## Saturday, 30 November 2013

### The trace distance between pure states

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 $||(\rho-\sigma)||_{Tr}=||v_\rho-v_\sigma||_2=1$.