# Really how can an observable quantity be equal to an operator?

A wave-function can be written as $$\Psi = Ae^{-i(Et - px)/\hbar}$$ where $E$ & $p$ are the energy & momentum of the particle. Now, differentiating $\Psi$ w.r.t. $x$ and $t$ respectively, we get \begin{align} \frac{\partial \Psi}{\partial x} &= \frac{i}{\hbar} p\Psi \\ \frac{\partial \Psi}{\partial t} &= -\frac{i}{\hbar}E\Psi \, . \end{align} The above equations can be written in suggestive forms \begin{align} p \Psi &= \left( \frac{\hbar}{i} \frac{\partial}{\partial x} \right) \Psi \\ E \Psi &= \left( i \hbar \frac{\partial}{\partial t} \right) \Psi \, . \end{align} Evidently the dynamical quantities momentum and energy are equivalent to the operators \begin{align} p &=\frac{\hbar}{i} \frac{\partial}{\partial x} \\ E &= i \hbar \frac{\partial}{\partial t} \, . \end{align}

quoted from Arthur Beiser's Concept of Modern Physics

How can observable quantities be equal to operators? You can't measure an "operator". Can anyone intuitively explain how momentum and energy are equal to operators?

• Just read on, if the text is any good, expectation values of operators should be discussed shortly. – ACuriousMind Jun 14 '15 at 19:54
• In particular, you'll see that the results of measurements will be eigenvalues of self-adjoint operators (which are real, so it makes sense that they are measurable). – Danu Jun 14 '15 at 19:56
• @user36790 A measured quantity is not the operator itself. The act of "measurement" of an observable for some state can be characterized mathematically by acting on the state vector with the corresponding operator for the observable. The postulates of QM then say that the value you measure is one of the eigenvalues of the operator. The actual thing you measure with your experimental apparatus is an eigenvalue. – JohnnyMo1 Jun 14 '15 at 20:24
• @user36790, in fact, the author does not say "equal", he says "equivalent". Another (more usual way) to express the same idea is to say that observables in QM are represented by self-adjoint operators. – Wildcat Jun 14 '15 at 20:25
• In a commutative setting, the observables are identified with real-valued functions; in a non-commutative setting they are identified with operators on some Hilbert space. This is a mathematical result (called Gel'fand-Naimark theorem) that holds under very reasonable assumptions on the set of observables. Therefore, since the observables of QM are experimentally seen (for what we know) to satisfy those assumptions and indeed be non-commutant, they are naturally represented mathematically as operators in a Hilbert space. – yuggib Jun 14 '15 at 20:26

## 4 Answers

Not equal, but equivalent, in the sense that they have the same effect on the wavefunction in question.

More precisely, the book is using a slight abuse of terminology. Taking momentum as an example, it's not really the case that the dynamical quantity of momentum is equivalent to the operator $\frac{\hbar}{i}\frac{\partial}{\partial x}$, because a number can't actually be equivalent to an operator. A number is a thing all on its own, whereas an operator is something that needs to be applied to something else to have any meaning. But the operation of multiplying the wavefunction by the wave's momentum (a number) is equivalent to the operation of taking the derivative and multiplying by $\frac{\hbar}{i}$. In mathematical language: $$\frac{\hbar}{i}\frac{\partial}{\partial x}\psi = p\psi\tag{1}$$ whereas, strictly speaking, we can't say this: $$\frac{\hbar}{i}\frac{\partial}{\partial x} = p$$ At least, not in the way you're thinking about the notation.

What we normally do in quantum physics is always think about quantities as operators. That is, if you write just $p$ in an equation, it's implicitly understood that this should be applied to some wavefunction. For example, if you write $$H = \frac{p^2}{2m}$$ what you really mean is $$H\psi = \frac{1}{2m}p (p \psi)$$

• Operators have spectra and they can be treated in the abstract, without having to resort to a particular representation, see Phoenix87 with his remark about C*-algebras. Unfortunately, none of that makes any difference to the beginner (if I recall my first lessons on QM). :-( – CuriousOne Jun 14 '15 at 21:27
• Can you tell me what the difference between equivalent & equal? – user36790 Jun 15 '15 at 5:13
• @user36790 in this context, "equivalent" means the objects can be manipulated in the same way (e.g. have the same effect) under a particular set of circumstances, whereas "equal" means they are the same thing, which in particular implies that they have the same effect under all possible circumstances. – David Z Jun 15 '15 at 5:31
• As you have mentioned, sir, that being equivalent means they can be manipulated in the same way under particular set of circumstances, can you tell where can I not manipulate them in the same way? I mean to say can you tell me of such a circumstance where the operator cannot be used in place of the quantity(they are equivalent,right)? – user36790 Jun 15 '15 at 7:15
• @user36790 in general, it's context-dependent. For this specific case, any time you have a wavefunction that is not an eigenfunction of the operator, the operator and the quantity will be different. $\frac{\hbar}{i}\frac{\partial}{\partial x}\psi$, if $\psi$ is not an eigenfunction of the momentum operator, will not be the same as $p\psi$ for any value of $p$. – David Z Jun 15 '15 at 14:58

The reason operators correspond to measured values has to do with what happens when you connect a measurement apparatus to the system under observation.

Suppose the Hamiltonian of the system by itself is $H_S$ and the Hamiltonian of the measurement apparatus by itself is $H_M$. When $M$ is physically connected to $S$, we get a additional "interaction" term $H_I$ in the Hamiltonian so that the complete Hamiltonian is

$$H = H_S + H_M + H_I \, .$$

Typically, $H_I$ is a product of two operators, one acting on $S$ and one acting on $M$. For example, if $A_S$ is an operator on $S$ and $B_M$ is an operator on $M$, then we might have $$H_I = A_S \otimes B_M \, .$$ If you don't know what $\otimes$ means just think of $A_S \otimes B_M$ as a list of two operators, the first of which acts on $S$ and the second of which acts on $M$. It turns out that when you have an interaction Hamiltonian like this and the measurement apparatus has many degrees of freedom which we don't have any specific information about, then the result of the interaction is that $S$ collapses to a random probability distribution of possible states, each of which has a different value of $A$. The possible values of $A$ are the eigenvalues of the operator $A_S$.

So, if the interaction Hamiltonian is $$H_I = x_S \otimes O_M$$ where $x_S$ is the position operator on $S$ and $O_M$ is an arbitrary operator on $M$, then the effect of connecting the measurement apparatus to the system is to collapse the system into a probability distribution over various states, each of which having a specific value of position.

• Two thumbs up... but will this actually help the OP to understand what's going on? – CuriousOne Jun 14 '15 at 21:21
• Total agreement about the frustration here. I also agree that having a quantum mechanical model of the measurement process is a very good step somewhere along the line. Not sure I really ever "understood" quantum mechanics, it's more that I have made peace with its structure because it is mathematically beautiful (who doesn't love a linear theory?) and self-consistent to a degree that few other theories (barring maybe thermodynamics and statistical mechanics) reach. – CuriousOne Jun 14 '15 at 21:37
• You are right, I hate statistical mechanics, but I still recognize the utility and structural consistency of tools that I hate. – CuriousOne Jun 14 '15 at 21:46
• @Nogueira: The question is really what upsets the novice most? Is is that the structure of the theory seems to drop out of nowhere without the guidance of spring loaded force gauges and pulleys (which are really poor ways of teaching classical mechanics that are nonetheless accepted by most students without questioning) or is it that it leaves us feeling standing in front of the Born rule curtain behind which we keep imagining a man that just won't let us in? In the end, no matter how we re-axiomatize QM, the gap between classical expectations and QM expectation values remains unbridged. – CuriousOne Jun 14 '15 at 21:50
• @Nogueira: Good for you! Not everybody can develop that level of intuition that quickly and/or has teachers who can instill these lessons. You are right, part of the problem is that we are teaching classical mechanics in ways which neglect the reality of the measurement process. My high school teacher did try to motivate some of that even while he was teaching mechanics, but I noticed how it completely flew by the other students. It is very easy to mistake idealized measurements with an interaction-free measurement process, which, of course, in QM is not the case any longer. – CuriousOne Jun 14 '15 at 22:09

As argued by von Neumann, the measuring process has many properties that resemble those found in the theory of operator algebras. For instance, if you have an instrument, you can measure something, say the length of a table, to get a certain value $x$ within experimental errors. What you can now do is relabel the ticks of your instrument according to a certain function $f$. If you measure the same quantity you will now find $f(x)$ within some other error. Without loss of generality you can assume that the outcome of a measurement lies in a compact subset of the real line. The fact that one inevitably has to deal with experimental errors restrict the set of admissible functions to those that are uniformly continuous on this compact set, hence continuous. What you then have is something that resembles the spectral mapping theorem, that is $$\sigma(f(O)) = f(\sigma(O))$$ where $O$ is an observable (note that I'm not assuming $O$ to be an operator yet, this will follow as a consequence of these heuristic considerations), and $f$ is the act of relabeling the ticks, whereas $\sigma(O)$ denotes for the time being all the possible outcomes of a measurement of $O$ (the so-called physical spectrum).

The way states are defined in this approach turns out to lead to a set, the set of admissible states, which has some mathematical properties. In particular this set, when equipped with a suitable topology, becomes compact and convex. The same holds for the state space (this name stems precisely from this fact) of a C*-algebra whereby one can identify the operation of measuring $O$ over a state $\omega$ as the evaluation of the linear functional $\omega$ over some "operator" $O$.

The above has then led to the axiomatisation of quantum (and classical as well, when one consider commutative settings only) mechanics in the following terms

Definition and axiom A physical system is a C*-algebra $A$ whose self-adjoint part is the set of observables and (a subset of) its state space is the set of all the physically admissible states for the system.

The "operators" that one usually deals with in quantum mechanics arise now as a consequence of the above axiom. A C*-algebra is an abstract object which becomes concrete when one considers a representation of it. Among all representations, the important ones are the irreducible representations. It turns out that a quantum mechanical systems with $n$ degrees of freedom that satisfy to the Heisenberg relations (i.e. the canonical commutation relations) generates a C*-algebra which is isomorphic to the C*-algebra of compact operators on an infinite dimensional separable Hilbert space. The theory of representation for such an algebra is such that there exists only one class of unitary equivalence of irreducible representation. Since one of these is the Schroedinger representation, this is essentially the unique one. The associated Hilbert space is $L^2(\mathbb R^n)$ with Lebesgue measure. This is how Dirac's formalism for quantum mechanics arises from just first principles, in a nutshell.

• Nothing in physics "arises from first principles". It's ALL based on measurements. What you are talking about is equivalent axiomatizations of the same theory, assuming that they are actually equivalent. – CuriousOne Jun 14 '15 at 21:15
• i don't follow this - are you saying that the Born's rule arises from a more fundamental, "first principle"? – innisfree Jun 14 '15 at 21:33
• Most of the "magic" of quantum mechanics as presented to physicists can be justified in a mathematically rigorous way. By first principle here I mean the assumption that a physical system is a C*-algebra. This, together with the theory of operator algebras, is enough to justify the mathematical structure of quantum mechanics (in particular I'm referring to the Dirac picture of his infinite dimensional Hilbert spaces and so on and so forth). – Phoenix87 Jun 14 '15 at 21:37
• Most of what theoretical physicists do ad-hoc can be rewritten in a more rigorous way, but none of what experimental physicists do has a counterpart in mathematics. I am not trying to ding mathematicians here who are trying to clean up our mess, Choquet-Bruhat's book, for instance, is actually one of my favorites, even though I don't understand most of it in detail. All I am saying is that which mathematical structures are approximated by parts of nature is nature's choice, not ours. – CuriousOne Jun 14 '15 at 21:44

Note that you only can identify the numbers with the operators when you hold a plane wave function. For an arbitrary wave functions, you can't identify the operator $(\hbar/i)(\partial / \partial x)$ with some number $p$, but you can hold the operator to have some significant meaning related to this numbers. If you apply $$\langle P\rangle = \int dx \, \psi^*(x)\frac{\hbar}{i}\frac{\partial}{\partial x}\psi(x)$$ when $\langle P\rangle$ is the expectation value for the measurement of the number $p$. For the position $x$ the operator is simple the function $f(x)=x$, or c-number $x$.

This is the Born principle of quantum mechanics! (can't be deduced, only understood, is a principle.).

Understanding the principle: the wave function $\psi (x)$ is some sort of a state of knowledge about the system. When the wave function is localized at some point, say $\delta (x)$, the Fourier transform of this function is simple $1$, means that the integration above presents as: $$\int dp p$$ then, the distribution probability of the momentum is spread over all values equally. The localized wave function don't have any knowledge about the momentum. So, Born principle met naturally the Heisenberg principle when we defined the momentum operator as $(\hbar/i)(\partial/\partial x)$ acting in functions over space.