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In the following notes from an MIT OCW course, Zweibach claims that energy eigenstates are not necessarily normalized. https://ocw.mit.edu/courses/physics/8-05-quantum-physics-ii-fall-2013/lecture-notes/MIT8_05F13_Chap_01.pdf

A solution $\psi(x)$ associated with an energy $E$ is called an energy eigenstate of energy $E$. [...] We do not impose the requirement that $ψ(x)$ be normalizable. This would be too restrictive. There are energy eigenstates that are not normalizable. Momentum eigenstates of a free particle are also not normalizable. Solutions for which $ψ$ is not normalizable do not have a direct physical interpretation, but are very useful: suitable superpositions of them give normalizable solutions that can represent a particle.

However, this page implies that if you perform an energy measurement, the system collapses into the energy eigenstate corresponding to the resultant measured value. http://physics.gmu.edu/~dmaria/590%20Web%20Page/public_html/qm_topics/superposition/superposition.html

Result of measurement [in QM Statefunction]: The state [$\Psi$] is destroyed. The system falls to one of the eigenstates jn after measurement.

I think that this means that those eigenstates exist but cannot be observed. Is this correct? If they aren't observed, what is their significance?

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At the bottom of page 5 and top of page 6 in the linked notes, Zweibach says

For simplicity we discuss the case when the spectrum is denumerable so that, as above, we can label the states and energies with the integer label $n$. In general a potential $V(x)$ can result in a spectrum that contains a discrete part and a continuous part. The discrete part is denumerable but the continuous part is not. The formulae we will write below require some modification when there spectrum contains a continuous part (emphasis mine). The eigenstates of the continuum spectrum are not normalizable.

You'll notice that the link from GMU says

The superposition principle states that a statefunction $|\Psi\rangle$ can be expanded as a linear combination of the normalized eigenstates $|\varphi_n\rangle$ of a particular operator that constitute a basis of the space occupied by $|\Psi\rangle$. For the discrete case (emphasis mine):

$$|\Psi\rangle = \sum_{n=0}^\infty b_n|\varphi_n\rangle$$

The GMU site never explicitly treats the continuous case, and as a result, their description of the effects of measurement needs to be modified.

Ultimately, the resolution of this problem lies in the fact that because measurements have finite resolution, measurement of an observable which has a continuous spectrum (like position, momentum, or, in the case of the free particle on a line, energy) never returns a single real number, but rather some interval $I\subset \mathbb R$. As a result, the system is projected not into a single (non-normalizable) eigenstate, but rather into a (normalizable) superposition of those eigenstates. The language is technical, but the Wiki article on the position operator might be helpful; for a more visual picture of how position measurement works, see my answer here.

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Good question!

You need to make a distinction between QM with countable number of Hilbert space dimensions (even better finite) like spin systems or the harmonic oscillator, and QM with uncountable (continuous) dimensions like particles in free space. In the case of countable dimensions it is possible to normalize the basis states. In the uncountable case the useful "basis" states are not normalizable (but they are useful). In the countable case you can measure an energy eigenstate and get that precise state after measurement. In the continuous (uncountable) case you can only measure things with finite precision and the post measurement state is some state that consists of all the states that fall within the margin of error. Those are normalizable states but they are never exact eigenstates because exact eigenstates in this cases are not real (still, very useful). These details are more mathematically involved and textbooks on quantum mechanics choose to strategically dodge them for more streamline exposition.

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  • $\begingroup$ This doesn't really have anything to do with the underlying Hilbert space, which is the same for the harmonic oscillator and the free particle - the key point is that some operators have purely continuous spectra. $\endgroup$ – J. Murray Jul 27 at 1:59
  • $\begingroup$ I agree, thank you for pointing that out. I was thinking about how to make this argument as straight forward as possible but I cut one corner to many. $\endgroup$ – oleg Jul 27 at 16:18

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