Does a position operator exist?

In summary, the conversation discusses the existence of a position operator and its action on wavefunctions. It is established that a position operator does exist, and it gives a position eigenfunction when operated on. However, this only applies to the correct wavefunction (Dirac delta function), and for other wavefunctions, the position operator does not give a position eigenfunction. It is also clarified that the momentum operator does not collapse the wavefunction, but instead gives a new wavefunction. The concept of a measurement is also discussed, and it is explained that the action of an operator on a wavefunction is not equivalent to a measurement.
  • #1
chrisphd
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Does a position operator exist?
 
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  • #2


What do you mean exist? I don't want to just blurt out "yes" to such a seemingly obvious question...
 
  • #3


Does a position operator acting on a wavefunction give a position eignenfunction?
 
  • #4


Yes, if the wavefunctions are eigenfunctions of the position operator ;)
 
  • #5


For the correct wavefunction (Dirac delta function), yes. For other wavefunctions, no.
 
  • #6


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.
 
  • #7


chrisphd said:
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?
 
  • #8


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.
 
  • #9


chrisphd said:
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 [tex]F'=p_{1}0.8a+p_{2}0.6b[/tex]. (And you'd still have to normalize that.)
 
  • #10


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.
 
  • #11


chrisphd said:
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.
 
  • #12


chrisphd said:
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.
 

Related to Does a position operator exist?

1. What is a position operator?

A position operator is a mathematical operator used in quantum mechanics to describe the position of a particle in space. It represents the location of a particle in three-dimensional space and is denoted by the symbol x.

2. Does a position operator exist in classical mechanics?

No, a position operator does not exist in classical mechanics. In classical mechanics, the position of a particle is described by its coordinates in space, whereas in quantum mechanics, the position is described by the position operator which is a mathematical abstraction.

3. How is the position operator related to the uncertainty principle?

The position operator is related to the uncertainty principle, which states that it is impossible to simultaneously determine the position and momentum of a particle with absolute certainty. The position operator plays a crucial role in this principle, as it represents the measurement of position, which can never be known with complete accuracy.

4. Can the position operator be observed or measured directly?

No, the position operator itself cannot be observed or measured directly. It is a mathematical concept used in quantum mechanics to describe the position of a particle. However, the position of a particle can be measured experimentally using physical instruments and techniques.

5. How is the position operator used in quantum mechanics?

The position operator is used in quantum mechanics to represent the position of a particle in three-dimensional space. It is an essential part of the mathematical framework used to describe the behavior of particles at the quantum level. The position operator is also used to calculate the probability of finding a particle at a certain location in space.

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