I've read several Q&A's regarding free particles and the associated wave packet in this website, but found the answer to my question nowhere. It's OK to attribute a Gaussian wave packet to the free particle in order to normalize it. The question is now what the uncertainty in momentum is. The start point to find the wave function is the equation: $$H \psi = E \psi$$

we find an exponential function which is not a proper solution, so we add up infinite number of these waves and make a wave packet that also has a width in momentum space.

But we know that writing the above equation means we have measured energy of the particle. So, it has already gone to an eigenvector of energy and its energy and momentum are definite, not uncertain. If we want to decrease the momentum uncertainty to zero, we have to choose a Dirac delta function as the distribution function in momentum space. What happens is that the wave packet breaks into a simple exponential function, what we had at the beginning. What am I missing?

  • 2
    $\begingroup$ No writing $H\psi=E\psi$ doesn't mean that you observed something. You are playing with equations not with highly expensive toys. $\endgroup$
    – Gonenc
    Jul 7, 2015 at 10:34

1 Answer 1


First of all I think I should clarify the confusion of the terminology "uncertainty in $x$". What we mean by uncertainty is simply the standard deviation of the observable defined by:

$$\Delta x= \sqrt{ \left< x^2 \right> - \left< x \right>^2}$$

where $<A>$ denotes the expectation value of the operator $A$ is some state $\psi$. You should be aware that I won't put any hats (^) on the operators, so when I write $x$, what I really mean is $\hat x$. Furthermore I'll drop all the $x$ (the variable not the operator) dependencies e.g. I'll write $\psi$ to indicate $\psi(x)$.

Note that all of the proper states ie square integrable functions, that solve the Schrödinger Equation, must satisfy the uncertainty relationship, given by

$$\Delta x \Delta p \ge \frac \hbar 2$$

so saying that the state is an "eigenvector of energy and its momentum is definite, not uncertain" is not true in general, though it is correct for this specific case.

Returning to the question of Gaussian wave packages, you should note that the equation

$$H\psi = E \psi$$

does not collapse the wave function. Take for example the state $\psi = a_1 \phi_1 + a_2\phi_2$, where $\phi_i$ are the energy eigenfunctions of the Hamiltonian $H$ with energy eigenvalue $E_i$ and $a_i \in \mathbb C$ satisfy $\sum_i |a_i|^2 = 1$. Now applying the Hamiltonian gives:

$$H\psi = a_1 H \phi_1 + a_2 H \phi_2 = a_1E_1\phi_1+ a_2 E_2\phi_2$$

Note that this describes a state having probability $|a_i|^2$ of having energy $E_i$ and it is certainly not an eigenstate of energy neither did we collapse the wave function by acting on it by the Hamiltonian. In particular applying the Hamiltonian to a wave function $\psi$ gives the probability density of the state $\psi$ having energy $E_i$.

The Gaussian wave package is just a particular superposition of energy eigenstates and everything I did above applies. However you should note that a Gaussian package is a superposition of infinitely many energy eigenfunctions, whereas I only took a superposition of 2 states in my example.

  • $\begingroup$ Thanks dear Gonenc. I have serious problem with understanding some points you mentioned. for instance, I guess, in your example, you missed phi_1 and phi_2 in the right-hand side. If so, it's not still an eigenvalue equation. What I say is that a state may be a1 phi_1 + a2 phi_2, just like your example, but when H acts on that, this state jumps into phi_1 or phi_2 and the measurement returns E1 or E2, each by a probability of |a1|^2 or |a2|^2. am I still wrong? $\endgroup$
    – N.S.
    Jul 7, 2015 at 15:03
  • $\begingroup$ @NavidSanayei: Both of what you said is correct. Note that if indeed you have an eigenvalue equation for energy, then the gaussian wave package is not a solution thereof. $\endgroup$
    – Gonenc
    Jul 7, 2015 at 15:05
  • $\begingroup$ In other words the gaussian wave package is not an eigenfunction of the hamiltonian but is a superposition of eigenfunctions. $\endgroup$
    – Gonenc
    Jul 7, 2015 at 15:10
  • $\begingroup$ yeah, that's it! the Gaussian wavefunction is not an eigenfunction of H, in other words it doesn't satisfy the first equation I wrote "H psi = E psi". Then how can we consider the particle to be a Gaussian wavefunction? $\endgroup$
    – N.S.
    Jul 7, 2015 at 15:42
  • $\begingroup$ If you think about it, nothing is in energy eigenstate and it is perfectly fine that a state is in superposition of energy eigenstates, like the gaussian wave function. $\endgroup$
    – Gonenc
    Jul 7, 2015 at 15:44

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