# Uncertainty principle once again

1. Mar 17, 2009

### Unkraut

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 $H=T+V=\frac{p^2}{2m}+\frac{kq^2}{2}$ and for the energy we have $E=H$. Because T and V are both nonnegative we obtain the bounds $E>T>0$ and $E>V>0$ for the kinetic and potential energy respectively.

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

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

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:

$H=\frac{p^2}{2m}+\frac{kq^2}{2}$

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

Uncertainty principle: $(\Delta p)^2(\Delta q)^2\geq \frac{\hbar^2}{4}$

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

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

we obtain: $(\Delta p)^2_0=\frac{\hbar}{2} \sqrt{km}$

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

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 $\Delta p$ and $\Delta q$ for p and q? Obviously he means that $|p|$ and $|q|$ must be greater than $\Delta p$ and $\Delta q$. But that totally makes no sense to me, because $\Delta p$ and $\Delta q$ 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 $\Delta p$.

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...

2. Mar 17, 2009

### StatusX

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. Mar 18, 2009

### Unkraut

Hi StatusX!

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 $|pq|<\frac{\hbar}{2}$ then we would know the values of p and q with uncertainties $\Delta p$ and $\Delta q$ that must also hold $\Delta p\Delta q<\frac{\hbar}{2}$. If the uncertainties were bigger we couldn't know $|pq|<\frac{\hbar}{2}$. 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. Mar 19, 2009

### StatusX

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.