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Suppose that a particle is in a linear momentum eigenstate $\mid p \rangle$ that is $\hat p \mid p \rangle=p \mid p \rangle$. According to the Heisenberg uncertainty principle we must have $p\neq 0$. I am not familiar with how detectors work, but they should have finite size. So if a detector detects a particle with zero momentum then the particle should remain in the detector. To not contradict the Heisenberg uncertainty principle the detector should have infinite size.

Now there is a reference frame where this particle has linear momentum equal to zero so in this frame the particle is represented by $\mid 0 \rangle $.

How can we resolve this apparently contradiction?

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  • $\begingroup$ If the particle is in a momentum eigenstate, then what does it mean for it to be "in a reference frame where this particle has momentum = 0"? $\endgroup$
    – Daddy Kropotkin
    Commented Mar 13, 2021 at 17:46
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    $\begingroup$ According to the Heisenberg uncertainty principle we must have $p\neq0$. Why? The uncertainty principle doesn't say anything about the values of position and/or momentum, only their "uncertainties". $\endgroup$
    – Philip
    Commented Mar 13, 2021 at 17:48
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    $\begingroup$ In no way HUP implies that $p \neq 0$. $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Commented Mar 13, 2021 at 17:51
  • $\begingroup$ @Philip Correct me if I am wrong. What do you mean is that the Heisenberg uncertainty principle tell us about our incapacity of knowing the position and momentum at the same time that is the position and the momentum exist but we are incapable to measure at the same time $\endgroup$
    – amilton moreira
    Commented Mar 13, 2021 at 17:54
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    $\begingroup$ There is a certain truth to linear momentum eigenstates not existing: They don't exist as elements of the Hilbert state of normalizable states, but can only be approximated by elements of that space. The uncertainty conundrum is seen to be solved when looking at the approximations: The position uncertainty goes appropriately to infinity as we approach zero momentum uncertainty. $\endgroup$
    – Sebastian Riese
    Commented Mar 13, 2021 at 18:07

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Well, they "exist" in our (physicists') imagination, as a limit, but they are not normalizable.

Adopting your notation for the null eigenstate of the momentum operator, $\hat p|0\rangle=0$, you may readily identify it as Dirac's celebrated standard ket, through which he defines his bra-ket formalism in Ch III.20 of his textbook. But, since $\langle p|0\rangle=\delta(p)$, your hyper-localized state has infinite norm, "$\delta(0)$". You may, if you wished, approximate it as the limit of a suitably normalized Gaussian in momentum space, as its width goes to zero. (Intelligent squeezed coherent states. )

  • Note it is translationally invariant, since the exponential of the momentum operator acting on it amounts to one.

The x-space wavefunction of it is a plane-wave of zero momentum, so, then, a constant, $\langle x|0\rangle= 1/\sqrt{2\pi \hbar}$, everywhere in space; also unnormalizable, and terminally delocalized: a flatline, translationally invariant. In your Gaussian approximation, it would be a "Gaussian" with infinite width.

But you may translate this state in momentum space by acting on it with the exponential of the position operator, $$ \exp (ip\hat{x}/\hbar ) | 0\rangle = |p\rangle ,$$ and further define, with Dirac, $$|x\rangle= \delta(\hat{x}-x) | 0\rangle \sqrt{2\pi \hbar} .$$

The uncertainty principle is fine as a freak limit, but you'd better reassure yourself with the Gaussians involving mutually inverse widths. As it stands, it is egregious nonsense, which should never have been written down,$\cancel{ \frac{\langle 0|\hat p^2 |0\rangle }{\langle 0|0\rangle } \frac{\int dx ~x^2 }{\int dx } }$. I didn't write this...

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  • $\begingroup$ You gave me a good answer theoretically but what about experimentally. According to your answer the wave function is well delocalized. But if we detect a particle with 0 momentum it will remain in the detector which is not delocalized $\endgroup$
    – amilton moreira
    Commented Mar 13, 2021 at 20:06
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    $\begingroup$ Experimentally? Detectors are so huge in $\hbar$ terms, that they are circumscribed by the uncertainty principle for sure. $\endgroup$
    – Cosmas Zachos
    Commented Mar 13, 2021 at 20:16
  • $\begingroup$ @ Cosmas Zachos What about if we could construct a detector small enough ? $\endgroup$
    – amilton moreira
    Commented Mar 13, 2021 at 20:38
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    $\begingroup$ Oh, conventionally we call detectors those macroscopic objects which couple to quantum states and decohere their wavefunctions. But you may keep thinking... $\endgroup$
    – Cosmas Zachos
    Commented Mar 13, 2021 at 20:39
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    $\begingroup$ Indeed, the detector itself is circumscribed by the uncertainty principle. It it is an intelligent squeezed state, it won't be accurate as a detector for detecting your state... $\endgroup$
    – Cosmas Zachos
    Commented Mar 13, 2021 at 20:49

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