Quantum state with well-defined position and momentum?

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Discussion Overview

The discussion centers around the existence of a quantum state that possesses well-defined position and momentum simultaneously, exploring the implications of the Heisenberg uncertainty principle and the nature of eigenstates in quantum mechanics.

Discussion Character

  • Debate/contested
  • Technical explanation

Main Points Raised

  • Some participants propose that if both position and momentum were in a simultaneous eigenstate, it would theoretically allow for their measurement without altering the wavefunction.
  • Others argue that such an eigenstate does not exist due to the Heisenberg uncertainty principle, which asserts that position and momentum cannot be precisely defined at the same time.
  • One participant mentions a misconception regarding the relationship between measurement and the uncertainty principle, emphasizing that it is a mathematical theorem derived from the properties of non-commuting operators.
  • Another participant notes that the non-commutativity of position and momentum operators prevents the construction of a simultaneous eigenstate, specifically in the same direction.
  • There is a claim that while there are states that minimize the uncertainty principle, no states exist where the uncertainty product is exactly zero.

Areas of Agreement / Disagreement

Participants generally agree that no simultaneous eigenstate exists for position and momentum due to the Heisenberg uncertainty principle, but there are varying interpretations and nuances regarding the implications and mathematical foundations of this principle.

Contextual Notes

Some discussions touch on the limitations of understanding related to the measurement process and the strict mathematical nature of the uncertainty principle, which may not be fully resolved in the conversation.

romsofia
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Well, I know if both the position and momentum are in a simultaneous eigenstate then, theoretically, we would be able to measure momentum and position without changing the wavefunction. But, is there such an eigenstate out there?

Any help is appreciated!

(sorry if the wording is awkward)
 
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romsofia said:
is there such an eigenstate out there?

No, because of the Heisenberg uncertainty principle.
 
jtbell said:
No, because of the Heisenberg uncertainty principle.

I've always thought they're would be maybe ONE eigenstate which could contain them both.

I just found a site proving it though: http://inst.eecs.berkeley.edu/~cs191/fa05/lectures/lecture13_fa05.pdf
(I never saw a proof, guess I have a flaw in my understanding of the principle!).

Thanks for answering though!
 
The eigenstates for position and momentum are noncommutative, hence the uncertainty principle.
 
What Chronos meant is that the position and momentum operators are non-commutative, therefore you cannot construct a simultaneous eigenstate.

One should mention that this is true only for position and momentum in the same direction.
 
There is a general misconception that this has something to do with measurement; that's not true. The HUP followes as a strict mathematical theorem from Hilbert space geometry. This demonstrates that for non-commuting operators the product of their uncertainties is always non-zero. This holds for arbitrary states, i.e. there are states minimizing the HUP, but there are no states for which the result is exactly zero!
 
romsofia said:
Well, I know if both the position and momentum are in a simultaneous eigenstate then, theoretically, we would be able to measure momentum and position without changing the wavefunction. But, is there such an eigenstate out there?

Any help is appreciated!

(sorry if the wording is awkward)

There's no common eigenstate. Not even one. If it were, it would violate the canonical commutation relations which are the cornerstone of quantum mechanics.
 

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