Heisenberg Uncertainty Principle problem

[K.QMUL]

Homework Statement

Explain, using the Heisenberg Uncertainty Principle, how classical physics is reached a a limit of quantum physics when (h-bar) tends to 0.

Homework Equations

ΔxΔp(x) ≥ (h-bar)/2

The Attempt at a Solution

The only reasonable answer I can formulate is the fact that when 'h-bar' is zero, the momentum/its kinetic energy is greater than or equal to zero. However in classical physics the kinetic energy of an object can be zero (if its at rest), whereas in Quantum physics, using the Heisenberg Uncertainty principle above, it cannot be zero.

Any help?

 

[carllacan]

You're almost there. Take note that the delta-x and delta-p in the uncertainty relation are not the position and the momentum of the particle, but the standard derivations of them, which measure the "uncertainty" of each variable.

Therefore if the right-hand-side is 0 there is actually no restriction on them, as the relation would only state that their product must be positive and that is always true (they are both always positive).

If the bar-h is not cero, though, we run into problems. Imagine you have a "fixed" delta-x and a delta-p that decreases over time (because we perform better experiments, for example). The product on the left-hand-side will decrease as well, and eventually it will be equal to the right-hand-side, so that another decreasing would "break" the rule, it would make it lower. According to the Uncertainty Principle one of two things should then occur: either delta-p would stop decreasing (we could not do a better experiment) or either delta-x would start increasing (experiments would show more accurate p measures but less acurate x measures).

I hope this helps.

 

[K.QMUL]

Thanks, that does help! But is my explanation complete if I consider it to be standard deviations? Is there anything else I am missing to explain?

 

[carllacan]

Thanks, that does help! But is my explanation complete if I consider it to be standard deviations? Is there anything else I am missing to explain?

Mmm, not quite. You say that in quantum mechanics momentum can't be zero. I don't see why neither momentum or its standard derivation could not be zero.

Let me put the relation on words: if delta-p or delta-x decrease the other one must increase (or viceversa) so that their product is always a minimum (namely h-bar/2). If that minimum is satisfied (because it is really little or even zero) they can take pretty much any value.

 

[K.QMUL]

Right, sorry, I forgot to put that in. I did write that down previously however. Thanks!