Solving the time dependent Schrödinger equation (here in 1D) with time independent potential: $$i\hbar\frac{\partial}{\partial t}\Psi(x,t)=\hat{H}\Psi(x,t),$$ using Ansatz: $$\Psi(x,t)=\psi(x)\phi(t),$$ and separation of variables, gives: $$\phi(t)=e^{-iE_nt/\hbar}$$ $\psi(x)$ are the eigenfunctions and $E_n$ the energy eigenvalues. With the superposition principle: $$\Psi(x,t)=\displaystyle\sum_n^{+\infty}c_n\psi_n(x)e^{-iE_nt/\hbar}$$ Assume that at $t=0$ the wave function was $\Psi_0$, then: $$\Psi(x,0)=\Psi_0=\displaystyle\sum_n^{+\infty}c_n\psi_n(x)$$ Assuming the system has domain $\Delta x$, we should then be able to determine the coefficients with Fourier: $$c_n=\frac{2}{\Delta x}\int_{\Delta x}\Psi_0\psi_n(x)dx$$

Is this true? If so, is it a 'practical' way of determining $c_n$?

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    $\begingroup$ Except for the issue with complex conjugation, this is really the only way for an arbitrary wavefunction. $\endgroup$
    – Javier
    Nov 16, 2016 at 21:42

1 Answer 1


I would say this is not always true (the eigenfunctions are not necessarily real). You should use the condition of orthonormality for the eigenfunctions (if it holds, otherwise use the relevant coefficients). The orthonormality conditions contain complex conjugates of the eigenfunctions.

EDIT: (11/16/2016) You have an orthonormality condition for the eigenfunctions: $<\psi_m|\psi_n>=\delta^m_n=\int^{+\infty}_{-\infty}\psi_m^*\psi_n dx$. Therefore, if $\Psi_0=\sum_n c_n\psi_n$, then $<\psi_m|\Psi_0>=\int_{-\infty}^{+\infty}\psi_m^*\Psi_0 dx=\int^{+\infty}_{-\infty}\psi_m^*\sum_n c_n\psi_n dx=\sum_n c_n\delta^m_n=c_m$. Therefore, $c_m=\int_{-\infty}^{+\infty}\psi_m^*\Psi_0 dx$.

  • $\begingroup$ Can you elaborate slightly? $\endgroup$
    – Gert
    Nov 16, 2016 at 18:20
  • $\begingroup$ @Gert: Please see the EDIT. $\endgroup$
    – akhmeteli
    Nov 16, 2016 at 21:28

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