# Simultaneous measuring of two operators

• Caulfield
In summary, the conversation discusses the concept of observables and corresponding operators in quantum mechanics. It is stated that for particles to have definite values of both observables a and b at the same time, the operators A and B must commute. This is explained by the fact that measuring an observable collapses the particle's state into an eigenstate of that observable. However, if the operators do not commute, the particle's state will be disturbed and cannot have definite values of both observables at the same time. This is further supported by a theorem on non-commuting observables.
Caulfield
Hello everybody.

I am new here, and also new to quantum mechanics. This is the question to which I can't answer neither in mathematical nor physical way.

a,b → observables (like position and momentum)
A,B → corresponding operators.

"It is possible for particles to be in a state of definite a and b at the same time only if the corresponding operators A and B commute."

Why?

When you measure something, you collapse its state into an eigenstate of whatever you are measuring. The state remains undisturbed if it is already an eigenstate of whatever you are measuring. This is why a particle in an eigenstate of A is said to have a definite value of A. A particle can be in an eigenstate of both A and B only if A and B commute.

The other way to say it is that if you measure an observable A on a particle in a state, you will get a particular result with certainty only if the state is an eigenstate of A. This is why we say an eigenstate of A has a definite value of A. Again, a particle can only be in an eigenstate of both A and B only if A and B commute.

Last edited:
if they don't commute, for example: [A,B]=-iħ

A(ψ)=a(ψ)

BA(ψ)=Ba(ψ)

-iħA(B(ψ))=aB(ψ)

A(B(ψ))=(i/ħ)aB(ψ)

but A(ψ)=aψ, and (i/ħ) is not equal to a.

This means we have two different wave functions as eigenfunctions of the operator A. (So eigenstate is disturbed)

How this imply that B(ψ) is not equal to bB(ψ)? (that is, how does it imply that b is not definite?)

Let ψ be an eigenstate of A and of B.

Aψ=aψ

Bψ=bψ

[A,B]ψ
=ABψ-BAψ
=Abψ-Baψ
=bAψ-aBψ
=baψ-abψ
=0

So if ψ is an eigenstate of A and B, then A and B will commute. (It may depend on the subspace, and some other conditions, which I don't remember.)

Caulfield said:
"It is possible for particles to be in a state of definite a and b at the same time only if the corresponding operators A and B commute."

Its a theorem on non commuting observables:
http://www.pa.msu.edu/~mmoore/Lect23_HeisUncPrinc.pdf

Also any good book on QM will detail it - eg see section 8.4 - Ballentine - Quantum Mechanics - A Modern Development.

But seeing why commuting observables have no issues isn't that hard. Basically it means they have the same set of eigenvectors, which roughly translates to they are really the same observation in disguise - note to the more knowledgeable reading this - I said ROUGHLY. Technically such are said to be compatible.

Thanks
Bill

## 1. What is simultaneous measuring of two operators?

Simultaneous measuring of two operators refers to the process of measuring two different physical quantities at the same time. It allows for the comparison and analysis of the relationship between the two quantities.

## 2. What are some examples of simultaneous measuring of two operators?

Some examples of simultaneous measuring of two operators include measuring the voltage and current in an electrical circuit, measuring the position and velocity of an object in motion, and measuring the wavelength and frequency of a wave.

## 3. How is simultaneous measuring of two operators useful in scientific research?

Simultaneous measuring of two operators is useful in scientific research because it allows for the collection of more comprehensive data. It can reveal patterns and relationships between two quantities that may not be apparent when measured separately.

## 4. What are some techniques used for simultaneous measuring of two operators?

Some techniques used for simultaneous measuring of two operators include using sensors, data loggers, and oscilloscopes. These tools can measure and record multiple quantities simultaneously, making it easier to analyze and compare the data.

## 5. What are the limitations of simultaneous measuring of two operators?

One limitation of simultaneous measuring of two operators is that it can be challenging to ensure that both quantities are measured accurately at the same time. This requires careful calibration and synchronization of the measuring equipment. Additionally, measuring more than two operators simultaneously can be even more complex and may require specialized equipment.

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