Consider the position wavefunction $|Ψ\rangle$, which can be written as a linear combination of the eigenstates $|Ψ_n\rangle$ of the position operator:

$$|Ψ\rangle = c_1|Ψ_1\rangle + c_2|Ψ_2\rangle + ... c_n|Ψ_n\rangle + ... $$

(where the $c_n$ are the probability amplitudes, according to Born's rule).

If we perform the measurement of the position through the position operator $\hat{x}$, the wavefunction collapses to only one of these eigenstates $|Ψ_n\rangle$, and the operator's equation yields an eigenvalue which corresponds to the result of the measurement, i.e. the position of the wavefunction:

$$\hat{x}|Ψ_n\rangle = x_n|Ψ_n\rangle. $$

But how is it possible that particle is exactly at $x_n$, as this equation is saying? $|Ψ\rangle$ has collapsed and now it's only $|Ψ_n\rangle$, and $|Ψ_n\rangle$ is located at position $x_n$ according to the operator's equation. Heisenberg's uncertainty principle clearly states that a particle can't be measured to be located in a point (aka, in only one location) since that would imply maximum certainty on $x_n$, so $\Delta x=0$. From the principle:

$$\Delta x \Delta p \geq \frac{h}{4\pi}. $$

If we plug in $\Delta x=0$, the inequality is not satisfied. This would have been avoided if the wavefunction collapsed to a superposition of more than one eigenstate.

  • $\begingroup$ Note that '\rangle' is better to use than '>'. $|\psi\rangle$ vs. $|\psi>$. Also is there a reason you are using a discrete rather than continuous basis for the position basis? $\endgroup$ Jul 13, 2021 at 19:14
  • $\begingroup$ Related question: Position of the wavefunction's collapse $\endgroup$ Jul 13, 2021 at 19:18
  • $\begingroup$ @BioPhysicist n could be $\in R$ as well, wouldn't that make it continuous? :) $\endgroup$ Jul 13, 2021 at 19:27
  • 1
    $\begingroup$ No, because you can't write out the list of the elements of an uncountably infinite set like that. A countable list of real numbers is still discrete. $\endgroup$ Jul 13, 2021 at 19:32
  • $\begingroup$ See for example this excellent post. $\endgroup$ Jul 13, 2021 at 19:34

3 Answers 3


The position basis is a continuous one:

$$|\psi\rangle=\int_{-\infty}^\infty\psi(x)|x\rangle\,\text dx$$

If you make a position measurement, the mathematical idealization is that the state collapses to just one eigenstate $|x_0\rangle$, so that $\psi(x)$ is a Dirac Delta function $\psi(x)=\delta(x-x_0)$. Note that this can still be consistent with the HUP by taking a function with a finite spread in position space $\Delta x>0$ and then taking limits $\Delta x\to0$ and $\Delta p\to\infty$ such that $\Delta x\Delta p=a\geq\hbar/2\pi$ remains constant.

For a more realistic position measurement, your state just collapses to a state centered at $x=x_0$ in position space with some finite width $\Delta x>0$. This can be described as a superposition of position states.

  • $\begingroup$ The part that confuses me is the Δ𝑥Δ𝑝=𝑎≥ℏ/2𝜋 : how can it be so if Δ𝑥→0 and Δ𝑝→∞? wouldn't that work only for certain functions? i mean, how can we know if this product 𝑎 really is bigger than ℏ/2𝜋, as opposed to smaller? $\endgroup$ Jul 13, 2021 at 19:49
  • $\begingroup$ @NegentropySeeker If that product is smaller then you're not working with a physically realizable state. Plus note that I said take the limit. We do the same thing to "make" an ideal dipole. Take two charges $q$ and $-q$ that are separated a distance $d$, then take the limit $q\to\infty$ and $d\to0$ while $qd=p$ stays constant: you have yourself an ideal dipole. $\endgroup$ Jul 13, 2021 at 19:57
  • $\begingroup$ Thank you. I still need the clarification on a certain point, which is really crucial for my question; in your past answer there are some sentences that hit the nail right in the head. You say: "If you were to make a position measurement of your system, then the wavefunction collapses to a state with definite position". Now since the collapse hypothesis states that the wavefunction collapse to an eigenstate, this state with definite position you mention is an eigenstate. [continues...] $\endgroup$ Jul 13, 2021 at 20:13
  • $\begingroup$ [contin...] You then say: "Now, the above is an idealization. States with definite position are not physical. So yes, there will be some inherent uncertainty, and the collapsed wavefunction will be a spike of finite width centered at the location the particle was observed to be at." Wouldn't this mean that the wavefunction has collapsed to a combination of the "states with definite position", aka to a combination of eigenstates? (since in the previous part of the comment we said that "a state with definite position" is an eigenstate) $\endgroup$ Jul 13, 2021 at 20:13
  • $\begingroup$ [contin... 3/3] the operator equation only gives one eigenvalue (one position) for each eigenstate, so if the wavefunction is spread out over more positions (aka more eigenvalues) wouldn't that mean that it has collapsed to more eigenstates? $\endgroup$ Jul 13, 2021 at 20:18

A position eigenstate is a state with a definite position $x_0$, i.e.,


Its position uncertainty is indeed $\Delta x=0$ as you point out. The momentum and position representations of a state are related by a Fourier transform, i.e.,

$$ \mathcal{F} (\delta(x-x_0)) = \exp(- ix_0p/\hbar) $$

This is a state with maximum uncertainty in momentum because the probability density is uniform over $p$, i.e., $|\exp(-2 \pi ix_0p/\hbar)|^2=1$. In other words, $\Delta p=\infty$ so that $\Delta x \Delta p=0 \cdot\infty$, which is an indeterminate quantity that can be finite and non-zero.

Edit (which is inspired by the comments of Jakob and the OP and an elaboration of the answer by BioPhysicist)

To be more concrete and rigorous, consider the Gaussian wavefunction represented both in position and momentum space

$$\psi(x)= \dfrac{A}{\sqrt{a}} \exp(-\dfrac{x^2}{a^2})$$

$$\psi(p)=C \exp(-\dfrac{a^2 p^2}{4 \hbar^2})$$

where $A$ and $C$ are normalization constants. This state satisfies the minimum uncertainty condition: $\Delta x \Delta p=\hbar/2$. By letting $a \to 0$, we can make $\psi(x)$ approach $\delta(x)$ arbitrarily closely, and $\psi(p)$ approach 1 (which is the Fourier transform of $\delta(x)$) while maintaining $\Delta x \Delta p=\hbar/2$ for all $a$ along the way.

  • $\begingroup$ that makes sense, but i still have one doubt: if the indeterminate quantity can be finite and non-zero, how do we make sure that it's still bigger than h/4π? $\endgroup$ Jul 13, 2021 at 19:52
  • $\begingroup$ Related. $\delta(x)$ is not an element of $L^2(\mathbb{R})$ and not a physical state. $\endgroup$ Jul 13, 2021 at 19:57
  • $\begingroup$ To OP and @Jakob : Please check my edit. $\endgroup$
    – Omar Nagib
    Jul 13, 2021 at 21:06
  • $\begingroup$ @Jakob You're right, and I have ignored these subtleties in my answer. I have made an edit to give a more rigorous answer. $\endgroup$
    – Omar Nagib
    Jul 13, 2021 at 21:14
  • $\begingroup$ since we are taking the limit to 0, and not tackling directly the delta function, would you say that this means that the position wavefunction collapsed to a superposition of eigenstates instead to just one eigenstate? (an eigenstate Ψ𝑛 needs to be an unphysical delta function since Ψ𝑛 has only one eigenvalue 𝑥𝑛, according to the eigenvalue equation 𝑥̂|Ψ𝑛⟩=𝑥𝑛|Ψ𝑛⟩ ) $\endgroup$ Jul 13, 2021 at 21:22

In your last equation, you say "If we plug in $Δx=0$, we get $0≥h/4\pi$, which is not possible." You are getting the $0$ in the LHS there because you are assuming $\Delta p$ remains finite. But it does not. If somehow this highly idealized situation could become a reality and $x$ was known precisely with no uncertainty whatsoever, $\Delta p$ would indeed be infinite, and it would be "infinite enough" so that Heisenberg's relation would remain fulfilled. An infinite uncertainty might seem counter-intuitive or even impossible but, well, so is a $\Delta x = 0$.

  • $\begingroup$ yes i modified that part earlier, prior to your comment. but wouldn't the inequality still be unsatisfied? originally, if we are considering the position eigenstate (the delta function) Δ𝑥 Δp are not limits to $0$ and to infinite , they actually are $0$ and infinite. The inequality could be satisfied if we just approximate the delta function via a combination of multiple eigenstates, and take the limits $\endgroup$ Jul 13, 2021 at 21:41
  • $\begingroup$ In math, there is never "actually are" 0 and infinite when you are calculating a product between the two, because you have to know if the infinite is "infinite enough" to produce an inifinity, a finite, or a zero when the product is taken. (And mathematicians out there please excuse my sloppy language.) And the only way to know this is to take a limit somehow. I would suggest you look at the expressions Omar gives in his answer and look at what happens when $a\rightarrow \infty$. $\endgroup$ Jul 13, 2021 at 23:28
  • $\begingroup$ But i'm talking about the case in which the wavefunction $Ψ$ collapsed to a single eigenstate $Ψ_n$. The eigenvalue equation shows that the location of $Ψ_n$ is perfectly known to be at $x_n$, so $\Delta x$ is exactly 0: the single eigenstate is a delta distribution at $x_n$, not an approximation to it (after all the limit of a function around a point is not guaranteed to be the value of the function at the point). If, otherwise, $Ψ$ actually collapsed to a superposition of more eigenstates, you could approximate the delta function as much as you want, and you could take the limit to 0 $\endgroup$ Jul 14, 2021 at 10:55
  • $\begingroup$ Well, doesn't matter. It still remains that only way to do this sort of calculation is to take a limit. If you have found a case where a limit can't be taken, then the answer is simply "We don't know". But maybe this will help: So $\Delta x$ collapses to zero. Maybe I'm misinterpreting, but you seem to be assuming that $\Delta p$ remains constant, or at least finite. Why assume that? The measurement will change both $\Delta x$ and $\Delta p$. That's the uncertainty principle: You can't do a measurement that will change only one of them. And $\Delta p$ will become infinite. $\endgroup$ Jul 14, 2021 at 12:38
  • $\begingroup$ i agree that Δ𝑝 is infinite. my point is that in order to take a limit, and in order to approach the delta distribution of the particle's location (without actually being the delta distribution, just an approximation to it) you don't have anymore a single position eigenstate: you have at least a superposition of more than one states, since a position eigenstate is exactly the delta distribution (recall that the limit of a function around a point is not guaranteed to be the value of the function at the point) $\endgroup$ Jul 14, 2021 at 21:01

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