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

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.

Writing $$|b^n\rangle := |b\rangle^{\otimes n}$$ for $$b \in \{0,1\}$$, we have $$$$\rho_2 - \rho_1 = \frac{1}{2}|0^n\rangle \langle 0^n| - \frac{1}{2}|0^n\rangle \langle 1^n| - \frac{1}{2}|1^n\rangle \langle 0^n| -\frac{1}{2}|1^n\rangle \langle 1^n| \tag{1}$$$$ Regardless of the value of $$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$$-dimensional subspace spanned by the elements in Eq. (1), it will only have two nonzero eigenvalues. These eigenvalues are computed from the matrix $$$$\frac{1}{2}\begin{pmatrix} 1 & -1 \\ -1 & -1 \end{pmatrix} \begin{array}{c} \leftarrow|0^n\rangle \\ \leftarrow|1^n\rangle \end{array} \tag{2}$$$$ where the kets with the arrows identify the subspace $$\text{span}(|0^n\rangle, |1^n\rangle)$$ of our Hilbert space in which we know the eigenvectors must live (not that this is relevant for computing the eigenvalues).
• ... regarding the eigenvalues, this is more or less the Hadamard matrix, up to a minus sign, so $\pm1/\sqrt{2}$. Commented Mar 31 at 18:34
We can answer this question in the general case. Let $$\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$$.