Does a position operator exist?

[chrisphd]

Does a position operator exist?

 

[Pengwuino]

What do you mean exist? I don't want to just blurt out "yes" to such a seemingly obvious question...

 

[chrisphd]

Does a position operator acting on a wavefunction give a position eignenfunction?

 

[Pengwuino]

Yes, if the wavefunctions are eigenfunctions of the position operator ;)

 

[PhaseShifter]

For the correct wavefunction (Dirac delta function), yes. For other wavefunctions, no.

 

[chrisphd]

Yes, if the wavefunctions are eigenfunctions of the position operator ;)

And if the wavefunction was not an eigenfunction of the position operator, one would expect that the wavefunction will transform into an eigenfunction of the position operator, when operated on.

 

[Pengwuino]

And if the wavefunction was not an eigenfunction of the position operator, one would expect that the wavefunction will transform into an eigenfunction of the position operator, when operated on.

No... why would it?

 

[chrisphd]

No... why would it?

Suppose you have a wavefunction that is a linear combination of momentum eigenfunctions. ie, F = 0.8a + 0.6b where F is the wavefunction and a and b are momentum eigenfunctions. When the momentum operator, O, acts on F, the wavefunction F will collapse into either state a, state b with probabilites of 0.64 and 0.36 respectively. So I assumed the same process would occur with the position operator.

 

[PhaseShifter]

Suppose you have a wavefunction that is a linear combination of momentum eigenfunctions. ie, F = 0.8a + 0.6b where F is the wavefunction and a and b are momentum eigenfunctions. When the momentum operator, O, acts on F, the wavefunction F will collapse into either state a, state b with probabilites of 0.64 and 0.36 respectively.

No, the momentum operator does not collapse the wavefunction.

You just get a new wavefunction F'=p_{1}0.8a+p_{2}0.6b. (And you'd still have to normalize that.)

 

[chrisphd]

Mathematically that is what you get when you apply the momentum operator to the wavefunction F. However, I was actually thinking about physically measuring the momentum of the wavefunction, in which the wavefunction then collapses into a or b. So my question is then, when the position of a wavefunction is measured, will the wavefunction likewise collapse into a position eigenfunction just as it would if it was a momentum measurement.

 

[PhaseShifter]

Mathematically that is what you get when you apply the momentum operator to the wavefunction F. However, I was actually thinking about physically measuring the momentum of the wavefunction, in which the wavefunction then collapses into a or b. So my question is then, when the position of a wavefunction is measured, will the wavefunction likewise collapse into a position eigenfunction just as it would if it was a momentum measurement.

In principle it would, but unless it's trapped in an infinitely deep and infinitesimally narrow potential well, that eigenfunction won't be a stationary state, and will immediately evolve into something else. A wavefunction corresponding to an exactly defined position you would also have infinite uncertainty in momentum.

Essentially it would require a potential resembling the limiting case of an infinite square well, where the length of the well approaches zero.

 

[xepma]

No... why would it?

Suppose you have a wavefunction that is a linear combination of momentum eigenfunctions. ie, F = 0.8a + 0.6b where F is the wavefunction and a and b are momentum eigenfunctions. When the momentum operator, O, acts on F, the wavefunction F will collapse into either state a, state b with probabilites of 0.64 and 0.36 respectively. So I assumed the same process would occur with the position operator.

I think you are mixing things up here. The action of a measurement is not given by acting with an operator on the state. The simplest interpretation of a measurement is that the |coefficients|^2 represent the probability of a certain outcome, and that after that measurement the wavefunction has collapsed. But that does not involve the action of the operator itself. The action of any (physical) operator on a wavefunction is never a collapse. This also goes for the momentum operator.