## An interesting result in functional analysis

Theorem. For any bounded operators A and B on a complex Hilbert space H, it is impossible that AB – BA = I, where I is the identity operator on H.

To establish the theorem, we note some elementary results concerning the spectrum of a bounded operator on H.

Definition. For any bounded operator A on H, the spectrum of A is given by $sp(A) = \{ \lambda \in \mathbb{C} : A - \lambda I \text{ is not invertible} \}$.

Proposition 1. For any bounded operator A, sp(A)  is a nonempty closed subset of the closed disk with radius $||A||$.
Proof: [Next time. The relatively more difficult part is that the spectrum is never empty.]

Proposition 2. For any bounded operators A and B, $sp(AB) \cup \{0\} = sp(BA) \cup \{0\}$.
Proof: Suppose c is nonzero such that AB -cI is invertible. It suffices to consider the case c = 1. Let C be the inverse of AB – I. We claim that BCA – I is the inverse of BA – I. Indeed,

(BA – I)(BCA – I) = BABCA – BA – BCA + I =  B(AB – I)CA – BA + I = BA – BA + I = I

and similarly (BCA -I)(BA – I) = I.

Proof of Theorem: Suppose AB – BA = I for some bounded operators A and B on H. Then by Proposition 2, for any positive integer n, if both sp(AB) and sp(BA) contain 0 or do not contain 0 then

$sp(AB) = sp(I + BA) = 1 + sp(BA) = 1 + sp(AB) = 2 + sp(AB) = \cdots = n + sp(AB);$

and if 0 lies in precisely one of sp(AB) and sp(BA), say in sp(BA),  then

$sp(AB) = 1 + sp(BA) = 1 + (sp(AB) \cup \{0\}) = (1 + sp(AB)) \cup \{1\} = (2 + sp(AB)) \cup \{1, 2\} = \cdots = (n + sp(AB)) \cup \{1,2,\dots,n\}.$

Therefore sp(AB) or sp(BA) is unbounded, contradicting Proposition 1.

A consequence of the theorem is that the Heisenberg commutation relation in quantum mechanics cannot be implemented using bounded operators.

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### 8 Responses to An interesting result in functional analysis

1. KK Kwong says:

For the finite dimensional case, taking trace on both sides and use the fact that
$tr(AB)=tr(BA)$ will prove the result. Is there any elementary proof like that in the general case?

It would be nice too if you could explain a little bit about quantum mechanics too.

2. Ken Leung says:

The general proof is to consider the spectrum of AB and BA. I’ll elaborate the post later on when I’m free. For the quantum mechanical part, I’m still learning this and perhaps I’ll write something about it.

• kingleunglee says:

It is interesting. I try finding such operator before…

3. michaelngelo says:

I thought there is such pair of matrices too, and tried to search about it in a Gilbert Strang’s book(I can’t find a hard copy, just found an old version online) after seeing this post. It seems that what the book said is not so correct…

4. KKK says:

Can you give an example of $A, B$ such that $0\in sp(AB)$ but $0\notin sp(BA)$?

• KKK says:

Hmm… seems I’ve asked an embarrassingly stupid question. Just find $A, B$ such that $AB$ is invertible but $BA$ is not. For example $l^2=\{(a_1, a_2, \cdots): \sum |a_i|^2<\infty\}$ with $B$ being the right translation $(a_1, a_2, \cdots)\mapsto (0, a_1, a_2, \cdots)$ and $A$ being the left translation $(a_1, a_2, \cdots)\mapsto (a_2, a_3, \cdots)$.

5. honleungdydx says:

About the last sentence, maybe Ken can give a brief introduction on unbounded operators and how to use them.

• Ken Leung says:

Well, I still haven’t stated learning unbounded operators. Perhaps later I’ll talk about the last point when I read more about that issue. I guess those who study PDE often encounter unbounded operators!