Schrödinger Eq: Questions on Mass, Uncertainty & Pi

[Sruli]

Some questions on Schrödingers equation (se)...

1) what does the m in se refer too the equation describes a wave function not a particle so how can we associate a mass with it.

2) the equation has to have fixed numbers ie it can't have any numbers with uncertainies yet shouldn't mass be subject to Heisenberg's uncertainty principle??

3) this applies to other physics as well...it has pi in the equation yet pi goes on for ever how does it make sense to assign a number that goes on for ever to a physical quantity as surely that must have a limit?

 

[jerromyjon]

Hi and welcome to PF!

1) I'm not comfortable discussing the math yet, as I'm more of a concept kind of guy. Perhaps someone else can help with that...

2) You might be surprised what tricks you can do with the math, but that has nothing to do with Heisenberg's uncertainty principle. That merely states that you can only accurately MEASURE a particle's position but the momentum will be uncertain, or accurately MEASURE a particle's momentum and the position will be uncertain.

3) You only need to be as precise as required for your purpose. If all you want to do is get an estimate of the circumference of a circle of 1m diameter, 3.14 is fine, if however you want to get the circumference of the observable universe in meters, and all you have is the diameter, you would certainly need several more decimal places to get a reasonable result.

 

[Sruli]

Hi and welcome to PF!

1) I'm not comfortable discussing the math yet, as I'm more of a concept kind of guy. Perhaps someone else can help with that...

2) You might be surprised what tricks you can do with the math, but that has nothing to do with Heisenberg's uncertainty principle. That merely states that you can only accurately MEASURE a particle's position but the momentum will be uncertain, or accurately MEASURE a particle's momentum and the position will be uncertain.

3) You only need to be as precise as required for your purpose. If all you want to do is get an estimate of the circumference of a circle of 1m diameter, 3.14 is fine, if however you want to get the circumference of the observable universe in meters, and all you have is the diameter, you would certainly need several more decimal places to get a reasonable result.

Sorry I meant the energy time uncertainty relation and if you don't measure the mass how do you know what to put in the equation

 

[jerromyjon]

The time-energy uncertainty? What else would you use for mass?

 

[strangerep]

1) what does the m in se refer too the equation describes a wave function not a particle so how can we associate a mass with it.

Well, the deeper answer is that it's really associated with a "representation of the Galilei group of transformations" (I'm restricting to the nonrelativistic case here). See, e.g., Wiki Galilean transformation, and possibly its representation theory.

But from your questions, I suspect these references might be too advanced. The Galilean transformations correspond to the familiar motions you can perform in space and time, i.e., rotation, displacement, velocity boost, etc. For such a group of transformations, there are quantities called "Casimir operators", which remain unchanged under these transformations. Mass is one of the Casimirs of the Galilean group -- it doesn't change under any of these transformations. The idea in QM is to represent these transformations as operators on an (abstract) Hilbert space of wave functions. The Hilbert space for wave functions corresponding to mass=$m_1$, say, is distinct from the Hilbert space wave functions corresponding to mass=$m_2$ (where $m_1 \ne m_2$). The Schrodinger equation is just an expression of how time evolution proceeds when represented in terms of such wave functions and the Hilbert space containing them.

I'd normally recommend Ballentine's QM textbook for this sort of question, but perhaps a more elementary text is needed here. Others may have suggestions.

2) the equation has to have fixed numbers ie it can't have any numbers with uncertainies yet shouldn't mass be subject to Heisenberg's uncertainty principle??

Quantum uncertainties are associated with noncommuting operators. E.g., for the Galilean case, the operators corresponding to position and momentum do not commute, and hence lead to a position-momentum uncertainty principle. But mass commutes with all the operators in the Galilei group, hence we don't get such an uncertainty principle involving mass.

3) this applies to other physics as well...it has pi in the equation yet pi goes on for ever how does it make sense to assign a number that goes on for ever to a physical quantity as surely that must have a limit?

OK, I guess you haven't yet studied the mathematical concept of real numbers? They're well understood mathematically, and thousands of people use them every day all over the world. As for comparing any predictions arising from the SE with experiment, well sure, all real-world experiments are necessarily of finite accuracy. So the theoretical predictions (which may indeed involve a non-terminating real number) need only be compared with experimental results up to the level of accuracy possible with that experiment's equipment.

As for time-energy uncertainty, that gets trickier since one must use the Poincare group instead of the Galilei group to analyze the relativistic case. But mass (i.e., invariant mass in the sense of special relativity) is still a Casimir of the Poincare group.

HTH.

 

[jtbell]

1) what does the m in se refer too the equation describes a wave function not a particle so how can we associate a mass with it.

The wave function is used for calculating the probbility of detecting a particle at various locations. We use the mass of that particle, e.g. the mass of the electron if we're talking about the SE for the electron in a hydrogen atom.

 

[bhobba]

Just to elaborate on Strangerep's excellent explanation, which is the correct, although unfortunately advanced, answer; it turns out if you chug through the math you end up with a quantum equation p = Mv where M is some constant and v is the quantum mechanical velocity observable. This looks strikingly like the definition of momentum in classical physics. You can in fact make it the same by introducing a change of units so m = hbar M - where hbar is a constant that simply changes units and m the mass. Hence you end up with hbar p = mv. This means hbar p is the analogue of classical momentum. hbar, called the reduced plank constant, is h/2pi where h is Planks constant. Planks constant is simply something that allows us to work in out usual units. Its nothing deep and mysterious like some accounts suggest.

Thanks
Bill