# I A bar magnet and magnetic moment in a magnetic field

#### hokhani

Why a magnet bar would become aligned with the external magnetic field while a magnetic moment would precess around the external magnetic field?

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Staff Emeritus
How is this quantum mechanics? And where are you getting this from?

Staff Emeritus
Your question is still extremely vague. Are you asking why a classical magnet behaves differently than a quantum mechanical one? Or are you asking why precession exists in the first place?

#### hokhani

Your question is still extremely vague. Are you asking why a classical magnet behaves differently than a quantum mechanical one? Or are you asking why precession exists in the first place?
As you said, my main question is "why a classical magnet behaves differently than a quantum mechanical one"; If a classical magnetic moment were in an external magnetic field, it becomes aligned with the magnetic field or would have a precession around the magnetic field (I also don't know the difference between these two as raised in my first post here) while the spin state would not change in an external magnetic field. For example, the state $|S_x+,t_0>=1/\sqrt(2)|S_z+>+1/\sqrt(2)|S_z->$ in the magnetic field $\vec B=B_0 \hat z$ changes as $|S_x+; t>=exp(-i\alpha t) |S_x+,t_0>$ which means that the spin state hasn't changed in the magnetic field.
In summary, to have an intuitive understanding of the behavior of spin in a magnetic field, I simulate the $|S_x+>$ as a magnetic bar that its N-pole directed in the x-direction. But I can not obtain a correct picture, even classically.

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Staff Emeritus
why a classical magnet behaves differently than a quantum mechanical one
A. Quantum mechanics is not classical mechanics
B. If we had a million units of hbar of angular momentum, like a macroscopic object might well, the behavior would be closer.

#### vanhees71

Gold Member
I still don't get the question. A classical magnetic moment doesn't behave so differently from the quantum one (at least in the sense of Ehrenfest's theorem).

Take the classical magnetic moment. As the most familiar example take a compass needle in the Earths' magnetic field. Make the $x$ axis pointing north and use polar coordinates. Thereby note that the geographical north pole of the earth is by convention its magnetic south pole!

$$L=\frac{I}{2} \dot{\varphi}^2 + B \mu \cos \varphi.$$
Here $I$ is the moment of inertia of the needle through its rotation axis. Then you get by the Euler-Lagrange equation
$$I \ddot{\varphi}=\dot{J} =-B \mu \sin \varphi,$$
which is the equation of motion of a pendulum, and thus indeed the static stable solution is $\varphi=0$.

The equation also shows that in vector notation the torque is given by
$$\vec{\tau}=\vec{\mu} \times \vec{B}.$$

#### hutchphd

I still don't get the question. A classical magnetic moment doesn't behave so differently from the quantum one (at least in the sense of Ehrenfest's theorem).
I believe the source of confusion of the OP is the angular momentum involved. Precession requires external torque and "large" angular momentum. For a macroscopic bar magnet the onboard angular momentum causing the magnetization is negligible. Not so for a proton.

#### vanhees71

Gold Member
What do you mean by "onboard angular momentum"?

In classical theory the magnetic moment of a current distribution is related to the angular momentum of the corresponding charge fluid by
$$\vec{\mu}=\frac{1}{2} \int_{\mathbb{R}^3} \mathrm{d}^3 x \vec{x} \times \vec{j}(\vec{x}).$$
If now the current consists of charged particles carrying a charge $q$ and the charge density (number of charges per unit volume) is $n$ we have $\vec{j}=q n \vec{v}$, where $\vec{v}$ is the fluid velocity (Eulerian description). Thus we have
$$\vec{\mu} = \frac{q}{2} \int_{\mathbb{R}^3} \mathrm{d}^3 x n \vec{r} \times \vec{v} = \frac{q}{2m} \vec{L},$$
where $\vec{L}$ is the orbital angular momentum of the moving charges.

If there is an external magnetic field $\vec{B}$ around, one can show that the force on the current distribution is approximately
$$\vec{F}=\vec{\nabla} (\vec{m} \cdot \vec{B}),$$
and the torque
$$\vec{\tau}=\vec{\mu} \times \vec{B}.$$
The latter equation can be rewritten as an equation of motion for $\vec{\mu}$ by using the above relation of it to angular momentum
$$\dot{\vec{L}}=\vec{\tau}=\vec{\mu} \times \vec{B} \; \Rightarrow \; \dot{\vec{\mu}} = \frac{q}{2m} \vec{\mu} \times \vec{B}.$$
This means that the magnetic moment precesses around the direction of $B$ with the (classical) Larmor frequency
$$\omega_{\text{Larmor}}=\frac{qB}{2m}.$$
Now for free Dirac particles it turns out that the magnetic moment is related to the spin by
$$\vec{\mu}=\frac{q}{2m} g \vec{s}, \quad g \simeq 2.$$
As we have seen from the classical consideration for the orbital angular momentum we have $g=1$. Thus in general you have some gyrofactor which is not easy to calculate but can be measured.

"A bar magnet and magnetic moment in a magnetic field"

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