In the quantum mechanics case of the infinite square well, the general solution to the Schrodinger equation is a linear combination of solutions with definite energy states. When you measure the particle, it will take one of these energy values.

I am now looking at the case of the free particle and I see that (because the stationary states are not normalizable) a free particle cannot exist in a stationary state. In his textbook, Griffiths states "there is no such thing as a free particle with a definite energy".

I'm struggling to see what this actually means; if you measure the energy of a free particle you will still get a value right? And won't this energy necessarily correspond to one of the infinite solutions for the free particle?


4 Answers 4


if you measure the energy of a free particle you will still get a value right? And won't this energy necessarily correspond to one of the infinite solutions for the free particle?

No, when you measure the energy of a (not necessarily free) particle you find an interval of values due to the finite precision of every measurement apparatus.

If you know that the energy is discrete (when the system is not free) and the precision of the instrument is smaller than the distance of two consecutive levels, then you can argue that you found a definite value of the energy and that the post measurment state is represented by the corresponding (proper) eigenstate of the energy observable.

When instead the energy is continuous, then you cannot argue anything in addition to that the energy stays in some $[a,b]$. In particular the energy is not definite.

A problem arises here concerning the post measurment state. The abstract idea is stated within the Luders-von Neumann postulate. It states that, if the state before measurment is the normalized wavefunction $\psi$ and the outcome interval is $[a,b]$, then the post-measurment state is $P^{(H)}_{[a,b]} \psi$ (up to normalization). Here, $P^{(H)}_{[a,b]}$ is the orthogonal projector associated to $[a,b]$ of the spectral measure of the energy observable $H$, the unique satisfying $$H = \int_{\mathbb{R}}\lambda dP^{(H)}_{\lambda}\:,$$ that is the general extension of the eigenvector decomposition $$H = \sum_{n} E_n P^{(H)}_{E_n}\:, \quad P^{(H)}_{E_n} = |\psi_{E_n}\rangle\langle \psi_{E_n}|$$ when the spectrum of $H$ is a point spectrum of energies $E_1\leq E_2 \leq \cdots$.

This general assumption is nowadays sometimes considered a bit too ideal and a better idea on the post-measurment state relies on a more concrete description of the measurement apparatus in terms of quantum operations (mathematically speaking, POVMs and their Kraus decompositions).

These observations can be ascribed to every observable with a continuous part of the spectrum. It is true that it is comfortable to deal with various types of plane waves which mimic definite energy (and also momentum) states. But these objects are not normalizable wavefunctions, so that they do not define physical quantum states. However they are useful limit ideas fruitfully exploited in various computations.

  • $\begingroup$ How do we know the energy of discrete energy levels exactly? $\endgroup$
    – Xfce4
    Sep 30, 2021 at 15:43

It is not that "there is no such thing as a free particle with a definite energy" but rather "there is (really) no such thing as a (completely) free particle".

Because the free particle is not normalised, asuuming it existed, getting one value for the energy out of an infinite set of values is undefined, the probability of getting the outcome of a measurement $A_i$ out of a measurement would be $~ 1/\infty = 0$. Simply, the measurement of $A_i$ is not defined. One exception would actually the the momentum measurement, as plane waves have a definite momentum and therefore energy, but then their position is completely uniform in space (becuase of uncertainty) and the probability of finding it "somewhere" becomes $0$. If, on the other hand, you combine several plane waves with different momentum, then measuring the actual momentum becomes also hard, unless each wave is weighted differently in the sum (then you get more well-defined wave-packets).

So free particle simply can not exist, they are a just theoretical description, very useful for other predictions but still purely theoretical in essence!

A more realistic view is a particle confined in a very big potential well (the particle-in-a-box model or infinite potential well, where infinite referes to the depth). Such a particle can then be normalised and we can extract a well-defined probability density from it.

Remember that for a 1D infinite well of length $L$ the energy spectrum is given by $$E_n = {n^2\pi^2\hbar^2\over 2mL^2}$$ so that $$\Delta E = E_{n+1}-E_n = (n+1){\pi^2\hbar^2\over 2mL^2}$$ which depends on $L$.

Thus in a "big" well ($L\gg 1$) you basically get a continuous spectrum ($\Delta E \ll 1$) but each $E_n$ is well defined. In the limit $L\to \infty$ you get the free particle but you can not define $E_n$ anymore and thus such a particle can not exist.


Short answer: physically no plane waves really exist. Mathematically, we use them all the time. If we are desperate for normalization, we normalize them by particle flux, rather than the total probability.

Let me first deviate into discussing photons, snce this situation has been extensively coveregd on this site: in order to have a truly plane wave we have to be able to observe it during an infinitely long time and at infinitely long lengths. This is impossible: firstly, because our observations are necessarily made over finite time and distance, which impose the finite width on the photon spectrum. Moreover, a photon has been emitted at some time that is at a finite distance from now (certainly not earlier than the creation of the unievrse), which again imposes finite width of the spectrum.

The situation is exactly the same for particles, although it might appear less avident. Our ability to observe plane waves is equally limited, and they necessarily interact with their surroundings. Yet, in many problems we can neglect these effects as small, and use the plane waves (typically as a Fourier expansion).


If a particle is somewhere in a very large volume then it is described by wave function normalisation factor that tends to zero. Because it is limited to a finite volume it will have a band width. This bandwidth can be made arbitrarily small. Perhaps Griffiths intends to state this fact.

Plane waves are useful as basis functions and to describe particle bundles that are wide compared to a relevant length scale. For example, for atomic transitions one can use a plane wave for the EM field because at the transition energy the EM wavelength is much larger than the scale of the electronic orbital. This leads to Fermi's golden rule.


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