Eigenfunctions of a Hermitian operator are orthogonal if they have different eigenvalues. Because of this theorem, we can identify orthogonal functions easily without having to integrate or conduct an analysis based on symmetry or other considerations.
If a1 and a2 in are not equal, then the integral must be zero. This result proves that nondegenerate eigenfunctions of the same operator are orthogonal. If the eigenvalues of two eigenfunctions are the same, then the functions are said to be degenerate, and linear combinations of the degenerate functions can be formed that will be orthogonal to each other.
Since the two eigenfunctions have the same eigenvalues, the linear combination also will be an eigenfunction with the same eigenvalue. Degenerate eigenfunctions are not automatically orthogonal but can be made so mathematically.
The proof of this theorem shows us one way to produce orthogonal degenerate functions. If two operators commute, then they can have the same set of eigenfunctions. For example, the operations brushing-your-teeth and combing-your-hair commute, while the operations getting-dressed and taking-a-shower do not.
Simply assuming that the boundary conditions give sufficiently strongly vanishing near infinity or have periodic behavior allows an operator to be Hermitian in this extended sense if. In order to prove that eigenvalues must be real and eigenfunctions orthogonal , consider.
Assume there is a second eigenvalue such that. Now multiply 4 by and 6 by. But because is Hermitian, the left side vanishes. If eigenvalues and are not degenerate, then , so the eigenfunctions are orthogonal.
If the eigenvalues are degenerate, the eigenfunctions are not necessarily orthogonal. Now take. The integral cannot vanish unless , so we have and the eigenvalues are real. For a Hermitian operator ,. Given Hermitian operators and ,. Because, for a Hermitian operator with eigenvalue ,.
Therefore, either or. But iff , so. This means that , namely that Hermitian operators produce real expectation values. Every observable must therefore have a corresponding Hermitian operator.
It only takes a minute to sign up. Connect and share knowledge within a single location that is structured and easy to search. Position, momentum, energy and other observables yield real-valued measurements. The Hilbert-space formalism accounts for this physical fact by associating observables with Hermitian 'self-adjoint' operators. The eigenvalues of the operator are the allowed values of the observable. Since Hermitian operators have a real spectrum, all is well. However, there are non-Hermitian operators with real eigenvalues, too.
Consider the real triangular matrix:. Obviously this matrix isn't Hermitian, but it does have real eigenvalues, as can be easily verified. Why can't this matrix represent an observable in QM? What other properties do Hermitian matrices have, which for example triangular matrices lack, that makes them desirable for this purpose? See also this Phys. SE post. If you want to see something different, there are actually a few articles by Carl Bender developing quantum mechanics formulated with parity-time symmetric operators.
He shows that some Hamiltonians are not Hermitian, yet they have real eigenvalues and seem to represent valid physical systems. If you think about it, the requirement that your operator is parity-time symmetric makes more sense physically than hermiticity. In a later article, his quantum mechanics approach was proven to be equivalent to the standard one where operators are hermitian.
To give an answer that is a little more general than what you're asking I can think of three reasons for having hermitian operators in quantum theory:. And unitary Lie group representations come with a lie algebra of hermitian operators. This structure if efficiently represented by a hermitian operator that comes with an eigenstructure that matches these requirements precisely.
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