# Quantum physics normalization

1. Apr 26, 2014

### kdlsw

I do understand the probability of a wave function ψ is given by ∫ψ* ψ d3x, which after normalization is equal to 1. However, I then saw the following, ∫ψn* ψmmn

Here is my understanding, the discussion is about discrete eigenfunction and value as expressed in m and n, ψn and ψm are two discrete wave functions, and with the orthonormal method, they are arranged to be ∫ψn* ψm=1 when m=n and 0 when m≠n.

What I don't understand is, doesn't such form equal to ∫ψ* ψ d3x ? Where the complex conjugate and normal wave function part comes from the same ψ. What is the point of discussing the complex conjugate of a different wave function ψn* in the first place? Are ψn ψm just two different states of the same wave function ψ?

Thanks a lot

2. Apr 26, 2014

### Matterwave

$\psi$ in the first equation is a general wave function. It can describe some particle somewhere. The $\psi_n$ are a set of orthonormal basis functions. Usually they are the eigenfunctions of some important operator (quite often the Hamiltonian). They provide a convenient way in which to express your general wave function:

$$\psi=\sum_n c_n\psi_n$$

Any $\psi$ can be expressed this way provided that the $\psi_n$ form a complete basis of the Hilbert space. It's useful that the basis functions are ortho-normal so that taking inner products is easy.

3. Apr 26, 2014

### kdlsw

Thanks a lot, but if I may ask a bit more, what is the meaning of having ∫ψn* ψm if they are just basis functions?

4. Apr 26, 2014

### Matterwave

What do you mean by "meaning of having..."? I'm not sure what your question is so I don't know how to answer adequately. Could you maybe rephrase it? The basis functions $\psi_n$ are called orthonormal if:

$$\int \psi_n^*(x)\psi_m(x) dx=\delta_{n,m}\quad \forall n,m$$

There's nothing special happening here. The fact that the basis functions are orthonormal means you can get $c_n$ really easily:

$$\int \psi^*(x)\psi_n(x)dx=\int\sum_m c_m^*\psi_m^*(x)\psi_n(x)dx=\sum_m c_m^*\delta_{m,n}=c_n^*$$

5. Apr 26, 2014

### kdlsw

Got it, Thanks!