Uncertainty principle once again

In summary, Nolting provides a greater lower bound for the energy of the harmonic oscillator than I do. I don't understand why, and he provides a solution that I don't understand.
  • #1
Unkraut
30
1
Hi!
I've just begun reading a textbook about quantum mechanics (by Wolfgang Nolting - a german book). And before the real quantum mechanics stuff has actually started I am already stuck at the first two small exercises at the end of the introductory chapter (despite the solutions being included in the book). They are about the Heisenberg uncertainty principle.

1. "Use the uncertainty principle to determine the lower bound for the energy of the [classical] harmonic oscillator."

My solution:
The Hamiltonian is given by [itex]H=T+V=\frac{p^2}{2m}+\frac{kq^2}{2}[/itex] and for the energy we have [itex]E=H[/itex]. Because T and V are both nonnegative we obtain the bounds [itex]E>T>0[/itex] and [itex]E>V>0[/itex] for the kinetic and potential energy respectively.

[itex]E>T=\frac{p^2}{2m}>0 \Rightarrow \sqrt{2mE}>|p|>0 \Rightarrow \Delta p < 2 \sqrt{2mE}[/itex]
[itex]E>V=\frac{kq^2}{2}>0 \Rightarrow \sqrt{2\frac{E}{k}}>|q|>0 \Rightarrow \Delta q < 2 \sqrt{2\frac{E}{k}}[/itex]
Thus, for a certain energy E the momentum and position would have to be confined to the ranges [itex]\Delta p < 2 \sqrt{2mE}[/itex] and [itex]\Delta q < 2 \sqrt{2\frac{E}{k}}[/itex] simultaneously.

By the Heisenberg principle we obtain then:
[itex]\frac{\hbar}{2}<\Delta p\Delta q<8E\sqrt{\frac{m}{k}}[/itex]
[itex]\Rightarrow E > \frac{\hbar}{16}\sqrt{\frac{k}{m}}=\frac{\hbar}{16}\omega[/itex].

This is certainly a lower bound for the energy due to the Heisenberg principle. But I cannot guarantee mathematically that it is the greatest lower bound that can be obtained from the Heisenberg principle. And obviously, it is not. Because the solution in my book states otherwise. The problem is that I don't understand it. It goes as follows:

Nolting's solution:

[itex]H=\frac{p^2}{2m}+\frac{kq^2}{2}[/itex]

It must hold: [itex]E \geq \frac{(\Delta p)^2}{2m}+\frac{k(\Delta q)^2}{2} ~ ~ ~ ~ (*)[/itex]

Uncertainty principle: [itex](\Delta p)^2(\Delta q)^2\geq \frac{\hbar^2}{4}[/itex]

Thus follows: [itex]E \geq \frac{(\Delta p)^2}{2m}+\frac{\hbar^2 k}{8(\Delta p)^2}[/itex]

From [itex]\frac{dE}{d(\Delta p)^2}=0=\frac{1}{2m}-\frac{\hbar^2 k}{8(\Delta p)^4}[/itex]

we obtain: [itex](\Delta p)^2_0=\frac{\hbar}{2} \sqrt{km}[/itex]

Insert this to the unequation for E: [itex]E\geq\frac{\hbar}{4}\frac{\sqrt{km}}{m}+\frac{\hbar}{4}\frac{k}{\sqrt{km}}=\frac{\hbar}{2}\sqrt{\frac{k}{m}}=\frac{\hbar}{2}\omega[/itex]

So, Nolting obtained a greater lower bound for the energy than I did. He pwn3d me. But how did he do that?
I just don't understand (*). How can he simply substitute the [itex]\Delta p[/itex] and [itex]\Delta q[/itex] for p and q? Obviously he means that [itex]|p|[/itex] and [itex]|q|[/itex] must be greater than [itex]\Delta p[/itex] and [itex]\Delta q[/itex]. But that totally makes no sense to me, because [itex]\Delta p[/itex] and [itex]\Delta q[/itex] are only the uncertainties, not total values for the position and momentum and I don't see why p and q should not be allowed to be smaller than their respective uncertainties.
The rest is okay for me though. I can comprehend that. He inserts the least possible uncertainty for q into the unequation for E and then minimizes E in terms of [itex]\Delta p[/itex].

I remember that this same problem already bugged me when we had quantum mechanics in school. And now, seven years later, when I'm trying to learn it on a university level I still get stuck with it and haven't gained any better understanding for it.

The same problem I have with the simple statement that "if one confines a particle to a very tiny area the momentum will become huge". Why is that? Heisenberg's principle only states that the uncertainty for the momentum will become huge so we don't have a clue anymore about the size of the momentum. How can we then be so certain that the impulse will be so huge? It seems to be totally unlogic for me.
I use to blame phyicists for their fuzzy reasoning. But I'm not sure if that would be appropriate in this case. It seems to me that I am the only person in the world having this problem in understanding. At least I could not find a solution by asking Google. So I'm a little afraid no one will see what my problem is. But I hope someone does and can explain the solution to me...
 
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  • #2
You shouldn't think of that argument as a rigorous proof of a lower bound for the ground state energy. It's really only an order of magnitude calculation. The idea is that the typical position/momentum for the particle is of the same order of magnitude as the uncertainty in position/momentum. It's just a coincidence that he gets the best possible lower bound (ie, the exact ground state energy).

For example, if I added a small dip in the potential near q=0, the ground state energy would go down, but that calculation would give the same result, which would thus be incorrect as a lower bound.
 
  • #3
Hi StatusX!

Thanks for your answer and sorry for my getting back late.

I don't understand what you mean by: "It's a coincidence that he gets the best possible lower bound". That's probably because I don't even know what a ground state is.

Anyway, after due examination I finally understand what I didn't understand before:

If we knew that [itex]|pq|<\frac{\hbar}{2}[/itex] then we would know the values of p and q with uncertainties [itex]\Delta p[/itex] and [itex]\Delta q[/itex] that must also hold [itex]\Delta p\Delta q<\frac{\hbar}{2}[/itex]. If the uncertainties were bigger we couldn't know [itex]|pq|<\frac{\hbar}{2}[/itex]. It's as simple as that. And that's why it is a valid argument to substitute the deltas into the energy equation.

Thank you anyway for trying to help me. It was as I expected: that no one would know what my problem is. And of course I now don't really see why I didn't understand it earlier.
 
  • #4
Ok... but like I said, don't take this argument too seriously. For example, it's meaningless to say |x|<a for some a, as the particle (in the ground state) has some non-zero probability of being anywhere. It's just that it's most likely to be found in the region where |x|<a. The argument is just a rough estimate.
 

What is the Uncertainty Principle?

The Uncertainty Principle, also known as Heisenberg's Uncertainty Principle, is a fundamental concept in quantum mechanics that states that it is impossible to know both the exact position and momentum of a particle at the same time.

Why is the Uncertainty Principle important?

The Uncertainty Principle is important because it sets a limit on our ability to measure and predict the behavior of particles at the quantum level. It also has implications for our understanding of the nature of reality and the limitations of our knowledge.

Who discovered the Uncertainty Principle?

The Uncertainty Principle was first proposed by German physicist Werner Heisenberg in 1927 as part of his work on quantum mechanics. However, the concept had been hinted at by earlier physicists, such as Niels Bohr and Max Planck.

What are some real-life examples of the Uncertainty Principle?

While the Uncertainty Principle is most commonly applied to particles at the subatomic level, it also has applications in other areas. For example, it can help explain the limitations of certain technologies, such as MRI machines, and the unpredictability of weather patterns.

Is the Uncertainty Principle absolute?

No, the Uncertainty Principle is not an absolute limitation. It is based on the limitations of our current understanding and measurement techniques. As our technology and knowledge continue to advance, it is possible that we may find ways to better measure and predict the behavior of particles, potentially challenging the principles of uncertainty.

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