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A Understanding spin precession in a magnetic field

  1. Mar 11, 2017 #1
    I have some trouble understanding the concept of spin precession in an external field from a quantum mechanics viewpoint. Hopefully someone will be willing to enlighten me.

    Consider a single spin ½ particle. In the absence of an external field, the projection of the spin angular momentum on any arbitrarily defined quantization axis can only take on two specific values, i.e. -½ (spin down) and +½ (spin up). Let us define a quantization axis z and further suppose we have prepared this particle in a particular eigenstate, say the +½ state. In other words, this is equivalent to saying that the particle has a definite projection angular momentum on the z-axis equal to ħ/2.

    We now turn on or place this particle in an external and constant magnetic field B which is aligned with our chosen quantization axis (z-axis). In certain (for e.g. nuclear magnetic resonance (NMR)) material that I've come across online, it is usually said the particle will now begin to precess about the B field, i.e. the z-axis.

    My questions are as follows:

    1. Will a particle prepared in the +½ Sz eigenstate precess?

    When referring to material on the quantum mechanics of precession, one usually defines the spin ½ particle with an arbitrary orientation in space (see for e.g.)


    In summary, it is something typically along the lines of defining the particle in a quantum state:
    Ψ = cosθ|1> + sinθ|2> where θ is the angle between the spin angular momentum vector and the z-axis that the B field is applied along.

    It can then be shown that the expectation value of the various projections of the magnetic dipole moment resembles that of precession about the z-axis at a precession frequency equivalent to the Larmor frequency.

    My reasoning is that a particle prepared in +½ Sz eigenstate has a projection that is aligned with the z-axis, so θ=0 and therefore no precession should occur. Is this correct?

    In the absence of an external field, the energies for the pair of eigenstates (spin up and spin down) are degenerate and so the time dependency within the Schrodinger equation only amounts to a constant phase factor. However, upon application of the magnetic field along the z-axis, the phase factors for each eigenstate are different. My second question is:

    2. For the same conditions as described above: (i.e. a particle prepared in the +½ Sz eigenstate, with a B field applied along the z-axis), how does this spin ½ particle evolve with time? Since it is in a stationary state, does it ever flip-flop between eigenstates at the Larmor frequency?

    Any help would be most appreciated. Thanks!
     
  2. jcsd
  3. Mar 11, 2017 #2
  4. Mar 11, 2017 #3
    Hi,

    I'd like to warn you that english is not my mother tongue so I may not understand what you write or make some mistakes, sorry about that. For your first question, I think that the B field is usually represented as a vector but this vector is only the result of the attraction, in this picture https://www.dropbox.com/s/9tjd5zk1oh0kgpz/Capture.PNG?dl=0 B's vector point up but the attaction comes from left and right, I think that's the reason why spins always precess. For your second I think you might misunderstand what stationary state is : https://en.wikipedia.org/wiki/Stationary_state

    Also I'm on this forum for a specific question and this topic just got the right subject so allow me to borrow it please :) I'm currently studying spin for my memoir and I came across a site that says that spin particle in +½ eigenstate will always be more numerous than the -½ ones when affected by B , the only explanation is "Blotzmann distribution". Could someone please explain that to me ?

    Here is the quote for those who speak French :
    Dans un champ magnétique B0, la proportion de spins dans le sens de B0 (parallèles) est supérieure à celle des spins orientés dans le sens contraire (antiparallèles) : statistique de Boltzmann. L’écart de population entre spin "parallèles" et "antiparallèles" est proportionnel à l’amplitude du champ magnétique principal.
     
  5. Mar 11, 2017 #4
    Jilang and Dinofou, thanks for both your responses and the material.

    Jilang, I am still not entirely convinced that precession will take place for the condition that I've specified. In the link that you provided, it says on slide 18 that the spin precession frequency is independent of orientation. I agree with that statement in general but I am not sure if it applies to the specific case where the particle is an eigenstate, i.e. θ=0 or for that matter, θ=180°. Can someone please enlighten me?

    Dinofou, when a magnetic field is applied to an ensemble of atoms, the energy levels are split due to the field, i.e. the spin +½ eigenstate has a different energy compared with the the spin -½ eigenstate. This energy difference is proportional to the strength of the magnetic field. And under equilibrium conditions, because of this energy difference, there will be a difference in the number of atoms in each of these levels based on Boltzmann statistics.

    Try searching for Zeeman effect (which will tell you more about the energy difference that occurs due to the application of an external B field) and Maxwell Boltzmann statistics (which will tell you more about the difference in the number of atoms in the spin -½ vs +½ levels.). Hope this helps.
     
  6. Mar 11, 2017 #5
    Tainty, don't get hung upon on the the eigenstates, they are just a relative phenomena.
     
  7. Mar 11, 2017 #6
    Thanks a lot ! This is a lot more clear, I'm glad I could make myself understood. I hope you'll get your answers but as you can see I'm new to this so I can't really help.
     
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