## 7.13 *Remarks 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 **partial inner product with |x\rangle.**
It is first defined on the product vectors

For example, the partial inner product of

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

Consider a single-qubit channel induced by the action of the

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

We can calculate the Kraus operators

The unitary action of the gate when the state of the target qubit is fixed at

The evolution of the control qubit alone can be expressed in the Kraus form as

As we can see, the diagonal elements of **coherences**) disappear.
The two Kraus operators, *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:

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

These two apparently very different physical processes correspond to the same quantum channel:

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

### 7.13.5 Cooling down

We can show that the process of cooling a qubit to its ground state, described the map

### 7.13.6 Unchanged reduced density operator

Show that, for any operator

*(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

### 7.13.8 Pauli twirl

Show that randomly applying the Pauli operators

### 7.13.9 Depolarising channel

The most “popular” Pauli channel is the **depolarising channel**

(For

### 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

### 7.13.11 Complete positivity of a certain map

Let *positive*, and the range for which it is *completely positive*.

### 7.13.12 Toffoli gate

Consider the Toffoli gate

Express

### 7.13.13 Duals

We say that **dual** of a linear map

- Show that, if
\mathcal{E} is trace preserving, then\mathcal{E}^\star is unital. - 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

(For example, if we are interested in the component

### 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

To show this, we start with the spectral decomposition of

This observation leads to a way of relating *all* convex decompositions of a given density operator: let

### 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

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, ofU , i.e.*without*also taking the transpose).The swap operation,

S\colon|i\rangle|j\rangle\mapsto|j\rangle|i\rangle , can be expressed as^{125}\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

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

### 7.13.18 Rotating Kraus operators

Mathematically speaking, Kraus operators

### 7.13.19 No pancakes

Consider a single qubit operation which causes the

Explain why we cannot physically implement such a map.

We write

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