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Quantum physics normalization

  1. Apr 26, 2014 #1
    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. jcsd
  3. Apr 26, 2014 #2

    Matterwave

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    ##\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.
     
  4. Apr 26, 2014 #3
    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?
     
  5. Apr 26, 2014 #4

    Matterwave

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    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:

    $$\int \psi^*(x)\psi_n(x)dx=\int\sum_m c_m^*\psi_m^*(x)\psi_n(x)dx=\sum_m c_m^*\delta_{m,n}=c_n^*$$
     
  6. Apr 26, 2014 #5
    Got it, Thanks!
     
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