# Finding measurements in non-Hermitian operators

I know how the measurement postulate in quantum mechanics works, in regard to hermitian operators, but what if an operator is non-hermitian? Can i apply the following reasoning?

If an operator is represented by a non-hermitian matrix, i know that i can't apply the same postulate for hermitian matrices because the eigenvalues might not be real, and the eigenvalues corresponding to different eigenvalues might not be orthogonal, but if when i try to find the eigenvalues of this matrix, i find that some are real, and to those real eigenvalues, some of them have eigenvectors that are orthogonal to eachother. Given a eigenstate of a system, can i say that the possible values of a measurement of that operator in that state are the eigenvalues corresponding to each orthogonal eigenvector that i found and that each probability is the sum of the inner products of the eigenstate with the eigenvectors corresponding to a eigenvalue?

Or is there some other procedure to find expected values and probabilities of a non hermitian operator?

Many texts say that an observable should be represented by a hermitian operator. That is sufficient, but not necessary. More generally, we can use any operator that can be expressed as a linear combination of mutually commuting projection operators. Such an operator is called a normal operator. A normal operator $$N$$ is most simply characterized by the fact that it commutes with its own adjoint: $$N^*N = NN^*$$. Examples of normal operators include hermitian operators, individual projection operators, and unitary operators.

Here's an example to illustrate this idea. If $$P_1,P_2,...$$ are mutually orthogonal projection operators, then the operator $$A = a_1 P_1 + a_2 P_2 + \cdots$$ is a normal operator for any choice of (possibly complex) coefficients $$a_k$$. The eigenvectors of this operator are mutually orthogonal because the projection operators $$P_k$$ are. If the coefficients are real, then $$A$$ is self-adjoint (hermitian). But what really matters in quantum theory is the projection operators $$P_k$$. These are what determine the various possible outcomes of the measurement and the relative frequencies of those outcomes. The coefficients $$a_k$$ are just convenient labels for the outcomes, making it possible to define statistics like mean values and standard deviations.

Using only self-adjoint operators is sufficient, because allowing complex coefficients is only allowing a more general way of labelling the various implied projection operators. Nature doesn't care how we label things.

In the first paragraph, I said "mutually commuting projection operators", which is more general than "mutually orthogonal projection operators." The latter implies the former, but not conversely. The former is needed in order to include observables like the position operator in non-relativistic quantum mechanics, which does not have (normalizable) eigenvectors. However, it still implicitly defines projection operators like $$P\psi(x)=\begin{cases} \psi(x)&\text{ if }x\in R\\ 0 &\text{otherwise}, \end{cases}$$ where $$R$$ is some region of space. We can think of the usual position operator $$X$$ as a convenient single-operator representation of this whole algebra of mutually commuting projection operators. It's the projection operators that we use in the measurement postulates. The fact that any normal operator implicitly defines such a set of mutually commuting projection operators is the subject of the spectral decomposition theorem.

Given any observable $$A$$, if $$P$$ is one of the projection operators that it implicitly defines (through the spectral decomposition theorem), then a measurement of $$A$$ will result in a state $$|\psi'\rangle$$ that satisfies either $$P|\psi'\rangle = |\psi'\rangle$$ or $$(1-P)|\psi'\rangle=|\psi'\rangle$$. (I'm not trying to advocate any particular interpretation of quantum theory here; I'm just trying to be concise.) In terms of the state $$|\psi\rangle$$ prior to the measurement, the relative frequencies of these two possible outcomes are $$\psi(P)$$ and $$\psi(1-P)$$, respectively, using the abbreviation $$\psi(\cdots)\equiv\frac{\langle\psi|\cdots|\psi\rangle}{\langle\psi|\psi\rangle}.$$ The point here is that we don't need to worry if $$A$$ doesn't have a complete set of (normalizable) eigenstates. As long as $$A$$ is a normal operator, we can still use the corresponding projection operators to make useful predictions, because each of the projection operators has (normalizable) eigenvectors. As long as they all commute with each other, we can think of this whole set of projection operators as a bunch of mutually compatible observables, each of which has only two possible outcomes.

• A caveat (see Weinberg vol.1 p.52): a symmetry transformation is of form $U=1+i \epsilon T$ with T Hermitian. Then Weinberg says "Indeed, most (and perhaps all) of the observables of physics, arise in this way from symmetry transformations." This is the deepest reason I have read why quantum textbooks restrict to Hermitian observables. – Bruce Greetham Nov 4 '18 at 9:33
• Actually a normal operator can uniquely be decomposed into a linear combination of two mutually commuting selfadjoint operators, so selfadjoint operators are also necessary...Normal operstors viewed as observables are nothing but a pair of compatible (seladjoint) observables. – Valter Moretti Nov 4 '18 at 14:03
• Necessity is due to the fact that the spectrum must be real if it represents the outcomes of measurements. A normal operator with real spectrum is selfadjoint necessarily. – Valter Moretti Nov 4 '18 at 14:07
• @ValterMoretti You made a good point that any normal operator can be written as a linear combination of commuting selfadjoint operators. But the essence of an observable is its projection-valued measure (PVM), not its spectrum. We are free to label the outcomes with complex numbers, or even with words. If words are used, then we can't encode the labels into a single operator, but using a single operator to represent an observable is only for convenience. We could define an observable to be a PVM, with no coefficients, and we'd still be doing standard quantum theory. – Chiral Anomaly Nov 4 '18 at 17:45
• See also this answer of mine for my general viewpoint physics.stackexchange.com/questions/116595/… which I suspect to be quite close to yours. – Valter Moretti Nov 4 '18 at 18:01