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How do we find the probability density as a function of (x,t), if the wavefunction is expressed as an infinite superposition of eigenstates? When the wavefunction is expressed as a superpostion of merely two eigenstates we end up with some overlap term when we compute the density function. How do we deal with a superposition of more than two eigenstates?

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    $\begingroup$ Why do you think the process is any different? Compute $\rho = \lvert \psi \rvert^2$. It doesn't matter what form $\psi$ is given in. $\endgroup$
    – ACuriousMind
    Jun 12, 2015 at 12:46

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In the case of an infinite superposition of eigenstates it becomes more complicated but we can still write a general expression for it.

If $$\psi(x) = \sum_{n=0}^\infty a_n \phi_n(x)$$ where the $\phi_n$ are the eigenstates of the Hamiltonian. The time dependent wavefunction will look like: $$\Psi(x,t) = \sum_{n=0}^\infty a_n \phi_n(x) T_n(t)$$ where $T_n = e^{-i E_n t/\hbar}$. So the probability density will be the absolute square of this: $$\rho(x,t) = \left|\Psi(x,t)\right|^2 = \sum_{m=0}^\infty\sum_{n=0}^\infty a_n a_m^* \phi_n(x)\phi_m^*(x) T_n(t)T_m^*(t) \\= \sum_{m=0}^\infty\sum_{n=0}^\infty a_n a_m^* \phi_n(x)\phi_m^*(x) \exp\left(i(E_m-E_n)t/\hbar\right)$$

This looks pretty hideous, I'll admit!

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  • $\begingroup$ Right, that's what I got. But how do you work with a thing like that? For example, if I want to calculate so simple a thing as the expected value of the position, I end up with an ugly integral. $\endgroup$
    – user36850
    Jun 13, 2015 at 19:45
  • $\begingroup$ Sadly there are a lot of ugly integrals in quantum mechanics, although at least some of them can be simplified in some way, for example you may have a degenerate system where $E_n=E_m$ for some $m,n$ which will kill the exponential term, or you may be able to use orthogonality of states as well $\endgroup$
    – danimal
    Jun 13, 2015 at 22:49
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In that case, you have a sum of overlaps between a pair of functions having same eigenvalue index. The cross overlap will be zero because of the orthogonality of basis set which is very important.

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you can make any state if you properly choose bases as,

$|\Psi> = \alpha|0> + \beta |1>$

where $|0>$ and $|1>$ are assumed as the complete bases. In this case, $\alpha$ and $\beta$ are the probability amplitude to observe 0 or 1 respectively.

Similarly if the system can be written by a continuous bases, like space $|x>$, any state can be represented as,

$|\Psi> = \int{\rm d}x\Psi(x)|x>$.

The probability amplitude one may find a particle (state) at $x$ is $<x|\Psi> = \Psi(x)$. This is the wave function.

I hope this helps!

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  • $\begingroup$ The probability amplitude is $\lvert \psi(x) \rvert^2$, not the wavefunction $\psi(x)$. $\endgroup$
    – ACuriousMind
    Jun 12, 2015 at 12:47
  • $\begingroup$ Thanks the comment to my anser. You should check the definition of the probability and the probability amplitude. $\endgroup$
    – rhtica
    Jun 12, 2015 at 13:35
  • $\begingroup$ I'm not talking about the probability amplitude - I'm talking about the probability density function, which is the wavefunction squared. And I'm asking about a system that is a superposition of more than two states - see the answer by danimal. $\endgroup$
    – user36850
    Jun 13, 2015 at 19:47
  • $\begingroup$ Sorry for my misunderstanding. Danimal is correct. And if you want to make the bases orthogonal (to make the equation simpler), you can follow the Gram-Schmidt procedure. $\endgroup$
    – rhtica
    Jun 14, 2015 at 11:02

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