- #36
pmb_phy
- 2,952
- 1
Its unfortunate that you feel the need to read things into my posts which are not there. However in this case you do demonstrate a lack of understanding of the uncertainty principle. This is very common even for people who have studied quantum mechanics, even at the college level.Fredrik said:This is at least the third time in a short time that you have claimed that something trivial and obviously correct in one of my posts is wrong, and then started talking to me as if you think I just started to learn this stuff yesterday.
I see no reason to start making accusations such as this. In case you're unaware of it this kind of response is insulting. I know very well what I'm saying. I can only infer what you mean by what you say. This material was already discussed recently in another thread of the same title. Its for these reasons I've asked you certain questions, i.e. so that from your responses I can determine exactly what you are are trying to say rather than trying to deduce it from what you posts. However if you find that you're unable to respond without this poor attitude then I will not respond to your posts and will place you on my igore list. I don't have the time nor the patience for this kind of nonsense. Your post will be reported to the powers that be in this case. The moderators do not allow posts which are problematic (harassment, fighting, or rude) and unappropriated.For some reason you don't seem to do a lot of thinking before you reply to one of my posts.
Then please explain the difference between them as you understand them. Why do you think I asked you where you got that idea?? It sure can't be found in any text in quantum mechanics. It does not appear as if you understand the difference from what you've written in your posts, hence my question.It's absurd that you assume that I'm confusing the thing I called dx with the [itex]\Delta x[/itex] in the uncertainty relation.
Your posts have a very combative attitude. This forum is moderated so that this kind of thing doesn't happen.This should be good... prove it.
Answer - Simple. First off there is no reason to assume that one can't measure something like position exactly. So in what follows it will be assumed that x is exact. As an example consider an electron moving in the x-direction and for which the wave function initially has a Gaussian shape. Nowe measure the position. Repeat this process and for each value of position record how many times the electron was measured to be at this location. Repeat this process. In theory this will establish a probability density. In practice one cannot carry out infinite amounts of measurements. Note that in each case you start with the same state, i.e. the one described by the Gaussian function. After you have this data then you can determine the "spread" of the values of x. The standard deviation (sd) is the measure of this "spread" of values. Now calculate the standard deviation. The value of the sd is precisely identical to [itex]\Delta x[/itex]. I.e. they are two different words which describe the exact same thing.
Were you aware that the [itex]\Delta x[/itex] is identical to the standard deviation? Were you unaware that the value of [itex]\Delta x[/itex] is determined from the wave function and has nothing to do with how position is measured?
One last question so that I know we are talking about the same thing - Please post the exact mathematical definition of [itex]\Delta x[/itex] and why you believe that it has anything to do with how precise a measurement of something like position is made. Thanks.
Pete
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