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hokhani

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In summary, the spin magnetic moment of a proton is related to the spin by the gyrofactor g, which is not easily calculable.

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hokhani

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Vanadium 50

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

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hokhani

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I want to raise the spin magnetic moment.Vanadium 50 said:How is this quantum mechanics?

Please see http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/larmor.htmlVanadium 50 said:where are you getting this from?

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hokhani

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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.Vanadium 50 said:

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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Vanadium 50

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hokhani said: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.

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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!

Thus the Lagrangian reads

$$L=\frac{I}{2} \dot{\varphi}^2 + B \mu \cos \varphi.$$

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

$$\vec{\tau}=\vec{\mu} \times \vec{B}.$$

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hutchphd

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I believe the source of confusion of the OP is the angular momentum involved. Precession requires external torquevanhees71 said: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).

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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 is a permanent magnet that has a north and south pole, and produces a magnetic field. It is typically in the shape of a rectangular bar and is made of a ferromagnetic material, such as iron or steel.

The magnetic moment of a bar magnet is a measure of its strength and orientation of its magnetic field. It is a vector quantity, meaning it has both magnitude and direction.

A bar magnet will experience a force when placed in a magnetic field. The north pole of the magnet will be attracted to the south pole of the field, and vice versa. The strength of the force depends on the strength of the magnet and the magnetic field.

Yes, the magnetic moment of a bar magnet can be changed by applying an external magnetic field or by heating the magnet above its Curie temperature. This can cause the alignment of the magnetic domains within the magnet to change, altering its magnetic moment.

The magnetic moment of a bar magnet can be measured using a device called a magnetometer. This instrument measures the strength and direction of the magnetic field produced by the magnet, which can then be used to calculate the magnetic moment using the formula M = B x A, where B is the magnetic field strength and A is the area of the magnet's cross-section.

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