## What happens when two operators commute?

If two operators commute, then they can have the same set of eigenfunctions. If two operators commute and consequently have the same set of eigenfunctions, then the corresponding physical quantities can be evaluated or measured exactly simultaneously with no limit on the uncertainty.

**What does it mean for two things to commute?**

Frequency: The definition of commute means to travel between home and work, or to change one thing for another. For example, numbers commute under addition, which is a commutative operation. Generally, any two operators H and G commute if their commutator is zero, i.e. HG − GH = 0.

### What is the commutator of two operators?

The Commutator of two operators A, B is the operator C = [A, B] such that C = AB − BA. If the operators A and B are scalar operators (such as the position operators) then AB = BA and the commutator is always zero.

**Do commuting operators share Eigenstates?**

So commuting observables share the same eigenstates if the associated eigenvalues are non-degenerate.

#### What is the physical significance of two operators commute?

If two operators commute then both quantities can be measured at the same time, if not then there is a tradeoff in the accuracy in the measurement for one quantity vs. the other.

**What are simultaneous eigenfunctions?**

The simultaneous eigenfunctions of L2 and Lz are the spherical harmonics Ylm(θ, φ) and the simultaneous eigenfunctions of S2 and Sz are |SMs〉 with S = 1 and Ms = 1,0, − 1. From: Atoms and Molecules, 1978.

## What can you conclude when you are informed that two operators commute?

**What does it mean to commute for a job?**

A commute is a journey you take from home to work and back again. You might enjoy your subway commute because it gives you lots of time to read. Your commute is your trip to work, and the verb commute describes making that trip — like your preference to commute by public bus.

### Are quantum operators commutative?

Let u and v be two physical quantities, and let uop and vop be the associated quantum operators. Two particular quantum operators for which the relation (5.1) is true for every function φ are called commutative operators, and the corresponding physical quantities are called commutative entities. …

**Are commutators associative?**

In other words, an algebra M is commutant-associative if the commutant, i.e. the subalgebra of M generated by all commutators [A, B], is an associative algebra. …

#### Do commuting operators share eigenvectors?

Commuting matrices do not necessarily share all eigenvector, but generally do share a common eigenvector. Let A,B∈Cn×n such that AB=BA. There is always a nonzero subspace of Cn which is both A-invariant and B-invariant (namely Cn itself).

**How do you find if two operators commute?**

To determine whether two operators commute first operate A ^ E ^ on a function f (x). Then operate E ^ A ^ the same function f (x). If the same answer is obtained subtracting the two functions will equal zero and the two operators will commute.on If two operators commute, then they can have the same set of eigenfunctions.

## Do the two operators of a wave function commute?

The two operators do not commute. Because the difference is zero, the two operators commute. Operators are very common in Physical Chemistry with a variety of purposes. They are used to figure out the energy of a wave function using the Schrödinger Equation.

**How do you find the commutator of two observables?**

The commutator of two observables A and B with operators ˆA and ˆB is defined to be, [ˆA, ˆB] = ˆAˆB − ˆBˆA A commutator is a mathematical construct that tells us whether two operators commute or not. Suppose A corresponds to a dynamic observable A and B corresponds to the dynamic observable B.

### How do two operators \\hat a and \\hat b$ commute?

Let us first restate the mathematical statement that two operators $\\hat A$ and $\\hat B$ commute with each other. It means that $$\\hat A \\hat B – \\hat B \\hat A = 0,$$ which you can rearrange to $$\\hat A \\hat B = \\hat B \\hat A.$$