# I Eigenstate of two observable operators

#### Kavorka

Let's say you have two operators A and B such that when they act on an eigenstate they yield a measurement of an observable quantity (so they're Hermitian). A and B do not commute, so they can't be measured simultaneously. My question is this: You have a matrix representation of A and B and you find their eigenvalues and corresponding eigenstates (eigenvectors), let's say they each have a basis of 3 eigenstates. My understanding is that if their eigenstate basis is not identical, they are not simultaneously measurable quantities and thus do not commute. What if A and B, which don't commute share a 1 eigenstate in common out of their 3 states? Does that mean that the two observables are simultaneously measurable when acting on that eigenstate? What if you had a mixed state, not a singular state, would there then be a probability of the observables being simultaneously measurable and a probability that they won't be? This wouldn't make sense to me in terms of non-commuting operators, but I'm not sure.

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#### hilbert2

Gold Member
Sometimes non-commuting operators can have a common eigenstate "by accident", for example in the radial-symmetric s orbital states of a hydrogen atom all the components of orbital angular momentum are zero at the same time, while usually you can't measure the angular momentum components simultaneously.

#### Kavorka

Sometimes non-commuting operators can have a common eigenstate "by accident", for example in the radial-symmetric s orbital states of a hydrogen atom all the components of orbital angular momentum are zero at the same time, while usually you can't measure the angular momentum components simultaneously.
So you would be able to measure the eigenvalues of the non-commuting operators simultaneously if they acted on the eigenstate they share?

#### hilbert2

Gold Member
Yes, you can measure all the components of orbital angular momentum of a hydrogen atom ground state as many times as you want and you always get the result zero and the state of the atom doesn't change in any way (at least in an idealized situation).

#### Kavorka

Yes, you can measure all the components of orbital angular momentum of a hydrogen atom ground state as many times as you want and you always get the result zero and the state of the atom doesn't change in any way (at least in an idealized situation).
Does it being 0 make it a special case, or it can be any eigenvalue?

Also, if you had a mixed state, would that mean the probability of simultaneous measurement would be the product of the probabilities of finding the corresponding eigenvalue for each operator? Like if A had a 50% chance of yielding the eigenvalue of the eigenstate it shares with B, and B had a 30% chance of yielding that eigenvalue, then there would be a chance of simultaneous measurement of 15%? In effect, non-commuting operators means that they are not ALWAYS simultaneously measurable, but they could be sometimes? How can that be true if the commutator is ALWAYS non-zero?

#### hilbert2

Gold Member
You can also play with block diagonal matrices like
\begin{bmatrix}
1 & 0 \\[0.3em]
0 & A \\[0.3em]
\end{bmatrix}

and

\begin{bmatrix}
1 & 0 \\[0.3em]
0 & B \\[0.3em]
\end{bmatrix}

where A and B are non-commuting hermitian matrices, and you'll see that they have at least one common eigenstate where both eigenvalues are 1.

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