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Nuclear magnetic resonance

  1. Jul 2, 2004 #1
    I've been reading about nuclear magnetic resonance, and one aspect is bothering me. (I haven't studied quantum physics yet.)

    Quantum mechanically, the proton has only 2 energy states: spin +1/2 spin and spin -1/2. In NMR, a pulse of electromagnetism is applied to alter the orientation of the proton, ie. in a classical sense, to alter the angle between the axis of the spinning proton and the direction of the external magnetic field. Texts write of angles of 90 degrees, 60 degrees, 30 degrees etc... But what do these various angles represent in a quantum sense? I thought the proton can only be in one of 2 different states (+1/2 and -1/2)???
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  3. Jul 3, 2004 #2
    To be very brief (but hopefully not too simplistic for being on the run...):

    The magnetization that is manipulated during NMR experiments is due to the difference in populations of the two states. Under ambient conditions where most NMR experiments are conducted, there is a slight excess of protons in the lower energy state of the two, with the remainder in the higher energy state (remember Boltzmann!). In the vectorial representation that has evolved for doing basic NMR theory, that relates to a vector pointing "up." When you irradiate the sample enough to invert the populations, you now have a vector pointing "down" due to a 180 degree pulse. When you irradiate the sample enough to equilibrate the populations, you have a vector pointing perpendicular to the original system, the result of a 90 degree pulse.

    Things are of course more complicated for a spin-1 nucleus or above, as well as for systems of coupled spins, but that's another post for another day.
  4. Jul 4, 2004 #3


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    Yes, this means that the state space of a proton is a 2-dimensional hilbert space, that is spanned by two orthogonal "vectors" (usually called kets in Dirac notation). One corresponds to "spin up" and the other to "spin down" along the *z-axis*. Let us call them |z+> and |z->. But you could take two OTHER orthogonal basis vectors (which are then of course linear combinations of |z+> and |z->). Funny thing is, they then correspond to a "spin up" and a "spin down" state along another axis. So you can associate a specific direction to each thinkable linear combination of |z+> and |z->, or, if you want to, to each state vector in the state space.
    Read up the very first chapter of Sakurai for an intuitive explanation.

    It turns out that the quantum mechanical evolution of a ket which corresponds to a certain orientation in space (theta,phi) at time t0, and which evolves in a B-field along the z-axis, results after a time t-t0 in a ket which has the same orientation in space as would have had a classical magnetic top. Of course, because it is quantum mechanics, if you are going to MEASURE the spin along z or x or whatever axis, you'll allways find spin up or a spin down result, with probabilities given by the superposition of the two states that are aligned along the measuring axis, of the state you've been measuring.

  5. Jul 4, 2004 #4
    To make sure I am following correctly... The component of the magnetization vector along the z axis represents the balance between spin up and spin down protons. A vector pointing in the +z direction represents an excess of low energy protons, and a vector pointing in the -z direction represents an excess of high energy protons. If one irradiates with a 90 degree pulse, then the 2 populations are balanced in numbers, and the "z" component becomes zero.

    So what happens if we irradiate with a pulse that is > 180 degrees? I would expect that we move from an excess of low energy protons at the start, to a balance of high and low energy protons at 90 degrees, to an excess of high energy protons at 180 degrees, but then what? What happens at 181 degrees? For example, what if we irradiate with 90 degrees, then before longitudinal relaxation has occurred, we give another 180 degree pulse, ie. a total of > 180 degrees?
  6. Jul 5, 2004 #5
    With a pulse of 181 degrees, you'd have a population inversion and then starting your way back to the typical Boltzmann distribution. Conversely, one could just apply a 179 degree pulse and end up with the same distribution. If you think about this simplified model, you can either be facing up, down, or somewhere in the middle. Any intermediate that can be accomplished by a 190 degree pulse can also be done with a 170 degree pulse, or what have you. If you'll think about a spherical coordinate system, the angle phi (well, at least how I was originally taught in calculus) only goes from 0 to 180 degrees (pi) since that's all that is needed to specify that coordinate. So a rotation of 225 degrees is equivalent to a 135 degree pulse going in the opposite direction. Does this make sense?

    If you have a 90 degree pulse followed by a 180 pulse (along the same axis), you're simply rotating the magetization 270 degrees (or 90 degrees in the opposite direction from the original 90 degree pulse).
    Last edited: Jul 5, 2004
  7. Jul 6, 2004 #6
    I understand how a 180 degree pulse would cause a magnetization vector aligned with the external magnetic field to rotate into anti-parallel alignment with the external magnetic field.

    But using a 181 degree pulse rather than a 180 degree pulse, ie adding MORE energy, how do we get a magnetization vector that points only to 179 degrees? Doesn't this imply that adding energy has caused the protons to take on less energetic states. My understanding is that if the magnetization vector is pointing up, we have an excess of low energy protons. Then, by irraditating the protons, we shift the balance from an excess of high energy protons to an excess of low energy protons. But once we pass 180 degrees, how does ADDING ADDITIONAL energy shift the protons to a LOWER energy state? (Thanks for all your help.)
  8. Jul 6, 2004 #7
    Let me try this again.

    The important thing to remember is that the magnetization vector is tracing a path along a circle. So, for sake of an example, let's say we do a 181 degree pulse that takes the magnetization from along the positive z-axis along the y-z plane, down to right along the positive y-axis and finally down and around to one degree past the negative z-axis in the y-z plane. So to do this would in fact require more energy than to simply do a 179 degree pulse in the opposite direction where the magnetization vector goes down, to right along the negative y-axis, and finally to one degree short of the negative z-axis. This idea is actually revisited in certain calibration schemes - one wishes to determine the pulse length for a 360 degree rotation (from which you can determine the 90 time and such), which should put the magnetization (nearly) back at equilibrium and should require a very small recycle delay.

    I hope that clears things up.
  9. Jul 10, 2004 #8
    I wasn't so sure where your discussion was going; hope this helps.

    In nmr, the ensemble of nuclei is subject to a constant magnetic field Ho (~10 Tesla), which in the case above is taken along the +z direction. The perturbation is a radio frequency (rf) magnetic field H1 (~10 Gauss) at the resonance frequency, where for the examples above H1 is along the +x (perpendicular to z), and causes the expectation value of the nuclear magnetization operator to rotate in the y-z plane. One then needs the sign of the nuclear magnetic moment to know if it is a clockwise or counterclock rotation.

    Now the end state of the nuclear ensemble after a 181 pulse of H1 along the +x is the same as that of a 179 along the -x (this is of course neglecting any relaxation effects on the population within the time it takes to rotate 2 deg, which in real nmr experiments is totally neglegible). However, the problem with this discussion is that the absolute phase of a single pulse is arbitrary, +/- x etc. Only relative phases with respect to other pulses in a pulse sequence are meaningful (given an nmr setup with quadrature detection, of course!).

    So yes the 181 pulse has added a fraction more rf energy than the 179 pulse, but the excited state of the nuclear ensemble after is the same. Anyway, this state is not a lower energy state, but an excited energy state. In terms of an end temperature of the ensemble, a 180 pulse causes a negative nuclear spin temperature, while a 90 pulse causes an effective infinite temperature. These are effective temperatures since they are not at equilibrium with the lattice, but in a sense, negative temperature is "hotter" than an infinite one. The lattice then cools off the nuclear spin system with a characteristic relaxation time T1 after the rf pulse.
    Last edited: Jul 10, 2004
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