} u^*, 4u+2v] + [\frac{1}{2} u^* - v^*, -2v] = 2

and hence there exists a bounded operator {a} such that

\displaystyle  [\frac{1}{2} u^*, 4u+2v] = 1+a; \quad [\frac{1}{2} u^* - v^*, -2v] = 1-a.

Moving now to the Banach algebra of {2 \times 2} matrices with entries in {B(H)} (which can be equivalently viewed as {B(H \oplus H)}), a short computation then gives the identity

\displaystyle  \left[ \begin{pmatrix} \frac{1}{2} u^* & 0 \\ a & \frac{1}{2} u^* - v^* \end{pmatrix}, \begin{pmatrix} 4u+2v & 1 \\ 0 & -2v \end{pmatrix} \right] = \begin{pmatrix} 1 & v^* \\ b & 1 \end{pmatrix}

for some bounded operator {b} whose exact form will not be relevant for the argument. Now, by Neumann series (and the fact that {u,v} have unit operator norm), we can find another bounded operator {c} such that

\displaystyle  c + \frac{1}{2} v c u^* = b,

and then another brief computation shows that

\displaystyle  \left[ \begin{pmatrix} \frac{1}{2} u^* & 0 \\ a & \frac{1}{2} u^* - v^* \end{pmatrix}, \begin{pmatrix} 4u+2v & 1 \\ vc & -2v \end{pmatrix} \right] = \begin{pmatrix} 1 & v^* \\ 0 & 1 \end{pmatrix}.

Thus we can express the operator {\begin{pmatrix} 1 & v^* \\ 0 & 1 \end{pmatrix}} as the commutator of two operators of norm {O(1)}. Conjugating by {\begin{pmatrix} \varepsilon^{1/2} & 0 \\ 0 & \varepsilon^{-1/2} \end{pmatrix}} for any {0 < \varepsilon \leq 1}, we may then express {\begin{pmatrix} 1 & \varepsilon v^* \\ 0 & 1 \end{pmatrix}} as the commutator of two operators of norm {O(\varepsilon^{-1})}. This shows that the right-hand side of (3) cannot be replaced with anything that blows up faster than {\varepsilon^{-2}} as {\varepsilon \rightarrow 0}. Can one improve this bound further?