An hermitian operator problem

Question

It is possible to have two hermitian operators $A$ et $B$, with :

$B^2 = \mathbb{I}d$

$[A,B] = i * \mathbb{I}d$

where $i$ is the usual (complex) square root of $(-1)$, and $\mathbb{I}d$ is the identity operator ?

(I think that A is necessarily not bounded, due to the last condition)

If it is possible, may we exhibit a explicit representation of these operators $A$ and $B$ ?

You are correct that $A$ would have to be unbounded because $B$ is bounded (e.g., see here). But I don't know the answer. — Jonas Meyer, Jan 05 '12 at 11:04
For what it's worth, replacing $B$ with $\frac{B+1}{2}$ and $A$ with $2A$, the problem is the same if we ask whether we can have $B^2=B$, i.e., $B$ is an orthogonal projection. — Jonas Meyer, Jan 05 '12 at 11:24
@JonasMeyer I did not follow you, if I take $B = (B'+1)/2$, it gives $B'² = 3 \mathbb{I}d - 2 B'$ and not $B'² = B'$ ??? — Trimok, Jan 05 '12 at 11:48
@Trimok: he meant the opposite way around: $[(B+1)/2]^2 = (B^2 + 1)/4 + 2B / 4 = (B+1)/2$. And $[2A,(B+1)/2] = [A,B] = i$. So you can ask the equivalent question of $C^2 = C$ and $[D,C] = i$. — Willie Wong, Jan 05 '12 at 11:56
BTW, in view of that first comment by @Jonas perhaps you can drop the (linear-algebra) and (matrices) tags and replace by (functional-analysis)? — Willie Wong, Jan 05 '12 at 12:01

Nikolaj-K · Accepted Answer · 2012-01-05T13:07:19.347

5

The commutator $[A,B]$ is proportional to $\mathbb{I}$ and therefore commutes with everything. So $$A-BAB=AB^2-BAB=[A,B]B=B[A,B]=BAB-B^2A=BAB-A$$ or $$A=BAB.$$ But then $$[A,B]=AB-BA=(BAB)B-BA=BA-BA=0.$$ Btw. from this it follows that the result generalizes to all commutators with $$[A,B]=f(B).$$

edited Jan 05 '12 at 13:07

answered Jan 05 '12 at 12:46

Nikolaj-K

12,249

OK, nice, +1 for you – Trimok Jan 05 '12 at 13:02
1

@Trimok: Also, I suggest that you change the name of the question to "An idempotent operator problem". There is not so much hermitian about it, except that the commutator smells like quantum mechanics. – Nikolaj-K Jan 05 '12 at 13:20
Was your question motivated by a Putnam or IMO problem? – Nikolaj-K Jan 05 '12 at 18:21
It was motivated by a quantum mechanics interrogation (what is the representation of position and impulsion operators for a free particle), but, with some reflection , the question seems to be ill-defined. – Trimok Jan 05 '12 at 18:45
Which question? Anyway, you'll like this question. It's about the traps of infinitdimensional representations. – Nikolaj-K Jan 05 '12 at 19:21
OK, thank you very much. – Trimok Jan 05 '12 at 19:35
see also here – Trimok Jan 05 '12 at 19:59
k, thx as well. – Nikolaj-K Jan 06 '12 at 01:44

Geoff Robinson · Answer 2 · 2012-01-06T11:36:04.993

2

This is impossible. In fact, I will prove that if $A,B$ are operators in a real or complex algebra such that $B$ has finite multiplicative order ( that is, $B^{k} = I$ for some positive integer $k),$ then we can't have $AB -BA = cI$ for any non-zero scalar $c.$ For otherwise, an induction argument (due to Wielandt, and to be found in the link given in Jonas's comment) and which does not require any boundedness assumption, shows that we have $A(c^{-1}B)^{n} - (c^{-1}B)^{n}A = n(c^{-1}B)^{n-1}$ for every positive integer $n$. Taking $n =k$ yields a contradiction, since then the left side is $0,$ but the right side is non-zero.

An alternative approach is to note that if $AB - BA = I,$ then $B$ is not an algebraic element (when $B$ is Hermitian, this implies that the spectrum of $B$ must be infinite). By an algebraic operator, we mean an operator $T$ such that $f(T) =0$ for some monic polynomial $f(z) \in \mathbb{C}[z].$ Note that an algebraic operator has a unique minimum polyomial, that is a unique monic polynomial $p(z) \in \mathbb{C}[z]$ of least degree subject to $p(T) =0.$ For suppose that $B$ is algebraic, and let $q(z)$ be the minimum polynomial for $B,$ say of degree $r.$ Then $r >1$ since $AB \neq BA.$ Then we have $0 = Aq(B) - q(B)A = (r-1)B^{r-1} +$ (some linear combination of lower powers of $B$), so that $h(B) =0$ for some monic monic polynomial $h(z) \in \mathbb{C}[z]$ of degree $r-1,$ contrary to the fact that the minimum polynomial of $B$ has degree $r.$

edited Jan 06 '12 at 11:36

answered Jan 05 '12 at 12:05

Geoff Robinson

24,035

But A is not bounded, so the demonstration you made here does not work ??? – Trimok Jan 05 '12 at 12:51
It does work, but I have rewritten it in more generality in any case. – Geoff Robinson Jan 05 '12 at 17:02
Understood..., there is a clear contradiction for n=1 or n= 2, for instance. – Trimok Jan 05 '12 at 18:38
Sorry, but it seems impossible to put the flag "answer to the question" to several persons, which seems not very clever... – Trimok Jan 05 '12 at 18:46
Maybe it's an other question, but do you think the conclusion will be the same if B was bounded (but not a root of unity). – Trimok Jan 05 '12 at 19:07
I was wondering about that. I am not sure. The Wielandt-type argument does not seem to givie it immediately as far as I can see. – Geoff Robinson Jan 05 '12 at 20:31
@Trimok : I have now outlined an argument to show that when $B$ is Hermitian, its spectrum must at least be infinite. – Geoff Robinson Jan 06 '12 at 11:44
Thank you, I will study this. – Trimok Jan 09 '12 at 11:19

An hermitian operator problem

2 Answers2