Paradox of wavefunction collapse into an unphysical state

Question

"A measurement always causes the system to jump into an eigenstate of the dynamical variable that is being measured, the eigenvalue this eigenstate belongs to being equal to the result of the measurement."

— P.A.M. Dirac, The Principles of Quantum Mechanics

This is one of the postulates of quantum mechanics. However, there are some cases in which this statement leads to contradictions.

For example, we know that the eigenfunctions of the momentum operator (in 1D for simplicity)

$$\hat p = -i \hbar \frac{\partial}{\partial x}$$

are plane waves:

$$\psi_p(x) = A e^{ipx/\hbar}$$

These eigenfunctions are not normalizable and therefore are not acceptable as physical states.

If we try to apply the cited postulate to the momentum operator, we would therefore incur in a contradiction: the system cannot jump into an eigenstate of the momentum operator, because such an eigenstate would not be normalizable and therefore would not be a physical state.

This paradox is usually dismissed by saying that this line of reasoning applies to an ideal measurement, which cannot be realized in practice, and that for non-ideal measurement the situation is different. But this answer doesn't seem to be satisfying to me: although it makes sense, it is not clear what is the theoretical reason why an ideal measurement is not realizable.

There seem to be only two possible solutions to this paradox:

1. The cited postulate is wrong.
2. The momentum operator is somewhat ill-defined: for example, maybe we cannot just take its domain to be the set of all sufficiently regular (*) functions $f \in L^2(\mathbb R)$ as we usually do. In this case, maybe it is possible to give a definition of the momentum operator which agrees with the cited postulate.

What is a possible solution to this paradox?

PS: As far as I'm concerned, it is perfectly fine to answer that the solution is that an ideal measurement is not physically realizable in practice, but only if such a claim is backed up with rigorous theoretical arguments explaining why this is the case.

(*) Sometimes, the condition imposed is the absolute continuity of $f$, but I don't know if it can be relaxed.

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Updates

• Related questions and answers:

-Measurement of observables with continuous spectrum: State of the system afterwards (suggested by ACuriousMind). After some discussion, the author added a wonderful Addendum that maybe can be considered as an answer to this question.

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I found this article and this article (free download) which are about this exact problem, but they are quite technical and I still have to properly dig into them.

Answer 1

This answer is coming extremely late, but I think it's worth putting out a physical answer even if you requested a mathematical one.

First consider a position measurement. If you did an ideal position measurement, the wavefunction would collapse to a position eigenstate, which is unnormalizable and hence not a valid physical state. The resolution is that ideal position measurements don't actually exist, because real measurements have finite solution.

As a crude way to account for this, suppose your measurement can only return values that are integers, in some units. Then effectively you are measuring the operator $$\hat{x}_{\text{f}} = \lfloor \hat{x} \rfloor $$ where the floor function of an operator is defined elementwise: if $\hat{x} | x \rangle = x |x \rangle$, then $$\hat{x}_f | x \rangle = \lfloor x \rfloor |x \rangle.$$ As a result, all of the states $|x' \rangle$ for $x' \in [n, n+1)$ are degenerate eigenvectors of $\hat{x}_f$. All positions in this range correspond to the same measurement result $n$.

In the case of degeneracy, the collapse postulate is slightly generalized. The Hilbert space is now divided into orthogonal subspaces, one for each distinct eigenvalue of the operator measured. Upon measurement of that operator, the state vector collapses to its projection into one of these subspaces, with probability proportional to the magnitude squared of its projection. (This reduces to the postulate Dirac gives when all the subspaces are one-dimensional, i.e. no degeneracy.)

Therefore, if you have a state with wavefunction $\psi(x)$ and measure $\hat{x}_f$, you collapse it to $$\psi'(x) = \psi(x) \cdot \begin{cases} 1 & n \leq x < n+1 \\ 0 & \text{otherwise} \end{cases}$$ if your measurement result is $n$. This state is perfectly normalizable.

The exact same logic applies to $\hat{p}$, with the logic above occurring in Fourier space rather than real space, and it leads to the same conclusion: you get a final, normalizable state with finite width in Fourier space, which corresponds to a wavepacket in real space.

So it all works out... but is it "rigorous"?! Well, the point of quantum mechanics is to predict physical results. The purpose of formulating it in terms of postulates is just to give us a concrete basis for computing results, with the ultimate goal of matching to experiment. It is well known that if you take Dirac's stated postulates at face value, and use it to compute unphysical quantities, then you can get all kinds of mathematical contradictions -- his postulates are sloppy even by physicists' standards. We continue teaching them because they work for any experiment we can conceivably perform.

If you wanted to be "rigorous", you would do it by adding piles of extra postulates that amount to saying "you're not allowed to use $\hat{x}$ in the collapse postulate, but stuff like $\hat{x}_f$ is okay". But the experimental physicist will not care at all, because they've always known that there are limitations on what they can measure, fancy postulates or not.

Answer 2

The thing of it all is this:

There are two kinds of eigenfunctions of hermitian operators. The ones which admit discrete spectra (eigenvalues are separated from one another) and the other, continuous spectra (eigenvalues fill out an entire range). If the spectra is continuous, then they DO NOT represent possible wavefunctions (only a linear combination of them.....yes a gaussian wavepacket kind of thing may be normalizable). In case of momentum operators, $$\frac{\hbar}{i}\frac{\partial f_p(x)}{\partial x} = pf_p(x)$$....$f_p(x)$ is some momentum eigenfunction.... solving the above gives $$f_p(x) = Ae^{\frac{ipx}{\hbar}}$$ which is not a square integrable one.

But since momentum is an observable, we only take real values of p and use dirac orthonormality by $$\int_{-\infty}^{\infty}f^*_{p^\prime}(x)f_p(p)dx = |A^2|\int_{-\infty}^{\infty}e^\frac{(p^\prime-p)x}{\hbar}dx = |A^2|2\pi\hbar\delta(p-p^\prime)$$, and then picking $A=\frac{1}{\sqrt{2\pi\hbar}}$, we have $$\langle f_{p^\prime}|f_p\rangle = \delta(p-p^\prime)$$......

Now, this means that the eigen functions of momentum are sinusoidal (this itself is unrealizable since any TRUE or PERFECT sine wave has to extend from $-\infty$ to $\infty$)

But there is no such thing as a particle with definite momentum, courtesy heisenberg uncertainty principle........also implies that measurement cannot collapse a wavefunction to an eigenstate with a perfectly defined momentum

This is why we make a normalizable wavepacket with a narrow range of momenta.....to make the whole thing physically realizable. None of the eigenfunctions of $\hat p$ live in hilbert space but those with real eigenvalues (wavepackets) and which are dirac normalizable do. They (eigenfuntions of $\hat p$) do not represent possible physical states but are very useful in problems like scattering from a potential hill or a barrier.

Reference :- Griffiths, Introduction to Quantum Mechanics

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EDIT

Please do not upvote my answer as it does not completely address the concerns raised by OP (take a look at the comment section below this answer), ie, a formal treatment of 'wavepacket'(not wavefunction) collapse.....i am terribly sorry if invoking such a statement is wrong. At best my answer is partially complete.

Answer 3

Here I think is the resolution to the paradox first recall that the quote you gave above corresponds to $Hermitian $ operators. The key insight is the following:

The momentum operator is not a hermitian operator on the space of functions in which the plane waves are members and can be regarded as eigenfunctions of the momentum operator. On this extended domain the momentum operator does not correspond to any measurement.

For the momentum operators to be hermitian we would like to show that $ \int \phi^* P \psi = \int P^*\phi^* \psi $. Consider how we prove the momentum operator is Hermitian, we do the following calculation: $ \int \phi^*(-i \hbar \frac{d}{dx} \psi) dx = -i \hbar( \phi^* \psi |^{\infty}_{-\infty} - \int \frac{d}{dx}\phi^* \psi dx ) $. Notice that for the first term on the right hand side the function have to vanish at $\pm \infty $. If this happens then the momentum operator is equal to its hermitian adjoint and is therefore hermitian. This excludes the plane waves because they do not vanish at $\pm \infty $. So the momentum operator is not hermitian on the space of functions in which plane waves are members. Therefore the momentum operator does not correspond to any physical measurement in this space of functions.

I think the resolution has nothing to do with whether the measurement is ideal or not, whether the momentum operator is ill-defined or not. The momentum operator does not correspond to any measurement whether ideal or not if it acts on an extended domain that includes plane waves because it is not hermitian on this extended domain.