In this post, the relationship between two distance measures on density operators will be analyzed for a few specific cases. Namely, the trace distance and the Euclidean distance of two pure states will be studied.
The trace norm is defined as $||M||_{Tr}=Tr(\sqrt{M^\dagger M})$, and the trace distance between two density operators $\rho$ and $\sigma$ is given by $||\rho-\sigma||_{Tr}$.
Let's calculate an expression for the trace distance between $\ket{0}$ and $\cos(\theta)\ket{0}+\sin(\theta)\ket{1}$ as a function of $\theta$.
Let
\[
\rho=\ket{0}\bra{0}=\begin{pmatrix} 1&0\\0&0\end{pmatrix}
\]
and
\[ \begin{align*}
\sigma&=(\cos(\theta)\ket{0}+\sin(\theta)\ket{1})(\cos(\theta)\bra{0}+\sin(\theta)\bra{1})\\
\sigma&=\cos^2(\theta)\ket{0}\bra{0}+\cos(\theta)\sin(\theta)\ket{0}\bra{1}+\cos(\theta)\sin(\theta)\ket{1}\bra{0}+\sin^2(\theta)\ket{1}\bra{1} \\
\sigma&=\begin{pmatrix}\cos^2(\theta)&\cos(\theta)\sin(\theta)\\ \cos(\theta)\sin(\theta)&\sin^2(\theta)
\end{pmatrix}.
\end{align*}\]
In this case, since both $\rho$ and $\sigma$ are density operators, they are Hermitian. Thus, $(\rho-\sigma)^\dagger=(\rho-\sigma)$ implying that $(\rho-\sigma)^\dagger(\rho-\sigma)=(\rho-\sigma)^2$. Then
\[
\rho-\sigma=\begin{pmatrix}1-\cos^2(\theta)&-\cos(\theta)\sin(\theta)\\ -\cos(\theta)\sin(\theta)&-\sin^2(\theta)\end{pmatrix}=\begin{pmatrix}\sin^2(\theta)&-\cos(\theta)\sin(\theta)\\ -\cos(\theta)\sin(\theta)&-\sin^2(\theta)\end{pmatrix},
\]
from which it follows that
\[\begin{align*}
(\rho-\sigma)^2&=\begin{pmatrix}\sin^2(\theta)&-\cos(\theta)\sin(\theta)\\ -\cos(\theta)\sin(\theta)&-\sin^2(\theta)\end{pmatrix}\begin{pmatrix}\sin^2(\theta)&-\cos(\theta)\sin(\theta)\\ -\cos(\theta)\sin(\theta)&-\sin^2(\theta)\end{pmatrix}\\
&=\begin{pmatrix}\sin^4(\theta)+\cos^2(\theta)\sin^2(\theta)&-\sin^2(\theta)\cos(\theta)\sin(\theta)+\sin^2(\theta)cos(\theta)\sin(\theta) \\ -\sin^2(\theta)\cos(\theta)\sin(\theta) +\sin^2(\theta)cos(\theta)\sin(\theta)&\sin^4(\theta)+\cos^2(\theta)\sin^2(\theta)\end{pmatrix} \\
&=\begin{pmatrix}
\sin^2(\theta)&0 \\
0&\sin^2(\theta).
\end{pmatrix}
\end{align*}\]
Thus,
\[
\sqrt{(\rho-\sigma)^2}=\begin{pmatrix}
\sqrt{\sin^2(\theta)}&0 \\
0&\sqrt{\sin^2(\theta)}\end{pmatrix}
=\begin{pmatrix}
|\sin(\theta)|&0 \\
0&|\sin(\theta)|\end{pmatrix},
\]
and therefore
\[
||(\rho-\sigma)||_{Tr}=Tr\begin{pmatrix}
|\sin(\theta)|&0 \\
0&|\sin(\theta)|\end{pmatrix}=2|\sin(\theta)|.\]
As another example, lets now calculate an expression for the Euclidean distance between the two points in the Bloch sphere that correspond to the pure states $\ket{0}$ and $\cos(\theta)\ket{0}+\sin(\theta)\ket{1}$.
In the Bloch sphere representation, a general density matrix can be expressed as
\[
\frac{1}{2}(I+c_xX+c_yY+c_zZ),
\]
where $X$, $Y$, $Z$ are the Pauli matrices, and the vector $(c_x,c_y,c_z)$ gives the coordinates of the density matrix on the Bloch sphere. In such a representation, $\rho$ and $\sigma$ can be expressed as follows:
\[\begin{align*}
\rho=&\frac{1}{2}(I+Z),\\
\sigma=&\frac{1}{2}(I+2\cos(\theta)\sin(\theta)X+\cos(2\theta)Z),
\end{align*}\]
so that the
coordinate vectors for the Block sphere representations of $\rho$ and $\sigma$ are given by, respectively,
\[\begin{align*}
v_{\rho}=&(0,0,1) \\
v_{\sigma}=&(2\cos(\theta)\sin(\theta),0,\cos(2\theta)).
\end{align*}\]
Then the Euclidean distance between $\rho$ and $\sigma$ on the Block sphere is given by
\[\begin{align*}
||v_\rho-v_\sigma||_2&=\sqrt{(-2\cos(\theta)\sin(\theta))^2+(1-\cos(2\theta))^2}\\
&=\sqrt{4\cos^2(\theta)\sin^2(\theta)+4\sin^4(\theta)} \\
&=\sqrt{4\sin^2(\theta)(\cos^2(\theta)+\sin^2(\theta))}\\
&=\sqrt{4\sin^2(\theta)}\\
&=2|\sin(\theta)|.
\end{align*}\]
Hence, $||(\rho-\sigma)||_{Tr}=||v_\rho-v_\sigma||_2=2|\sin(\theta)|$, and the two distance measures agree.
Now we'll repeat the calculations done above for these two states: $\rho=\begin{pmatrix} 1/2 &0 \\ 0 & 1/2 \end{pmatrix}$ and the pure state $\cos(\theta)\ket{0}+\sin(\theta)\ket{1}$ as a function of $\theta$.
As calculatedabove, the density matrix corresponding to the state $\cos(\theta)\ket{0}+\sin(\theta)\ket{1}$ is given by
\[
\sigma:=\begin{pmatrix}\cos^2(\theta)&\cos(\theta)\sin(\theta)\\ \cos(\theta)\sin(\theta)&\sin^2(\theta)
\end{pmatrix}
\]
Calculating the relevant matrices in order to determine the trace distance $||(\rho-\sigma)||_{Tr}=Tr(\sqrt{(\rho-\sigma)^2})$ yields
\[\begin{align*}
\rho-\sigma=\begin{pmatrix}\frac{1}{2}-\cos^2(\theta)&\cos(\theta)\sin(\theta)\\ \cos(\theta)\sin(\theta)&\frac{1}{2}-\sin^2(\theta)
\end{pmatrix},
\end{align*}\]
and after some further calculation
\[ \begin{align*}
(\rho-\sigma)^2&=\begin{pmatrix}\frac{1}{2}-\cos^2(\theta)&\cos(\theta)\sin(\theta)\\ \cos(\theta)\sin(\theta)&\frac{1}{2}-\sin^2(\theta)
\end{pmatrix}\begin{pmatrix}\frac{1}{2}-\cos^2(\theta)&\cos(\theta)\sin(\theta)\\ \cos(\theta)\sin(\theta)&\frac{1}{2}-\sin^2(\theta)
\end{pmatrix}\\
&=\begin{pmatrix}
\frac{1}{4}&0\\
0&\frac{1}{4}
\end{pmatrix},
\end{align*}\]
which implies that
\[
\sqrt{ (\rho-\sigma)^2}=\begin{pmatrix}
\sqrt{ \frac{1}{4}}&0\\
0&\sqrt{\frac{1}{4}}
\end{pmatrix}=\begin{pmatrix}
\frac{1}{2}&0\\
0&\frac{1}{2}
\end{pmatrix}
\]
Therefore,
\[
||(\rho-\sigma)||_{Tr}=Tr(\sqrt{(\rho-\sigma)^2})=Tr\begin{pmatrix}
\frac{1}{2}&0\\
0&\frac{1}{2}
\end{pmatrix}=1
\]
is the trace distance between the two state $\rho$ and $\sigma$.
Now, to calculate the Euclidean distance between the two states in the Block sphere representation observe that
\[ \begin{align*}
\rho=&\frac{1}{2}I,\\
\sigma=&\frac{1}{2}(I+2\cos(\theta)\sin(\theta)X+\cos(2\theta)Z),
\end{align*}\]
so that the
coordinate vectors for the Block sphere representations of $\rho$ and $\sigma$ are given by, respectively,
\[\begin{align*}
v_{\rho}=&(0,0,0) \\
v_{\sigma}=&(2\cos(\theta)\sin(\theta),0,\cos(2\theta)).
\end{align*}\]
The Euclidean distance is then given by
\[ \begin{align*}
||v_\rho-v_\sigma||_2&=\sqrt{(-2\cos(\theta)\sin(\theta))^2+(-\cos(2\theta))^2}\\
&=\sqrt{4\cos^2(\theta)\sin^2(\theta)+\cos^2(2\theta)} \\
&=\sqrt{4\cos^2(\theta)\sin^2(\theta)+(\cos^2(\theta)-\sin^2(\theta))^2}\\
&=\sqrt{2\cos^2(\theta)\sin^2(\theta)+\cos^4(\theta)+\sin^4(\theta))^2}\\
&=\sqrt{2\cos^2(\theta)\sin^2(\theta)+\cos^4(\theta)+\sin^4(\theta))^2}\\
&=\sqrt{(\cos^2(\theta)+\sin^2(\theta))^2}\\
&=\sqrt{1^2}\\
&=1,
\end{align*}\]
and so the trace distance of $\rho$ and $\sigma$ agrees with their Euclidean distance on the Block sphere since $||(\rho-\sigma)||_{Tr}=||v_\rho-v_\sigma||_2=1$.
The trace norm is defined as $||M||_{Tr}=Tr(\sqrt{M^\dagger M})$, and the trace distance between two density operators $\rho$ and $\sigma$ is given by $||\rho-\sigma||_{Tr}$.
Let's calculate an expression for the trace distance between $\ket{0}$ and $\cos(\theta)\ket{0}+\sin(\theta)\ket{1}$ as a function of $\theta$.
Let
\[
\rho=\ket{0}\bra{0}=\begin{pmatrix} 1&0\\0&0\end{pmatrix}
\]
and
\[ \begin{align*}
\sigma&=(\cos(\theta)\ket{0}+\sin(\theta)\ket{1})(\cos(\theta)\bra{0}+\sin(\theta)\bra{1})\\
\sigma&=\cos^2(\theta)\ket{0}\bra{0}+\cos(\theta)\sin(\theta)\ket{0}\bra{1}+\cos(\theta)\sin(\theta)\ket{1}\bra{0}+\sin^2(\theta)\ket{1}\bra{1} \\
\sigma&=\begin{pmatrix}\cos^2(\theta)&\cos(\theta)\sin(\theta)\\ \cos(\theta)\sin(\theta)&\sin^2(\theta)
\end{pmatrix}.
\end{align*}\]
In this case, since both $\rho$ and $\sigma$ are density operators, they are Hermitian. Thus, $(\rho-\sigma)^\dagger=(\rho-\sigma)$ implying that $(\rho-\sigma)^\dagger(\rho-\sigma)=(\rho-\sigma)^2$. Then
\[
\rho-\sigma=\begin{pmatrix}1-\cos^2(\theta)&-\cos(\theta)\sin(\theta)\\ -\cos(\theta)\sin(\theta)&-\sin^2(\theta)\end{pmatrix}=\begin{pmatrix}\sin^2(\theta)&-\cos(\theta)\sin(\theta)\\ -\cos(\theta)\sin(\theta)&-\sin^2(\theta)\end{pmatrix},
\]
from which it follows that
\[\begin{align*}
(\rho-\sigma)^2&=\begin{pmatrix}\sin^2(\theta)&-\cos(\theta)\sin(\theta)\\ -\cos(\theta)\sin(\theta)&-\sin^2(\theta)\end{pmatrix}\begin{pmatrix}\sin^2(\theta)&-\cos(\theta)\sin(\theta)\\ -\cos(\theta)\sin(\theta)&-\sin^2(\theta)\end{pmatrix}\\
&=\begin{pmatrix}\sin^4(\theta)+\cos^2(\theta)\sin^2(\theta)&-\sin^2(\theta)\cos(\theta)\sin(\theta)+\sin^2(\theta)cos(\theta)\sin(\theta) \\ -\sin^2(\theta)\cos(\theta)\sin(\theta) +\sin^2(\theta)cos(\theta)\sin(\theta)&\sin^4(\theta)+\cos^2(\theta)\sin^2(\theta)\end{pmatrix} \\
&=\begin{pmatrix}
\sin^2(\theta)&0 \\
0&\sin^2(\theta).
\end{pmatrix}
\end{align*}\]
Thus,
\[
\sqrt{(\rho-\sigma)^2}=\begin{pmatrix}
\sqrt{\sin^2(\theta)}&0 \\
0&\sqrt{\sin^2(\theta)}\end{pmatrix}
=\begin{pmatrix}
|\sin(\theta)|&0 \\
0&|\sin(\theta)|\end{pmatrix},
\]
and therefore
\[
||(\rho-\sigma)||_{Tr}=Tr\begin{pmatrix}
|\sin(\theta)|&0 \\
0&|\sin(\theta)|\end{pmatrix}=2|\sin(\theta)|.\]
As another example, lets now calculate an expression for the Euclidean distance between the two points in the Bloch sphere that correspond to the pure states $\ket{0}$ and $\cos(\theta)\ket{0}+\sin(\theta)\ket{1}$.
In the Bloch sphere representation, a general density matrix can be expressed as
\[
\frac{1}{2}(I+c_xX+c_yY+c_zZ),
\]
where $X$, $Y$, $Z$ are the Pauli matrices, and the vector $(c_x,c_y,c_z)$ gives the coordinates of the density matrix on the Bloch sphere. In such a representation, $\rho$ and $\sigma$ can be expressed as follows:
\[\begin{align*}
\rho=&\frac{1}{2}(I+Z),\\
\sigma=&\frac{1}{2}(I+2\cos(\theta)\sin(\theta)X+\cos(2\theta)Z),
\end{align*}\]
so that the
coordinate vectors for the Block sphere representations of $\rho$ and $\sigma$ are given by, respectively,
\[\begin{align*}
v_{\rho}=&(0,0,1) \\
v_{\sigma}=&(2\cos(\theta)\sin(\theta),0,\cos(2\theta)).
\end{align*}\]
Then the Euclidean distance between $\rho$ and $\sigma$ on the Block sphere is given by
\[\begin{align*}
||v_\rho-v_\sigma||_2&=\sqrt{(-2\cos(\theta)\sin(\theta))^2+(1-\cos(2\theta))^2}\\
&=\sqrt{4\cos^2(\theta)\sin^2(\theta)+4\sin^4(\theta)} \\
&=\sqrt{4\sin^2(\theta)(\cos^2(\theta)+\sin^2(\theta))}\\
&=\sqrt{4\sin^2(\theta)}\\
&=2|\sin(\theta)|.
\end{align*}\]
Hence, $||(\rho-\sigma)||_{Tr}=||v_\rho-v_\sigma||_2=2|\sin(\theta)|$, and the two distance measures agree.
Now we'll repeat the calculations done above for these two states: $\rho=\begin{pmatrix} 1/2 &0 \\ 0 & 1/2 \end{pmatrix}$ and the pure state $\cos(\theta)\ket{0}+\sin(\theta)\ket{1}$ as a function of $\theta$.
As calculatedabove, the density matrix corresponding to the state $\cos(\theta)\ket{0}+\sin(\theta)\ket{1}$ is given by
\[
\sigma:=\begin{pmatrix}\cos^2(\theta)&\cos(\theta)\sin(\theta)\\ \cos(\theta)\sin(\theta)&\sin^2(\theta)
\end{pmatrix}
\]
Calculating the relevant matrices in order to determine the trace distance $||(\rho-\sigma)||_{Tr}=Tr(\sqrt{(\rho-\sigma)^2})$ yields
\[\begin{align*}
\rho-\sigma=\begin{pmatrix}\frac{1}{2}-\cos^2(\theta)&\cos(\theta)\sin(\theta)\\ \cos(\theta)\sin(\theta)&\frac{1}{2}-\sin^2(\theta)
\end{pmatrix},
\end{align*}\]
and after some further calculation
\[ \begin{align*}
(\rho-\sigma)^2&=\begin{pmatrix}\frac{1}{2}-\cos^2(\theta)&\cos(\theta)\sin(\theta)\\ \cos(\theta)\sin(\theta)&\frac{1}{2}-\sin^2(\theta)
\end{pmatrix}\begin{pmatrix}\frac{1}{2}-\cos^2(\theta)&\cos(\theta)\sin(\theta)\\ \cos(\theta)\sin(\theta)&\frac{1}{2}-\sin^2(\theta)
\end{pmatrix}\\
&=\begin{pmatrix}
\frac{1}{4}&0\\
0&\frac{1}{4}
\end{pmatrix},
\end{align*}\]
which implies that
\[
\sqrt{ (\rho-\sigma)^2}=\begin{pmatrix}
\sqrt{ \frac{1}{4}}&0\\
0&\sqrt{\frac{1}{4}}
\end{pmatrix}=\begin{pmatrix}
\frac{1}{2}&0\\
0&\frac{1}{2}
\end{pmatrix}
\]
Therefore,
\[
||(\rho-\sigma)||_{Tr}=Tr(\sqrt{(\rho-\sigma)^2})=Tr\begin{pmatrix}
\frac{1}{2}&0\\
0&\frac{1}{2}
\end{pmatrix}=1
\]
is the trace distance between the two state $\rho$ and $\sigma$.
Now, to calculate the Euclidean distance between the two states in the Block sphere representation observe that
\[ \begin{align*}
\rho=&\frac{1}{2}I,\\
\sigma=&\frac{1}{2}(I+2\cos(\theta)\sin(\theta)X+\cos(2\theta)Z),
\end{align*}\]
so that the
coordinate vectors for the Block sphere representations of $\rho$ and $\sigma$ are given by, respectively,
\[\begin{align*}
v_{\rho}=&(0,0,0) \\
v_{\sigma}=&(2\cos(\theta)\sin(\theta),0,\cos(2\theta)).
\end{align*}\]
The Euclidean distance is then given by
\[ \begin{align*}
||v_\rho-v_\sigma||_2&=\sqrt{(-2\cos(\theta)\sin(\theta))^2+(-\cos(2\theta))^2}\\
&=\sqrt{4\cos^2(\theta)\sin^2(\theta)+\cos^2(2\theta)} \\
&=\sqrt{4\cos^2(\theta)\sin^2(\theta)+(\cos^2(\theta)-\sin^2(\theta))^2}\\
&=\sqrt{2\cos^2(\theta)\sin^2(\theta)+\cos^4(\theta)+\sin^4(\theta))^2}\\
&=\sqrt{2\cos^2(\theta)\sin^2(\theta)+\cos^4(\theta)+\sin^4(\theta))^2}\\
&=\sqrt{(\cos^2(\theta)+\sin^2(\theta))^2}\\
&=\sqrt{1^2}\\
&=1,
\end{align*}\]
and so the trace distance of $\rho$ and $\sigma$ agrees with their Euclidean distance on the Block sphere since $||(\rho-\sigma)||_{Tr}=||v_\rho-v_\sigma||_2=1$.