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Electron precession and flip contridiction?

  1. May 18, 2012 #1

    edguy99

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    In a Leonard Susskind video lecture at http://www.youtube.com/watch?v=VtBRKw1Ab7E&feature=relmfu at 8:30 minutes he talks about an electron entering a strong magnetic field will precess, in fact does a demo of the precession with a stick. He states that the precession will end with radiated energy from the electon.

    Later on around 17:00 he talks about the setup of an electron at 45 degrees in a strong magnet, then sending the electron through a magnet at 0 degrees. He is very emphatic that only one photon of a very specific wavelength (and no other radiated energy) may or may not be detected depending on whether the electon was "flipped".

    This seems to be a contridiction. Clearly, based on the first part of the lecture, the electron will start to precess when it enters the second magnetic setup, but where is the photon (radiated energy) associated with the ending of the precession? Where is the evidence the electron has stopped precessing?
     
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  3. May 18, 2012 #2

    tiny-tim

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    hi edguy99! :smile:
    but at about 14:00 he says that so far he's been talking about a classical particle, and that an electron doesn't behave like that (ie it doesn't precess and gradually give off radiation until it aligns with the field) … it only behaves as if it's aligned exactly with or exactly against the field

    (since the magnet measures the spin of the electron, this is the well-known feature of quantum theory, that if you measure an electron's spin in a particular direction, it will be found to be either in or against that direction)
     
  4. May 18, 2012 #3

    edguy99

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    Thanks for the note.
    Does QM predict NO precession for the electron, proton and neutron in a magnetic field?
     
  5. May 19, 2012 #4

    tiny-tim

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    YES … the electron spin can only be exactly in or opposite any direction that we measure it

    since it cannot be at any angle other than 0 or 180°, THERE IS NOTHING TO PRECESS!​

    susskind at 8:30 was talking about an ordinary bar magnet … it spins at an angle, that angle precesses, and radiation eventually reduces the angle to 0

    shortly after that, he talks about electrons, and specifically about "preparation" and "detection"

    we can only detect electrons at 0 or 180°

    he doesn't actually say so (in the short part i've watched), but until we detect the spin,
    it is not at an "undetected angle", therefore capable of undergoing precession,
    it is in a superposition of two states, 0 and 180°, neither of which precesses!

    study this CERN webpage on electron spin resonance … http://project-physicsteaching.web....h/experiments/electron-spin-resonance-qrg.pdf

    the classical magnet precesses in a "primary" magnetic field, and a "secondary" field changes the angle of precession

    the electron spin aligns with or against the "primary" magnetic field (or is in a superposition of those two states), and a "secondary" electromagnetic field (ie a photon) changes ("flips") the alignment

    no precession! :smile:

    (btw, "The Bohr–van Leeuwen theorem … makes magnetism in solids solely a quantum mechanical effect and means that classical physics cannot account for diamagnetism, paramagnetism or ferromagnetism.")
     
  6. May 19, 2012 #5
    Hmm, I thought that neutrons precess quite nicely. This is used in spin-echo techniques.

    http://en.wikipedia.org/wiki/Neutron_spin_echo

    or am I missing something essential here?
     
  7. May 19, 2012 #6
    Here is the part I don't get:

    In a magnetic field, spin-up and spin-down are the eigenstates of energy, and their energies differ by 2mu_Bohr B.

    To precess, the neutron spin has to be at an angle with the magnetic field, i.e. it has to be a superposition of spin-up and spin-down states. That is easy to do for each and any spin direction you what.

    What bothers me is the fact that such a wave function is not an Eigenstate of the Hamiltonian. Does that mean that it does not have a well-defined energy? And if so, is that a problem?

    Just looking at the wave function, the relative phase of the spin-up and spin-down components would vary with time. This corresponds to an evolution of <S_x> and <S_y> with time (with B || z), i.e. precession.
     
  8. May 19, 2012 #7

    tiny-tim

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    i've always assumed that the "precession angle" of a neutron was just another name for the "angle" between the spin-up and spin-down states in a superposition

    and that what was precessing was not an angle in actual space but in "phase space",

    so that a magnetic field could cause a bunch of neutrons to get out of phase with each other, and then the resonance trick would be to get them back in phase, as evidenced by maximising the detected signal

    ie if a neutron is (cosθ<+> + sinθ<->)eiωt, then it's ωt that "precesses", and no angle away from the axis is involved :confused:

    can anyone help? :smile:
     
  9. May 19, 2012 #8

    jtbell

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    The magnitude of the electron's intrinsic angular momentum ("spin") is

    $$S = \sqrt{s(s+1)}\hbar = \sqrt{\frac{1}{2}\left(\frac{1}{2}+1\right)}\hbar = \frac{\sqrt{3}}{2}\hbar$$

    Its component along whatever direction that we measure it in is

    $$S_z = \pm \frac{1}{2}\hbar$$

    Therefore it has a fixed polar angle with respect to the direction of measurement:

    $$\theta = \cos^{-1} \left(\frac{S_z}{S}\right)$$

    which is either 54.74 or 125.26 degrees, corresponding to "spin up" and "spin down." The azimuthal angle ##\phi## around the direction of measurement is indefinite because the QM operators for Sx, Sy and Sz do not commute with each other. Only one of the three can have a definite value at any time. Many introductory books model this indefinite ##\phi## as a "precession" of the angular momentum vector around the measurement direction. This is misleading because ##\phi## is simply indefinite. It does not have a value that is changing at some rate ##\omega## (angular velocity of precession).
     
    Last edited: May 19, 2012
  10. May 19, 2012 #9
    jtbell: Yes, I know that, but that does not answer the question.
    tiny-tim: <+> and <-> have different energies, and therefore different ω...

    Let's start from the beginning. I think we all agree that

    [itex]i \hbar \frac{\partial}{\partial t} \left| \Psi \right> = H \left| \psi \right>[/itex]

    so that

    [itex] i\hbar \frac{\partial}{\partial t} \left \uparrow \right> = \omega \left| \uparrow \right> \Longrightarrow \left|\uparrow\right>(t) = \exp(-i \omega t) \left| \uparrow \right>(t=0) \\
    i \hbar \frac{\partial}{\partial t} \left \downarrow \right> = -\omega \left| \downarrow \right> \Longrightarrow \left|\downarrow\right>(t) = \exp(i \omega t) \left|\downarrow\right>(t=0)[/itex]

    where [itex]\omega = \mu_0 B / \hbar[/itex]

    Using the Pauli matrices, we can write the spin operators as

    [itex]
    S_x = \frac{1}{2} \hbar \left( \left| \uparrow \right> \left< \downarrow \right| + \left| \downarrow \right> \left< \uparrow \right| \right) \\
    S_y = \frac{1}{2} \hbar \left(-i \left| \uparrow \right> \left< \downarrow \right| +i \left| \downarrow \right> \left< \uparrow \right| \right) \\
    S_z = \frac{1}{2} \hbar \left( \left| \uparrow \right> \left< \uparrow \right| - \left| \downarrow \right> \left< \downarrow \right| \right)
    [/itex]

    Take the inital state as [itex]\left|\psi\right>(t=0) = \frac{1}{\sqrt{2}}\left( \left| \uparrow \right> + \left| \downarrow \right> \right)[/itex]

    [itex] \left<\psi \right| S_x \left|\psi\right>(t=0) = \frac{1}{2}\hbar\\
    \left<\psi \right| S_y \left|\psi\right>(t=0) = 0\\
    \left<\psi \right| S_z \left|\psi\right>(t=0) = 0 [/itex]
    The spin is oriented along the real space x-axis. If you measure along the +x-axis, you will always find [itex]+\frac{1}{2}\hbar[/itex]. If you measure along the y or z axis, you will find [itex]\pm\frac{1}{2}\hbar[/itex], but zero on average.

    Now insert the time-dependant wave function... (short break, overdose of itex)
     
  11. May 19, 2012 #10
    ... and you get

    [itex]
    \left<\psi\right| S_x \left| \psi\right> = \frac{1}{2}\hbar \cos(2\omega t) \\
    \left<\psi\right| S_y \left| \psi\right> = -\frac{1}{2}\hbar \sin(2\omega t) \\
    \left<\psi\right| S_z \left| \psi\right> = 0
    [/itex]

    I am a bit surprised by the "-" for <S_y>, so probably I have an odd number of wrong signs :-p

    In any case, I interpret this a precession of the (observable!) expectation value of the spin vector about the z-axis=magnetic field axis.

    By picking other combinations of up and down, you can make the vector expectation value point in any direction you want. It will always precess about the z-axis.

    So again the question: Am I making some fundamental error here?
     
  12. May 19, 2012 #11

    jtbell

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    (emphasis added)

    OK, I'll buy that, with the understanding that any description of what a particular electron is "really doing" before we make an observation is a matter of interpretation. Any attempt to impute specific "hidden" values to the quantities Sx, Sy and Sz simultaneously, for a particular electron, would probably land us in the swamps of the Bell's Theorem arguments that flare up here periodically.
     
  13. May 19, 2012 #12
    I agree.

    Interestingly, for each time you can find a direction cos(2ωt) x - sin(2ωt) y along which the expectation value will be [itex] +\frac{1}{2} \hbar[/itex]. Basically this means that the relative phase of the up and down components of the wave function is an observable.

    Since at that particular time the neutron/electron/proton is in an eigenstate of that particular mesurement, the measurement does not even perturb the wave function. If you measure something else, like the z-component only, then of course you perturb the wave function.

    Note also that I make no claims about the "missing" angular momentum, i.e. the difference between |S|^2 and |S_z|^2 that you pointed out above.

    I don't think that this is in conflict with Bell's theorem.

    For arguments sake, take up and down as basis vectors, but apply the magnetic field along x. Then eigenstates of H will be (up ± down)/√2 with eigenvalues ±1/2. Once you measure S_x=±hbar/2, the wave function will remain in that eigenstate and subsequent measurements will give the same value - the wave function remains unchanged, and in particular the relative phase between the up and down basis vectors remains constant.
    After all, this is just a change of coordinate system away from the usual case with B || z and up and down as basis vectors.
     
  14. May 19, 2012 #13

    edguy99

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    You are correct that at 8:30 he is talking about a bar magnet, but at 8:54 he clearly says the "electron" will precess. At times he talks about the "classical picture" and later the "qm" picture, but does not, that I can see, make any claims one way or the other about "precession" in the "qm" picture.

    I have always felt that qm built on the work of people like Rabi and Larmour, not contridicting it. If you dont have "precession" then what the heck is larmour frequency other then a measure of the rate of precession? If you dont have larmour frequency, how do you explain the workings of a NMR machine?
     
  15. May 19, 2012 #14

    edguy99

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    If you would like to talk about Bell's theorem please start a new thread, it is not the topic here.
     
  16. May 19, 2012 #15

    edguy99

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    Consider for a moment the picture:
    spinvec.gif
    taken from http://hyperphysics.phy-astr.gsu.edu/hbase/spin.html

    If you assume the electron simply "precesses" up till 45 degrees (and hence has an "average" direction of only up or down in an external magnetic field such as stern-gerlach), it also helps a person to understand why you only have to flip the electron 90 degrees to flip it up or down. One would expect the energy difference between up and down to be as illustrated.
     
  17. May 19, 2012 #16

    tiny-tim

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    that "electron" was a mistake, susskind immediately corrects it to "bar magnet"

    eventually, at 14:15 he dismisses everything he's said so far …
    "now i've told you completely the wrong story for a real electron, this is not the way it works, this is not the way it works for a genuine electron, something else happens and i'll tell you now what it is …"​
    … and at no point after that does he talk about precession
    i don't think that's intended to be a physical picture

    anyway, it's not 45°, that's a 1:√2:√3 triangle! :wink:
    larmour frequency is the frequency of the photon

    yes i see it as related to the angular velocity of the phase of the electron … i suppose you could regard that as a half-plane about the field axis, related to a physical spin, but i don't see where precession comes into it (which involves a spin axis not along the field axis)
     
  18. May 19, 2012 #17

    edguy99

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    Please have a look at http://en.wikipedia.org/wiki/Larmor_precession

    Quoting: "In physics, Larmor precession (named after Joseph Larmor) is the precession of the magnetic moments of electrons, atomic nuclei, and atoms about an external magnetic field..."
     
  19. May 19, 2012 #18

    edguy99

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    I would disagree with that, but it does not really matter. The original question posted is what I am trying to resolve.

    I agree, thats why I asked the question.
     
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