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Is there any physical significance of operator norm/spectral norm of a hermitian operator?

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Well, the operator norm of any normal - and therefore Hermitian - operator is equal to the spectral radius. Thus it gives the least upper bound on the magnitude of the largest eigenvalue - colloquially, the size of the largest observable quantity. Of course, many important operators on various Hilbert spaces in QM are not bounded, so there is not terribly much physical insight from this reasoning to be gained in those cases.

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I know this is a somewhat irrelevant technical remark, but in the case of operators on infinite dimensional Hilbert spaces, isn't it the case that there may not be a largest eigenvalue (say if there is an accumulation point in the spectrum like with the hydrogen atom), so (being rather unfamiliar with this stuff) I would imagine that it would instead give some sort of a bound on the spectrum instead of directly giving the magnitude of a largest eigenvalue? –  joshphysics Feb 20 '13 at 0:59
    
@joshphysics You're right of course - the spectral radius is the supremum of eigenvalue radii, and even the most well-behaved operators may fail to achieve it. I've improved the wording. –  Chris White Feb 20 '13 at 1:44
    
Cool thanks for the clarification and the nice answer. –  joshphysics Feb 20 '13 at 2:59
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