Extended phase graph

In the Bloch equations, within a voxel each spin along the dimension ${\displaystyle z}$ can be described by a magnetization vector ${\displaystyle [M_{x}(z),M_{y}(z),M_{z}(z)]^{T}}$ where every element is real valued.^[6] But ${\displaystyle M_{x}(z)}$ and ${\displaystyle M_{y}(z)}$ can also be described by the complex number ${\displaystyle M_{+}(z)=M_{x}(z)+iM_{y}(z)}$.^[7] Using a similarity matrix ${\displaystyle S}$ we can convert between the real valued basis and the complex transverse magnetization basis:

$${\displaystyle {\begin{aligned}&S={\begin{bmatrix}1&i&0\\1&-i&0\\0&0&1\end{bmatrix}},&\mathbf {m} (z)={\begin{bmatrix}M_{+}(z)\\M_{-}(z)\\M_{z}(z)\end{bmatrix}}=S{\begin{bmatrix}M_{x}(z)\\M_{y}(z)\\M_{z}(z)\end{bmatrix}}\\\end{aligned}}}$$

where ${\displaystyle M_{-}}$ is the complex conjugate of ${\displaystyle M_{+}}$. To perform a rotation on the complex magnetization vector ${\displaystyle \mathbf {m} }$ due to an RF pulse with angle ${\displaystyle \alpha }$ and phase ${\displaystyle \phi }$, we first apply a change of basis with ${\displaystyle S^{-1}}$, then apply standard cartesian rotation matrices ${\displaystyle R_{z}(\phi )}$, ${\displaystyle R_{x}(\alpha )}$,^[8] then another change of basis with ${\displaystyle S}$:^[9]

$${\displaystyle {\begin{aligned}&R_{z}(\phi )={\begin{bmatrix}\cos \phi &-\sin \phi &0\\\sin \phi &\cos \phi &0\\0&0&1\end{bmatrix}},\;\;R_{x}(\alpha )={\begin{bmatrix}1&0&0\\0&\cos \alpha &-\sin \alpha \\0&\sin \alpha &\cos \alpha \end{bmatrix}}\\&\\&T_{\phi }(\alpha )=SR_{z}(\phi )R_{x}(\alpha )R_{z}(-\phi )S^{-1}\\&\\&T_{\phi }(\alpha )={\begin{bmatrix}\cos ^{2}{\frac {\alpha }{2}}&e^{2i\phi }\sin ^{2}{\frac {\alpha }{2}}&-e^{i\phi }\sin \alpha \\e^{-2i\phi }\sin ^{2}{\frac {\alpha }{2}}&\cos ^{2}{\frac {\alpha }{2}}&ie^{-i\phi }\sin \alpha \\-{\frac {i}{2}}e^{-i\phi }\sin \alpha &{\frac {i}{2}}e^{i\phi }\sin \alpha &\cos \alpha \end{bmatrix}}\\&\\&\mathbf {m} ^{+}(z)=T_{\phi }(\alpha )\mathbf {m} (z)\end{aligned}}}$$

Where ${\displaystyle \mathbf {m} ^{+}}$ denotes the magnetization vector after the rotation. In the EPG representation we perform a Fourier decomposition of each element of the complex magnetization vector:^[2]

$${\displaystyle {\begin{aligned}&M_{+}(z)=\sum _{k=-\infty }^{\infty }F_{+}(k)e^{i2\pi kz}\;\Leftrightarrow \;F_{+}(k)=\int _{-0.5}^{0.5}M_{+}(z)e^{-i2\pi kz}dz\\&M_{-}(z)=\sum _{k=-\infty }^{\infty }F_{-}(k)e^{i2\pi kz}\;\Leftrightarrow \;F_{-}(k)=\int _{-0.5}^{0.5}M_{-}(z)e^{-i2\pi kz}dz\\&M_{z}(z)=\sum _{k=-\infty }^{\infty }Z(k)e^{i2\pi kz}\;\Leftrightarrow \;Z(k)=\int _{-0.5}^{0.5}M_{z}(z)e^{-i2\pi kz}dz\\\end{aligned}}}$$

If ${\displaystyle F_{+}(k)=0}$, ${\displaystyle F_{-}(k)=0}$, and ${\displaystyle Z(k)=0}$ when ${\displaystyle k>n}$ or ${\displaystyle k<-n}$. We can write the summations as being from ${\displaystyle k=-n}$ to ${\displaystyle n}$. We can also write the Fourier coefficients as a vector valued function:

$${\displaystyle {\begin{aligned}&\mathbf {f} (k)={\begin{bmatrix}F_{+}(k)\\F_{-}(k)\\Z(k)\end{bmatrix}},&\mathbf {m} (z)=\sum _{k=-n}^{n}\mathbf {f} (k)e^{i2\pi kz}\\\end{aligned}}}$$

When performing a rotation, because ${\displaystyle T_{\phi }(\alpha )}$ is constant across ${\displaystyle k}$ it can enter the sum:

$${\displaystyle {\begin{aligned}&\mathbf {m} ^{+}(z)=T_{\phi }(\alpha )\mathbf {m} (z)=\sum _{k}T_{\phi }(\alpha )\mathbf {f} (k)e^{i2\pi kz}\\&\mathbf {f} ^{+}(k)=T_{\phi }(\alpha )\mathbf {f} (k)\end{aligned}}}$$

It can be seen in the equation above that the rotation matrix can be applied directly to the Fourier coefficients to produce updated Fourier coefficients.^[1]

A gradient in the z dimension that creates one twist from ${\displaystyle z=-0.5}$ to ${\displaystyle z=0.5}$ can be applied with the matrix:

$${\displaystyle {\begin{aligned}&T(z)=SR_{z}(2\pi z)S^{-1}={\begin{bmatrix}e^{i2\pi z}&0&0\\0&e^{-i2\pi z}&0\\0&0&1\end{bmatrix}}\\\end{aligned}}}$$

This matrix is independent of ${\displaystyle k}$ so it can enter the Fourier sum:^[10]

$${\displaystyle {\begin{aligned}&\mathbf {m} ^{+}(z)=T(z)\mathbf {m} (z)=T(z)\sum _{k}\mathbf {f} (k)e^{i2\pi kz}=\sum _{k}T(z)\mathbf {f} (k)e^{i2\pi kz}\\\end{aligned}}}$$

Looking at each element of ${\displaystyle \mathbf {m} ^{+}}$:

$${\displaystyle {\begin{aligned}&M_{+}^{+}(z)=\sum _{k}e^{i2\pi z}F_{+}(k)e^{i2\pi kz}=\sum _{k}F_{+}(k)e^{i2\pi (k+1)z}\\&M_{-}^{+}(z)=\sum _{k}e^{-i2\pi z}F_{-}(k)e^{i2\pi kz}=\sum _{k}F_{-}(k)e^{i2\pi (k-1)z}\\&M_{z}^{+}(z)=M_{z}(z)\\\end{aligned}}}$$

Performing a substitution with ${\displaystyle {\hat {k}}=k+1}$ for the first element we get:

$${\displaystyle {\begin{aligned}&M_{+}^{+}(z)=\sum _{{\hat {k}}=-n+1}^{n+1}F_{+}({\hat {k}}-1)e^{i2\pi {\hat {k}}z}\\\end{aligned}}}$$

If ${\displaystyle F_{+}(n)=0}$ and ${\displaystyle F_{+}(-n-1)=0}$ the above summation is equal to:^[11]

$${\displaystyle {\begin{aligned}&M_{+}^{+}(z)=\sum _{{\hat {k}}=-n}^{n}F_{+}({\hat {k}}-1)e^{i2\pi {\hat {k}}z}\\\end{aligned}}}$$

and it can be seen that ${\displaystyle F_{+}^{+}(k)=F_{+}(k-1)}$. By the same logic ${\displaystyle F_{-}^{+}(k)=F_{-}(k+1)}$. So gradients in the EPG formalism simply shift the indices of the Fourier coefficients.^[12][1]

The relaxation and recovery matrices given by the Bloch equations for a real valued magnetization vector ${\displaystyle \mathbf {v} }$ are:^[7] $${\displaystyle {\begin{aligned}&E={\begin{bmatrix}e^{-t/T2}&0&0\\0&e^{-t/T2}&0\\0&0&e^{-t/T1}\end{bmatrix}},&\mathbf {r} ={\begin{bmatrix}0\\0\\M_{0}(1-e^{-t/T1})\end{bmatrix}}\\&\\&\mathbf {v} ^{+}=E\mathbf {v} +\mathbf {r} \end{aligned}}}$$

Substituting with ${\displaystyle \mathbf {v} =S^{-1}\mathbf {m} }$:

$${\displaystyle {\begin{aligned}&S^{-1}\mathbf {m} ^{+}=ES^{-1}\mathbf {m} +\mathbf {r} \\&\mathbf {m} ^{+}=SES^{-1}\mathbf {m} +S\mathbf {r} \end{aligned}}}$$

${\displaystyle S}$ and ${\displaystyle E}$ are commutative so ${\displaystyle SES^{-1}=ESS^{-1}=E}$. Also ${\displaystyle S\mathbf {r} =\mathbf {r} }$, so:

$${\displaystyle {\begin{aligned}&\mathbf {m} ^{+}(z)=E\mathbf {m} (z)+\mathbf {r} =E\sum _{k}\mathbf {f} (k)e^{i2\pi kz}+\mathbf {r} =\sum _{k}E\mathbf {f} (k)e^{i2\pi kz}+\mathbf {r} \\\end{aligned}}}$$

${\displaystyle \mathbf {r} }$ is a constant vector but we can still Fourier decompose it to ${\displaystyle \mathbf {r} (z)=\sum _{k}\mathbf {g} (k)e^{i2\pi kz}}$ where ${\displaystyle \mathbf {g} (k)=\mathbf {r} }$ if ${\displaystyle \mathbf {k} =0}$ else ${\displaystyle \mathbf {g} (k)=\mathbf {0} }$, so:^[13]

$${\displaystyle {\begin{aligned}&\mathbf {m} ^{+}(z)=\sum _{k}E\mathbf {f} (k)e^{i2\pi kz}+\sum _{k}\mathbf {g} (k)e^{i2\pi kz}=\sum _{k}(E\mathbf {f} (k)+\mathbf {g} (k))e^{i2\pi kz}\\\end{aligned}}}$$

So it can be seen that to apply a relaxation, you can multiply every Fourier coefficient with the relaxation matrix ${\displaystyle E}$ and then add the recovery vector ${\displaystyle \mathbf {r} }$ to the 0th coefficient.^[14]

The total transverse signal in the voxel is obtained by integrating over ${\displaystyle M_{+}(z)}$:

$${\displaystyle {\begin{aligned}&M_{+total}=\int _{-0.5}^{0.5}\sum _{k=-n}^{n}F_{+}(k)e^{i2\pi kz}\;dz=\sum _{k=-n}^{n}(F_{+}(k)\int _{-0.5}^{0.5}e^{i2\pi kz}dz)=\sum _{k=-n}^{n}F_{+}(k){\frac {1}{i2\pi k}}(e^{i\pi k}-e^{-i\pi k})\\\end{aligned}}}$$

so when ${\displaystyle k}$ is nonzero, ${\displaystyle F_{+}(k)}$ is multiplied by zero and doesn't end up in the sum. If ${\displaystyle k=0}$ the inner integral is 1. So the total transverse signal in the voxel is just ${\displaystyle F_{+}(0)}$. By the same logic the total longitudinal signal is ${\displaystyle Z(0)}$.^[11]

It can be proven that ${\displaystyle F_{+}(k)=F_{-}(-k)^{*}}$:^[13][15]

$${\displaystyle {\begin{aligned}&F_{-}(k)=\int M_{+}(z)^{*}e^{-i2\pi kz}dz\\&F_{-}(-k)=\int M_{+}(z)^{*}e^{i2\pi kz}dz=\int (M_{+}(z)e^{-i2\pi kz})^{*}dz=F_{+}(k)^{*}\\\end{aligned}}}$$

and so in practice only coefficients with ${\displaystyle k\geq 0}$ are stored and if a gradient occurs that would shift a coefficient to having a negative index, instead that coefficient is conjugated and stored at a non-negative index.

In practice a discrete number of Fourier coefficients are tracked. A matrix can be set up with ${\displaystyle n+1}$ frequencies tracked like so:^[14]

$${\displaystyle {\begin{bmatrix}F_{+}(0)&F_{+}(1)&...&F_{+}(n)\\F_{-}(0)&F_{-}(1)&...&F_{-}(n)\\Z(0)&Z(1)&...&Z(n)\\\end{bmatrix}}}$$

This matrix can be multiplied by the RF pulse rotation matrix ${\displaystyle T_{\phi }(\alpha )}$, or multiplied by the relaxation matrix ${\displaystyle E}$ and then have ${\displaystyle {\textbf {r}}}$ added to the first column, to update the coefficients. A gradient that adds one twist along the z dimension will result in the first row being shifted right by one, and the second row being shifted left. Then ${\displaystyle F_{+}(0)}$ can be set to ${\displaystyle F_{-}(0)^{*}}$.^[16]

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