2

I read the following claim in Slichter's popular book, Principles of Magnetic Resonance (after Fig. 4.3, it's p100 in the newest version.). Despite the title, the author claims it in a quite general manner in terms of common quantum mechanics.

Angular momentum changing in multiples of $\hbar$ is compelling only for a complete system, e.g., an electron and another magnetic moment. Division of angular momentum change between the parts of a coupled system doesn't have to be in integral units of $\hbar$.

Any clarification or examples?

xiaohuamao
  • 3,621
  • Even Bohr didn't know why angular momentum should be quantized. – Geremia Dec 18 '14 at 04:11
  • Can you give a reference to that claim? I have no idea what a "non-complete system" is supposed to be. It's not a term I have ever come across in physics. The quantization of angular momenta is a consequence of the rotational symmetry of space. See Lubos Motl's answer here: http://physics.stackexchange.com/q/22806/. He does point out that systems without this property can have properties that look similar to angular momenta but without having sharp eigenvalues. A good example may be quasi-crystals. – CuriousOne Dec 18 '14 at 07:07
  • I see...while I can't seem to look up the page with Fig 4.3, I would assume that what is meant is that the expectation value of the angular momentum can take on non-quantized values. Since the book is about magnetic spin resonance, you have to keep in mind that the single spin/angular momentum is rarely of concern. What we measure in these experiments is an expectation value, which is really the sum of signals form many nuclear spins. The expectation value behaves in many aspects like a magnetized spinning top, even though the individual spins can only give us quantized signals. – CuriousOne Dec 18 '14 at 08:02
  • @CuriousOne No, this claim is not about expectation value. The page is 100. My apology. – xiaohuamao Dec 18 '14 at 08:06
  • I am sorry, but I can't seem to access page 100. Can you give us a sufficiently long quote? – CuriousOne Dec 18 '14 at 08:30
  • @CuriousOne Updated. – xiaohuamao Dec 18 '14 at 08:41
  • Interesting... I have no idea what the author could mean by that, at least not without the context. An electron is not a divisible system, certainly not at the level of nagnetic resonance. A magnetic moment is not even a part of the system, but a property of the system... If we couple multiple electrons or nuclear spins, then the energy structure becomes more complicated (which is why magnetic resonance can be used for chemical structure determination), but it will still change its angular momentum in integer units in an electromagnetic field, because the em field is quantized by bosons. – CuriousOne Dec 18 '14 at 08:47

1 Answers1

2

The difference between the whole system properties and its constituents can be explained on two-particle system. Consider positronium (electron+positron) in the state $l=1$. The quantum number $l$ describes the relative motion of constituents. This state has a magnetic moment, which belongs to the whole system.

But when you consider the angular momentum operator of one particle in this system, for example, that of the electron, it is uncertain: it is equal to $\mathbf{l}/2+\mathbf{\delta}$, where $\mathbf{l}$ is the total angular momentum and $\mathbf{\delta}$ is a fluctuating part. The latter part fluctuates due to coupling to the other particle. Similarly, the positron angular momentum operator equals $\mathbf{l}/2-\mathbf{\delta}$ and is uncertain too. The expectation value of $(l_z)_{\rm{e}}$ is $\hbar/2$, but it is not an eigenvalue!

The sum of operators is not fluctuating thanks to compensation of fluctuating parts. The same reasoning is valid for the "magnetic moments" of constituents.

Generally a coupled constituent is in a mixed state, so its quantum numbers have certain expectation values, but have no eigenvalues.

Vladimir Kalitvianski
  • 13,885