# Observable: possible outcome of measurement vs (linear) transformation

One of the postulates of quantum mechanics is that every physical observable corresponds to a Hermitian operator $H$, that the possible outcomes of the measurements are eigenvalues of the operator, and that after a measurement the state collapses into the eigenfunction corresponding to the observed eigenvalue.

Some authors state that this is the reason quantum mechanics is "built on" linear algebra. However from a linear algebra standpoint this postulate seems strange. In linear algebra operators are linear transformations that take vectors to vectors; hence I would expect an operator also be a linear transformation, physically realised by a black box that takes in a particle in a state $\psi$ and emits the particle in the transformed state $H \psi$.

In fact, if I look at the creation operator (from say an electron in a harmonic oscillator potential), this clearly can't correspond to an observable because it's not hermitian. I can imagine the black box to be implemented by a machine that fires a photon into the electron. I can also imagine operators that for example rotate the angular momentum of an electron without measuring it. Can I create a physical apparatus that "implements" the position operator (in position space, it takes a state $\psi(x)$ and returns $x \psi(x)$)? Why does QM use operators in these two different senses and how do I distinguish them?

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Hermitian operators have real eigenvalues and this is why they correspond to physical measurable quantities. All other operators like creation operator are not hermitian. All operators are just linear transformations from Hilbert space to itself. –  Andyk Dec 18 '11 at 16:26
The title is misleading and has nothing to do with two questions you asked on the last two lines... –  Andyk Dec 18 '11 at 19:29

One distinction I find it useful to make is between operators that are essentially measurement operators and operators that are essentially part of the mathematics that is used to construct a state. The operator $\hat x$ is a measurement operator, but we also need a state to tell us what measurement results we will observe. To construct one particular state, the conventional ground state of the simple harmonic oscillator, we introduce another operator, the annihilation operator $\hat a$ and write (not worrying about constant factors) $\hat x=\hat a+\hat a^\dagger$. We can use $\hat a$ to define/construct an object $\left|0\right>$ for which $\hat a\left|0\right>=0$. This object lets us construct the expectation values of any function of $\hat x$ in the state $f(\hat x)\rightarrow\left<0\right|f(\hat x)\left|0\right>$, using the commutation relations $[\hat a,\hat a^\dagger]=1$.

At this point, we have to introduce a moderately high level of mathematics, which lets us use the vacuum state (or any other state that we can create using other mathematics) as input to the Gelfand-Naimark-Segal construction of a Hilbert space, but the essential fact at the elementary level is as I've already noted, that we can construct the expectation values of any function of $\hat x$ (and we can extend the algebra of functions to include functions of both $\hat x$ and $\hat p=\mathrm{i}(\hat a^\dagger-\hat a)\;$) in the vacuum state, using the commutation relations $[\hat a,\hat a^\dagger]=1$.

The SHO is especially simple because there are only bound states, so that we can have $\hat x$ be self-adjoint by (abstract) construction as it is here, but the same distinction can be helpful elsewhere --- between operators that are introduced as mathematical analogues of measurements and operators that are introduced as mathematical tools for constructing states. I find this distinction helpful in quantum field theory, for example, because the SHO is foundational there.

I hope this is not too abstract. It's not a distinction that I have seen made as explicitly as this in textbooks, so use it carefully. The dichotomy between states and observables is not often emphasized as much as I think it might be, however it's manifest in the well-known expression of the expected value of an observable in a given state as a trace of the product of an observable and a density operator, $\mathsf{E}=\mathsf{Tr}\bigl[\hat A\hat\rho\bigr]$, which to me suggests that we think about whether various operators are used more to construct the $\hat A$'s or more to construct the $\hat\rho$'s.

Almost immediately an EDIT, to engage with your Question slightly more: All the above is built on abstract linear algebras, however the construction of a state can alternatively be put in terms of the construction of a representation of the linear algebra --- which in elementary mathematics is one and the same as the linear algebra itself, all done as "matrices", but a representation of a linear algebra requires considerably more structure than is required to construct an abstract linear algebra.

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I think that your problem is the question of collapse. Collapse of wavefunction is somewhat far from all the mathematical structure of quantum mechanics and was devised by Von Neumann as a further add-on to the formalism. The understanding of it in the framework of quantum mechanics formalism is something that, so far, has singled out a possible successful framework in the idea of decoherence.

The idea of collapse can be seen as somewhat at the boundary when one understands that the following process

$$|\psi\rangle=\sum_nc_n|n\rangle\rightarrow |k\rangle$$

lies somewhat far from a linear formalism.

Much of the research work today in the area of the interpretation of quantum mechanics is rooted in a way or another into a clear framework to fit the bill for the collapse.

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The question was: "Can I create a physical apparatus that "implements" the position operator?" –  Andyk Dec 18 '11 at 19:30
The collapse postulate is sometimes called the projection postulate and a projection operator is linear. What you wrote down, for a fixed $k$, is linear if you bear in mind that phase factors don't have any physical significance so we could alter what you wrote down to $c_k\vert k \rangle$ without altering its physical significance. Yet, although it is linear, it's not unitary, and that is significant, but even so, what you have done is only a partial answer to the OP. –  joseph f. johnson Jan 17 '12 at 3:22

We work with representations of operators in a space (vector, tensor or higher order). The operator is an abstract object that has no meaning unless it acts on a space (basis). A linear operator can be defined as a homomorphism with (an) additional compatibility rule(s). There is nothing forbidding Quantum Mechanics to work in the higher level tensor space.

Also, there is no fundamental rule that forbids outcome of measurements to be in a field other than the Real field. It just turns out to be so and since we don't seem to have a device that can measure abstract fields, we have to stick it in there as a postulate.

There is no measurement problem if we simply realize that the best measurement we can obtain is only good up to an isomorphism. In my view, the postulate you stated is superfluous and should not have a place in modern QM.

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this is quite inadequate as an answer. –  joseph f. johnson Dec 31 '11 at 7:47

Well actually to be more general and precise, the physical observables correspond to " self - adjoint operators" not " Hermitian operators". If I recall well my functional analysis lessons, in infinite dimensional Hilbert spaces every self - adjoint operator is Hermitian but not every Hermitian is necessarily Self -adjoint ! So only in finite dimensional Hilbert spaces the two concepts may be taken as meaning the same thing. I guess physicists are not that worried about rigour, hence they take the two concepts as synonymous. Also physicists use the term " linear operator " as synonymous of " linear transformation ", in fact most of times they even omit the term "linear " and only use the term " operator ".

Observation : Also I think the use of Hilbert spaces and self-adjoint operators in quantum mechanics is rather historically motivated than anything fundamental. Indeed the motivation behind the Hilbert space formalism was that, Schrodinger employed the space $L^2(\mathbb{R})$ of square integrable complex functions to formulate his wave mechanics, whereas Heisenberg employed the space $l^2$ of square - summable sequence of complex numbers to formulate his matrix mechanics. Now von Neumann realized that $L^2(\mathbb{R})$ and $l^2$ have something deep in common. That deep thing is the underlying mathematical structure of a separable complex Hilbert space. Hence the unification of the two formalisms into a single abstract mathematical formalism was born by employing an abstract separable complex Hilbert space.

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This isn't an answer, it's a critique of the question. –  Colin K Dec 18 '11 at 19:12
Thanks for the reply, fixed! I wanted to post it as a comment under my answer. I have added it under "Observation " in my answer. –  Serifo Blade Dec 18 '11 at 19:31

QM uses Hermitian (okay, technically « self-adjoint ») operators for one sort of thing, observables, but uses many other operators for other purposes. For example, unitary operators might be symmetry operators such as mirror reversal or charge conjugation, or, more importantly, the time-evolution of a system is given by a unitary operator.

Now, your notion of black box is going to apply to one type but not the other type. Any real black box means the time evolution of some system, so it will always be a unitary operator. But observables are not unitary operators, so there is no very direct relation between measurement and black boxes. You should avoid thinking that an operator has to be a black box in Physics, even though mathematicians like to teach functions and operators by imagining it is a black box that takes your input, like $x$, and emits its ouptut, like $x^2$.

What I am saying has a great deal to do with Prof. Morgan's answer, and points made by other posters too, but is more focussed on the essentials, and I hope is much clearer.

In particular, applying an observable such as « multiplication by $x$ » to a vector such as $\psi(x)$ has no very direct physical meaning, and is not a physical process. It has, in a way, an indirect physical significance, but has no relation to the time evolution of a system and so cannot be physically implemented.

The only way to tell « the difference », as you call it, i.e., to tell whether a transformation of $\psi(x)$ is a physically implementable transformation or a mathematical, conceptual one, is to find out if there is a Hamiltonian $H$ such that someone in the family of unitary operators $e^{itH}$ gives the transformation you are interested in studying.

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