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"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.


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.

-Quantum mechanics - measuring position.

  • Related articles:

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.

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    $\begingroup$ You probably would have seen this one, but still i am posting on the off chance ........anyway, the only difference between your question and the one in the link is that of position and momentum eigenfuntions $\endgroup$ – Prasad Mani Oct 30 '16 at 9:15
  • $\begingroup$ @PrasadMani Yes, I've seen that question, but I think that mine is a bit more specific: I'm interested in a rigorous formal solution of this paradox. Thanks anyway. $\endgroup$ – valerio Oct 30 '16 at 9:46
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    $\begingroup$ Have a look at this question and answer (possible duplicate!) - the postulate you cite is for an observable with discrete non-degenerate spectrum and needs to be replaced by the Lüders-von Neumann axiom in general. $\endgroup$ – ACuriousMind Oct 30 '16 at 15:37
  • $\begingroup$ @ACuriousMind That answer is interesting indeed, but somehow not completely satisfactory because it is not clear what this $\delta$ should be exactly. Maybe I should take a closer look at POVMs and the formalism of Kraus operators, as the author of the answer suggests in one of the comments. I studied them in a quantum information course, but not so in-depth. In the end, it all comes down to a rigorous mathematical formalization of the measurement process, and as far as I know there is not only one accepted theory... $\endgroup$ – valerio Oct 30 '16 at 22:39
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    $\begingroup$ An experimentalist's pov: the ideal measurement is not measurable even theoretically because ANY measurement involves potentials of some sort. Potentials generate discrete eignevalues. To see the particle represented by the plane wave , it should interact and interaction implies a potential which will have eigenfunctions appropriate to the problem under study. $\endgroup$ – anna v Nov 3 '16 at 5:53
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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


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.

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  • $\begingroup$ Yes, I've read Griffith's book. So basically your answer is that the cited postulate does not apply to the momentum operator and that when we measure momentum in real life we obtain a wavepacket with a narrow range of momenta. This explanation could be correct, but it is not complete: how exactly does the collapse to such a wavepacket happen? How can we formally treat it? Also, isn't it possible that the problem is with our definition of momentum operator to begin with? $\endgroup$ – valerio Oct 30 '16 at 11:21
  • $\begingroup$ Actually, i dont think its application excludes $\hat p$; if it bothers you, just do fourier transform, go to momentum space, you will end up with basically the same problem; with position operator given by $i\hbar\frac{\partial}{\partial x}$ and momentum as $\delta(p-p^\prime)$ The spread in wavefunction after collapse comes naturally out of the nature of operators in quantum mechanics and when we measure $\hat x$ in position basis, we dont claim it to be at a position $x^\prime$ from the eigenstate $\delta (x-x^\prime)$ with arbitrary accuracy; its within the limit of heisenberg uncertainty. $\endgroup$ – Prasad Mani Oct 30 '16 at 12:13
  • $\begingroup$ Now of course you could proceed and ask (at least i would), what happens in case of energy eigenstates? Since they are normalizable in most cases (not free particle or such), do they have a spread in the measured energy after the wavefunction collapse or should we specify it by also mentioning the TIME it took to measure that energy and thereby put an estimate on the spread of the energy? This is the extent of my knowledge, i am sorry. But i am looking forward to more refined and better answers to your question as well as the one in this comment! $\endgroup$ – Prasad Mani Oct 30 '16 at 12:16
  • $\begingroup$ I wrongly said that position operator in momentum space is $i\hbar\frac{\partial}{\partial x}$.....read that as $i\hbar\frac{\partial}{\partial p}$ $\endgroup$ – Prasad Mani Oct 30 '16 at 12:19
  • $\begingroup$ "But there is no such thing as a particle with definite momentum, courtesy heisenberg uncertainty principle" Heisenberg's uncertainty of principle just says there's no such thing as a particle with both definite position and definite momentum, it doesn't say that neither position nor momentum can be definite. $\endgroup$ – Keshav Srinivasan Oct 31 '16 at 8:25
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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.

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  • $\begingroup$ But it is precisely because plane waves are not contained in $L^2$ that we can say that momentum is an hermitian operator: in your calculation, neither $\psi$ nor $\phi$ can be a plane wave, because they are not contained in $L^2$. $\endgroup$ – valerio Nov 4 '16 at 22:00
  • $\begingroup$ @valerio92 The point is that we crucially depended on the fact the we were dealing with function in $L^2$ to conclude the momentum operator was hermitian. If the momentum operator acts on plane waves or the space that includes plane waves, it is not hermitian and therefore can't correspond to any measurement. $\endgroup$ – Amara Nov 4 '16 at 22:38
  • $\begingroup$ @valerio92 On the domain of $L^2$ the momentum operator is hermitian but not on the extended domain which includes plane waves. This is my point. $\endgroup$ – Amara Nov 4 '16 at 22:51
  • $\begingroup$ Yes, but who said that it must be hermitian in such an extended domain? From a formal point of view, it will simply be an operator without an eigenfunction, because the solution of the eigenvalue problem is a plane wave, which is not contained in $L^2$. [PS actually the proper domain of $p$ is $\{\psi \in L^2(\mathbb R) : \psi' \in L^2(\mathbb R)\}$, but this is just a detail in the present discussion] $\endgroup$ – valerio Nov 4 '16 at 22:56
  • $\begingroup$ @valerio92 it must be hermitian on the extended domain because you want to talk about measurement. If you want to talk about measurement quantum mechanics says you must have a hermitian operator and then Dirac's statement follows. But on the extended domain you do not have a hermitian operator all you have is an operator that has plane waves as eigenfunctions.That is not good enough $\endgroup$ – Amara Nov 5 '16 at 0:37

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