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I am new to Quantum Mechanics. I read the uncertainty principle - it says there are pairs of physical quantities which can't both be determined with certainty for a particle.

My question is does the same apply to a system of particles - the nucleus for example ? Can we determine both the position and momentum of a nucleus (containing more than one protons) with certainty?

Any help would be appreciated.

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One way to look at this is that if the uncertainty principle applied to fundamental particles but not to compound systems, then it would be too good to be true in a sense, because it would be the ultimate microscope. For instance, if we wanted to search for substructure inside electrons, we wouldn't need to do high-energy particle physics experiments; we could just infer that they were fundamental from the fact that they obey the uncertainty principle. In reality, compound systems obey the same rules as fundamental ones. In particular, note that protons and neutrons are themselves compound. – Ben Crowell May 20 '13 at 15:29
@Panx: If you like Siva's answer, you should not just accept it but also upvote it. Usually people upvote first, then wait a while before accepting, because you never know if the answer will turn out to be wrong, or a better one will come along. – Ben Crowell May 20 '13 at 15:32
@BenCrowell I know, but I dont have enough reputation to upvote currently :| – Panx May 20 '13 at 15:33
up vote 1 down vote accepted

The uncertainty principle applies to any quantum system, and is way more general than just single particle examples. It is defined for any pair of operators (physical quantities) $A$ and $B$, with the system in a state $|\psi\rangle$ $$ \Delta A \; \Delta B \geq \frac{\hbar}{2} \langle \psi| [A, B] |\psi \rangle$$ Note: The constant factor ($\frac{1}{2}$ here) varies in different derivations, depending on how exactly you define $\Delta A$ and $\Delta B$, but the essence is the same.

In the case of simple quantum systems, you could take $A$ to be the position operator and $B$ to be the momentum operator. In your case, it seems like you would like to consider the whole nucleus as an effective particle and apply these operators on it's wavefunction/state. Sure, you could do that, and you'll get an uncertainty relation from that.

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Good answer. But note that in practical terms, e.g., the de Broglie wavelength of an alpha particle emitted by alpha decay is very short compared to the size of an alpha particle. Therefore you're nowhere near saturating the limit imposed by the uncertainty principle, and the center-of-mass motion is classical to a good approximation. – Ben Crowell May 20 '13 at 15:26
Thanks. That helped :) – Panx May 20 '13 at 15:31

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