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?
3 Answers
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. $\endgroup$– user36850Jun 13, 2015 at 19:45
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$\begingroup$ 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 $\endgroup$– danimalJun 13, 2015 at 22:49
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.
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
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$\begingroup$ Thanks the comment to my anser. You should check the definition of the probability and the probability amplitude. $\endgroup$– rhticaJun 12, 2015 at 13:35
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$\begingroup$ 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. $\endgroup$– user36850Jun 13, 2015 at 19:47
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$\begingroup$ 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. $\endgroup$– rhticaJun 14, 2015 at 11:02