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

Question

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?

Answer 1

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.

Answer 2

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 $$H\psi = \frac{1}{2m}p (p \psi)$$

Answer 3

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 system with $n$ degrees of freedom that satisfies 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 representations 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 only one. In other words, any other representation, such as the Heisenberg, or the Dirac representation, are all equivalent to each other and to the Schroedinger representation. 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.

Answer 4

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.