Commuting Operators: When 2 Operators Don't Commute

• Fe-56
In summary, the conversation discusses the concept of operators A, B, and C and their commutation in Hilbert space. When two operators commute, they have the same basis in (Hilbert) space. However, if they do not commute, their basis will be different. This only happens when there is degeneracy in the eigenvalues of the operators. An exception is if there is a state for which applying all three operators gives zero. This concept is seen in the example of {\vec L}^2, L_x, and L_z. The conversation also clarifies that this is not a homework assignment.
Fe-56
hello...this might look very stupid but I am totally confused...

Let have operators A, B, C. Let [A,B]=[A,C]=0 and [B,C] not 0...

When two operators commute, they have the same base in (Hilbert) space.
So base in A representation is the same as in the B representation and also
the basis in A and C representations are the same.

But if B and C do not commute, their basis are different.

There has to be a fatal error in here
thanx

Fe-56 said:
hello...this might look very stupid but I am totally confused...

Let have operators A, B, C. Let [A,B]=[A,C]=0 and [B,C] not 0...

When two operators commute, they have the same base in (Hilbert) space.
So base in A representation is the same as in the B representation and also
the basis in A and C representations are the same.

But if B and C do not commute, their basis are different.

There has to be a fatal error in here
thanx

As far as I know, this happens only when there is degeneracy.

Let's say that the eigenstates of A are degenerate. Then if [A,B]=0, it means that the eigenstates of B can be written as some linear combination of the eigenstates of A. If [A,C]= 0, the same thing applies for the eigenstates of C (they can be written as a linear combination of the eigenstates of A).

Consider a set of eigenstates of A all corresponding to the same eigenvalue of A, say $A \psi_n(x) = a \psi(x)$ . Then it will be possible to write some of the eigenstates of B (corresponding possibly to different eigenvalues of B) as some linear combinations of these $\psi_n(x)$. And it will be possible to write some of the eigenstates of C as a different linear combination of the eigenstates of A (still corresponding to the same degenerate eigenvalue $a$). So A and B share a common set of basis states, A and C share a common basis of states but B and C don't.

An exception is if there is a state for which applying A, B or C all gives zero. Then all three operators may share the same eigenstate even if they don't all commute.

This is what happens, say, with ${\vec L}^2$ and $L_x$ and $L_z$ for example. We have $[ {\vec L}^2,L_x] = [{\vec L}^2,L_z]=0$ but $[L_x,L_y] \neq 0$ .

Last edited:
:!)

YES, that's it! :!)

so...there must be degeneracy in all eigenvalues of A, yes?

thankx a lot...

...and this was not any homework, someone replaced it from QT

1. What are commuting operators?

Commuting operators are operators in quantum mechanics that can be applied in any order and still produce the same result. In other words, the order in which the operators are applied does not matter.

2. Why is it important to study commuting operators?

Studying commuting operators allows us to better understand the fundamental principles of quantum mechanics and how different operators interact with each other. This knowledge can then be applied in various quantum systems and experiments.

3. What happens when two operators do not commute?

When two operators do not commute, the order in which they are applied matters and can result in different outcomes. This is known as non-commutativity and is a fundamental concept in quantum mechanics.

4. Can commuting operators be used to simplify calculations?

Yes, commuting operators can be used to simplify calculations in quantum mechanics. Since the order of application does not matter, we can rearrange the operators in a way that makes the calculations easier.

5. Are commuting operators always observable?

No, commuting operators are not always observable. In order for an operator to be observable, it must be Hermitian and have a complete set of eigenstates. Commuting operators do not necessarily fulfill these requirements.

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