## 7.13Remarks and exercises

### 7.13.1 Partial inner product

The tensor product structure brings with it the possibility to do “partial things” beyond just the partial trace. Given \mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}}, any vector |x\rangle\in\mathcal{H}_{\mathcal{A}} defines an anti-linear map \mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}}\to\mathcal{H}_{\mathcal{B}} called the partial inner product with |x\rangle. It is first defined on the product vectors |a\rangle\otimes|b\rangle by the formula |a\rangle\otimes|b\rangle \longmapsto \langle x|a\rangle|b\rangle and then extended to other vectors in \mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}} by linearity. Similarly, any |y\rangle\in\mathcal{H}_{\mathcal{B}} defines a map \mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}}\to\mathcal{H}_{\mathcal{A}} via |a\rangle\otimes|b\rangle \longmapsto |a\rangle\langle y|b\rangle

For example, the partial inner product of |\psi\rangle=c_{00}|00\rangle+c_{01}|01\rangle+c_{10}|10\rangle+c_{11}|11\rangle\in\mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}} with of |0\rangle\in\mathcal{H}_{\mathcal{A}} is \langle 0|\psi\rangle = c_{00}|0\rangle + c_{01}|1\rangle and the partial inner product of the same |\psi\rangle with |1\rangle\in\mathcal{H}_{\mathcal{B}} is \langle 1|\psi\rangle = c_{01}|0\rangle + c_{11}|1\rangle.

### 7.13.2 The “control” part of controlled-NOT

Consider a single-qubit channel induced by the action of the \texttt{c-NOT} gate. Recall that the unitary operator associated with the \texttt{c-NOT} gate can be written as U = |0\rangle\langle 0|\otimes\mathbf{1}+ |1\rangle\langle 1|\otimes X where is X is the Pauli \sigma_x gate (i.e. the \texttt{NOT} gate). Let us step through the following simple circuit:

This time we are interested in the evolution of the control qubit: the control qubit will be our system, and the target qubit, initially in a fixed state |0\rangle, will play the role of an ancilla.

We can calculate the Kraus operators A_i = (\mathbf{1}\otimes\langle i|) U (\mathbf{1}\otimes|0\rangle) which we simply write as A_i=\langle i|U|0\rangle, for i=0,1. We see that \begin{aligned} A_i = \langle i|U|0\rangle &= \langle i| (|0\rangle\langle 0|\otimes\mathbf{1}+ |1\rangle\langle 1|\otimes X) |0\rangle \\&= |0\rangle\langle 0|\langle i|\mathbf{1}|0\rangle + |1\rangle\langle 1|\langle i|X|0\rangle \\&= |i\rangle\langle i| \end{aligned} We can also check the normalisation condition: A_0^\dagger A_0 + A_1^\dagger A_1 = |0\rangle\langle 0| + |1\rangle\langle 1| =\mathbf{1}.

The unitary action of the gate when the state of the target qubit is fixed at |0\rangle can be written as \begin{aligned} |\psi\rangle|0\rangle \longmapsto & A_0|\psi\rangle|0\rangle + A_1|\psi\rangle|1\rangle \\=& |0\rangle\langle 0||\psi\rangle|0\rangle + |1\rangle\langle 1||\psi\rangle|1\rangle \\=& \langle 0|\psi||\rangle 0\rangle|0\rangle + \langle 1|\psi||\rangle 1\rangle|1\rangle \end{aligned} which is a familiar \texttt{c-NOT} entangling process: if |\psi\rangle=\alpha_0|0\rangle+\alpha_1|1\rangle then |\psi\rangle|0\rangle becomes \alpha_0|0\rangle|0\rangle+\alpha_1|1\rangle|1\rangle.

The evolution of the control qubit alone can be expressed in the Kraus form as \begin{aligned} \rho \longmapsto \rho' &= A_0\rho A_0^\dagger + A_1\rho A_1^\dagger \\&= |0\rangle\langle 0|\rho|0\rangle\langle 0| + |1\rangle\langle 1|\rho|1\rangle\langle 1| \\&= \rho_{00}|0\rangle\langle 0| + \rho_{11}|1\rangle\langle 1|. \end{aligned} Then, in the matrix form, if the initial state of the control qubit is |\psi\rangle=\alpha_0|0\rangle+\alpha_1|1\rangle, we get \begin{bmatrix} |\alpha|_0^2 & \alpha_0\alpha_0^\star \\\alpha_0^\star\alpha_1 & |\alpha_1|^2 \end{bmatrix} = \rho \longmapsto \rho' = \begin{bmatrix} |\alpha_0|^2 & 0 \\0 & |\alpha_1|^2 \end{bmatrix}.

As we can see, the diagonal elements of \rho survive, and the off-diagonal elements (the coherences) disappear. The two Kraus operators, A_0=|0\rangle\langle 0| and A_1=|1\rangle\langle 1|, define the measurement in the standard basis, and so you may think about this operation as being equivalent to measuring the control qubit in the standard basis and then just forgetting the result.

### 7.13.3 Surprisingly identical channels

Let us now compare two single qubit quantum channels: \mathcal{A}(\rho)=\sum_k A_k\rho A^\dagger_k, defined by the Kraus operators \begin{aligned} A_1 = |0\rangle\langle 0| &= \begin{bmatrix}1&0\\0&0\end{bmatrix} \\A_2 = |1\rangle\langle 1| &= \begin{bmatrix}0&0\\0&1\end{bmatrix} \end{aligned} and \mathcal{B}(\rho)=\sum_k B_k\rho B^\dagger_k, defined by the Kraus operators \begin{aligned} B_1 = \frac{\mathbf{1}}{\sqrt{2}} &= \frac{1}{\sqrt{2}}\begin{bmatrix}1&0\\0&1\end{bmatrix} \\B_2 = \frac{Z}{\sqrt{2}} &= \frac{1}{\sqrt{2}}\begin{bmatrix}1&0\\0&-1\end{bmatrix}. \end{aligned}

We are familiar with the first channel from the previous example: it performs the measurement in the standard basis, but doesn’t reveal the outcome of the measurement. The second channel chooses randomly, with equal probability, between two options: it will either let the qubit pass undisturbed, or apply the phase-flip Z.

These two apparently very different physical processes correspond to the same quantum channel: \mathcal{A}(\rho)=\mathcal{B}(\rho) for any \rho. Indeed, you can check that B_1=(A_1+A_2)/\sqrt{2} and B_2=(A_1-A_2)/\sqrt{2}, whence \begin{aligned} \mathcal{B}(\rho) &= B_1\rho B_1^\dagger + B_2\rho B_2^\dagger \\&= \frac{1}{2} (A_1+A_2)\rho (A_1+A_2)^\dagger + \frac{1}{2} (A_1-A_2)\rho (A_1-A_2)^\dagger \\&= A_1\rho A_1^\dagger + A_2\rho A_2^\dagger \\&= \mathcal{A}(\rho). \end{aligned}

You can also check that the two channels can be implemented by the following two circuits:

### 7.13.4 Independent ancilla

Another way to understand the freedom in the operator-sum representation is to realise that, once the system and the ancilla cease to interact, any operation on the ancilla alone has no effect on the state of the system.

That is, the two unitaries U and (\mathbf{1}\otimes R)U (where R acts only on the ancilla) describe the same channel, even though the Kraus operators E_k=\langle e_k|U|e\rangle for the latter are \begin{aligned} F_k &= \langle e_k|(\mathbf{1}\otimes R)U|e\rangle \\&= \sum_j \langle e_k|R|e_j\rangle\langle e_j|U|e\rangle \\&= \sum_j R_{kj}E_j \end{aligned} Indeed, the unitary evolution (\mathbf{1}\otimes R) U gives \rho\otimes|e\rangle\langle e| \longmapsto \sum_{kl} E_k \rho E_l^\dagger \otimes R|e_k\rangle\langle e_l| R^\dagger and the subsequent trace over the environment gives \begin{aligned} \operatorname{tr}_E \sum_{kl} E_k \rho E_l^\dagger \otimes R|e_k\rangle\langle e_l| R^\dagger &= \sum_{kl} E_k \rho E_l^\dagger \langle e_l| R^\dagger R|e_k\rangle \\&= \sum_{k} E_k \rho E_k^\dagger. \end{aligned}

### 7.13.5 Cooling down

We can show that the process of cooling a qubit to its ground state, described the map \mathcal{E}(\rho)=|0\rangle\langle 0|, is a quantum channel. Indeed, the set of Kraus operators is |0\rangle\langle 0| and |0\rangle\langle 1|, and all Bloch vectors are mapped to the Bloch vector representing state |0\rangle\langle 0|.

### 7.13.6 Unchanged reduced density operator

Show that, for any operator \rho on \mathcal{H}_\mathcal{A}\otimes\mathcal{H}_\mathcal{B}, and, for any operator R on \mathcal{H}_\mathcal{B}, we have \operatorname{tr}_\mathcal{B} \left[(\mathbf{1}\otimes R) \rho (\mathbf{1}\otimes R^\dagger)\right] = \operatorname{tr}_\mathcal{B} \rho. That is, the reduced density operator \rho_\mathcal{A}=\operatorname{tr}_\mathcal{B} \rho is not affected by R.

(Hint: show this for operators \rho which are tensor products \rho=A\otimes B and then extend the result to any operator \rho.)

### 7.13.7 Order matters?

We know that, given a fixed state of the environment, the unitaries U and (\mathbf{1}\otimes R)U (where R acts only on the environment) define the same quantum channel. Is the same true for U and U(\mathbf{1}\otimes R)? That is, do these two unitaries define the same quantum channel as one another?

### 7.13.8 Pauli twirl

Show that randomly applying the Pauli operators \mathbf{1}, X, Y, and Z, with uniform probability, to any density operator \rho of a single qubit results in the maximally mixed state \frac{1}{4} \mathbf{1}\rho\mathbf{1}+\frac{1}{4} X\rho X + \frac{1}{4} Y\rho Y + \frac{1}{4} X\rho Z = \frac{1}{2}\mathbf{1}.

### 7.13.9 Depolarising channel

The most “popular” Pauli channel is the depolarising channel \rho\longmapsto (1-p)\rho + \frac{p}{3}\left(X\rho X+Y\rho Y+Z\rho Z\right). In the depolarising channel, a qubit in state \rho remains intact with probability 1-p, or is otherwise transformed with one of the Pauli operators X, Y, and Z, each chosen randomly with probability p/3. Show, using the Pauli twirl or otherwise, that we can rewrite the depolarising channel as \rho \longmapsto \rho' = \left(1-\frac{4}{3} p\right) \rho + \frac{4}{3}p\frac{1}{2}\mathbf{1}.

(For p\leqslant\frac{3}{4} we can thus say that the channel either does nothing or, with probability \frac{4}{3}p, throws away the initial quantum state and replaces it by the maximally mixed state.)

### 7.13.10 Depolarising channel and the Bloch sphere

It is also instructive to see how the depolarising channel acts on the Bloch sphere. An arbitrary density matrix for a single qubit can be written as \frac{1}{2}(\mathbf{1}+\vec{s}\cdot\vec{\sigma}), where \vec{s} is the Bloch vector, and \vec{\sigma}=(\sigma_x,\sigma_y,\sigma_z) is the vector of Pauli matrices. The depolarising channel maps this state to \frac{1}{2}\left[ \mathbf{1}+ \left(1-\frac{4}{3}p\right)\vec{s}\cdot\vec{\sigma} \right]. The Bloch vector shrinks by a factor of 1-\frac{4}{3}p. This means that, for p\leqslant\frac{3}{4}, the Bloch sphere contracts uniformly under the action of the channel; for p=\frac{3}{4}, the sphere is contracted to a single point at its centre; and for \frac{3}{4}\leqslant p\leqslant 1, the Bloch vector is flipped, and starts pointing in the opposite direction.

### 7.13.11 Complete positivity of a certain map

Let \mathcal{E} be the linear map on single a qubit defined by \begin{aligned} \mathcal{E}(\mathbf{1}) &= \mathbf{1} \\\mathcal{E}(\sigma_x) &= a_x\sigma_x \\\mathcal{E}(\sigma_y) &= a_y\sigma_y \\\mathcal{E}(\sigma_z) &= a_z\sigma_z \end{aligned} where a_x, a_y, and a_z are some fixed real numbers. Using the Choi matrix of \mathcal{E}, determine the range of x, y, z for which the map \mathcal{E} is positive, and the range for which it is completely positive.

### 7.13.12 Toffoli gate

Consider the Toffoli gate

Express \rho' as a function of \rho in the Kraus representation.

### 7.13.13 Duals

We say that \mathcal{E}^\star\colon\mathcal{B}(\mathcal{H})\to\mathcal{B}(\mathcal{H}') is the dual of a linear map \mathcal{E}\colon\mathcal{B}(\mathcal{H})\to\mathcal{B}(\mathcal{H}') if \operatorname{tr}\mathcal{E}^\star (X)Y = \operatorname{tr}X\mathcal{E}(Y) for any operators X and Y in \mathcal{B}(\mathcal{H}).

1. Show that, if \mathcal{E} is trace preserving, then \mathcal{E}^\star is unital.
2. Show that, if \mathcal{E}=\sum_i E_i\cdot E_i^\dagger, then \sum_i E^\dagger_i\cdot E_i is an operator-sum decomposition of \mathcal{E}^\star.

### 7.13.14 Trace, transpose, Choi

Let \mathcal{E}\colon\mathcal{B}(\mathcal{H})\to\mathcal{B}(\mathcal{H}'), and let d=\dim\mathcal{H} and d'=\dim\mathcal{H}'. Show that, for any (d\times d) matrix X and any (d'\times d') matrix Y, \operatorname{tr}\mathcal{E}(X)Y = \operatorname{tr}\widetilde{\mathcal{E}} (X^T\otimes Y).

(For example, if we are interested in the component \mathcal{E}(X)_{ij}=\langle i|\mathcal{E}(X)|j\rangle, then we can take Y=|j\rangle\langle i|.)

### 7.13.15 Purifications and isometries

All purifications of a density operator are related by an isometry acting on the purifying system. That is, if \rho is a density operator on \mathcal{H}, and |\psi_A\rangle\in \mathcal{H}\otimes\mathcal{H}_\mathcal{A} and |\psi_B\rangle\in\mathcal{H}\otimes\mathcal{H}_B (\dim\mathcal{H}_\mathcal{A}\leqslant\dim\mathcal{H}_\mathcal{B}) are two purifications of \rho, then |\psi_B\rangle=(\mathbf{1}\otimes V)|\psi_A\rangle.

To show this, we start with the spectral decomposition of \rho \rho = \sum_i p_i|i\rangle\langle i| and note that \begin{aligned} |\psi_A\rangle &= \sum_i \sqrt{p_i} |i\rangle\otimes|a_i\rangle \\|\psi_B\rangle &= \sum_i \sqrt{p_i} |i\rangle\otimes|b_i\rangle \end{aligned} define an isometry V=\sum_i |b_i\rangle\langle a_i| satisfying the desired equation.

This observation leads to a way of relating all convex decompositions of a given density operator: let \{p_k,|\psi_k\rangle\} and \{q_l,|\phi_l\rangle\} be convex decompositions of a density operator \rho; then there exists an isometry V such that these two decompositions \rho = \sum_{k=1}^n|\widetilde{\psi}_k\rangle\langle\widetilde{\psi}_k| = \sum_{l=1}^m|\widetilde{\phi}_l\rangle\langle\widetilde{\phi}_l|, (where n\geqslant m, and |\widetilde{\psi}_k\rangle=\sqrt{p_k}|\psi_k\rangle and |\widetilde{ \phi}_l\rangle=\sqrt{q_l}|\phi_l\rangle) are related: |\widetilde{\psi}_k\rangle = \sum_{l} V_{kl} |\widetilde{\phi}_l\rangle.

### 7.13.16 Tricks with a maximally entangled state

A maximally entangled state of a bipartite system can be written, using the Schmidt decomposition, as \begin{aligned} |\Omega\rangle &= \frac{1}{\sqrt d}\sum_i |i\rangle|i\rangle \\|\Omega\rangle\langle\Omega| &= \frac{1}{d} \sum_{ij}|i\rangle\langle j|\otimes|i\rangle\langle j| \end{aligned} Each subsystem is of dimension d, and all the Schmidt coefficients are equal. Here are few useful tricks involving a maximally entangled state.

• If we take the transpose in the Schmidt basis of |\Omega\rangle, then \langle\Omega|A\otimes B|\Omega\rangle = \frac{1}{d}\operatorname{tr}(A^T B).

• Any pure state of the bipartite system |\psi\rangle=\sum_{ij} c_{ij}|i\rangle|j\rangle can be written as (C\otimes\mathbf{1})|\Omega\rangle = (\mathbf{1}\otimes C^T)|\Omega\rangle. This implies that (U\otimes\overline{U})|\Omega\rangle=|\Omega\rangle (where \overline{U} denotes the matrix given by taking the complex conjugate, entry-wise, of U, i.e. without also taking the transpose).

• The swap operation, S\colon|i\rangle|j\rangle\mapsto|j\rangle|i\rangle, can be expressed as125 \begin{aligned} S &= d |\Omega\rangle\langle\Omega|^{T_{\mathcal{A}}} \\&= d \sum_{ij} \big(|i\rangle\langle j|\big)^T\otimes|i\rangle\langle j| \\&= d \sum_{ij} |j\rangle\langle i|\otimes|i\rangle\langle j|. \end{aligned} This implies that \operatorname{tr}[(A\otimes B)S] = \operatorname{tr}AB and that (A\otimes\mathbf{1})S = S(\mathbf{1}\otimes A).

### 7.13.17 Trace preserving and partial trace

Show that \mathcal{E} is trace preserving if and only if \operatorname{tr}_\mathcal{B}\widetilde{\mathcal{E}} = d\mathbf{1}. (Here the partial trace is over the second system in our definition \widetilde{\mathcal{E}} =\mathcal{I}\otimes\mathcal{E}|\Omega\rangle\langle\Omega|).

(Hint: show that \operatorname{tr}\mathcal{E}(|i\rangle\langle j|=\delta_{ij}.)

### 7.13.18 Rotating Kraus operators

Mathematically speaking, Kraus operators E_k are vectors in a dd'-dimensional Hilbert space, with the Hilbert–Schmidt inner product \operatorname{tr}E^\dagger_k E_l. So, our intuition tells us, in order to describe any quantum channel with \dim\mathcal{H}=\dim\mathcal{H}'=d, we should need no more than d^2 Kraus operators. We can pick an orthonormal basis of operators \{B_i\} and express each Kraus vector in this basis as E_k=\sum c_{ki} B_i (where i=1,\ldots,d^2 and k=1,\ldots,n, with n possibly much larger than d^2). This gives us \begin{aligned} \rho \longmapsto & \sum_{ij} B_i\rho B^\dagger_j \left(\sum _k c_{ki}c^\star_{kj}\right) \\=& \sum_{ij} B_i\rho B^\dagger_j C_{ij} \end{aligned} The matrix C_{ij} is positive semidefinite, and hence unitarily diagonalisable: C_{ij}=\sum_k U_{ik} d_k U^\dagger_{kj} for some unitary U and some d_k\geqslant 0. We can then unitarily “rotate” our operator basis and use the C_k=\sum_j U_{jk}B_j \sqrt{d_k} as our new Kraus operators.

### 7.13.19 No pancakes

Consider a single qubit operation which causes the z-component of the Bloch vector to shrink while preserving the values of the x- and y-components. Under such an operation, the Bloch sphere is mapped to an oblate spheroid which touches the Bloch sphere along its equator.

Explain why we cannot physically implement such a map.

1. We write X^{T_{\mathcal{A}}} to mean the partial transpose over \mathcal{A}, i.e. T\otimes\mathcal{I}.↩︎