# Orthogonality condition

1. Mar 5, 2013

### pluspolaritons

What does it mean when we say that two functions are orthogonal (the physical meaning, not the mathematical one)? I tried to search for the physical meaning and from what I read, it means that the two states are mutually exclusive. Can anyone elaborate more on this?

Why do we impose orthogonality condition in many of the problems like infinite square well, harmonic oscillator etc?

2. Mar 5, 2013

### Jorriss

We don't impose it (usually). It's a natural property of solutions to certain types of differential equations. If you'd like to read more about it, look up sturm-liouville theory.

3. Mar 5, 2013

### ModusPwnd

Functions are not physical objects, they are mathematical objects. I dont think you should expect physical meaning out of mathematical objects.

Otherwise, I would use the analogies to 3-D space and the dot product along with the definition of inner product to build up an intuition about them.

edit - doh! I see you posted this in quantum section rather than math... My first critique is thus not that relevant...

Last edited: Mar 5, 2013
4. Mar 5, 2013

### pluspolaritons

@ModusPwnd - I should have written 'wavefunctions' instead of 'functions'.

Anyway, I found a pretty good website that helps to explain this a bit:
http://quantummechanics.ucsd.edu/ph130a/130_notes/node140.html

So from what I understand,

In QM, if we want our observable to be real and measurable, the operator has to be Hermitian. And the eigenfunctions of Hermitian operators are always orthogonal as shown on the website. Thus, in other words if we want our eignenvalues/measurables to be real, we require that the eigenfunctions to be orthogonal.

Orthogonal in where? Orthogonal in Hilbert space, the space where all the wavefunctions live. So back to first question. What does it mean to have 2 wavefunctions orthogonal? I think it means that the two wavefunctions will have real-valued measurables. If the wavefunctions are orthogonal, they can't be linear combination of each other. Otherwise I think they will give the same eigenvalues.

Is this correct?

And in the proof in the website I given, if a_1=a_2, then is the inner product of psi_1 and psi_2 zero? I thought the whole point is to prove that the inner product of psi_1 and psi_2 is zero, but why does the website go on and find the linear combination of the two wavefunctions that will give a zero inner product? What about the inner product of psi_1 and psi_2 itself when a_1=a_2, is it still zero?

5. Mar 6, 2013

### Fredrik

Staff Emeritus
Hermitian operators are what we use to mathematically represent measuring devices that always return a real number as the result of the measurement. We don't need any other measuring devices to turn the mathematics into a theory of physics, or to test the theory, so non-hermitian operators are usually ignored in QM.

I'm not sure what you mean by that.

The physical significance of orthogonality is that if f and g are orthogonal wavefunctions, and the system is in the state f, then the probability that a measurement will put the system in state g is zero.

First they prove that any two eigenvectors that correspond to two different eigenvalues are orthogonal. Then they ask and partially answer the question, what happens if two eigenvectors have the same eigenvalue? The partial answer is that the two eigenvectors span a 2-dimensional subspace, and there exists an orthogonal basis for that subspace. The reason why this is interesting is that you will often need to use that given a hermitian operator A, there's an orthonormal basis for the Hilbert space that consists of eigenvectors of A.

Last edited: Mar 6, 2013
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