2 added 2 characters in body; edited title

# Why is addition of operatorsobservables in quantum mechanics commutative?

I am no expert in the field. I hope the question is suitable for MO.

Background/Motivation

I once followed a quantum mechanics course aimed at mathematicians. Instead of the usual motivations coming from experiment at the turn of the 19th century, the following argument (more or less) was given to show that the QM formalism is in some sense unavoidable.

When one does physics, he is interested in measuring some quantity on a given state of the universe. The quantity (say the speed of a particle) is defined experimentally by the tool used to do the measure. We define such an instrument, with a given measure unit, an observable. So for every state and every observable we get a real number.

We can now define a sum and a product of observables. These are obtained by performing the two measures and then adding or multiplying their values. Similarly we can define scalar multiplication. These operations are then associative, but there is no reason why they should be commutative, since performing the first measure can (and indeed does) change the state of the universe. For some reason I cannot understand, anyway, addition is assumed commutative. I also see no reason why multiplication should distribute over addition. We can now also consider observables with complex values, by linearity.

At this point observables form an $\mathbb{R}$-algebra. We intoduce a norm it as follows. The norm of an observable is the sup of the absolute values of the quantities which can be measured. Every instrument will have a limited scale, so this is a real number. By definition this is a norm. Moreover it satisfies $\|A B \| \leq \|A\| \| B \|$. We can now formally take the completion of our algebra and obtain a Banach algebra.

Finally we define an involution * on our algebra by complex conjugation of observables. This yields a Banach * -algebra, and the third assumption which is mysterious to me is that the $C^*$ identity holds.

Finally we can use the Gelfand-Naimark theorem to represent the given algebra as an algebra of operators on a Hilbert space. If this turns out to be separable, it is isomorphic to $L^2(\mathbb{R}^3)$ and we recover the classical Schrodinger formalism.

The problems

In this approach I see three deductions which seems arbitrary: addition is commutative, multiplication is distributive and the $C^*$ identity holds. Is there any kind of hand-waving which can jusify these? In particular

Why is addition of operators observables commutative, while multiplication is not?

1

# Why is addition of operators in quantum mechanics commutative?

I am no expert in the field. I hope the question is suitable for MO.

Background/Motivation

I once followed a quantum mechanics course aimed at mathematicians. Instead of the usual motivations coming from experiment at the turn of the 19th century, the following argument (more or less) was given to show that the QM formalism is in some sense unavoidable.

When one does physics, he is interested in measuring some quantity on a given state of the universe. The quantity (say the speed of a particle) is defined experimentally by the tool used to do the measure. We define such an instrument, with a given measure unit, an observable. So for every state and every observable we get a real number.

We can now define a sum and a product of observables. These are obtained by performing the two measures and then adding or multiplying their values. Similarly we can define scalar multiplication. These operations are then associative, but there is no reason why they should be commutative, since performing the first measure can (and indeed does) change the state of the universe. For some reason I cannot understand, anyway, addition is assumed commutative. I also see no reason why multiplication should distribute over addition. We can now also consider observables with complex values, by linearity.

At this point observables form an $\mathbb{R}$-algebra. We intoduce a norm it as follows. The norm of an observable is the sup of the absolute values of the quantities which can be measured. Every instrument will have a limited scale, so this is a real number. By definition this is a norm. Moreover it satisfies $\|A B \| \leq \|A\| \| B \|$. We can now formally take the completion of our algebra and obtain a Banach algebra.

Finally we define an involution * on our algebra by complex conjugation of observables. This yields a Banach * -algebra, and the third assumption which is mysterious to me is that the $C^*$ identity holds.

Finally we can use the Gelfand-Naimark theorem to represent the given algebra as an algebra of operators on a Hilbert space. If this turns out to be separable, it is isomorphic to $L^2(\mathbb{R}^3)$ and we recover the classical Schrodinger formalism.

The problems

In this approach I see three deductions which seems arbitrary: addition is commutative, multiplication is distributive and the $C^*$ identity holds. Is there any kind of hand-waving which can jusify these? In particular

Why is addition of operators commutative, while multiplication is not?