# Quantum state with well-defined position and momentum?

1. Oct 4, 2011

### romsofia

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)

2. Oct 4, 2011

### Staff: Mentor

No, because of the Heisenberg uncertainty principle.

3. Oct 4, 2011

### romsofia

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!).

4. Oct 4, 2011

### Chronos

The eigenstates for position and momentum are noncommutative, hence the uncertainty principle.

5. Oct 5, 2011

### Matterwave

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.

6. Oct 5, 2011

### tom.stoer

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!

7. Oct 5, 2011

### dextercioby

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.