How can the collapse of the wavefunction be compatible with Heisenberg uncertainty principle?

Question

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.

Answer 1

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.

Answer 2

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

$$\delta(x-x_0)$$

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.

Answer 3

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 $\Delta x = 0$.