What's the trace distance between $|0\rangle^{\otimes n}$ and $\frac{1}{\sqrt{2}}\big(|0\rangle^{\otimes n} + |1 \rangle^{\otimes n} \big)$?

Question

I'm trying to figure out the trace distance between the states $\rho_1$ and $\rho_2$, where $$ \begin{align}\rho_1 &= (|0\rangle \langle 0|)^{\otimes n}\,,\\ \rho_2 &= \dfrac{1}{2}(|0\rangle^{\otimes n} + |1 \rangle^{\otimes n}) (\langle 0|^{\otimes n} + \langle 1 |^{\otimes n})\,. \end{align}$$

However, I am not sure how to proceed.

I know that we can calculate the trace distance using the sum of the absolute values of the eigenvalues of $\rho_1-\rho_2$, but it is not at all obvious what that is to me.

Answer 1

Writing $|b^n\rangle := |b\rangle^{\otimes n}$ for $b \in \{0,1\}$, we have \begin{equation} \rho_2 - \rho_1 = \frac{1}{2}|0^n\rangle \langle 0^n| - \frac{1}{2}|0^n\rangle \langle 1^n| - \frac{1}{2}|1^n\rangle \langle 0^n| -\frac{1}{2}|1^n\rangle \langle 1^n| \tag{1} \end{equation} Regardless of the value of $n$, this matrix basically looks the same: It is a $2^n\times 2^n$ matrix with a $1/2$ in the top left corner, and $-1/2$ in the other three corners. Since this matrix only acts nontrivially on a $2$-dimensional subspace spanned by the elements in Eq. (1), it will only have two nonzero eigenvalues. These eigenvalues are computed from the matrix \begin{equation} \frac{1}{2}\begin{pmatrix} 1 & -1 \\ -1 & -1 \end{pmatrix} \begin{array}{c} \leftarrow|0^n\rangle \\ \leftarrow|1^n\rangle \end{array} \tag{2} \end{equation} where the kets with the arrows identify the subspace $\text{span}(|0^n\rangle, |1^n\rangle)$ of our Hilbert space in which we know the eigenvectors must live (not that this is relevant for computing the eigenvalues).

I don't see an "easy way" to get the eigenvalues of Eq. (2), but you can now either solve for them by hand or use computer software (Wolfram Alpha), and the trace distance will follow immediately.

Answer 2

We can answer this question in the general case. Let $\mathbb{P}_\psi\equiv|\psi\rangle\!\langle\psi|$ and $\mathbb{P}_\phi\equiv|\phi\rangle\!\langle\phi|$ be two arbitrary pure states. We want to compute $\|\mathbb{P}_\psi-\mathbb{P}_\phi\|_1$, where $\|A\|_1\equiv \operatorname{tr}\sqrt{A^\dagger A}=\operatorname{tr}|A|$. Equivalently, $\|A\|_1$ is the sum of the singular values of $A$, or the sum of the absolute values of the eigenvalues of $A$ when $A$ is normal or Hermitian.

In other words, we are interested in the eigenvalues of $\mathbb{P}_\psi-\mathbb{P}_\phi$. To this end, observe that regardless of the underlying space dimension, this is effectively a 2x2 matrix, because it only acts nontrivially on the space spanned by $|\psi\rangle$ and $|\phi\rangle$. Consider then what the matrix looks like in an orthonormal basis for this subspace containing $|\psi\rangle$. That is, we represent the matrix in the orthonormal basis $\{|\psi\rangle, |\psi_\perp\rangle\}$ where $$|\psi_\perp\rangle \equiv N (|\phi\rangle - |\psi\rangle \langle\psi|\phi\rangle)$$ with $N$ suitable renormalisation constant. The idea is of course that $\langle \psi|\psi_\perp\rangle=0$ and $\operatorname{span}(\{|\psi\rangle,|\psi_\perp\rangle\})=\operatorname{span}(\{|\psi\rangle,|\phi\rangle\})$. In this basis, we have $$\mathbb{P}_\psi-\mathbb{P}_\phi = \begin{pmatrix}1- F & -\langle\psi|\phi\rangle \langle\phi|\psi_\perp\rangle \\ -\langle\psi_\perp|\phi\rangle \langle\phi|\psi\rangle & F-1 \end{pmatrix} = \begin{pmatrix}1-F & -\sqrt{F(1-F)} e^{i\alpha} \\ -\sqrt{F(1-F)} e^{-i\alpha} & F-1 \end{pmatrix},$$ where $F\equiv |\langle\psi|\phi\rangle|^2$ and $\alpha\in\mathbb{R}$ is some phase. It follows that the eigenvalues are $$\lambda_\pm = \pm\sqrt{(1-F)^2 + F(1-F)} = \pm \sqrt{1-F}.$$ We conclude that $$\|\mathbb{P}_\psi-\mathbb{P}_\phi\|_1 = 2 \sqrt{1-F}.$$ In the particular case at end, we have $F=1/2$.

See also Is the trace distance upper bounded by the Euclidean distance? for some applications of this result.