Magnetic dipole in magnetic field

[Niles]

Hi

The energy U of a magnetic dipole in an external magnetic field is given by
$$U = -\mu \cdot B$$
so the energy is zero when they are perpendicular and maximal when they are antiparallel. This makes very good sense intuitively. Quantum-mechanically we have that
$$\mu = -m_Fg_F \mu_B$$
so U becomes
$$U = \mu_Bg_Fm_FB,$$
which is just the Zeeman shift of an atom. My questions is on how these two different scenarios - quantum and classical - are related.

The first relation states that the particle gains energy due to the torque exerted on it by B. However a Zeeman shift of an atom is - how I have understood it - basically not related to that the atom gains enegy. It just means that its internal levels are shifted. So it is not intuitive to me how the magnetic field "imparts" energy onto the particle in the second relation.

I hope my question is clear.


Niles.

 

[mfb]

The first relation states that the particle gains energy due to the torque exerted on it by B.

Only if it can rotate in that direction. The transition between the corresponding quantum mechanical energy levels is not really a rotation, but has a similar effect: it is a change of µ, which changes the energy.

 

[Niles]

Only if it can rotate in that direction. The transition between the corresponding quantum mechanical energy levels is not really a rotation, but has a similar effect: it is a change of µ, which changes the energy.

Thanks. When you say that it is a change of μ, then you are referring to that m_F is changed in
$$\mu = -m_Fg_F \mu_B$$
?


Niles.

 

[mfb]

As the other two are constant... right ;).

 

[sardar]

Hello Niles,

Thank you for your question. The relationship between the classical and quantum descriptions of a magnetic dipole in a magnetic field can be quite complex and confusing, so I will try to explain it as simply as possible.

In classical mechanics, a magnetic dipole is treated as a small bar magnet with a north and south pole. When placed in an external magnetic field, the dipole experiences a torque, which causes it to align either parallel or antiparallel to the field. This alignment corresponds to the minimum and maximum energy states, respectively.

In quantum mechanics, the magnetic dipole is described as an atomic or subatomic particle with a magnetic moment. This magnetic moment can be thought of as the spin of the particle, which can have two possible orientations: spin up or spin down. When placed in an external magnetic field, the energy levels of the particle split into two, corresponding to the two spin orientations. This is known as the Zeeman effect.

Now, the energy expression in the quantum case, U = μBgFmF, may seem similar to the classical expression, but there are a few key differences. Firstly, the μ in the quantum case is the magnetic moment of the particle, not the dipole moment as in the classical case. Secondly, gF is the Landé g-factor, which takes into account the quantum mechanical effects of the particle's spin and orbital angular momentum. And finally, mF is the quantum number representing the particle's spin orientation.

So, to answer your question, the two scenarios are related in that they both describe the energy of a magnetic dipole in an external magnetic field. However, the classical description is a simplified version of the quantum description, which takes into account the quantum mechanical effects of the particle's spin. In the quantum case, the energy is not being imparted onto the particle, but rather the energy levels are being split due to the interaction with the external magnetic field.

I hope this helps clarify the relationship between the classical and quantum descriptions of a magnetic dipole in a magnetic field. Let me know if you have any further questions. Keep exploring the fascinating world of magnetism and quantum mechanics!