Consider a free particle with hamiltonian $\hat{H}=\frac{\hat{p}^2}{2m}$ and propagator $\hat{U}(t) = e^{-\frac{i}{\hbar}\hat{H}t}$:

we can compute the time evolution of a position wavefunction as:

$$ \langle x|\psi(t)\rangle = \int_{-\infty}^\infty dp\langle x|p\rangle \langle p | \psi(t)\rangle = ... = \sqrt{\frac{m}{2\pi\hbar i t}}\int_{-\infty}^\infty dx'\psi(x',t=0)e^{i\frac{(x-x')^2}{2\hbar t}m} $$

However, when trying to conceptually play with this equation, I stumble on a problem: By the postulates, after a position measurement on the free particle, its state collapse into $|x_0\rangle$ and has position wavefunction: $\langle x|x_0\rangle = \delta(x-x_0)$. Intuitively, one expects that the time evolution of a measured free particle disperses the wavefunction again into a gaussian, until further measurement or interaction. However, the time evolution of the dirac delta, doesn't seem to yield such a result:

By setting $|\psi(t=0)\rangle = |x_0\rangle$, and then trying to time evolve this initial state, we get:

$$ \psi(x,t) = \sqrt{\frac{m}{2\pi\hbar i t}}\int_{-\infty}^{\infty}dx'\delta(x'-x_0)e^{i\frac{(x-x')^2}{2\hbar t}m} = \sqrt{\frac{m}{2\pi\hbar i t}}e^{i\frac{(x-x_0)^2}{2\hbar t}m} $$

But this is an unphysical! As the position probability density $|\psi(x,t)|^2 = \frac{m}{2\pi\hbar t}$. Which is constant over all space...


2 Answers 2


This question reveals that the basis states of the position and momentum operators are not really proper states. They are completely non-normalizable, and they thus do not represent a valid probability density. They are a useful mathematical concept as in the Dirac formulation of state-space quantum mechanics, but not a particularly valid physical one. This is also reflected in the fact that if you were to prepare a quantum system in an eigenstate of the position operator then the standard deviation of the position would be 0, in flagrant violation of the Heisenberg uncertainty principle that $\sigma_x \sigma_p \geq \frac{\hbar}{2}$. This means that no physical system can ever be prepared in an exact eigenstate of the position operator (or the momentum operator for that matter). Instead, you could consider preparing the system in some initial Gaussian state with a narrow spread over position. This would allow you to properly propagate the dynamics and observe the "spreading" phenomenon that you desired.

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    $\begingroup$ okay but still the basic postulate of QM states that when measuring a position x0, the states collapses in the eigenstate x0. i do get that an easy counter argument to this is that no apparatus has infinite precision, but then in natural physical interactions, does that mean that a system is never ever in a position eigenstate? $\endgroup$
    – cmatteo
    Commented Nov 19, 2023 at 14:22
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    $\begingroup$ Collapse is not a particularly well-understood phenomenon. Since the HUP is perfectly general, I am more inclined to believe that the collapse we see is just the uncertainty in position becoming rather small. As you have shown here, there is really nonsensical way to actually put a system into a position eigenstates. $\endgroup$ Commented Nov 19, 2023 at 14:30
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    $\begingroup$ @cmatteo: A better formulation of quantum observables replaces the eigenstate-operator business with projector operators/answer operators, or projection-valued measures (PVMs). These are much more intuitive to how you are thinking of measurements, because each operator now literally stands for the receipt of a given answer to the measurement question and is an actual legible mathematical operation to update the state accordingly (up to normalization, at least). $\endgroup$ Commented Nov 20, 2023 at 5:46
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    $\begingroup$ In particular, the PVM for position basically looks like a whole bunch of projector operators corresponding to receiving "yes"/affirmative answers to questions of the form "Is the particle located within $R$?" for each and every open-set region of space, $R$. There is then no projector operator for single points, because single points are not open sets in the usual Euclidean topology. And thus, when using this formalism, we don't need to worry about special, nonexistent eigenvectors of position or the like. $\endgroup$ Commented Nov 20, 2023 at 5:49

What could be the physical meaning of "setting $|\psi(t=0)\rangle = |x_0\rangle$"? Assume that this means to "see", using light, that the particle is, with certainty, in a (nearly) exact location. That light would then have (nearly) zero wavelength and thus (nearly) infinite energy.

The interaction with that (nearly) infinite energy light disturbs the particle so greatly that it can (nearly) instantly end up (nearly) anywhere, over all space, as you say. Is the extremely quick loss of the knowledge of the position ("constancy" of the spatial distribution) really that surprising or unphysical, given your (already rather tall) assumptions?

You could construct a different apparatus, of course. But measurements are never gentle. Especially if they are too precise.


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