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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?

Cosmas Zachos
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amilton moreira
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  • 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"? – Daddy Kropotkin Mar 13 '21 at 17:46
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    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". – Philip Mar 13 '21 at 17:48
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    In no way HUP implies that $p \neq 0$. – GiorgioP-DoomsdayClockIsAt-90 Mar 13 '21 at 17:51
  • @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 – amilton moreira Mar 13 '21 at 17:54
  • Linked. – Cosmas Zachos Mar 13 '21 at 18:01
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    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. – Sebastian Riese Mar 13 '21 at 18:07
  • @CosmasZachos I have withdrawn my comment. – my2cts Mar 13 '21 at 21:15

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$\require{cancel}$

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...

Cosmas Zachos
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  • 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 – amilton moreira Mar 13 '21 at 20:06
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    Experimentally? Detectors are so huge in $\hbar$ terms, that they are circumscribed by the uncertainty principle for sure. – Cosmas Zachos Mar 13 '21 at 20:16
  • @ Cosmas Zachos What about if we could construct a detector small enough ? – amilton moreira Mar 13 '21 at 20:38
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    Oh, conventionally we call detectors those macroscopic objects which couple to quantum states and decohere their wavefunctions. But you may keep thinking... – Cosmas Zachos Mar 13 '21 at 20:39
  • @ Cosmas Zachos What i mean is you said that Detectors are so huge in ℏ terms, that they are circumscribed by the uncertainty. If we could construct a detector small enough than would this pose any restriction to a particle to have zero momentum – amilton moreira Mar 13 '21 at 20:46
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    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... – Cosmas Zachos Mar 13 '21 at 20:49
  • @ Cosmas Zacho Would you give me any reference for your last comment? – amilton moreira Mar 13 '21 at 20:53
  • Oh, I liked an answer in my answer... – Cosmas Zachos Mar 13 '21 at 20:56
  • ok thank you for your time – amilton moreira Mar 13 '21 at 20:58