Canonical momentum and the uncertainty principle

[proton]

I'd kiss you guys...good thing this is a forum!

 

*dances with glee*

 

I still need to do some research and take a few math classes, but I am not hopeless it understand it seems.

 

joy.

I'm going to try an experiment here. There is a debate on terminology in another thread and I want to try something here to see what happens.

 

You know that the uncertainty principle relates uncertainty in momentum to the uncertaintly in position, right? What you don't learn until advanced courses on quantum mechanics the momentum this refers to is actually what is called "cannonical momentum" (aka conjugate momentum). Its different than the momentum p = mv (small p) that you learn about in basic physics. Suppose a charged particle is moving in a magnetic field. The cannonical momentum p has the value

 

(B-45) p = mv + qA

 

where A is known as the magnetic vector potential. A graduate text by Cohen Tannoudji explains this (page 225) as follows

Care must be taken not to confuse p (the momentum of a particle, also called conjugate momentum of r) with mv (the mechanical momentum of the particle): the difference between these two quantities appears clearly in (B-45). ...However it is the conjugate momentum p and not the mechanical momentum mv which becomes in quantum mechanics the operator P which satisfies the canonical commutation relations.

Since canonical momentum is a rather advanced topic and probably not known to most visitors here I'm curious as to whether such a fact is of interest to anyone, i.e. would you want to know that p is a function of the magnetic field through the quantity A? Thanks for your input.

 

Mod note: moved to its own thread

Edited July 4, 2009 by swansont
add mod note - moved

[GutZ]

Proton:

 

For me not so much because I don't know when that would be relevant.

 

when does the cannonical momentum become crucial to add?

[proton]

Proton:

 

For me not so much because I don't know when that would be relevant.

 

when does the cannonical momentum become crucial to add?

What do you mean "crucial to add"? If you mean when it is important to know what quantity appears in the equations then, for example, when its a charged particle moving in a magnetic field. In that case the momentum the HUP refers to is cannonical momentum and depend on the magnetic field. As I said, this is an advanced topic. But I wished I knew about if when I first learned quantum and not when I studied it as an upper classman

[GutZ]

Yeah it's a bit much for me lol.

 

btw lol (laugh out loud) I say it all the time and forget not everyone is down with internet lingo.

[proton]

Yeah it's a bit much for me lol.

 

btw lol (laugh out loud) I say it all the time and forget not everyone is down with internet lingo.

I read an article on this. If there is no magnetic field then p = mv and [math]\Delta p_x \Delta p_y = 0[/math] which means that [math]\Delta v_x \Delta v_y = 0[/math]. When there is a uniform magnetic field present then this becomes

 

[math]\Delta v_x \Delta v_y > e\hbar/2m^2c |<B_z>|[/math]

 

This would not be apparent if one didn't know how p is actually defined.

[Xittenn]

(B-45) p = mv + qA

 

where A is known as the magnetic vector potential. A graduate text by Cohen Tannoudji explains this (page 225) as follows

 

Since canonical momentum is a rather advanced topic and probably not known to most visitors here I'm curious as to whether such a fact is of interest to anyone, i.e. would you want to know that p is a function of the magnetic field through the quantity A? Thanks for your input.

 

I read an article on this. If there is no magnetic field then p = mv and [math]

\Delta p_x \Delta p_y = 0

[/math] which means that [math]

\Delta v_x \Delta v_y = 0

[/math]. When there is a uniform magnetic field present then this becomes

 

[math]

\Delta v_x \Delta v_y > e\hbar/2m^2c |<B_z>|

[/math]

 

This would not be apparent if one didn't know how p is actually defined.

 

I would love to understand and am very interested in what you just said. I'm finding it hard to put it together though!

 

Is this, in effect, a mathematical statement of the uncertainty created by trying to measure a particle? Where attenuating the frequency of the magnetic field for momentum decreases the accuracy of the measurement taken by the same device for position and vice versa; with respect to plancks constant of coarse......

 

Thanks for your input Proton! It's those little bits I don't know to look for that often hold me back from seeing the big picture.

Edited July 4, 2009 by buttacup

[proton]

I would love to understand and am very interested in what you just said.

I'm both encouraged and surprised by your interest. It's heart warming to see such a keen interest in the physics as your displaying here. I'm also surprised because I would have thought that most people wouldn't concern themselve with this. I'm pleasantly surprised.

Is this, in effect, a mathematical statement of the uncertainty created by trying to measure a particle?

The idea that uncertainty is created by trying to measure a particle is an urban myth. It is more accurate to say that uncertainty is intrinsically inherent in the system rather than related to measurement. E.g. ideally, no matter how you choose to measure the properties of the system, the uncertainty remains unaffected and is dependant only on the state of the system. This means that once [math]\Psi[/math] is determined [math]\Delta x[/math] is determined. Or to be more precise, once you tell me what [math]\Psi[/math] is I can tell you the value of [math]\Delta x[/math] and I don't need to know anything about how you intend on measuring the postion.

Where attenuating the frequency of the magnetic field for momentum decreases the accuracy of the measurement taken by the same device for position and vice versa; with respect to plancks constant of coarse......

The A above refers to a classical uniform magnetic field. There is no frequency associated with it. The associated quantity B_z represents z-component of the magnetic field determined by A.

Thanks for your input Proton! It's those little bits I don't know to look for that often hold me back from seeing the big picture.

Your very welcome. I admire that attitude. Shows courage in the face of the scary quantum world!

 

One of the rewards of posting here is the satisfaction I get from seeing people enjoy physics and knowing I was able to help. It kind of makes up for those arguements with the resident thought police.

---

Merged post follows:

Consecutive posts merged

I forgot to mention that the HUP relation for position and momentum are unaltered in this case.

Edited July 4, 2009 by proton

[Xittenn]

What does the merged part mean?

[proton]

What does the merged part mean?

 

I had forgotten to mention the HUP so when I came back and added that comment it merged the two posts together. I think I was too late to edit it and that's why I had to quote myself.