The time evolution of the one-dimensional quantum mechanical free particle ($V(x) = 0$ $\forall x$) is described by the following Schroedinger equation

$ -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} = i\hbar\frac{\partial \Psi}{\partial t} $

and the requirement that $\int_{-\infty}^{+\infty}|\psi(x, t)|^2dx$ converges. Since $V(x)$ is time-independant, we have, equivalently that

$\Psi(x, t) = \psi(x) \cdot e^{\frac{iEt}{\hbar}}$

$-\frac{\hbar^2}{2m}\frac{d^2 \psi}{dx^2} = E \psi$

Where E represents the allowed values of the energy. Following the treatment and notation scheme of Griffiths' introductory textbook, we appear to arrive at the following allowed states for the free particle:

$\Psi_{k}(x, t) = A e^{i(kx - \frac{\hbar k^2}{2m}t)}$

$E_k = \frac{\hbar^2 k^2}{2m}$

$k \in \mathbb{R}$

until we discover that no choice of the constant $A$ can meet the additional requirement placed on $\Psi_{k}(x, t)$. The problem is circumvented by the introduction of a wave packet; a superposition of the uncountably infinite "eigenstates" $\Psi_{k}(x, t)$ which becomes an integral over all values of $k$

$\Psi(x, t) = \int_{-\infty}^{+\infty}\phi(k) \cdot e^{i(kx - \frac{\hbar k^2}{2m}t)}dx$.

which, for suitable $\phi(k)$ is a well-behaved wave function with a well defined expectation value and uncertainty for the energy. My questions then, are the following:

1) What DOES one see when they measure the energy of a free particle?

2) If a measurement returns $\frac{\hbar^2 k^2}{2m}$ for some $k$, then surely the measurement has collapsed $\Psi(x, t)$ onto $\Psi_{k}(x, t)$, but would this not contradict the observation that $\Psi_{k}(x, t)$ is not an allowed state?

3) Do repeated measurements return the same energy? If not, do the results differ with mean given by $<\Psi|\hat{H}|\Psi>$ and spread given by $\sigma_\hat{H}$?

4) Can we even measure the energy of a free particle?

Thank-you for reading and sorry for the length of my question and/or if it seems inane.

  • $\begingroup$ It's part of the system around here to accept an answer if someone answers your question to your satisfaction. If the existing answers aren't what you need, you can comment on those answers asking for more information or clarification. $\endgroup$
    – DanielSank
    Apr 29, 2014 at 23:34

1 Answer 1


This is an excellent set of questions.

The basic thing to realize here is that a wave function described by $Ae^{i(kx - \omega t)}$, where here $\omega \equiv E/\hbar = \hbar k^2 \ 2m$, extends with equal weight through all of the entire universe. These waves are called "plane waves". Because they are of infinite extent these wave functions are

  • Not normalizable in the usual sense, and
  • Not possible to create in any experiment.

Already we can answer your first question

  • Q1: What DOES one see when they measure the energy of a free particle?
  • A1: You can't ever get a wave function that is actually proportional to $e^{i(kx - \omega t)}$. Whatever the physical size of your experiment is limits the boundary in which the particles live. There are always boundaries. If the boundaries are large compared to the wave length of the particle then you can get wave functions that behave almost as if they were plane waves. You could have a wave function like $\Psi(x,t) = \phi(x)e^{i(kx-\omega t)}$ where $\phi(x)$ is an envelope with a large extent. If $\phi(x)$ extends over a large region of space, then this wave function is almost a plane wave. In this case, the Fourier transform of $\phi(k)$ will be a very narrow function in $k$-space. The values of energy you might get as the result of a measurement are just given by the probability weight $\left|\phi \left( k=\sqrt{\frac{2mE}{\hbar^2}} \right) \right|^2$, so if $\phi(k)$ is a narrow function, then the probability distribution of energies you can observe is narrow.

Your second question actually isn't really a quantum mechanics question. To answer it, we have to make an observation about the relationship between energy and time. At the beginning of your question you wrote


which is quite correct for a wave function of definite energy. What you can see is that just as with our discussion of how definite $k$ means your wave functions extends over all space, definite $E$ means your wave function extends over all time. A result of this is that any measurement you construct to measure $E$ for any finite amount of time will have an uncertainty bounded by $\Delta E\Delta t \gtrsim \hbar$. This is entirely similar to the classical relation for waves $\Delta \omega \Delta t \gtrsim 1$ through the quantum relation $E=\hbar \omega$.

  • Q2: If a measurement returns $\hbar^2 k^2/2m$ for some $k$, then surely the measurement has collapsed $\Psi(x,t)$ onto $\Psi_k(x,t)$, but would this not contradict the observation that $\Psi_k(x,t)$ is not an allowed state?
  • A2: Any measurement lasting a finite amount of time has an intrinsic uncertainty in the measurement of the energy. This is not a quantum thing, it is a classical thing, basically coming from the fact that you can't measure a frequency to infinite precision with a finite time measurement.

  • Q3: Do repeated measurements return the same energy? If not, do the results differ with mean given by $\langle\Psi|\hat{H}|\Psi\rangle$ and spread given by $\sigma_H$?

  • A3: Repeated measurements in the case you've described will not give the same result. Indeed, you will be drawing from a distribution with mean $\langle \Psi| \hat{H} | \Psi \rangle$ and width $\sigma_H$. That width comes precisely from the uncertainty relation we discussed leading up to question 2.

  • Q4: Can we even measure the energy of a free particle?

  • A4: At this point I think you get the picture.

Now, after all that strictly theoretical business we need to look at some numbers so you get a practical feel for how these constraints actually come into play. Suppose we have a particle in a one-dimensional box. The energy levels in that case are

$E_n = n^2 h^2 / 8mL^2$

where $m$ is the mass and $L$ is the length of the box. The energy difference between the first and second levels is $\Delta E_{1,2} = 3h^2 / 8mL^2$. Let's suppose this particle is an electron and see how much time we need to measure in order to distinguish these two levels. We get

$t \gtrsim \hbar / \Delta E_{1,2} = \frac{\hbar 8 m L^2}{3h^2} = \frac{4mL^2}{\pi3h}$

Now we have to pick a length of the box. Let's just put in a meter. Taking that and the mass of the electron we wind up with

$t(L=1\,\textrm{meter}) \gtrsim 10\,\textrm{minutes}$.

This is a very long time for typical experiments with single electrons. The reason is that with a box that big the energy levels are very close together, so distinguishing them takes a long time. For a more realistic length scale of $L=1\mu\textrm{m}$ you get

$t(L=1\mu\textrm{m}) \gtrsim 0.5\,\textrm{nanoseconds}$.

That's pretty darn fast! That means that for the right set of parameters where the energy levels are well-separated, this energy-time uncertainty relation imposes only pretty weak constraints. On the other hand, there are real life experiments going on these days where time scales of nanoseconds are commonplace and this sort of thing really matters.

I hope that was helpful. If something is not clear please just ask.

  • $\begingroup$ Thank-you! I really appreciate the level of detail in your answer. It helps a lot! $\endgroup$ Jul 17, 2014 at 23:01
  • $\begingroup$ @SeanMacfoy: Thank you for saying so. Positive responses like yours encourage good answers :) $\endgroup$
    – DanielSank
    Jul 22, 2014 at 21:40

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