Can you cancel a function out of a differential equation?

[bubblewrap]

Main Question or Discussion Point

I saw this in http://en.wikipedia.org/wiki/Momentum_operator From equation 4 to 5, it seems that a function is canceled out from the partial derivatives, is this possible?

 

[Stephen Tashi]

Why do you think a function was "canceled" out?

 

[DEvens]

Are you referring to the place where they say that the partial derive of psi gives p psi, that it suggests the partial derivative is the momentum operator?

They did not "cancel out" the function. The operation there was not "it therefor follows" but rather "it suggests." Getting from the one equation to the other has more support than they have given there. But it's quite a bit more complicated than the typical wiki article.

 

[bubblewrap]

Are you referring to the place where they say that the partial derive of psi gives p psi, that it suggests the partial derivative is the momentum operator?

They did not "cancel out" the function. The operation there was not "it therefor follows" but rather "it suggests." Getting from the one equation to the other has more support than they have given there. But it's quite a bit more complicated than the typical wiki article.

Then how is the momentum operator defined?

 

[timthereaper]

Try this post on StackExchange for a decent cursory explanation: http://physics.stackexchange.com/questions/77457/what-is-the-momentum-operator

 

[Khashishi]

An operator isn't part of the same kind of algebra as equations. In line 4, the expressions on each side of the equals are both values. In line 5, the expressions represent operations (or functions on functions), not values. It's kind of an abuse of notation. But in generalized algebra, you can make anything you want into an expression, as long as you know what you are doing.

In general, you can't cancel a function out of a derivative like that. It sort of works in this case because quantum mechanics is linear, which is something that came out of experiment and can't be derived mathematically.