Does $\sigma_x\sigma_p = 0 \cdot \infty$ after a measurement of particle position?

Question

I feel this question has an obvious answer that I should have been able to find independently, but I've searched for a while now it hasn't clicked.

When position is measured, the uncertainty of the resulting delta spike's position is $0$, and the momentum of the particle, determined by its wavelength, has therefore infinite uncertainty. (Alternatively, $\sigma_x=\infty$ and $\sigma_p=0$ after measuring momentum.) Wouldn't each of these give the uncertainty relation

\begin{align*} \sigma_x\sigma_p &\geq \frac{\hbar}{2} \\(0)(\infty) &\geq \frac{\hbar}{2} \end{align*} ?

Does $0 \times \infty$ even make sense? Is it just defined to be $\geq \frac{\hbar}2$ here so the uncertainty principle is obeyed?

The usual application of the uncertainty principle, as far as I know, is as a constraint before a measurement is actually made, with respect to the variation in the expectation values of observables. Once you've made a measurement, does the uncertainty principle simply no longer apply?

Answer 1

When position is measured, the uncertainty of the resulting delta spike's position is 0

This notion is the root of the problem. Quantum states which are actually eigenstates of the position operator are mathematically pathological and also completely unphysical.

Some math tools

Consider a one dimensional system. Suppose $\{|x\rangle \}$ is an orthonormal basis for the vector space of possible quantum states. Then we should be able to expand any general state $|\Psi \rangle$ in terms of this basis

$$|\Psi\rangle = \int_x |x\rangle \Psi(x) \, dx. \qquad (1) $$

In this equation, just think of $\Psi(x)$ as coefficients of an expansion of the vector $|\Psi\rangle$ in the basis $\{|x\rangle \}$. If this expansion is taken seriously so that $\Psi(x)$ are the coefficients of the expansion, and if the basis really is orthonormal, then it should be the case that

$$\langle y | \Psi \rangle = \Psi(y) . \qquad (2) $$

Stuffing (2) into (1) gives

$$|\Psi \rangle = \int_x |x\rangle \langle x | \Psi \rangle \, dx = \left( \int_x |x\rangle \langle x | \, dx \right) | \Psi \rangle $$

which suggests that

$$\int_x |x \rangle \langle x | \, dx = 1 \qquad (*) $$

Remember equation $(*)$ forever, as it is crucial.

Ok now let's start with (2) and stuff (1) into it:

$$ \begin{align} (2) &\qquad \langle y | \Psi \rangle &= \Psi(y) \\ \text{use }(1) &\qquad \langle y | \int_x |x \rangle \Psi(x) \, dx &= \Psi(y) \\ &\qquad \int_x \langle y | x \rangle \Psi(x) \, dx &= \Psi(y) \end{align} $$

This equation shows that $\langle y | x \rangle = \delta(y-x)$. This is just a formal proof what you already know that a position eigenstate is a delta function.

The eigenstates aren't normalized

Let $| p\rangle$ be a momentum eigenstate with eigenvalue $p$. You may know that $\langle x | p \rangle = e^{ipx/\hbar}$. In other words, the wave function in the position basis of a momentum eigenstate is $e^{ipx/\hbar}$. Using this we can try to find the norm of $|p\rangle$

$$ \begin{align} \langle p | p \rangle &= \langle p | \left( \int_x | x \rangle \langle x | \, dx \right) \left( \int_y | y \rangle \langle y| \, dy \right)| p \rangle \\ &= \left( \int_x \langle p | x \rangle \langle x | \, dx\right) \left( \int_y | y \rangle \langle y | p \rangle \, dy \right) \\ &= \int_{x,y}e^{-ipx/\hbar}e^{ipy/\hbar} \langle x | y \rangle \, dx \, dy \\ &= \int_{x,y} e^{ip(y-x)/\hbar}\delta(y-x) \, dx \, dy \\ &= \int_x 1 \, dx \end{align} $$

This is a problem. The integral of 1 over all possible $x$ is definitely not a finite number. This is a major problem because our bases are supposed to be normalized. It turns out that with a continuous set of eigenvalues, like position and momentum, you can't come up with normalizable basis states [1]. Note also that the momentum wave function $e^{ipx/\hbar}$ has infinite uncertainty in position.

The point is that position and momentum eigenstates are pathological mathematically, so any relations (like the uncertainty relation) derived in general falls apart when applied to position and momentum eigenstates.

Answer to questions

Is $0 \times \infty$ just defined to be $\ge \hbar /2 $?

No. The uncertainty relation just doesn't work with position and momentum eigenstates because they are pathological.

The usual application of the uncertainty principle, as far as I know, is as a constraint before a measurement is actually made, with respect to the variation in the expectation values of observables. Once you've made a measurement, does the uncertainty principle simply no longer apply?

The relation $\sigma_x \sigma_p \ge \hbar / 2$ is just a statement about the shapes that the wave function can have, or equivalently about the possible results you can get by making measurements on an ensemble of quantum states. It holds before, during, and after the measurement!

In real life, you can't do a measurement which would collapse the state into a position eigenstate. It turns out that to do that you'd need an apparatus which is infinitely large in extent. In any actual experiment there's always a finite width of the position basis wave function after you do the measurement. This in turns translates to finite values for $\sigma_x$ and $\sigma_p$ and the uncertainty relation always holds.

Go back to your equation $0 \times \infty \ge \hbar / 2$. Instead of having a delta function in position, let's assume we have a sharp but finite peak, something like $\exp [ -x^2 / (2\sigma_x^2)]$ where $\sigma_x$ is very small. In that case, the wave function in momentum is $\exp[ -p^2 / (2\sigma_p^2)]$ where $\sigma_p = \hbar / 2 \sigma_x$. Note that $\sigma_p$ is big because $\sigma_x$ is small. This makes $0 \times \infty \ge \hbar/2$ more like

$$\text{big thing} \times \text{small thing} \ge \hbar / 2$$

which remains true unless you actually consider the crazy case where the position uncertainty actually goes to zero.

[1] We could have seen this by trying to compute the norm of a position state $|x\rangle$, but this involves the square of delta functions which can be intrinsically confusing.

Answer 2

As far as I understand it, your claim is not per se about measurement but more like "if I prepare a particle in an eigenstate of either position or momentum, doesn't it mean that I have product of uncertainties $0.\infty$?" isn't it?

One first problem with this claim as it is, is that the corresponding "wave functions" of these states are not $L^2(\mathbb{R})$ and therefore cannot really represent the physical state of a particle.

A second problem is how would you actually prepare, in a repeated manner, the same eigenstate of either position or momentum. The route you suggest seems to be "by measuring either the position or the momentum" of a particle that would be in any quantum state to begin with and rely on the wave collapse. The problem with that strategy is that, since in an unbounded system, $X$ and $P$ are continuous variables, then so are their corresponding probability distributions. One thing to know about continuous probability distributions is that they never give twice exactly the same result i.e. the probability to get exactly $X=x$ or $P=p$ say is rigorously zero (that's in substance why these states correspond to unphysical wave functions).

Now, if instead, you imagine a box with hard walls, you may be able to prepare the particle in a definite momentum state for instance as momenta take only discrete values. The only problem with that is that the hamiltonian does not strictly commute with the momentum operator and the momentum will change over time (as the particle can bounce repeatedly on the walls and change at the very least the direction of the momentum).