# Orthogonal Eigenfunctions

1. Jun 22, 2004

### ce124

Can anyone give me a physical interpretation of what orthogonal eigenfunctions are please? I understand the mathematical idea, the overlap integral, but I'm not clear about what it implies for the different states. At the moment the way I'm thinking of it is that the energy eigenfunctions of an orthonormal set are at right angles to one another in multidimensional eigenvalue space, ie the overlap integral can tell us the probability of being in a particular state but not of being in two states at once. How far off is this?

2. Jun 22, 2004

You dont have a problem with orthogonal eigenstates, you have it with states. You have a problem with the first postulate of QM itself. A phisical state can be discribed by a ket, and the ket can be expressed in terms of a combination of the kets of a complete base, a complete base of kets that in turn represent physical states. If the base is orthogonal everything is easier. TRy expressing a function of x in terms of the base {x^n), n=0,infinity (taylor expansion). It's much harder than doing it in terms of legendre polynomials, which are orthogonal.

3. Jun 27, 2004

### Eye_in_the_Sky

NOTE: The measuring device referred to below is one which can be used to perform a "measurement" of a nondegenerate observable whose spectrum is discrete.

The physical measuring device (see "NOTE" above) has an orthonormal basis of eigenfunctions associated with it (call this basis {|n> ; n = 1,2,...[finite or infinite]}). Whenever this device is used to perform a "measurement", one of these eigenfunctions is, so to speak, "selected" as the "result" of the "measurement".

Suppose that the quantum system is in the state |f> at the time just before the "measurement" occurs. Then, the probability that the "selected result" will be the eigenfunction |n> is given by the modulus-square of the overlap integral between |n> and |f>, namely,

P(n) = |<n|f>|^2 .

If the initial state |f> of the quantum system already happens to be one of those special eigenfunctions associated with the device, say |f> = |m>, then the probability P(n) above equals zero for n different from m, and equals unity for n = m. Physically, this means:

If the quantum system is already in an eigenstate of the observable just before the "measurement" occurs, then that eigenstate will be "selected" as the "result" with certainty.

Does this help?

4. Jun 28, 2004

### speeding electron

I'm not sure if this is the actual meaning of the question, but what I would like someone to explain is:
1) What you actually mean when you say that two wavefunctions are orthogonal. I know what this means in terms of vector functions but wavefunctions are scalars.
2) What the physical significance of the different eigenfunctions of an operators being orthogonal - in a kind of "eigenspace" or whatever you like to call it. I know people always stress that the mathematics of quantum mechanics can rarely be put into physically intuitive terms, but can this idea be explained in this way?

5. Jun 28, 2004

Staff Emeritus
Wave functions are vectors in Hilbert space* of two vectors is zero, so that any variation in the magnitude of one does not affect the magnitude of the other. Recall that the squared magnitude of a wave function at a value is interpreted as the probability of finding whatever observable you are desribing at that value, so the point of orthogonality is that the probabilities of the two wave functions don't interact.

* The function itself may be scalar valued, but it, the function, is an element of the vector Hilbert space. The values of Hilbert space range over not the scalars but the FUNCTIONS.

Last edited: Jun 28, 2004
6. Jun 30, 2004

### turin

The functions are "vectors" before you evaluate them with a particular argument. When you evaluate them with a particular argument, you get a component value, where the component is the argument.

One basis for this vector space is the domain of the function. Another (equally valid, sometimes) basis is a set of polynomials.