Question

in shankar the principle of quantum mechanics chapter1

i don't understand the particular sentence in the part discussing about simultaneously diagonalization of two hermitian operators.

Proof. Consider first the case where at least one of the operators is nondegener ate, i.e., to a given eigenvalue, there is just one eigenvector, up to a scale. Let us assume Ω is nondegenerate. Consider any one of its eigenvectors: Since [A, 2]-0, i.e., A/ω> is an eigenvector of Ω with eigenvalue ω'. Since this vector is unique up to a scale

what does this vector is unique up to a scale mean?

Answer 1

please discuss if you have any query about it.

As in the Given problem we operator ohm on the eigenstate Vertical-bar wi > and get ohm Vertical-bar wi > = wi Vertical-bar wi > Also,^ohm Vertical-bar wi > = wi^ Vertical-bar wi > Since [lambda, ohm] = 0 So^ohm =^ohm commute ohm^ Vertical-bar wi > = wi^ Vertical-bar wi > ohm^ Vertical-bar wi > = wi^ Vertical-bar wi > This means that for a single eigenvalues these is a given This means that for a given eigenvalues, there is just one eigenvalues. The vector is unique up to a scale means it have only one eigenvalues corresponding to one operator which is also known as non - degenerated only occur not generate again

Thankyou