- #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 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...
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...