Sign Convention For Momentum Operators

1. Jun 1, 2012

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.

2. Jun 1, 2012

3. Jun 2, 2012

Septim

Thanks for the answer by the way can this convention have something to do with our depiction of the wave function as $e^{i(kx-wt)}$? 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

Last edited: Jun 2, 2012
4. Jun 2, 2012

strangerep

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.

5. Jun 3, 2012

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 $\hat{T}_{\pm}(\xi)=\exp(\pm \mathrm{i} \xi \hat{p})$. 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

$$\hat{T}_{\pm}(\xi) |x \rangle=|x-\xi \rangle. \qquad (*)$$

For a general state $|\psi \rangle$ this gives for the translation operation on the position-wave function

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

For a infinitesimal displacement you have on the one hand

$$\psi'(x)=\psi(x+\delta \xi)=\psi(x)+\delta \xi \partial_x \psi(x).$$

On the other that's

$$\psi'(x)=(1 \pm \mathrm{i} \delta \xi \hat{p}) \psi(x).$$

Comparing the two latter equations gives

$$\hat{p} \psi(x)=\mp \mathrm{i} \partial_x \psi(x).$$

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

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

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, $x \rightarrow x'=x \pm \xi$.

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