# Quantum Mechanical Treatment of the Spin Echo

I'm trying to find the expectation value of the spin in the $$y$$ direction after applying a $$\pi/2$$-$$\pi/2$$ sequence of pulses. In Slichter's Principles of Magnetic Resonance, he does a $$\pi/2$$-$$\pi$$ pulse which makes sense because that will get you the largest free induction signal you could get. I understand everything about the derivation for $$\langle I_y\rangle$$. What was nice about that derivation was that the rotation about the $$X$$-axis operator simply inverts $$I_y$$ (and $$I_z$$) such that the time evolution operators T still have the initial Hamiltonian in the z direction: $$T(t,h_0) = e^{i\gamma h_0tI_z}$$. So just like what Slichter did, you end up with terms $$T^{-1}(t-\tau,h_0)\cdot T(\tau,h_0)$$ resulting in 1 at the time of the echo $$2\tau$$. Then the rest is applying the $$X$$ operator to $$I_y$$. But, when it comes to the $$\pi/2$$-$$\pi/2$$ sequence, the integral looks like (ignoring integral over all spins):

$$\int\psi^*(0)X^{-1}(\pi/2)T^{-1}(\tau,h_0)e^{i\gamma h_0(t-\tau)I_y}\cdot I_z\cdot e^{-i\gamma h_0(t-\tau)I_y}T(\tau,h_0)X(\pi/2)\cdot \psi(0)d\tau_I$$

Reducing the time evolution operators to $$1$$ for $$t=2\tau$$ is no longer trivial. So I tried making the analogy that the coordinate system has been rotated by $$\pi/2$$ about the $$y$$-axis, as well as the Hamiltonian. From that, you'd have:

$$I_{x^{'}} = e^{-iI_y\phi} I_z e^{iI_y\phi} = I_x\cos\phi+I_z\sin\phi$$

$$I_{z^{'}} = e^{-iI_y\phi} I_z e^{iI_y\phi} = -I_x\sin\phi+I_z\cos\phi$$

$$I_{y^{'}} = I_y, \quad \phi = \gamma h_0(t-\tau)$$

Then, replacing $$I_z$$ from the integral gives two terms because of the $$\cos$$ and $$\sin$$. I then made the assumption that $$h_0 \approx 0$$ since we're trying to get as close to resonance as possible but small inhomogeneities are preventing this. So $$\sin\phi \approx 0 ,\cos\phi \approx 1$$ reducing the integral to just one term with:

$$e^{-i\gamma h_0\tau I_z}I_ze^{i\gamma h_0\tau I_z} = I_z$$

$$\langle I_{y,tot}(t)\rangle = -N\int p(h_0)dh_0\int\psi^*(0)X^{-1}(\pi/2)I_zX(\pi/2)d\tau = -N\langle I_y(0)\rangle$$

I interpret this as there was not a perfect $$z$$ magnetization due to inhomogenities in $$H_0$$ before the first pulse. Any help would be greatly appreciated.

• just one comment (not taking on the whole line of reasoning) - for spins which are on-resonance time evolution does nothing, so for $\gamma=0$ a $\pi/2 - \tau - \pi/2$ sequence is like a $\pi$ pulse. – tesch1 Jul 16 at 14:25
• Spins on resonance should mean time evolution is simply rotations. Right? In the rotating frame, time evolution would be a rotation about the x axis. In the lab frame, it’d be rotations about the z axis with nutations? Also, I’ve abandoned the small h_0 limit. I proceeded to apply rotation sandwiches to spin operators. Now what I’m unsure about is how to integrate over all spins since I now have trig functions dependent on h_0. I can post more on this tomorrow when I’m at my computer – theads Jul 17 at 3:49
• Sort of... what I was saying is that a spin on resonance in the rotating frame does not appear to rotate at all under non-pulsed time evolution (the $\tau$ above). Shameless plug: maybe have a look at SpinDrops (spindrops.org) to see the effects of different pulses and offsets and plain time evolution, you can create your $\pi/2 - \tau - \pi/2$ sequence and change the offset for a single spin under Spin Parameters. – tesch1 Jul 18 at 23:37