# Can an operator without complete eigenstates be measured?

• lackrange
In summary, the conversation discusses the conditions for an operator to be considered an observable. It is stated that the operator must be Hermitian and have a complete set of eigenfunctions. There is a discussion about whether xp+px, which is Hermitian but does not have a complete set of eigenfunctions, is considered an observable. It is concluded that according to Griffiths, an observable must have a complete set of eigenfunctions, so xp+px would not be considered an observable.
lackrange
Say you have some operator A with an incomplete set of eigenstates, but the state of the system is such that it happens to be expressible as a sum (possibly infinite, or integral, whatever) of the eigenstates of A, and let's say the eigenvalues are real and whatever is necessary...we may assume that A is a dynamical variable. Can A be measured?

The justification that an observable must have complete eigenstates is that if you make a measurement corresponding to the observable on a system, the system will jump into one of the observables eigenstates. But why must any state be expressible as a sum of an operators eigenstates, rather than just the current one?

In order for an operator to represent an observable it must be Hermitian. Hermitian operators always have eigenfunctions that form a complete set, so in order for something to be an observable it must be Hermitian and that implies that its eigenfunctions are complete.

I believe there are subtleties. I don't think the operators are necessarily hermitian on the whole Hilbert space. xp+px is hermitian, but I don't believe it has a complete set of eigenstates. So maybe the operator A isn't defined on the whole Hilbert space..

I do not believe xp+px is considered an observable because you would not be able to measure an exact eigenstate due to the uncertainty principle.

You wouldn't be measuring x and p separately and then putting them together to get xp+px...you would just be making one measurement, xp+px.

Yes I understand this. I am saying that I do not believe that xp+px is measurable because it does not have a complete set of eigenstates. I used the uncertainty principle to offer an explanation as to why it may not be able to be measured.

Right. But that is my question. If the system happens to be in an eigenstate of xp+px (I am not claiming they are complete, there just needs to be one), then could you make a measurement of xp+px?

Sorry I confused myself, in Griffiths it is stated as an axiom that all observables with Hermitian operators have a complete set of eigenfunctions. If xp+px does not have a complete set, it is not observable.

## 1. Can an operator without complete eigenstates still be measured?

Yes, an operator without complete eigenstates can still be measured. The results of the measurement may not be as precise as when using an operator with complete eigenstates, but it is still possible to obtain meaningful information from the measurement.

## 2. What are complete eigenstates in relation to operators?

Complete eigenstates refer to a set of states in which an operator has a well-defined value. These states form a basis for the Hilbert space of the operator, and any state can be expressed as a linear combination of these complete eigenstates.

## 3. How does the presence or absence of complete eigenstates affect the measurement process?

The presence or absence of complete eigenstates affects the measurement process by determining the precision and accuracy of the measurement. Operators with complete eigenstates allow for more precise measurements, while operators without complete eigenstates may result in less accurate measurements.

## 4. Are there any real-world examples of operators without complete eigenstates being measured?

Yes, there are several examples of operators without complete eigenstates being measured in various fields of science. One example is the measurement of the spin of a particle, which does not have complete eigenstates due to the uncertainty principle.

## 5. Can operators without complete eigenstates be used in quantum computing?

Yes, operators without complete eigenstates can be used in quantum computing. However, they may not be as efficient or effective as operators with complete eigenstates, and may require more complex algorithms to obtain meaningful results.

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