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Posted

is the product of errors in position and momentum measurement independent of velocity of particle?can the value of product be reduced below the specified limit?is the value product under special conditions equal to zero also?Please help me in understanding the problem

Posted

Whether the uncertainty is velocity-dependent might depend on the experimental setup. IOW, I can't categorically say no, though I can't think of an example at the moment)

 

The product cannot be lowered below the limit and can't be made zero. One term can be made arbitrarily small, as long as the other is larger, to compensate.

Posted

sir

according to your reply product of errors follows the inverse variation law What happens when the particle is at rest

sincerely

sidharath

Posted (edited)

Neither product can "be" 0, but they can infinitely approach 0...what exactly is the difference, because there's all sorts of things in mathematics where you "approach" 0 but in the end you just end up treating it as 0, like instantaneous velocity using derivatives, h "approaches" 0, but the actual equation for the instantaneous velocity actually is when h is actually 0 after simplifying. I mean, why exactly couldn't you have a particle that as a infinitely uncertain momentum or position? Is it because you'd have to divide by 0? But then you can just take the limit of approaching 0 and get an answer anyway, so why bother saying it's impossible?

I mean when we measure a particle, let's only look at that one instant where we measure the particle. We know of a position, and nothing more, from that one measurement of a single point in space, we have 0 idea of what the momentum is, so couldn't I use 0 that way?

Edited by SamBridge
Posted

Neither product can "be" 0, but they can infinitely approach 0...what exactly is the difference, because there's all sorts of things in mathematics where you "approach" 0 but in the end you just end up treating it as 0, like instantaneous velocity using derivatives, h "approaches" 0, but the actual equation for the instantaneous velocity actually is when h is actually 0 after simplifying. I mean, why exactly couldn't you have a particle that as a infinitely uncertain momentum or position? Is it because you'd have to divide by 0? But then you can just take the limit of approaching 0 and get an answer anyway, so why bother saying it's impossible?

I mean when we measure a particle, let's only look at that one instant where we measure the particle. We know of a position, and nothing more, from that one measurement of a single point in space, we have 0 idea of what the momentum is, so couldn't I use 0 that way?

 

The product of position and (canonical) momentum uncertainty must always be greater than or equal to [math]\hbar /2[/math]. So if you make (for example) the uncertainty in position close to zero then the uncertainty in momentum has to be very large for their product to be above the required number. If you take the limit where uncertainty in position goes to zero then the uncertainty in momentum grows to infinity.

Posted

The product of position and (canonical) momentum uncertainty must always be greater than or equal to [math]\hbar /2[/math]. So if you make (for example) the uncertainty in position close to zero then the uncertainty in momentum has to be very large for their product to be above the required number. If you take the limit where uncertainty in position goes to zero then the uncertainty in momentum grows to infinity.

So then if I have a purely known position, such as when I measure a particle as a point, don't I have an infinite undetirmined momentum? Wouldn't I have absolutely no clue what the momentum could possibly be if all I know about the particle is a single point?

Posted

It doesn't really matter about how or with what you would measure it - Heisenburg is not merely about levels of precision in measurement in a lab or even thought experiment; you cannot even theorise a particle with an arbitrarily well known position without the uncertainty in the momentum creeping in. I am sure I remember Leonard Susskind in one of the Stamford Lecture series using the fact that an idea would lead to a precise position and momentum as prima facie evidence for dismissing the idea. That is to say that if your hypothesis requires (or produces a situation in which it is possible that) position and momentum to be known with the product of sdivs less than hbar over two - then your hypothesis is wrong (or the HP is wrong)

Posted

It doesn't really matter about how or with what you would measure it - Heisenburg is not merely about levels of precision in measurement in a lab or even thought experiment; you cannot even theorise a particle with an arbitrarily well known position without the uncertainty in the momentum creeping in. I am sure I remember Leonard Susskind in one of the Stamford Lecture series using the fact that an idea would lead to a precise position and momentum as prima facie evidence for dismissing the idea. That is to say that if your hypothesis requires (or produces a situation in which it is possible that) position and momentum to be known with the product of sdivs less than hbar over two - then your hypothesis is wrong (or the HP is wrong)

I'm not saying it would mathematically make sense, it obviously wouldn't unless you used an infinity, but just as a concept. Aren't particle's themselves suppose to the "point-lie objects" when measured? I haven't seen anything that says otherwise, you can "measure" particle to the extent as the Heisenberg principal states that you know more or less about the position and momentum, but I don't know where the term "point-like" came from then.

Posted

I'm not saying it would mathematically make sense, it obviously wouldn't unless you used an infinity, but just as a concept. Aren't particle's themselves suppose to the "point-lie objects" when measured? I haven't seen anything that says otherwise, you can "measure" particle to the extent as the Heisenberg principal states that you know more or less about the position and momentum, but I don't know where the term "point-like" came from then.

 

The position and momentum (or any conjugate variable) wave functions are Fourier transforms of each other. The transform of a delta function is one that is a constant — the same everywhere.

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