# Orthogonal Eigenfunctions (Landau Lifshitz)

1. Sep 15, 2011

### Master J

I've been reading QM by Landau Lifgarbagez, in which I've come across a statement I can't seem to get my head around.

It states (just before equation 3.6):

a_n = SUM a_m. INTEGRAL f_m. f_n. dq

( a_n is the nth coefficient, f_m is the mth eigenfunction of an operator, dq is the differential element.

It follows from the equation for determining the coefficients in a wavefunction composed of a linear sum of eigenfunctions of an operator:

a_n = INTEGRAL f_n. f. dq )

It then states that it is evident from this that the eigenfunctions must be orthogonal. I don't see how this is? I would like to understand this, as it would imply that orthogonality of eigenfunctions would fall directly out of the mathematics of QM!

2. Sep 15, 2011

### Bill_K

Landau and Lifgarbagez is really laborious to read - I think the English translation is partly to blame. The wording is heavily convoluted, making it hard to follow the argument.

Notice that they snuck in without explicitly saying so the assumption that all the eigenvalues fn are distinct. In fact if two eigenvalues coincide, you can choose linear combinations of the Ψn's that are not orthogonal, rendering the conclusion false.

3. Sep 16, 2011

### Master J

Is it assumed in QM that all eigenvalues f_n of a given operator ARE distinct? Why?

4. Sep 16, 2011

### Master J

Is it assumed in QM that all eigenvalues f_n of a given operator ARE distinct? Why?

5. Sep 16, 2011

### sweet springs

Hi, Master J.

Operators corresponds to physical variables, e.g. energy, position, and eigenvalues of operator corresponds to values of physical variables e.g. 0.12 Joule, 3.45 meter. Observed value should be distinctive.

Regards.

Last edited: Sep 16, 2011
6. Sep 16, 2011

### HallsofIvy

The orthogonality "falls out" from that definition of the coefficients.

In particular, if you look for the coefficient, an, in the expansion of fm, you get
$$an= \int fn fm dq$$
But, fm is itself a basis function, so fm is just 1 times fm plus 0 times all the other basis functions. That is
$$am= \int fm^2 dq= 1$$
[tex]an= \int fn fm dqa= 0[tex]
for n not equal to m.

7. Sep 16, 2011

### Bill_K

No, of course not. Eigenvalues can be degenerate. In a spherically symmetric potential for example, solutions proportional to Yℓm for m = -ℓ,... ,+ℓ will have the same energy.

8. Sep 16, 2011

### muppet

It's mentioned at the start of that section that really one should speak about "complete sets" of observables, but for simplicity it was assumed that we'd chosen an observable whose eigenstates formed a complete set by themselves, which would I think mean that it had no degeneracy in its spectrum.

For example, in the case of a spherically symmetric potential, the states Bill_K mentioned with different values of the quantum number m are orthogonal to each other.