Sign Convention For Momentum Operators

[Septim]

Greetings,

How do we decide on which sign to take when using the momentum operator? The question may be very simple but I need a push in the right direction.

Many thanks.

 

[strangerep]

How do we decide on which sign to take when using the momentum operator? The question may be very simple but I need a push in the right direction.

We want it to satisfy the canonical commutation relations.

http://en.wikipedia.org/wiki/Canonical_commutation_relations

 

[Septim]

Thanks for the answer by the way can this convention have something to do with our depiction of the wave function as [itex]e^{i(kx-wt)}[/itex]? I think the statement on the second page of the lecture notes available on the following URL suggests this if I did not misinterpret it. If we selected the complex conjugate of the function then would the momentum operator be the complex conjugate of the previous momentum operator ?

For lecture notes
http://www.phys.spbu.ru/content/File/Library/studentlectures/schlippe/qm07-03.pdf

 

[strangerep]

Thanks for the answer by the way can this convention have something to do with our depiction of the wave function as [itex]e^{i(kx-wt)}[/itex]? I think the statement on the second page of the lecture notes available on the following URL suggests this if I did not misinterpret it. If we selected the complex conjugate of the function then would the momentum operator be the complex conjugate of the previous momentum operator ?

For lecture notes
http://www.phys.spbu.ru/content/File/Library/studentlectures/schlippe/qm07-03.pdf

I don't like the sequence of ideas in those notes, which (imho) is rather back-to-front. The function you mentioned is better derived by considering eigenstates of the momentum operator. Try the early chapters of Ballentine's textbook for a better (again, imho) sequence of development of the ideas.

 

[vanhees71]

This issue also puzzles me whenever I prepare a lecture on the Galileo or Poincare group ;-)). It's just a convention, how you describe spatial translations in terms of its infinitesimal generators, i.e., using [itex]\hat{T}_{\pm}(\xi)=\exp(\pm \mathrm{i} \xi \hat{p})[/itex]. It's arbitrary whether to use the upper or the lower sign convention. The usual one is the + convention.

The action of the translation operator on a generalized position eigenvector is defined by

[tex]\hat{T}_{\pm}(\xi) |x \rangle=|x-\xi \rangle. \qquad (*)[/tex]

For a general state [itex]|\psi \rangle[/itex] this gives for the translation operation on the position-wave function

[tex]\psi'(x)=\langle x|\hat{T}_{\pm}(\xi) \psi \rangle=\langle \hat{T}_{\pm}^{\dagger}(\xi) x|\psi \rangle=\langle x+\xi |\psi \rangle=\psi(x+\xi).[/tex]

For a infinitesimal displacement you have on the one hand

[tex]\psi'(x)=\psi(x+\delta \xi)=\psi(x)+\delta \xi \partial_x \psi(x).[/tex]

On the other that's

[tex]\psi'(x)=(1 \pm \mathrm{i} \delta \xi \hat{p}) \psi(x).[/tex]

Comparing the two latter equations gives

[tex]\hat{p} \psi(x)=\mp \mathrm{i} \partial_x \psi(x).[/tex]

The commutation relations for position and momentum of course also differ by a sign,

[tex][\hat{x},\hat{p}]=\pm \mathrm{i}.[/tex]

As I said, the usual convention is the upper sign.

Of course you can also mix up the whole issue further by using the upper sign convention for the translation operator but a different sign in Eq. (*) on the right-hand side. This again depends on whether you consider the translation as an active or passive operation, i.e., whether you define the translation of the position coordinates with either sign, [itex]x \rightarrow x'=x \pm \xi[/itex].

As I said, that's all convention, and it's good to stick to one once and forever not to get confused.