Talk:Momentum operator

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Derivation is unclear

The derivation is not clear, for instance it never explains why there is an h in there, or what the imaginary unit is doing. Someone should put this on a more theoretical basis, I guess I could try. -2/06/09 —Preceding unsigned comment added by 76.174.157.167 (talk) 09:15, 6 February 2009 (UTC)

Merge with momentum space

I suggest to merge this article with Momentum space. 79.11.106.253 (talk) 09:13, 19 August 2008 (UTC)

Definition of i

There is a mistake in this article: ${\displaystyle i={\sqrt {-1}}}$ <-- this is wrong

The definition is actually ${\displaystyle i^{2}:=-1}$ otherwise this would happen: ${\displaystyle -1=i^{2}=i\cdot i={\sqrt {-1}}\cdot {\sqrt {-1}}={\sqrt {(-1)\cdot (-1)}}={\sqrt {1}}=1}$

Your reasoning is wrong. The definition is okay, it's your proof that contains the flaw. ${\displaystyle {\sqrt {-1}}\cdot {\sqrt {-1}}={\sqrt {(-1)\cdot (-1)}}}$ does not hold for general complex numbers and is subject to branch considerations. See Square_root#Notes AlfredR (talk) 18:53, 24 June 2009 (UTC)
Part of the problem is that there are two square-roots of -1, namely i and -i. Of course, none of this matters...it's just a throwaway equation helping people who already understand complex numbers to confirm the notation. People who are mystified will click the link to imaginary unit. :-) --Steve (talk)

<actually it is wrong that i = \sqrt{-1} , I do not know if the proof is wrong or not!

i = \sqrt{-1} is wrong. assuming this is true we have 1=1 => -1=-1 => -1/1=1/-1 => sqrt(-1)/sqrt(1)=sqrt(1)/sqrt(-1) => i/1=1/i => i^2=1 => -1=1 — Preceding unsigned comment added by Tst0 (talkcontribs) 06:40, 8 December 2010 (UTC)

The equation ${\displaystyle {\sqrt {A/B}}={\sqrt {A}}/{\sqrt {B}}}$ is not always true. It's true when A and B are real positive numbers, but it's not true in other cases. You used this equation above, which is how you got a contradiction.
This is all explained in the article square root. The equation ${\displaystyle i={\sqrt {-1}}}$ is correct according to the common, conventional definition of ${\displaystyle {\sqrt {x}}}$. --Steve (talk) 07:07, 8 December 2010 (UTC)

Definition

There needs to be a discussion of the domain of definition of the momentum operator. As it is, the definition of the momentum operator doesn't make sense. Isdatmaths (talk) 12:48, 10 March 2012 (UTC)

I agree; I tagged the article for jargon. Bearian (talk) 21:23, 12 March 2012 (UTC)
A definition of operators in general is found on the Operator (physics) page. This is just the one for momentum. I'm not sure how much more of a definition it needs. 147.188.248.114 (talk) 10:05, 29 March 2012 (UTC)
Hmm, I can't see any definition or reference to the definition of an unbounded operator on that page. Isdatmaths (talk) 10:53, 29 March 2012 (UTC)
Of course there will be "jargon/tech terms" in an article like this - you have to expect it. The article says what the operator is. Everything is linked. How much more understandable could this be made? It works the other way round also: readers should have some tolerance for what subjects are, and we can't fall into the trap of degenerating it by emphasizing trivialities...
By no means am I an expert, but usually when this operator is defined in books, there is no mention of the domain of definition - it is just stated "the momentum operator is ${\displaystyle {\mathbf {\hat {p}}}=-i\hbar \nabla }$" or words to that effect.
Surely there is room for improvement in the presentation (or other more subtle things) which will be picked out (and hopefully fixed) by more people in time, but as far as I can tell the article seems fine for now. Maschen (talk) 09:30, 16 August 2012 (UTC)

Positive versus negative derivative and the abuse of notation

I participated last year in an interesting conversation about some subtleties in how the momentum operator is defined: See Talk:Bra-ket_notation/Archive_1#Abuse_of_Notation.

In particular, the definition ${\displaystyle p_{x}=-i\hbar (d/dx)}$ is correct in the position basis, although the way it's used involves sort of an abuse of notation. In fact, ${\displaystyle p_{x}=+i\hbar (d/dx)}$ (with a different definition of (d/dx)) is arguably a more "correct" definition from the perspective of linear algebra. (Obviously no one disagrees about how the momentum operator acts on any particular state, I'm just talking about how best to write it down.)

There might be something here worth putting in the article, although I'm skeptical...it may be a lot of extra complication with little gain in understanding. --Steve (talk) 13:10, 16 August 2012 (UTC)

Maybe this is something that can be explained in momentum space (or soon position and momentum space) but indicated here? That way the subtlety in sign is to the reader's attention here leaving the details in an article about the position and momentum spaces and linear algebra, since ${\displaystyle p_{x}=\pm i\hbar (d/dx)}$ is for position space, ${\displaystyle x=\mp i\hbar (d/dp)}$ is for momentum space (not sure about the sign for x operator in p-space...). Maschen (talk) 14:37, 16 August 2012 (UTC)