In classical mechanics one thinks of points in phase space as states, all the while knowing that for the precise description of any real physical system one will inevitably have to deal with uncertainties and thus will have to talk about probability distributions on phase space. One assumes that one can measure observables to infinite precision although in practice this is never possible.

In quantum mechanics "states" of definite momentum (or position) aren't states - they are not represented in the Hilbert space. They are used a lot but always bearing in mind that one is actually considering very narrow peaks rather than delta distributions, for example. Also strictly speaking momentum measurements are impossible, only measurements of they type "Was the particle's momentum uppon measurement contained in the Borel set $A$?" The mathematical machinery (Gelfand triples) is in place but one does not physically allow the generalized states (many non-normalizable functions as well as distributions) to be states. Why would it be wrong do define objects as states of definite momentum to be actual quantum states?

I can think of the following arguments:

  1. One can think of the infinite space model as the limit (i.e. an idealization) of a very large "box with boundary conditions" where momenta are discrete. Here distributions used to correspond to actual states before the limit. A common argument is that it should be impossible to distinguish a system in a very large box from a system in infinite space. Why does this not imply that one should be allowed to think of definite momentum states as actual states?

  2. In such a generalized state most observables won't have a probability distribution for the values one could obtain measuring them (a definite momentum state will not have a position distribution). I don't see how this is a problem, since there are vectors in the Hilbert space representing actual states for which the same is true (them not being in the domain of definition of certain observables). Looking at examples there is no obvious reason to dismiss these states as pathological. Now domains of observables being dense, one can always find a state that is arbitrarily norm-close to some given state and in the domain of a collection of arbitrary observables. However, norm-closeness has nothing to do with "closeness of physical states" (a notion I'm not completely sure of, which should probably be defined in terms of observables. I assume the projective Hilber space topology actually fails do provide such a notion.) so this argument is void as well, furthermore it would also apply to delta distributions in a generalized sense (those can be approximated by test functions arbitrarily well).

  3. Making some general assumptions about statistics and defining properties of a propositional lattice that should discribe QM one can show that this is equivalent to the lattice of closed subspaces of some Hilbert space. This clearly rules out the distributions as states. However, this step is not completely transparent to me and the conclusion seems conceptually very far from where the argument starts. I feel there should be some more intuitive way to provide a connection between these aspects of quantum logic and there not being such states.

Does anyone know a good argument for this or can explain the mistakes in the ideas above? Any sources on this issue?


1 Answer 1


A position ket $\lvert x \rangle$ simply does not define a state in the abstract sense: In the abstract algebraic approach to QM, the fundamental object is the $C^\ast$-algebra of observables, and states are simply linear functionals on observables, i.e. rules that assign to every observable its expectation value. This is the most fundamental property of a quantum mechanical state - it tells you what you're going to measure. The position (or momentum) ket doesn't do that, its expectation value of momentum (resp. position) is ill-defined, therefore it is not a physical state.

You cannot escape this by considering only measurements of Borel sets of position and momentum - since the kets are inherently unnormalizable, trying to apply the Born rule simply fails: It tells us to take the projector $P_i$ and compute $$ \frac{\langle \psi \vert P_i \vert \psi\rangle}{\langle \psi \vert \psi\rangle},$$ but the denominator simply doesn't exist. That this is truly irredeemable you can see by considering a momentum eigenstate $\psi_p(x) = A\exp(\mathrm{i}px)$ in position space - the "probability" to detect it within the interval $[x_1,x_2]$ is $\int_{x_1}^{x_2} 1 \mathrm{d}x$, which is larger than 1 and therefore not a probability for large $x_2 - x_1$ regardless of $A$.

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    $\begingroup$ The C* algebra approach implicitly assumes taking finite spectral projections so there is a priori no momentum operator there. The borel sets were not about escaping anything but observing that in standard qm you can't do momentum measurements outputting a single value. One could transform each observable to the space where it is a multiplication operator and only assign values where expressions exist without needing to divide a norm. For unbounded observables many expectation values don't exist anyway in reasonable states. Why is this not an argument? $\endgroup$ Aug 14, 2017 at 10:46
  • $\begingroup$ @AdomasBaliuka An argument for what? As I said, the Born rule simply fails for these states: You cannot reasonably compute the probability to measure the position of the momentum eigenstate to be in an interval $[x_1,x_2]$. But such a position measurement is certainly something we can carry out in practice. So this state represents something that doesn't exist in reality, hence declaring it a "physical state" would be silly. $\endgroup$
    – ACuriousMind
    Aug 14, 2017 at 11:00
  • $\begingroup$ There are physical states whose expectation values for position exist but not the momentum expectation and vice versa. Why is this not an argument for them being unphisical? I can accept that expectations of projection operators are more fundamental and those exist for Hilbert space vectors so that may be convincing to dismiss generalized states. What are your thoughts on my first argument? That in a box one can talk of such states and the limit in which they lose their meaning should have no physical consequence? $\endgroup$ Aug 14, 2017 at 18:54
  • $\begingroup$ @AdomasBaliuka I'm not talking about expectation values now. The requirement that we can compute the probability to measure a particle at position $[x_1,x_2]$ is much more basic than the existence of position expectation values, and is guaranteed for every physical state (since physical states are $L^2(\mathbb{R}^d)$, but not for the position/momentum states. $\endgroup$
    – ACuriousMind
    Aug 14, 2017 at 19:06
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    $\begingroup$ This "answer" has quite a few mathematical technical errors. From the point of view of mathematics, there is nothing intrinsically unnormalizable about functionals like the Dirac delta function. And the probability to detect a momentum eigenstate in position space in an interval $[x_1,x_2]$ is always $0$, which is less than $1$. $\endgroup$ Nov 5, 2018 at 18:28

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