Example Sheet 3

March 17, 2018*

*Last modified 25-Aug-18

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Example sheet 3
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Exercise 1.

Given two qubit states \(\rho\) and \(\omega\) with Bloch vectors \(\vec r\) and \(\vec s\) respectively, show that \[ \|\rho-\omega\|_1 = \|\vec r - \vec s\|_2. \] The two norm of a vector \(\vec v = (v_x,v_y,v_z)\) is \(\|\vec v\|_2 = \sqrt{v_x^2 + v_y^2 + v_z^2}\).

We have \[ \rho = \frac{1}{2}(I+ \vec r \cdot \vec \sigma), \qquad \omega = \frac{1}{2}(I+ \vec s \cdot \vec \sigma), \] so \[ \|\rho - \sigma\|_1 = \frac{1}{2}\| (\vec r - \vec s)\cdot \vec \sigma\|_1 =\frac{1}{2} \left\| \begin{pmatrix} v_z & v_x - i v_y \\ v_x + i v_y & - v_z \end{pmatrix}\right\|_1 \] where \(\vec v = \vec r - \vec s\), just as from exercise 1 from example sheet 2. We can use that for any self-adjoint matrix \(A\), we have \(\|A\|_1 = \operatorname{tr}[|A|] = \sum_{j} |\lambda_j|\) where \(\lambda_j\) are the eigenvalues of \(A\). Thus, we simply need to calculate the eigenvalues of the above \(2\times 2\) matrix. We find the eigenvalues are \[ \lambda_\pm = \pm \sqrt{ v_x^2 + v_y^2 + v_z^2} = \pm \| \vec r - \vec s\|_2. \] Thus, \(\|\rho - \sigma\|_1 = \frac{1}{2}(|\lambda_-| + |\lambda_+|) = \|\vec r - \vec s\|_2\).

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Exercise 2.

One Fuchs and van de Graaf inequality.

  1. Show that for two pure qubit states \(\psi\) and \(\phi\), we have that \[ D(\psi,\phi) = \sqrt{1 - F(\psi,\phi)^2}. \] where \(D\) is the trace distance, and \(F\) the fidelity, defined in the lectures.
  2. Using this, show that for any two mixed qubit states \(\rho\) and \(\sigma\), \[ D(\rho,\sigma) \leq \sqrt{1- F(\rho,\sigma)^2}. \]
  1. For two pure states, we have \(F(\psi, \phi) = |\langle \psi|\phi \rangle|\). We may write \[ | \phi \rangle = \cos (\theta) | \psi \rangle + \sin(\theta) | \psi^\perp \rangle \](1) for some vector \(| \psi^\perp \rangle\) orthogonal to \(| \psi \rangle\). In these terms, \(F(\psi, \phi) = |\cos \theta|\). To calculate the trace distance, we want to simplify the form of \(\psi - \phi = | \psi \rangle\langle \psi | -| \phi \rangle\langle \phi |\). Using (1), we find \[ \phi = \cos^2(\theta)\, \psi + \sin (\theta)\cos(\theta) | \psi^\perp \rangle\langle \psi | + \cos (\theta)\sin(\theta) | \psi \rangle\langle \psi^\perp | + \sin^2 (\theta) \, \psi^\perp. \] In the basis \(\{| \psi \rangle,| \psi^\perp \rangle\}\), we find \[ \psi - \phi = \begin{pmatrix} 1 -\cos^2(\theta) & - \sin (\theta)\cos (\theta) \\ - \sin(\theta)\cos(\theta) & - \sin^2(\theta)\end{pmatrix} \] which has eigenvalues \(\pm |\sin\theta|\). Thus, \[ \frac{1}{2}\|\psi - \phi\|_1 = |\sin\theta|. \] Since \(\sqrt{1- \cos^2(\theta)} = |\sin\theta|\), we therefore have \[ D(\psi,\phi) = \sqrt{1- \cos^2(\theta)} = \sqrt{1- F(\psi,\phi)^2}. \] Remark. In fact, this proof holds for qudits.
  2. Let \(\rho = \rho_A\) have purification \(\phi_{AR}\) and \(\sigma = \sigma_A\) have purification \(\psi_{AR}\) such that \[ F(\rho_A,\sigma_A) = |\langle \phi_{AR}|\psi_{AR} \rangle| \] using Uhlmann’s theorem. Then \[ \frac{1}{2}\|\psi_{AR} - \phi_{AR}\|_1 = \sqrt{1 - F(\psi_{AR},\phi_{AR})} = \sqrt{1 - F(\rho_A,\sigma_A)}. \](2) Since the partial trace is a linear CPTP map, and therefore a quantum operation, we may use the monotonicity of the trace distance under quantum operations to find \[ \frac{1}{2}\|\psi_{AR} -\phi_{AR}\|_1 \geq \frac{1}{2}\|\rho_A-\sigma_A\|_1 \] which concludes the proof by (2).
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Exercise 3.

Show that for any two states \(\rho\) and \(\sigma\) and any unitary \(U\), \[ F(U \rho U^\dagger,U \sigma U^\dagger) = F(\rho,\sigma) \] by using the fact that for any positive semi-definite operator \(A\) and unitary \(U\), we have \(\sqrt{U A U^\dagger} = U \sqrt{A} U^\dagger\).

We have \[\begin{aligned} F(U\rho U^\dagger, U\sigma U^\dagger) &= \| \sqrt{U \rho U^\dagger}\sqrt{U \sigma U^\dagger}\|_1 \\ &= \| U \sqrt{\rho} U^\dagger U \sqrt{\sigma}U^\dagger\|_1\\ &= \| U \sqrt{\rho}\sqrt{\sigma}U^\dagger\|_1\\ &= \max_{V} | \operatorname{tr}[ V U \sqrt{\rho}\sqrt{\sigma}U^\dagger ]| \\ &= \max_{V} | \operatorname{tr}[ U^\dagger V U \sqrt{\rho}\sqrt{\sigma} ]| \\ &= \max_{UVU^\dagger} | \operatorname{tr}[ U^\dagger U V U^\dagger U \sqrt{\rho}\sqrt{\sigma} ]| \\ &= \max_{UVU^\dagger} | \operatorname{tr}[\sqrt{\rho}\sqrt{\sigma} ]| \\ &= \|\sqrt{\rho}\sqrt{\sigma}\|_1 \\ &= F(\rho,\sigma). \end{aligned}\] We use the characterization of the trace distance as a maximum over unitary operators \(V\), and along the way proved that the trace norm was invariant under unitary conjugation. We also used that maximizing over all \(V\) was the same as maximizing over all \(U V U^\dagger\), since \[ \{U V U^\dagger: V \text{ unitary}\} = \{ V : V \text{ unitary}\}. \]