If I want to calculate perturbed energy states in 2-fold degenerate case in Quantum Mechanics.

Assume the Hamiltonian is $ H=H_0+H^\prime$

Then we can calculate matrix elements of W : $ \langle a|H^\prime|b\rangle$, where a and b are eigenvectors which span the degenerate subspace.

And by solving the characteristic Eq of W, we can get pertured energy states.

My questions are:

(1) Should we always take $i$ and $j$ to be orthogonal? In Griffiths' book the derivation of time-independent theory assumed orthonormality.

(2) Can any orthonormal states in subspace be "good states" ? Because in his book $|\psi_a\rangle$ and $|\psi_b\rangle$ are arbitrary and indeed good states.

the derivation


1 Answer 1


Using orthonormal states in the degenerate subspace (and in general) simplifies enormously all computations. When we work with a self-adjoint operator (like $H$ and all their order by order pieces, $H_0, H_1, ...$), we make use of the spectral theorem to claim that there is a basis of orthonormal vectors for our Hilbert space consistent of eigenvectors of such operator.

Moreover, for a self-adjoint operator, we are guaranteed that eigenvectors with distinct eigenvalues are orthogonal (technically this would be true for normal operators in general). But even in degenerate subspaces, where we don't have orthogonality apriori, it is incredibly powerful to use an orthonormal basis (perhaps generated with an orthonormalization procedure).

As long as you fully span the degerenate subspace of interest, any orthonormal basis will be good enough in principle. But later in the book you'll see how some bases simplify computations notably, in particular those in which the perturbation is diagonalized. Physical intuition will also allow you to find these special bases in some cases.

You can try to reproduce alternative versions of the expressions for the order by order corrections without the use of orthonormal bases and you'll find out rapidly how everything becomes increasingly complicated.


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