# Eigenstate of two observable operators

• I
• Kavorka
In summary, non-commuting operators A and B in a quantum system have different eigenstate bases and cannot be measured simultaneously. However, they can have a common eigenstate "by accident", as demonstrated in the radial-symmetric s orbital states of a hydrogen atom. In this case, the eigenvalues of A and B can be measured simultaneously when acting on the shared eigenstate. This is a special case and does not apply to all eigenstates. In a mixed state, there is a probability of simultaneous measurement of A and B if they share an eigenstate, but this probability is not always guaranteed. The commutator being non-zero means that the operators are not always simultaneously measurable, but there are instances where they can be. This can
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.

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.

hilbert2 said:
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?

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).

hilbert2 said:
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?

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.

Nugatory

## What is an eigenstate of two observable operators?

An eigenstate of two observable operators is a quantum state that is simultaneously an eigenvector of both operators. This means that when one of the operators is measured, the state will collapse into one of its eigenstates and the corresponding eigenvalue will be observed.

## What is the significance of eigenstates of two observable operators?

Eigenstates of two observable operators are important because they allow for the simultaneous measurement of two compatible observables. This can provide information about the state of a system and allows for the prediction of future measurements.

## How are eigenstates of two observable operators related to quantum entanglement?

Quantum entanglement occurs when two or more particles are in a state of superposition, meaning their states are not determined until they are measured. In this case, the particles are described by a joint wave function, and the eigenstates of two observable operators correspond to the different possible outcomes of measurements on the entangled particles.

## Can an eigenstate of two observable operators change over time?

No, once a state is in an eigenstate of two observable operators, it will remain in that state unless acted upon by an external force. This is because the state is an eigenvector of both operators, meaning it is an exact solution to their corresponding equations of motion.

## How are eigenstates of two observable operators used in quantum computing?

Eigenstates of two observable operators are used in quantum computing as the basis for quantum algorithms. By manipulating the state of a quantum system using quantum gates, the system can be prepared in an eigenstate of two observable operators, allowing for the measurement of multiple observables simultaneously and enabling more efficient and accurate computations.

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