For a free particle, the plane wave $f(x,t)=e^{i(kx-\omega(k)t)}$ where $E=\bar{h}\omega$ and $p=\bar{h}k$. It satisfies time dependent Schrodinger equation.

But this wave function is not normalizable. So we can superpose these waves to form a normalizable solution to represent the probability of finding the particle using Fourier's theorem.
So we can define $\psi(x,t)=\int_{\infty}^{\infty}dk\phi(k)e^{i(kx-\omega(k)t)}$.
Generally, $\phi(k)$ is such that it has a peak at some $k_o$ and tend to 0 at other places. Then by stationary phase analysis, $\psi(x,t)$ also has a peak at $x=\frac{d\omega}{dk}\Big|_{k_o} t$.
For non-relativistic particle, $E=\frac{p^2}{2m}$ thus $\omega(k)=\frac{\bar{h}k^2}{2m}$
Thus $\psi(x,t)$ has peak at $x=\frac{\bar h k_o}{m}t$.

Schrodinger equation is a linear equation, so superposition with any weight $\phi(k)$ of plane waves satisfies it.
But as seen above for peaks at different $k$ in $\phi(k)$, we get peaks at different $x$ in $\psi(x,t)$. Thus the position of maximum probability of finding the particle is also not same for different $\phi(k)$.

So for a same system we are getting different wave functions depicting different probabilities of finding the particle. Isn't that contradictory? Or am I missing something?

  • $\begingroup$ Why you think they are the same systems? $\endgroup$ Feb 16, 2022 at 7:04
  • $\begingroup$ Thanks for the reply. Yes you are correct that they are not same system because k is related to momentum which is an observable. Suppose we have uncertainty in momentum say $\Delta k$ around some $k_o$. Thus we have different plane wave solutions $e^{i(kx-\omega t)}$superposed with different weights $\phi(k)$. But in my above question $\phi(k)$ is some sort of an arbitrary function we can have any form for $\phi(k)$. For same system we can choose different $\phi(k)$ and can get different $\psi(x,t)$? $\endgroup$
    – Iti
    Feb 16, 2022 at 7:18

1 Answer 1


Indeed there can be different solutions for the wave function of a single system, but this is in no way a contradiction. The Schrödinger equation only describes the time evolution of a certain state, but it makes no statement about what the initial state is. This state, the boundary condition, has to be defined separately. For example, if I tell you that I have an initial state $$ \psi(x,0) = \int dk ~\phi(k) e^{i kx}~, $$ and it evolves according to the Schrödinger equation, $$ i \hbar \partial_t \psi(x,t) = H \psi(x,t)~, \qquad H = \frac{p^2}{2m}~, $$ you know immediately that at the time $t > 0$ the system will be in the uniquely defined state $$ \psi(x,t) = \int dk ~ \phi(k) e^{i (kx - \omega t)}~. $$

This is no different in classical mechanics. Consider for example Newton's equation for a free particle: $$ m \ddot{\vec x} = m \vec a = \vec F = 0~. $$ All solutions are of the fom $x(t) = x(0) + v(0) t$, but both the initial position $x(0)$ and the initial velocity $v(0)$ can be chosen arbitrarily, leading to particles starting at different positions and moving with different velocities in different directions. What is different, is that Newton's equation is of second order with respect to time, so two boundary conditions are needed, in contrast to Schrödinger's equations, which is of first order with respect to time and only requires a single boundary condition.

  • $\begingroup$ Thank you so much for the reply. Basically we can recognise this expansion of $\psi(x,t)$ as Fourier expansion with the weight $\phi(k)$ as its Fourier transform. So if we have initial wavefunction $\psi(x,0)$, then we can determine $\phi(k)$ by taking its Fourier transform. So $\phi(k)$ is not an unknown. Thus we get unique wave packet. Am I correct? $\endgroup$
    – Iti
    Feb 16, 2022 at 8:13
  • 1
    $\begingroup$ @Iti Yes, and because the plain waves, which are momentum eigenstates, are also eigenstates of the Hamiltonian in the above case of no potential, this Fourier transform is independent of time. $\endgroup$
    – sim0
    Feb 16, 2022 at 8:30

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