Operators with a common set of eigenvectors commute ↩

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shorthands: {}
created: 2022-01-06 02:14:55
modified: 2022-01-10 04:13:04

Where is the commutator.

Proof

From

Follows that from an arbitrary vector

Using the commutator yields zero as well since it is linear:

So the commutator yields for any vector, which means that it is the zero operator:

See also

• Commuting operators have a common set of eigenvectors