# Atom magnetic moment caused by orbiting electron

In Nielsen and Chuang Quantum Computation and Quantum Information book section 1.5.1 describing the Stern-Garlach experiment, it says: "Hydrogen atoms contain a proton and an orbiting electron. You can think of this electron as a little 'electric current' around the proton. This electric current causes the atom to have a magnetic field;" I thought the view of an electron orbiting the nucleus is outdated, replaced by the model of a probabilistic cloud around the nucleus in modern quantum physics view.

Can anyone help me to understand the relationship between the two views today? (I majored in computer science, without much formal education in modern physics.) Thanks.

• What is more important, and it is written a of couple paragraphs later in the book, that orbital movement of an electron in a hydrogen atom does not create magnetic field, and the whole magnetic moment of a hydrogen atom is due to electron's spin. IMO, there are no "two views" you are asking of today. – kludg Nov 16 '19 at 9:18

I will try to link the two concepts in the following way. It is rough, descriptive, but contains the essential idea.

Here we are talking about the so-called orbital angular momentum (OAM) of an electron orbiting around the nuclei. In classical mechanics, OAM is defined as

$$\vec{L} \equiv \vec{r}\times\vec{p}$$

where $$\vec{r}$$ is the position of the electron, and $$\vec{p}$$ is the momentum. One could roughly think that the magnetic moment to be proportional to $$\vec{L}$$, namely $$\vec{M}\propto\vec{L}$$ given that we ignore the spin part (which are in fact equally important). Now in quantum mechanics, the OAM is given by pretty much the similar expression, except two things. First, $$\vec{r}$$ and $$\vec{p}$$ should be thought as operators now (see below) and second, the operators are sandwiched between wavefunctions. Thus

$$\vec{L} = \langle \psi | \vec{r}\times\vec{p} |\psi \rangle$$

Precisely what this means is that, if you want to calculate say the component $$L_z$$, then the expression is

$$L_z = i\int d^3r \psi^*(r) ( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} )\psi(r)$$

where we have written out the operator $$\vec{p}$$ explicitly as $$i\frac{\partial}{\partial \vec{r}}$$ for the purpose of calculation.

That said, the two point of view, one with orbiting electron and the other with probability cloud is not in conflict with each other. Rather, they are complementary in the sense that the former view is more intuitive for us to imagine the physics and write down equations, but when it comes down to actual calculations, we have to use the wavefunctions.

Yes. It is outdated, however, to explain something without sounding too abstract the old stationary orbiting solutions come handy.

• I'm not sure that the orbiting view is totally outdated. On page 44, 2nd paragraph, "the spin of the electron is posited to make an extra contribution to the magnetic dipole moment..." In Wikipedia en.wikipedia.org/wiki/Electron_magnetic_moment, spin, orbital and total magnetic dipole moments are discussed, though the details are beyond my understanding. – czwang Nov 16 '19 at 16:45

I chatted with my physicist friend. If I understand him correctly, the modern view certainly uses the probabilistic cloud description. However this does not contradict with the motion of electrons which causes orbiting current (and momentum), as long as the motion does not change the probabilistic distribution. In other words, we cannot view each electron with a deterministic orbit, but the whole probabilistic cloud can move around the nucleus without changing the distribution.

Please feel free to correct this understanding. Thanks.

Happy holidays.