Rotating Reference Frame of Spin-Half Systems

From Icamipedia

Spin-Half systems[1] are of great importance in science, since these systems are used widely across fields such as condensed matter physics, chemistry and molecular biology. Specifically, experimental techniques such as NMR (Nuclear Magnetic Resonance)[2] and EPR[3] (Electron Paramagnetic Resonance) are built upon the spin-half system, since the particles probed are the spin-half particles present in materials such as nuclei and the electron.

Much information can be gained from NMR and EPR, such as resonance peaks and relaxation effects. However, to obtain the information easily a rotating reference frame[4] is used. The reason being is that nuclei and electrons need to be polarized first in a static external magnetic field before one can induce resonance. Due to the presence of this static field, a rotating magnetic field is applied to obtain a resonance response. By using the rotating reference frame, the rotating magnetic field appears static, and therefore one can observe the spin-half particles precess around this newly established static field. xxx

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Derivation of the Rotating Reference Frame

Here follows a derivation of the rotating refernce frame for a quantum spin-half system.

Shrodinger's famous equation is as follows

$-\frac{\hbar}{i}\frac{\partial \psi(t)}{\partial t} = \mathcal{H} \psi(t)..........(1)$

where $\mathcal{H}$ is the Hamiltonian, $ψ(t)$ is the time dependent wavefunction and $\hbar$ is Planck's constant divided by $2π$ . This can be solved by recognizing that the equation is an ordinary differential equation for which a suitable form of the wavefunction is

$\psi (t) = e^{-(i/\hbar)\mathcal{H}t}\psi (0)..........(2)$

where we assumed that $\mathcal{H}$ is time independent. To demonstrate that this is a viable solution, just substitute eqn 2 into eqn 1.

Now, let us consider the Hamiltonian for a spin of a half in a static magnetic field

$\mathcal{H}=-\gamma\hbar B_{0}\hat{\sigma}_{z}..........(3)$

where $γ$ is the gyromagnetic ratio[5], $B 0$ is the static magnetic field in the z direction and $\hat{\sigma}_{z}$ is the Pauli matrix[6] in the z direction. Then the wavefunction as written in eqn 2 becomes

$\psi (t) = exp\left[-(i/\hbar)(-\gamma\hbar H_{0}\hat{\sigma}_{z})t\right]\psi (0)..........(4)$

$\psi (t) = exp\left[i\omega_{0}t\hat{\sigma}_{z}\right]\psi (0)..........(5)$

where we have defined $ω 0 = γ B 0$ the Lamor frequency[7].

The field $B 0$ causes the spin to precess around the z axis at an angular velocity of $-\gamma B_{0}\mathbf{\hat{k}}$ . So, we can look at the wavefunction in eqn 5 and see that it is the time evolved wavefunction for $ψ(0)$ in which the time evolution causes the wavefunction to rotate at a negative angle of $ω 0 t$ . So we have effectively taken the wavefunction $ψ(0)$ and put it into a rotating frame of reference.

We can now compute a matrix element for the operator $\hat{\sigma}_{x}$ in this rotating frame by

$\int \psi^{*}(t) \hat{\sigma}_{x}\psi (t) d\tau = \int \psi^{*}(0) exp\left[-i\omega_{0}t\hat{\sigma}_{z}\right] \hat{\sigma}_{x}exp\left[i\omega_{0}t\hat{\sigma}_{z}\right]\psi (0) d\tau..........(6)$

$\int \psi^{*}(t) \hat{\sigma}_{x}\psi (t) d\tau = \int \psi^{*}(0) \hat{\sigma}_{x'} \psi(0)d\tau..........(7)$

where $\hat{\sigma}_{x'}\equiv exp\left[-i\omega_{0}t\hat{\sigma}_{z}\right] \hat{\sigma}_{x}exp\left[i\omega_{0}t\hat{\sigma}_{z}\right]$ . Essentially we have gone from the Shrodinger picture[8] to the Heisenberg picture[9]. The operator has been moved in the rotation frame by letting the operator time evolve. Things to note however, the wavefunction $ψ(0)$ rotated in the negative sense, but if it is held fixed and the operator is moved, the operator must be rotated in the positive direction to yield the same effect.

Moving on, let us show that $\hat{\sigma}_{x'}$ can be related to $\hat{\sigma}_{x}$ through a rotation axes, instead of using the handwaving explanation beforehand. So let us consider a general rotation operation acting on our operator $\hat{\sigma}_{x}$

$f(\phi)=exp\left[-i\phi\hat{\sigma}_{z}\right] \hat{\sigma}_{x}exp\left[i\phi\hat{\sigma}_{z}\right] ..........(8)$

$\frac{df(\phi)}{d\phi} = e^{-i\phi\hat{\sigma}_{z}}\left(-i\hat{\sigma}_{z}\hat{\sigma}_{x}+i\hat{\sigma}_{x}\hat{\sigma}_{z}\right)e^{i\phi\hat{\sigma}_{z}}..........(9)$

$\frac{df(\phi)}{d\phi}= e^{-i\phi\hat{\sigma}_{z}}\hat{\sigma}_{y}e^{i\phi\hat{\sigma}_{z}}..........(10)$

where in the last line the commutation relations for the Pauli Matrices were used. Continuing

$\frac{d^{2}f(\phi)}{d\phi^{2}} = e^{-i\phi\hat{\sigma}_{z}}\left(-i\hat{\sigma}_{z}\hat{\sigma}_{y}+i\hat{\sigma}_{y}\hat{\sigma}_{z}\right)e^{i\phi\hat{\sigma}_{z}}..........(11)$

$\frac{d^{2}f(\phi)}{d\phi^{2}}= -e^{-i\phi\hat{\sigma}_{z}}\hat{\sigma}_{x}e^{i\phi\hat{\sigma}_{z}}..........(12)$

$\frac{d^{2}f(\phi)}{d\phi^{2}}= -f(\phi)..........(13)$

$\Rightarrow \frac{d^{2}f(\phi)}{d\phi^{2}} + f(\phi) = 0..........(14)$

which we recognize as a second order ODE. One solution that satisfies eqn 14 is

f (φ) = A cosφ + B sinφ..........(15)

with the coefficients $A$ and $B$ still to be determined. To determine the coefficients we determine the initial conditions and obtain

f (φ = 0) = A ..........(16)

f'(φ = 0) = B ..........(17)

Looking back at eqn's 8 and 9 one can immediately identify $A=\hat{\sigma}_{x}$ and $B=\hat{\sigma}_{y}$ . Therefore, we have the final result being

$\hat{\sigma}_{x'}=\hat{\sigma}_{x}\cos{\phi} + \hat{\sigma}_{y}\sin{\phi}..........(18)$

The above calculation can be done for the other dimensions as well with results

$\hat{\sigma}_{y'} = -\hat{\sigma}_{x}\sin{\phi} + \hat{\sigma}_{y}\cos{\phi}..........(19)$

$\hat{\sigma}_{z'} = exp\left[-i\phi\hat{\sigma}_{z}\right] \hat{\sigma}_{z}exp\left[i\phi\hat{\sigma}_{z}\right] = \hat{\sigma}_{z}..........(20)$

So clearly we have a frame rotating around the z axis if we use the rotation operators and rotate at angle $φ$ . If we let $φ = ω 0 t = γ B 0 t$ , we can see that what was deduced in the discussion around eqn's 5 to 7 has a mathematical basis.

In conclusion, the time evolution operator $\hat{U}=exp[i\omega_{0}t\hat{\sigma}_{z}]$ can be used to make a rotation reference frame rotating around the z axis at the Lamor frequency $ω 0$ .

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Use of reference frame in NMR

Here is given an example of how the rotation reference frame is used in NMR. Demonstrated is how the rotation reference frame is used to turn the rotating magnetic field into a static field around which the spin-half particle can precess.

In NMR, one applies a magnetic field along the z direction in order to polarize the spin of the nuclei in one's sample. To get a resonance response, one then applies a rotating magnetic field, giving one the total magnetic field of

$\vec{B}(t) = \mathbf{\hat{\imath}}B_{1}\cos{\omega_{z}t}+\mathbf{\hat{j}}B_{1}\sin{\omega_{z}t}+\mathbf{\hat{k}}B_{0}..........(21)$

For a spin of magnetic moment $\vec{\mu}$ , one can describe the system quantum mechanically using Schrodinger's equation by

$-\frac{\hbar}{i}\frac{\partial \psi}{\partial t} = -\hat{\vec{\mu}}\cdot\vec{B}(t)\psi..........(22)$

$=-\gamma\hbar\left[B_{0}\hat{\sigma}_{z}+B_{1}\left(\hat{\sigma}_{x}\cos{\omega_{z}t+\hat{\sigma}_{y}\sin{\omega_{z}t}}\right)\right]\psi ..........(23)$

since $\hat{\vec{\mu}}=\gamma\hat{\vec{J}}=\gamma\hbar\hat{\vec{\sigma}}$ . Looking at eqn 18, one can see that one can substitute it into eqn 23 to yield

$\mathcal{H} = -\gamma\hbar\left(B_{0}\hat{\sigma}_{z}+B_{1}exp\left[-i\omega_{z}t\hat{\sigma}_{z}\right] \hat{\sigma}_{x}exp\left[i\omega_{z}t\hat{\sigma}_{z}\right]\right)..........(24)$

. Thus we have effectively put the Hamiltonian into the rotating frame of reference, we have put the Hamiltonian into the Heisenberg Picture. However, it is far more advantageous to use the Schrodinger Picture, since this will yield us the resonance behaviour we are seeking. To accomplish this, let us use a rotation operator on the wavefunction $ψ$ as follows

$\psi ' = e^{i\omega_{z}t\hat{\sigma}_{z}}\psi..........(25)$

$\psi = e^{-i\omega_{z}t\hat{\sigma}_{z}}\psi ' ..........(26)$

Do note that the function $ψ$ and $ψ'$ are assumed to be time dependent. Using eqn 26 we can get the following result

$\frac{\partial \psi}{\partial t} = -i\omega_{z}\hat{\sigma}_{z}e^{-i\omega_{z}t\hat{\sigma}_{z}}\psi' + e^{-i\omega_{z}t\hat{\sigma}_{z}}\frac{\partial \psi '}{\partial t} ..........(27)$

. What we want to do now is to put Schrodinger's equation in terms of $p s i'$ in order to work the entire system in the rotation frame of reference. Therefore, we substitute eqn 26, eqn 27 and multiply both left hand sides with $e^{i\omega_{z}t\hat{I}_{z}}$ into eqn 23 to get

$e^{i\omega_{z}t\hat{\sigma}_{z}} (-\frac{\hbar}{i}) (-i\omega_{z}\hat{\sigma}_{z}e^{-i\omega_{z}t\hat{\sigma}_{z}}\psi' + e^{-i\omega_{z}t\hat{\sigma}_{z}}\frac{\partial \psi '}{\partial t}) = e^{i\omega_{z}t\hat{\sigma}_{z}} ( -\gamma\hbar H_{0}\hat{\sigma}_{z})e^{-i\omega_{z}t\hat{\sigma}_{z}}\psi'$

$+ e^{i\omega_{z}t\hat{\sigma}_{z}} (H_{1}exp\left[-i\omega_{z}t\hat{\sigma}_{z}\right] \hat{\sigma}_{x}exp\left[i\omega_{z}t\hat{\sigma}_{z}\right]) e^{-i\omega_{z}t\hat{\sigma}_{z}}\psi'..........(28)$

$\hbar\omega_{z}\hat{\sigma}_{z}\psi' - \frac{\hbar}{i}\frac{\partial \psi'}{\partial t} = \left(-\gamma \hbar H_{0} \hat{\sigma}_{z} -\gamma\hbar\hat{\sigma}_{x}\right)\psi'..........(29)$

$- \frac{\hbar}{i}\frac{\partial \psi'}{\partial t} = -\left[\hbar(\omega_{z}+\gamma H_{0})\hat{\sigma}_{z}+\gamma\hbar B_{1}\hat{\sigma}_{x}\right]\psi'..........(30)$

So we have put the Shrodinger equation into a rotating frame by using $ψ'$ , but we can recognize that we have an effective static magnetic field within this reference frame as $\mathbf{\hat{k}}\left(B_{0}+\frac{\omega_{z}}{\gamma}\right)+\mathbf{\hat{\imath}}B_{1}$ , the same as in our classical field.

Now, let the newly transformed Hamiltonian be $\mathcal{H}' = -\left[\hbar(\omega_{z}+\gamma B_{0})\hat{\sigma}_{z}+\gamma\hbar B_{1}\hat{\sigma}_{x}\right]$ . Since this Hamiltonian is time independent, it allows us to solve for $ψ'(t)$ in analogy to eqn 2

$\psi'(t) = e^{-(i/\hbar)\mathcal{H}'t}\psi'(0)..........(31)$

Finally, we can solve for the original wavefunction $ψ$ by

$\psi(t) = e^{-i\omega_{z}t\hat{\sigma}_{z}}e^{-(i/\hbar)\mathcal{H}'t}\psi'(0)..........(32)$

Note that at $t = 0$ , $ψ'(0) = ψ(0)$ , thus we have $\psi(t) = e^{-i\omega_{z}t\hat{\sigma}_{z}}e^{-(i/\hbar)\mathcal{H}'t}\psi(0)$ .

Let us now use our wavefunction to solve for what happens to the magnetic moment of a spin, when we hit the spin's resonance frequency with our external rotating magnetic field. Essentially, we solve for $\left\langle \mu_{z}(t)\right\rangle$ when we let $ω z = - γ B 0$ . The Hamiltonian $\mathcal{H}'$ reduces to $-\gamma\hbar B_{1}\hat{\sigma}_{x}$ . Thus

$\left\langle \mu_{z}(t)\right\rangle = \int \psi^{*}(t)\mu_{z}\psi(t)d\tau..........(33)$

$= \gamma \hbar\int \left[e^{-i\omega_{z}t\hat{\sigma}_{z}}e^{i\gamma B_{1}t}\psi(0)\right]^{*}\hat{\sigma}_{z} \left[e^{-i\omega_{z}t\hat{\sigma}_{z}}e^{i\gamma B_{1}t}\psi(0)\right]d\tau..........(34)$

Using the complex conjugate rules we can rewrite to obtain

$\left\langle \mu_{z}(t)\right\rangle = \gamma \hbar\int \psi^{*}(0) e^{-i\omega_{1}t\hat{\sigma}_{x}}e^{i\omega_{z}t\hat{\sigma}_{z}} \hat{\sigma}_{z} e^{-i\omega_{z}t\hat{\sigma}_{z}}e^{i\omega_{1}t\hat{\sigma}_{x}}\psi(0)d\tau..........(35)$

where $\omega_{1}\equiv\gamma B_{1}$ is the Lamor frequency around the x direction.

$\left\langle \mu_{z}(t)\right\rangle = \gamma \hbar \int \psi^{*}(0) e^{-i\omega_{1}t\hat{\sigma}_{x}} \hat{\sigma}_{z} e^{i\omega_{1}t\hat{\sigma}_{x}} \psi(0)..........(36)$

By noting similarities with eqn 18, we can write

$e^{-i\omega_{1}t\hat{\sigma}_{x}} \hat{\sigma}_{z} e^{i\omega_{1}t\hat{\sigma}_{x}} = -\hat{\sigma}_{y}\sin{\omega_{1}t}+\hat{\sigma}_{z}\cos{\omega_{1}t}..........(37)$

and thus substitute into eqn 36 to obtain

$\left\langle \mu_{z}(t)\right\rangle = \gamma\hbar\int \psi^{*}(0) \left(-\hat{\sigma}_{y}\sin{\omega_{1}t}+\hat{\sigma}_{z}\cos{\omega_{1}t}\right) \psi(0) d\tau..........(38)$

$\left\langle \mu_{z}(t)\right\rangle = -\left\langle \mu_{y}(0)\right\rangle \sin{\omega_{1}t} + \left\langle \mu_{z}(0)\right\rangle \cos{\omega_{1}t}..........(39)$

If, at $t = 0$ , the mean magnetic moment is pointing in the z direction, then there is no component in the y direction\footnote{In an experimental setup, one will first polarize the sample in the z direction, and then apply the rotating magnetic field in the x direction. Essentially, at $t = 0$ the mean magnetic moment in the z direction is experimentally realised.}, thus $\left\langle \mu_{y}(0)\right\rangle=0$ . The final equation of motion is therefore

$\left\langle \mu_{z}(t)\right\rangle = \left\langle \mu_{z}(0)\right\rangle\cos{\gamma B_{1}t}$

Basically, at resonance the magnetic moment precesses in the z-y plane around the magnetic field $B 1$ , if we use the rotating frame of reference.

To finish off the section, I will now attempt to do this using matrix mechanics. Firstly, remember the eigenstates of the operator $\hat{\sigma}_{z}$ to be $\left|+1/2\right\rangle$ and $\left|-1/2\right\rangle$ , where the numbers in the brackets indicate the spin state. These form a complete basis set for a spin 1/2 particle, therefore, I can write any arbitrary wavefunction $Ψ(t)$ as a linear combination of the eigenstates with coefficients. Since the wavefunction is a function of time and the eigenstates of the operator is not, it means that the coefficients must be functions of time, therefore

$\left|\Psi(t)\right\rangle = a(t)\left|+1/2\right\rangle + b(t)\left|-1/2\right\rangle$

Using this wavefunction in the rotating frame of reference at resonance, one can substitute into the Shrodinger equation to yield

$\frac{\hbar}{i}\left( \frac{d a(t)}{dt}\left|+1/2\right\rangle + \frac{d b(t)}{dt}\left|-1/2\right\rangle \right) = \left(\gamma\hbar B_{1} \hat{\sigma}_{x}\right)\left(a(t)\left|+1/2\right\rangle + b(t)\left|-1/2\right\rangle \right)$

Multiplying both sides with either $\left\langle +1/2\right|$ or $\left\langle -1/2\right|$ we obtain two equations

$\frac{\hbar}{i}\frac{da(t)}{dt} = \gamma\hbar B_{1}b(t)\left\langle +1/2\right|\hat{\sigma}_{x}\left|-1/2\right\rangle$

$\frac{\hbar}{i}\frac{db(t)}{dt} = \gamma\hbar B_{1}b(t)\left\langle -1/2\right|\hat{\sigma}_{x}\left|+1/2\right\rangle$

Since $\left\langle +1/2\right|\hat{\sigma}_{x}\left|-1/2\right\rangle = \left\langle -1/2\right|\hat{\sigma}_{x}\left|+1/2\right\rangle = \frac{1}{2}$ , we have a set of coupled differential equations. These can be solved as follows by putting them into a matrix and letting $γ B 1 = ω 1$ .

$\mathbf{y'} = \mathbf{A}\mathbf{y}$

$\left( \begin{matrix}{c} \frac{db(t)}{dt} \\ \frac{da(t)}{dt} \end{matrix} \right) = \left(\begin{matrix}{cc} 0 & \frac{i\omega_{1}}{2} \\ \frac{i\omega_{1}}{2} & 0 \end{matrix}\right)\left(\begin{matrix}{c} b(t) \\ a(t) \end{matrix}\right)$

We can guess the overall solution to be in the form of : $\mathbf{y}=\mathbf{x}e^{\lambda t}$ , then $\mathbf{y}' = \lambda\mathbf{x}e^{\lambda t} = \mathbf{Ax}e^{\lambda t}$ , which can be rewritten into an eigenvalue equation $\mathbf{Ax}=\lambda\mathbf{x}$ . Now we solve for the eigenvalues and vectors.

$det\left(\mathbf{A}-\lambda\mathbf{I}\right) = 0$

$\left|\begin{matrix}{cc} -\lambda & \frac{i\omega_{1}}{2} \\ \frac{i\omega_{1}}{2} & -\lambda \end{matrix}\right| = \lambda ^{2} - \left(\frac{i\omega_{i}}{2}\right)^{2} = 0$

$\lambda = \pm \frac{i\omega_{1}}{2}$

These values are then used to determine the eigenvectors, which are

$for~\lambda= -\frac{i\omega_{1}}{2}~,~ \mathbf{x}_{-\lambda} = \left(\begin{matrix}{c} 1 \\ -1 \end{matrix}\right)$

$for~\lambda= \frac{i\omega_{1}}{2}~,~ \mathbf{x}_{\lambda} = \left(\begin{matrix}{c} 1 \\ 1 \end{matrix}\right)$

Employing the superposition principle we can write a general solution such as

$\mathbf{y} = c_{1} \mathbf{x}_{-\lambda}e^{-i\omega_{1}t/2}+ c_{2}\mathbf{x}_{\lambda}e^{i\omega_{1}t/2}$

$\left(\begin{matrix}{c} b(t) \\ a(t) \end{matrix}\right) = c_{1}\left(\begin{matrix}{c} 1 \\ -1 \end{matrix}\right)e^{-i\omega_{1}t/2}+c_{2}\left(\begin{matrix}{c} 1 \\ 1 \end{matrix}\right)e^{i\omega_{1}t/2}$

To solve for initial conditions, we let $t=0$, which gives the following solutions

a (0) = c 1 + c 2

b (0) = - c 1 + c 2

$\Rightarrow c_{1}=\frac{b(0)-a(0)}{2}$

$\Rightarrow c_{2}=\frac{b(0)+a(0)}{2}$

This leads to final solutions for the coefficients

a (t) = a (0)cosω 1 t / 2 + i b (0)sinω 1 t / 2

b (t) = i a (0)sinω 1 t / 2 + b (0)cosω 1 t / 2

The coefficients $a (0)$ and $b (0)$ can also be solved for if one remembers two things at $t = 0$ , we have a magnetic field in the z direction, thus all the spins are in the z direction and that the wavefunction has to be normalized, thus $a (0) = 1$ and $b (0) = 0$ . Therefore, the final wavefunction is

$\left|\Psi(t)\right\rangle = \left(\cos{\omega_{1}t/2}\right)\left|+1/2\right\rangle + \left(i\sin{\omega_{1}t/2}\right)\left|-1/2\right\rangle$

Now we can determine the motion of the magnetic moment in the rotation frame. The spins in the x' frame thus becomes

$\left\langle \mu_{x'}\right\rangle = \left\langle \Psi(t)\right|\gamma\hbar \hat{\sigma}_{x} \left|\Psi(t)\right\rangle$

$= \frac{\gamma\hbar}{2} \left(\cos{\omega_{1}t/2}\right)\left(i\sin{\omega_{1}t/2}\right) + \frac{\gamma\hbar}{2} \left(\cos{\omega_{1}t/2}\right)\left(-i\sin{\omega_{1}t/2}\right)$

= 0

For the z direction in the rotation frame

$\left\langle \mu_{z}\right\rangle = \left\langle \Psi(t)\right|\gamma\hbar \hat{\sigma}_{z} \left|\Psi(t)\right\rangle$

$= \frac{\gamma\hbar}{2} \left(\cos{\omega_{1}t/2}\right)^{2} - \frac{\gamma\hbar}{2} \left(i\sin{\omega_{1}t/2}\right)\left(-i\sin{\omega_{1}t/2}\right)$

$= \frac{\gamma\hbar}{2} \cos^{2}{\omega_{1}t/2} -\frac{\gamma\hbar}{2}\sin^{2}{\omega_{1}t/2}$

$= \frac{\gamma\hbar}{2} \left(\frac{1+\cos{\omega_{1}t}}{2}\right) - \frac{\gamma\hbar}{2} \left(\frac{1-\cos{\omega_{1}t}}{2}\right)$

$= \frac{\gamma\hbar}{2}\cos{\omega_{1}t}$

Finally, for the y' direction

$\left\langle \mu_{y'}\right\rangle = \left\langle \Psi(t)\right|\gamma\hbar \hat{\sigma}_{y} \left|\Psi(t)\right\rangle$

$= \frac{\gamma\hbar}{2}\left(\sin{\omega_{1}t/2}\right)\left(\cos{\omega_{1}t/2}\right) + \frac{\gamma\hbar}{2}\left(\cos{\omega_{1}t/2}\right)\left(\sin{\omega_{1}t/2}\right)$