Suppose, for the sake of proof by contradiction, that | ψ ψ ⟩ ⟩ is also a right eigenstate of momentum, with constant eigenvalue p 0. If this were true, then one could write ( x ^ ^ − − x 0 I ^ ^ ) p ^ ^ | ψ ψ ⟩ ⟩ = ( x ^ ^ − − x 0 I ^ ^ ) p 0 | ψ ψ ⟩ ⟩ = ( x 0 I ^ ^ − − x 0 I ^ ^ ) p 0 | ψ ψ ⟩ ⟩ = 0. On the other hand, the above canonical commutation relation requires that [ x ^ ^ , p ^ ^ ] | ψ ψ ⟩ ⟩ = i ℏ ℏ | ψ ψ ⟩ ⟩ ≠ ≠ 0. This implies that no quantum state can simultaneously be both a position and a momentum eigenstate.