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

-
Hurriedly looking over your question, don't you really want to ask why addition of observables is commutative? I mean, we know why addition of operators is commutative, but that doesn't seem to be what you're asking. – Yemon Choi Feb 6 2010 at 11:48
Sorry, I edit it. – Andrea Ferretti Feb 6 2010 at 11:50
It seems I cannot edit the title: it gets changed here, but not in the main page. Of course addition of observables was what I had in mind. – Andrea Ferretti Feb 6 2010 at 11:52
I don't think that is how addition and multiplication are defined. In fact it is definitely not that way, since we know the answer for quantum mechanics. Consider the finite dimensional case. Observables are modeled by certain operators and the numbers you get from measurement are eigenvalues. But operators' addition is not defined as addition of their eigenvalues unless you can simultaneously diagonalize them. – MBN Feb 6 2010 at 16:43
@Georges I did not mean for that to come across as condescending, or imply that I am in any way so brilliant that I find this book simple. I simply meant, that reading the book in the way I did, it did not require the level of effort that serious reading normally does, and most of what I read of it, took place walking across campus, lying in bed, or in a coffee shop. I hope I didn't give the wrong impression. – B. Bischof Feb 6 2010 at 22:00
show 4 more comments

Your description of the structure of the algebra of observables isn't quite how I'm used to it being. Indeed, I believe that in the best algebraic descriptions of quantum mechanics, addition is a formal operation, rather than a physical operation as you've described. The best reference I know for this point of view is L.D. Faddeev and O.A. Yakubovskii, 2009, Lectures on Quantum Mechanics for Mathematics Students. I don't have my copy handy right now, so I will describe my memory of how they set up the algebra of observables.

The first thing to point out is that in the real world, there is no such thing as pure states. This has nothing to do with quantum mechanics, and everything to do with an experimenter's inability to perfectly measure the initial set-up. For your notion of "state" to make sense physically, it must be something like "repeatable initial set-up for an experiment". Once this is your notion of state, you are perfectly able to run your experiment 1000 times, make your measurements (each individual run may give a different answer, but you can look at the distribution), and process them as you want.

So really an observable assigns a probability distribution on $\mathbb R$ to each state. We now demand the following axiom: the (good) functions $\mathbb R \to \mathbb R$ act on the set of observables by composition. So if $X: \{\text{states}\} \to \{\text{probability distributions}\}$ is an observable, so is $X^2$: the probability that the observable $X^2$ assigns to an interval $[a,b]$ is the same as the probability that $X$ assigns to the interval $[\sqrt a,\sqrt b]$. In particular, suppose you compose your observable $X$ with a step function $\Theta(x - \xi)$, where $\xi \in \mathbb R$. Then the observable $\Theta(X-\xi)$ measures the whether the value of $X$ is more than $\xi$. Then you can check that the full distribution $X$ is recoverable from the knowledge of all the $\Theta(X-\xi)$. In particular, it's recoverable from the expected values of $\Theta(X-\xi)$ on each state. So to set up the algebra of observables, it's enough to know only the expectation values for observable at each state.

Now you should realize the following. The previous paragraph makes sense even for classical mechanics, and in fact is the correct formalism (as there are no pure states). But in quantum mechanics, it's worse than that. A definite state is one that gives a delta distribution for each observable. Classically, we believe that a sufficiently good experimenter can approximate definite states to whatever desired accuracy. But there is very good evidence that this fails in the quantum world: no matter what tools you use, there are absolute bounds preventing states from approximating definite states. So the language of distributions and expectations is absolutely necessary to formalize quantum mechanics, whereas in classical mechanics you could say that there are idealized definite states, observables are functions on definite states, and states are probability distributions on the space of definite states.

Finally, the question is how to assign algebraical operations to the collection of observables. And here I admit that I don't have a great answer. One possibility is simply to convolve probability distributions: this gives a commutative addition, for example. Then you could define a commutative associative multiplication by taking logs and adding and exponentiation, but my memory is that this does not distribute over addition in general. F&Y define a commutative nonassociative multiplication by $(X,Y) = \frac12\bigl((X+Y)^2 - X^2 - Y^2\bigr)$. Oh, right. The problem is the following: do you add, multiply, etc. the distribitions, or the expectation values? For addition, adding expectation values is the same as the usual convolution of distributions and then taking expectation. But for multiplication it is not. I don't remember what F&Y do, but I think it might at the level of expectation values.

-
Wait, so it is in theory possible to have pure states? I thought that QM was nondeterministic. – Harry Gindi Feb 6 2010 at 23:29
@HG: I left out a definition. Given two states $\mu, \nu$, you can mix them with proportion $p, 1-p$, where $0 \leq p \leq 1$ is some probability, by flipping a (classical) coin that lands Tails with probability $p$ and Heads with probability $1-p$, and depending on how the coin falls, set up your experiment in state $\mu$ or $\nu$. You can call the resulting state $p\mu + (1-p)\nu$. Then a pure state is a state $\alpha$ so that if $\alpha = p\mu + (1-p)\nu$, then either $\alpha = \mu$ or $\alpha = \nu$. Classically, pure states are the same as definite states, but this fails in QM. – Theo Johnson-Freyd Feb 6 2010 at 23:53
The failure of this is the "nondeterminacy" of QM. Whether you can actually set up a pure state (really you cannot), you can (presumably) approximate pure states with arbitrary accuracy. But in quantum mechanics, there are absolute bounds preventing you from approximating definite states: for any state, there is an observable that does not give a single $\delta$-distribution as output when it eats that state, i.e. it assigns a nontrivial probability distribution. – Theo Johnson-Freyd Feb 6 2010 at 23:56
Of course, in every theory, the mechanics is deterministic. In QM, and also in the classical world, "dynamics" or "mechanics" describes how the "value" (a probability distribution or equivalently an expectation) of an observable on a given state evolves over time. It's reasonable to insist that the $\mathbb R\to \mathbb R$ action on observables is time-independent; then in particular any theory set up the way I've described has linear, deterministic evolution at the level of probability distributions. – Theo Johnson-Freyd Feb 6 2010 at 23:59
@HG: Time evolution in QM is always deterministic. Any state, mixed or pure, evolves deterministically via the Schrodinger equation. The formalism I outlined above fits better with the Heisenberg picture (states don't evolve), in which case the observables evolve deterministically (and linearly). The only non-deterministic part is the measurements: the pairing between states and observables. In classical mechanics, the value of an observable on a pure state is determined. In quantum mechanics, even on a pure state the value of an observable is not determined. – Theo Johnson-Freyd Feb 7 2010 at 17:52
show 2 more comments

We may think of a state $\omega$ as a functional on the algebra of observables $\mathcal O$ which is interpreted as giving the expected value of each observable. With this in mind, it is natural to require $\omega$ to be linear (as well as two other usual properties, positivity and normalization).

Thus given two observables $A, B \in \mathcal O$, if we are going to have a sum $A + B$ it should be true that $\omega(A + B) = \omega(A) + \omega(B)$ for any state $\omega$. Since this gives the values of $A + B$ on every state, it suffices to define it. Since $B + A$ has the same values on every state, $B + A$ is the same observable.

On the other hand, there is no natural way to say what $\omega(AB)$ should be, since states (like expectation values) need not be multiplicative.

So: commutativity of observables reduces to commutativity of $\mathbb C$ since expectations are linear, but nothing analogous applies to multiplication.

This is based on what I've read in F. Strocchi, An Introduction to the Mathematical Structure of Quantum Mechanics.

Note that you can eventually interpret states as arising out of probability distributions, leading to Theo's comments.

Personally I am still a bit hazy on why we postulate a multiplication on observables at all, when (unlike the classical case) there is not a clear physical interpretation of what such an operation should mean. However, given the full structure of a $C^*$-algebra, one can show that the uncertainty principle (or the existence of complementary observables) requires noncommutative multiplication, et voila, you have quantum mechanics.

-
 I followed myself the course from Strocchi, and it is exactly his argument I'm trying to fill. Defining a state as a linear functional seems as arbitrary as defining it as a multiplicative linear functional. The question is: how con we derive these properties from the description of the process of measure alone? I have tried to explain the full reasoning in my question. – Andrea Ferretti Jul 28 2010 at 10:20 Basically, a state is the current situation of the universe, an observable is some physical quantity we have some instrument to measure. They have an obvious pairing: in a given state, apply your instrument and read the result. If you now define sum and multiplication of observables by doing the two measures one after the other, the result may depend on the order. And indeed it does for the multiplication. Why not for the addition? – Andrea Ferretti Jul 28 2010 at 10:23 If you say $\omega(A + B)$ is the expected value of preparing the universe in state $\omega$, then measuring $A$, then measuring $B$ in the state the universe was left in after the first measurement, then summing the results, then indeed $\omega(B + A)$ may be different. Instead, $A + B$ represents a single measurement we can do that results (in expectation) in the sum of $A$ and $B$. That is a more sensible requirement than the corresponding one for multiplication, since even if $A$ and $B$ are somehow linked (as they are in QM), expectation is still linear. – Jonathan L Long Jul 29 2010 at 8:41 As for a purely physical definition of addition of observables, I don't have one, and neither does Strocchi; he even notes that the introduction of the sum leads to an extension of the physically defined observables. The situation for multiplication seems even worse, so I am not sure how much more physical motivation can be given at the purely algebraic level. – Jonathan L Long Jul 29 2010 at 8:47 While Theo's comments are excellent, yours (Jonathan) get to the heart of the difference between addition and multiplication: states are expectation values, expectations values add, they don't multiply. (Obviously there is more to be said.) – Toby Bartels Jan 3 2012 at 17:54

The introduction you outlined is basically reminiscent of the old quantum mechanics, anyway in the approach you depicted it is the culmination and not the premise of the construction, and the comment was surely intended to be explanatory of the far origin of these choice. I now try to resume the history.

There are basically two approaches to mathematical quantum mechanics. The first one very complex and stratified in its development, but simple in the premise was discussed by John Von Neumann in a lot of papers after the "Foundation of quantum mechanics", the second one is basically conceveid to be an extension for the first, and is this second approach you are referring to: the GNS approach.

Anyway both of them are surely derived after an abstraction process very far beginning on the methods of classical mechanics, joint to the newest evidence from atomic and particle physics of the first quantum mechanics.

Just as in classical mechanics we define functions of observables dynamical quantity so the founders of quantum mechanics conceived it is possible in quantum mechanics, anyway we need to clarify in which sense this is possible and explanation isn't fully depleted from the naive extension of classical theory of the measure, based on real numbers, but it need of a clear axiomatic and this was furnished from John Von Neumann (and in some way from Heisenberg, Dirac, and Schroedinger before him formulated this axiomatic)

Anyway, just as in classical mechanics there is a notion of repeatibility and regularity, so there is in quantum mechanics. The true difference is in the outcome of the measures, deterministic in classical, probabilistic in quantum mechanics. So that measure processes are conceived deterministic in a statistical sense, and, for example, the component energies of the isotropic harmonic oscillator sums exactly in mean value, but the variance is zero if the considered states are eigenstates. Old quantum mechanics can be founded on few axioms about the measures and led Von Neumann, in a natural way to linear operators acting, like a non commutative algebra, on Hilbert spaces.

In order to grant correspondence principle we, following the founders of quantum mechanics, need to hypothesize the existence of intrinsically deterministically evolving observable, and just the measure process make the difference, because these dynamical "quantities" with respect to the measures doesn't appear as real numbers, this point was the first time realized some time after the Copenaghen interpretation was developed.

So they are assumed, after Heisenberg (speaking of non commutative numbers) and Jordan (speaking of matrices), and Schroedinger (speaking of operators acting on functional space of probability) all these three point of view were showed to be in a certain strict framework to be equivalent, from Dirac assuming they are algebraic elements obeying to canonical commutation relation generalizing the Poisson algebra.

In brief the Dirac point can be summarized in assuming an Hilbert space structure for the states, and in developing step by step a theory of observables compatible with the Copenaghen interpretation spirit and with the correspondence principle.

Anyway Von Neumann felt the need to obtain an axiomatic foundation based on more general operators algebras, and an axiomatic of measure, unifying from scratch the theoretical framework, in fact obtaining a more general theory with respect to the Heisenberg and Dirac theoretical "prejudices". The Von Neumann point was in fact based on the general representation theory in the geometrical framework of Banach operator algebras of operators in Hilbert space, and in particular on the CCR irreducible representation theory, but from this point the research of Von Neumann continued in search of an intrinsic point of view based on the geometry of observable.

After time and time was in fact recognized that part of quantum theory of measure is nothing else then a generalized probabilistic theory in a Banach algebra and the general setting of Gelfand Najmark Segal construction rebuild intrinsically the Hilbert spaces. Anyway the field extension of this setting is very problematic and a hierarchy of Hilbert spaces appears. Anyway in this way a circle is closed and a new loop is opened: in the GNS approach to quantum mechanics we postulate that operators are living in an abstract algebra, obeying familiar rules for an algebra with an involution (the * operation). Via Gelfand theorem the commutative case led to the algebra of complex valued continuous functions in an Hausdorf space, the spectrum of the algebra (which will led the ordinary numerical set of coordinates of classical mechanics), and more in general to a spectral theory, culminating in the GNS construction, which associate to a given linear form an Hilbert space and a representation for the algebra.

Anyway the true achievement of this approach is the net of algebras, that is very more general with respect to the Hilbert space interpretation of quantum mechanics,this achievement is useful in relativistic field theory and leads to very far reaching results firstly partially discovered by Von Neumann in some papers, and after then developed from Araky, Haag, Kastler. In this full setting is now possible to address in more precise terms the question of the cluster decomposition principle implicit in the deterministic evolutionary scheme, and the question of repeatability principle of classical and quantum mechanics, and to understand quantitatively something about the limitation, arising from the change of the state of the universe, to this principle, which can be espressed, for example, in term of a change of representation, becaused from the change of the linear form representing the thermokinetic state of "universe", without any change in the postulates of quantum field theory and the derived quantum mechanics. This is perhaps the perspective of the search about KMS theorem.

I'm not very satisfied from this resume, anyway I think you can correct and integrate it, and I hope to read and write something else more precise and delimited.

-
 This looks very interesting and covers a lot of ground. Would you object if I made some small edits for language? – Yemon Choi Feb 7 2010 at 0:14 It looks very interesting indeed, but sadly it does not touch in any way the original question. – Andrea Ferretti Feb 8 2010 at 19:56 @ YC sure you can edit it, I'm sure there are a lot of mistaken. Thanks in advance. – Tetis - Gianmarco Bramanti Feb 9 2010 at 20:59 @AF I think the points "touching" the very interesting question you pointed out are two: the choice of a algebra structure for operator is IMHO based on both: conservative approach with regard to the construction of quantum mechanics ( in order to safe a predictive theory, so that the effect of measure in the state of universe was at first neglected at all) an empiric base with respect to the measure of quantum quantities corresponding to additive classical quantities. E.g. the energies and spin components. Anyway in old QM these were partly "theorems" nowadays, at all, definitions. – Tetis - Gianmarco Bramanti Feb 9 2010 at 21:17

Since you raised your question I'm uncomfortable about some question, and I re-read Von Neumann, in order to becalm myself, anyway in Von Neumann the problem isn't solved in a deductive way neither is justified at all, only is asserted the additivity, as customary, between commutating operators (so there the correspondence principle is granted) and then the additivity is extnded to non commutating operators.

Anyway reflecting on the practical use of non commutating linear combination of operators I realize that the arguments of linear combination are generally elements of some Lie Algebra, and their powers, and I remember a point in the first book of Landau about classical mechanics that I like to quote:

Conservation laws.

"Not all integrals of motion have an equally relevant role in mechanics. Among these there are some whose invariance over time has an origin very deep, related to fundamental properties of space and time and that is their homogeneity and isotropy. These quantity, these conservative, have an important general property, they are additive, that is, their value for a system composed of several elements, whose interaction can be neglected, is equal to the sum of the values for each of the elements."

It seems just like Landau is mixign two unrelated points: the isotropy and the commutativity. In fact this isn't the additivity we are thinking to. Anyway there is an important point: and this is the role of simmetry and in mathematic the role of the Stone-Neumann Theorem, and the role of parallel transport, and in mathematic the role of gauging the space and time, so we can perahps re-connect the two point arised by Landau to the additivity of observables.

Generalizing, perhaps we need to relate an observable to an infinitesimal of a continous symmetry group: an elementar generator of lie algebra in order to justify additivity. Just handwaving.

-

This question has bothered me for a long time! Although I don't have an answer, I'd like to mention an approach that looks promising at first, but turns out not to work.

First, recall that in quantum mechanics, you can think of a "state" as a way of preparing a physical system. Theo Johnson-Freyd pointed out in a comment that if you have two states $\rho$ and $\sigma$, you can construct a state that intuitively deserves to be called $\tfrac{1}{2}(\rho + \sigma)$:

Flip a fair coin. If the coin comes up heads, prepare the system in state $\rho$. If the coin comes up tails, prepare the system in state $\sigma$.

This state deserves the name $\tfrac{1}{2}(\rho + \sigma)$ because if $\rho[X]$ is the expectation value of the observable $X$ for a system prepared in state $\rho$, and $\sigma[X]$ is the expectation value of $X$ for a system prepared in state $\sigma$, the expectation value of $X$ for a system prepared in state $\tfrac{1}{2}(\rho + \sigma)$ should be $\tfrac{1}{2}(\rho[X] + \sigma[X])$, by the laws of classical probability.

Now, what happens if we use the same trick to define the sum of two observables? Given two observables $X$ and $Y$, let's define $X + Y$ to be the observable:

Flip a fair coin. If the coin comes up heads, measure $X$ and double the result. If the coin comes up tails, measure $Y$ and double the result.

The laws of classical probability tell us that if $\rho[X]$ and $\rho[Y]$ are the expectation values of $X$ and $Y$ for a system prepared in state $\rho$, the expectation value of $X + Y$ for a system prepared in state $\rho$ should be $\rho[X] + \rho[Y]$, just as you would hope.

Here's where things go pear-shaped. Given an observable $Z$, it makes sense to define $Z^2$ to be the observable:

Measure $Z$ and square the result.

So what's the expectation value of $(X + Y)^2$ for a system prepared in state $\rho$? The laws of classical probability tell us that it's $\rho[X^2] + \rho[Y^2]$. In the formalism of quantum mechanics, however, $X$ and $Y$ are operators and $\rho$ is a linear functional on the operator space, so

$\rho[(X + Y)^2] = \rho[X^2] + \rho[Y^2] + \rho[XY + YX]$.

If $\rho[XY + YX]$ is nonzero, this formula disagrees with the expectation value for $(X + Y)^2$ that follows from our definitions of $X + Y$ and $Z^2$, according to the laws of classical probability!

In practice, it's not hard to find observables $X$ and $Y$ for which $\rho[XY + YX]$ can be nonzero. For example, let $X$ and $Y$ be the x-spin and z-spin of a spin-1 particle, represented by the operators

$X = \frac{1}{\sqrt{2}}\left[\begin{array}{ccc}0&1&0\\1&0&1\\0&1&0\end{array}\right],\qquad Y = \left[\begin{array}{ccc}1&0&0\\0&0&0\\0&0&-1\end{array}\right].$

-
 'Given an observable Z, it makes sense to define Z2 to be the observable Measure Z and square the result' - I think this is where that argument falls over, and not just on a quantum but even a classical level. I expect to be able to define the observable X+Y readily simply because linearity is a 'plausible' thing to have, but I see no reason to believe that I'll be able to ascribe any sort of invariant meaning to Z2 in general. – Steven Stadnicki Jul 28 2010 at 17:12 @Steven Stadnicki: To me, given an observable $Z$ and a function $f \colon \mathbb{R} \to \mathbb{R}$, it seems totally natural to define $f(Z)$ as the observable you measure by measuring $Z$ and then applying $f$ to the result. What do you mean by an "invariant meaning," and why doesn't this definition give an "invariant meaning" to $f(Z)$ for any $Z$ and $f$? – Vectornaut Jul 28 2011 at 23:05 On the contrary: in quantum mechanics as it exists, measuring $Z$ and squaring the result does measure $Z^2$, but flipping a coin (then making an appropriate measurement and doubling the result) does not measure $X + Y$. – Toby Bartels Jan 3 2012 at 17:58

If I correctly understand, you are misinterpreting the meaning of the product and sum of observables.

When you say "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."

This cannot possibly describe the usual sum A+B and product AB of operators. For the product, it is not even hermitian unless A and B commute. Agreed, A+B is hermitian, but the spectrum of A+B does not contain the result of the sum of a measurement of A followed by a measurement of B (in either way), again unless A and B commute. For a counter-example take $A=\pmatrix{1& 0\cr 0&-1}$ and $B=\pmatrix{0&1\cr 1&0}$.

I hope I correctly understood your question.

-
 Well, the point of my question is in big part metamathematical. That is, the explanation I received was meant as a motivation to study quantum mechanics via C*-algebras, and a posteriori, via the Gelfand-Naimark-Segal theorem, via operator theory. I admit that the explanation does not convince me completely, but I think it has some points. The reason why I asked here was trying to fill the gaps. Instead you assume that observable are represented by operators, which is meant to be the conclusion – Andrea Ferretti Jan 26 2012 at 22:14