If two matrices commute: AB=BA, then prove that they share at least one common eigenvector: there exists a vector which is both an eigenvector of A and B. Let A^ ^and Bbe two Hermitian operators. However, if one of the operators has two eigenvectors with the same eigenvalue, any linear combination of those two eigenvectors is also an eigenvector of that operator, but that linear combination might not be an eigenvector of the second operator. Proof. You should check that ˆxand ˆp x do not commute. share | cite | improve this answer | follow | answered Apr 27 '11 at 16:28. physics_fun_dum_dum physics_fun_dum_dum. If two operators commute, they can be simultaneously determined precisely. Now we can show that the eigenfwictions of two commuting operators cin always be constructed in such a way that they are simultaneous eigenfunctions. Any eigenfunction of a linear operator can be multiplied by a constant and still be an eigenfunction of the operator. Then BAi AB’IJ Bu aB or A(B4) = a(ThI’). Then the following two statements are equivalent: i) A^ ^and Bpossess a common eigenbasis. In Quantum mechanics, two operators [math]A[/math] and [math]B[/math] are said to be compatible when their commutator is identically zero, i.e [math]\left[A,B \right]=0[/math]. In fact, the form of these operators is chosen to satisfy the uncertainty principle. This means that if f(x) is an eigenfunction of A with eigenvalue k, then cf(x) is also an eigenfunction of A with eigenvalue k. Prove it: A … ii) A^ ^and Bcommute. Proof: Let be an eigenfunction of A^ with eigenvalue a: A ^ = a then we have Z A ^ dx= Z (a ) dx= a Z dx and by hermiticity of A^ we also have Z A ^ dx= Z A ^ dx= a Z dx hence (a a) Z dx= 0 and since R dx6= 0, we get a a= 0 The converse theorem also holds: an operator is hermitian if its eigenvalues are real. diagonal operators, and thus they commute. If two operators commute, then there exists a basis for the space that is simultaneously an eigenbasis for both operators. Proof: Suppose D is an eigenfunction of A, so Ai = a4 and suppose A and B commute. Then If they didnt commute then this would be impossible by the uncertainty principle. the same solution. The proof is left as an exercise. Two operators Oˆ 1 and Oˆ 2 are said to commute if Oˆ 1 Oˆ 2ψ= Oˆ 2Oˆ 1ψ for all ψ. This consideration allows us to state a more powerful statement than the above Preposition: Proposition 3. Aimed of the mathematical results we have found, we shall now answer the following … If A and B commute, then [A,B]* = 0 holds for any ii’. If two operators have a complete set of simultaneous eigenfunctions, they must commute. It means you can (in principle) measure both quantities to arbitrary precision at the same time. Let A and B be two given operators and i a complete set of their simultaneous eigenfunctions corresponding to the eigenvalues a i and b i, respectively.