Chapter 8

Exercises

Exercise 8.1 (Unitary evolution as a quantum operation)

Pure states evolve under unitary transforms as \(\ket{\psi}\mapsto\,U\ket{\psi}\). Show that, equivalently, we may write \(\rho\mapsto\calE(\rho)\equiv\,U\rho\,U^\dagger\), for \(\rho=\proj{\psi}\).

Proof

Preparation

This is a really basic question and it should actually be obvious if you really understand the bra-ket notation (beware however that it might appear to be obvious if you merely feel familiar with it).

To give a meaningful solution beyond "it is obvious" we take a step back and recall what the bra-ket notation actually means. In quantum mechanics we always work on (complex) Hilbert spaces. A Hilbert space is a (complex) vector space \(\calH\) equipped with a sesquilinear form, which has to obey certain axioms to be mentioned soon. We denote it by

\[ (\ket{\psi}, \ket{\varphi}) . \]

A sesquilinear form is a mapping \(\calH\times\calH\to\CC\) which is linear in the second argument and conjugate-linear in the first argument. For a Hilbert space we require in addition that (positivity)

\[ \forall \ket{\psi} \in \calH\backslash\{0\}: \; (\ket{\psi}, \ket{\psi}) > 0 , \]

and that (symmetry)

\[ \forall \ket{\psi}, \ket{\varphi} \in \calH : \; (\ket{\psi},\ket{\varphi}) = \overline{(\ket{\varphi},\ket{\psi})} . \]

(The bar means complex conjugation.) Such a sesquilinear form is called a scalar product. This is what makes a complex Hilbert space, a complex vector space equipped with a scalar product. There are also real Hilbert spaces but we do not need them here. Note that we are always talking about finite dimensional vector spaces. In case of infinite dimensional vector spaces a completeness axiom is required in addition. (In the finite dimensional case this axiom follows automatically from the completeness axiom of real (and hence complex) numbers).

In the framework of Dirac's bra-ket notation the kets are just the elements of the Hilbert space. We already used this in the above introduction. The bras are elements of the dual space \(\calH^*\) which is just the space of linear maps \(\calH\to\CC\). For any \(\ket{\psi}\in\calH\) an example of such a map is given by

\[ \ket{\varphi} \mapsto (\ket{\psi}, \ket{\varphi}) . \]

In fact, one can prove that any element of \(\calH^*\) (that is, any linear map \(\calH\to\CC\)) can be written like this for a unique \(\ket{\psi}\). This is called the canonical isomorphism between \(\calH\) and \(\calH^*\) (it is called canonical since we could define it without knowing anything concrete about our Hilbert space, in other words it is definable in the most abstract terms). Note that the canonical isomorphism \(\iota\) is conjugate linear, that is, \(\iota(\lambda\ket{\psi})=\bra{\psi}\lambda^*\).

This canonical isomorphism actually motivates the bras. A bra \(\bra{\psi}\in\calH^*\) is the linear map given by

\[ \bra{\psi} : \; \ket{\varphi} \mapsto \sprod{\psi}{\varphi} := (\ket{\psi}, \ket{\varphi}) . \]

Having said that, terms like

\[ \bra{\psi} U \ket{\varphi} \]

are the same as

\[ \bra{\psi} U \ket{\varphi} = (\bra{\psi}\circ U) \ket{\varphi} = (\ket{\psi}, U\ket{\varphi}) , \]

where \(\circ\) denotes the composition of functions (\(U\) and \(\bra{\psi}\) as linear maps are both functions of course). And of course, the bra-ket notation also makes the above introduced notation for the scalar product obsolete since basically the bra and the ket are the left and right half of the scalar product.

The actual proof

Let us return to the actual exercise. We use the above notation \((\ldots,\ldots)\) for the scalar product to make the derivation clearer. Let \(U\) be a unitary operator on \(\calH\) and let

\[ \rho = \rho_{\ket{\psi}} = \proj{\psi} . \]

The task is to calculate \(\rho_{U\ket{\psi}}\) and to show that it is equal to \(\calE(\rho_{\ket{\psi}})\). Note that \(\rho\) is a linear map \(\calH\to\calH\) given by

\[ \rho \ket{\varphi} = \ket{\psi} \sprod{\psi}{\varphi} = (\ket{\psi}, \ket{\varphi}) \ket{\varphi} . \]

Hence

\begin{align*} \rho_{U\ket{\psi}} \ket{\varphi} &= (U\ket{\psi}, \ket{\varphi}) U\ket{\psi} \\ &= (\ket{\psi}, U^\dagger\ket{\varphi}) U\ket{\psi} \\ &= \bra{\psi} U^\dagger \ket{\varphi} U\ket{\psi} \\ &= U\ket{\psi} \bra{\psi} U^\dagger \ket{\varphi} \\ &= U\rho_{\ket{\psi}} U^\dagger \ket{\varphi} . \end{align*}

The first equality follows from the definition of \(\rho_{\ket{\psi}}\). The second one follows from the definition of the (hermitian) adjoint operator. The third one is the definition of the bras. The fourth equality just moves the complex number \(\bra{\psi}U\ket{\varphi}\) to the right. The last equation is again the definition of \(\rho_{\ket{\psi}}\).

Since the above eqation holds for all \(\varphi\) we have

\[ \rho_{U\ket{\psi}} = U\rho_{\ket{\psi}} U^\dagger . \]

QED.

Remark

There is another slightly deeper way to see this result. Let us denote the above mentioned canonical isomorphism by

\[ \iota(\ket{\psi}) = \bra{\psi} . \]

It is not hard to show and extension of the conjugate-linearity property of the canonical isomorphism:

\[ \iota(U\ket{\psi}) = \iota(\ket{\psi}) U^\dagger = \bra{\psi} U^\dagger . \]

In fact, the proof is even shorter than the proof of the claim of the exercise. That is, the canonical way to associate a linear mapping \(\calH^*\to\calH^*\) with \(U\) is the right composition with \(U^\dagger\). From this the claim of the exercise follows too.

Exercise 8.2 (Measurement as a quantum operation)

Recall from Section 2.2.3 (on page 84) that a quantum measurement with outcomes labeled by \(m\) is described by a set of measurement operators \(M_m\) such that \(\sum_mM_m^\dagger\,M_m=I\). Let the state of the system immediately before the measurement be \(\rho\). Show that for \(\calE_m(\rho)=M_m\rho\,M_m^\dagger\), the state of the system immediately after the measurement is

\[ \frac{\calE_m(\rho)}{\trace{\calE_m(\rho)}} . \]

Also show that the probability of obtaining this measurement result is \(p(m)=\trace{\calE_m(\rho)}\).

Solution

Clearly the exercise assumes that we only know the postulates of quantum mechanics for pure states as detailed in section 2.2.3 and not the generalization to mixed states as detailed in 2.4.1. Otherwise the exercise would be a trivial restatement of the evolution and measurement postulates.

Let us first assume that \(\rho=\proj{\psi}\) is a pure state (just expressed as a density matrix). In order to compute the trace of an operator \(\sigma\) we can take any orthonormal basis \(\{e\}\) and have \(\trace{\sigma}=\sum_{e}\bra{e}\sigma\ket{e}\). Let \(\sigma=\proj{\nu}\) with a not necessarily normalized non-zero \(\ket{\nu}\). Without loss of generality we may assume that \(\ket{\nu}/\norm{\nu}\) is part of our chosen ONB (this follows from the Gram-Schmidt procedure). Hence

\[ \trace{\sigma} = \sum_{e} \bra{e}\proj{\nu}\ket{e} = \sprod{\nu}{\nu} \sprod{\nu}{\nu} / \norm{\nu}^2 = \norm{\nu}^2 . \]

Given a state \(\ket{\psi}\) the post measurement state after measuring \(m\) is \(M_m\ket{\psi}/\norm{M_m\ket{\psi}}\). From the reasoning in exercise 1 (that \(U\) was a unitary operator didn't really matter there) we deduce that the corresponding density matrix is

\[ \frac{M_m \rho M_m^\dagger}{\norm{M_m\ket{\psi}}^2} . \]

Using the equation involving \(\sigma\) (with \(\ket{\nu}=M_m\ket{\psi}\)) we see that \(\norm{M_m\ket{\psi}}^2=\trace{M_m\rho\,M_m^\dagger}\). Hence the post measurement state is

\[ \frac{M_m \rho M_m^\dagger}{\trace{M_m\rho M_m^\dagger}} . \]

The probability to measure \(m\) (and to obtain this state) is

\[ p(m) = \norm{M_m\ket{\psi}}^2 = \trace{M_m \rho M_m^\dagger} . \]

This concludes the exercise for pure states. Let \(\rho\) be a mixed state now. This means that there is a decomposition

\[ \rho = \sum_j p_j \proj{\psi_j} \]

with strictly positive coefficients \(p_j\) such that \(\sum_jp_j=1\). The ensemble interpretation says that having such a mixed state means that, upon measurement, one of the pure states \(\ket{\psi_j}\) is taken at random with probability \(p_j\) and then the measurement proceeds as in the pure state case. It is not a-priori clear that this is well defined since the decomposition of \(\rho\) into pure states is not unique (c.f. Theorem 2.6 on page 103). But let us ignore this for the moment. We will see that the obtained probabilities and the post measurement states are independent of the concrete decomposition, hence the problem solves itself.

According to the ensemble interpretation the probability to measure \(m\) is

\[ p(m) = \sum_j p(m|j) \, p_j = \sum_j \trace{M_m\proj{\psi_j} M_m^\dagger} \, p_j = \trace{M_m \rho M_m^\dagger} . \]

Here

\[ p(m|j) = \norm{M_m \ket{\psi_j}}^2 = \trace{M_m \proj{\psi_j} M_m^\dagger} \]

is the conditioned probability to measure \(m\) if we already know that the pure state \(\psi_j\) was chosen. Suppose we know that we measured \(m\). What is the (mixed) state in that case? We already know that with (conditioned) probability \(p(j|m)\) the post measurement state is

\[ \frac{M_m\proj{\psi_j} M_m^\dagger}{p(m|j)} . \]

But this is an ensemble of pure states! Hence, using \(p(m|j)p_j=p(m\cap\,j)=p(j|m)p(m)\), the corresponding mixed state is

\[ \sum_j p(j|m) \frac{M_m\proj{\psi_j} M_m^\dagger}{p(m|j)} = \frac{1}{p(m)} \sum_j p_j M_m\proj{\psi_j} M_m^\dagger = \frac{M_m\rho M_m^\dagger}{p(m)} = \frac{M_m\rho M_m^\dagger}{\trace{M_m\rho M_m^\dagger}} . \]

This concludes the exercise. But let is briefly mention that in case we do not obtain the outcome of the measurement the post measurement state is

\[ \sum_{m,j} p(m|j) p_j \, \frac{M_m \proj{\psi_j} M_m^\dagger}{p(m|j)} = \sum_m M_m \rho M_m^\dagger . \]

Remark

Together with the conclusions from the actual exercise we can draw the following additional conclusion: If now we somehow obtain the measurement result than this sum "collapses" into one of its summands (just normalized), with probability \(\trace{M_m\rho\,M_m^\dagger}\) for each of them. That is, the measurement can be decomposed into two parts: First the state changes to

\[ \calE(\rho) = \sum_m \calE_m(\rho) \]

and then (if we obtain the measurement result) the state "collapses" into \(\calE_m(\rho)/\trace{\calE_m(\rho)}\) for one of the \(m\) with probability \(\trace{\calE_m(\rho)}\). So, the first part of the measurement can be expressed in terms of quantum operations. But not the second part (since it is not deterministic and cannot give rise to a mathematical function on density matrices).

Exercise 8.3

Our derivation of the operator-sum representation implicitly assumed that the input and output spaces for the operation were the same. Suppose a composite system AB initially in an unknown quantum state \(\rho\) is brought into contact with a composite system CD initially in some standard state \(\ket{0}\), and the two systems interact according to a unitary interaction \(U\). After the interaction we discard systems A and D, leaving a state \(\rho'\) of system BC. Show that the map \(\calE(\rho)=\rho'\) satisfies

\[ \calE(\rho) = \sum_k E_k \rho E_k^\dagger , \]

for some set of linear operators \(E_k\) from the state space of system AB to the state space of system BC, and such that \(\sum_kE_k^\dagger\,E_k\).

Proof

Basically we do the same as the book for the special case of equal input and output space. According to the exercise we have

\[ \calE(\rho) = \ptrace{AD}{U \rho \otimes \proj{0} U^\dagger} . \]

Let \(\{e_k\}\) be an orthonormal basis of AD. Hence

\[ \calE(\rho) = \sum_k \bra{e_k} U \rho \otimes \proj{0} U^\dagger \ket{e_k} . \]

Thus, defining \(E_k=\bra{e_k}U\ket{0}\), we have

\[ \calE(\rho) = \sum_k E_k \rho E_k^\dagger . \]

That is, \(\calE\) has the required form. We only have to show the completeness property \(J:=\sum_kE_k^\dagger\,E_k=I_{AB}\). Note that \(J=J^\dagger\) is self-adjoint and positive-definite. Let \(\rho\) be arbitrary and consider

\[ 1 = \ptrace{BC}{\calE(\rho)} = \sum_k \ptrace{BC}{E_k\rho E_k^\dagger} = \sum_k \ptrace{AB}{E_k^\dagger E_k \rho} = \ptrace{AB}{J^\dagger \rho} . \]

Here we used the well-known cyclicity property of the trace (which also holds for (matching) rectangular matrices). Thus

\[ \forall \rho : \; \trace{(J-I)^\dagger \rho} = 0 . \]

Here the quantification runs over all density operators (hermitian, positive definite, trace 1).

Remark: In case you wonder why I take the adjoint of \(J\). We have that \(S,T\mapsto\trace{S^\dagger\,T}\) turns the matrix space \(\CC^{n\times\?n}\) into a complex Hilbert space. See e.g. the definition of a Hilbert-Schmidt operator on wikipedia (it is about infinite dimensional spaces but just note that in finite dimensions every operator is a Hilbert-Schmidt operator). For this exercise it does not directly help us since the quantification above is not over all \(\rho\in\CC^{n\times\?n}\). But still nice to know 😉.

Considering all rank-1 density matrices \(\rho=\proj{\psi}\) we deduce

\[ \forall \ket{\psi} : \; \trace{(J-I)^\dagger \proj{\psi}} = \bra{\psi} (J-I)^\dagger \ket{\psi} = 0 . \]

(The first equality can either be seen by the cyclicity property of the trace or by using an ONB which involves \(\psi\) (but normalized) as one of its elements) But this can only be true (for all normalized \(\ket{\psi}\)) if \(J=I\). QED.

Exercise 8.4 (Measurement)

Suppose we have a single qubit principal system, interacting with a single qubit environment through the transform

\[ U = P_0 \otimes I + P_1 \otimes X \]

where \(X\) is the usual Pauli matrix (acting on the environment), and \(P_0\equiv\proj{0}\), \(P_1\equiv\proj{1}\) are projectors (acting on the system). Give the quantum operation for this process, in the operator-sum representation, assuming the environment starts in the state \(\ket{0}\).

Solution

Note that \(U\) is nothing else than the CNOT gate. Let us label the environment by E. One way to define the operation elements is

\begin{align*} E_0 &= \bra{0_E} U \ket{0_E} = P_0 \sprod{0_E}{0_E} + P_1 \bra{0_E}X\ket{0_E} = P_0 , \\ E_1 &= \bra{1_E} U \ket{0_E} = P_0 \sprod{1_E}{0_E} + P_1 \bra{1_E}X\ket{0_E} = P_1 . \end{align*}

(The \(\ket{0_E}\) to the right of \(U\) comes from the initial state of the environment.) Hence

\[ \calE(\rho) = P_0 \rho P_0 + P_1 \rho P_1 = \begin{bmatrix} \rho_{00} & 0 \\ 0 & \rho_{11} \end{bmatrix} . \]

That is, this operation acts like the first part of a measurement. It describes the state after a measurement in the Z-basis if nobody tells us the measurement outcome. Once we get the outcome the state "collapses" to one of the two summands (properly normalized of course). The respective probabilities are \(\rho_{00}\) (for outcome \(0\)) and \(\rho_{11}\) (for outcome \(1\)).

Exercise 8.5 (Spin flips)

Just as in the previous exercise, but now let

\[ U = \frac{1}{\sqrt{2}} \left( X \otimes I + Y \otimes X \right) . \]

Give the quantum operation for this process, in the operator-sum representation.

Solution

First of all let us make sure that \(U\) really is unitary. We have \(U^\dagger=U\), hence

\[ U^\dagger U = U^2 = \frac{1}{2} \left( X^2 \otimes I + Y^2 \otimes X^2 + (XY+YX)\otimes X \right) = I . \]

Hence \(U\) is indeed unitary. One way to define the operation elements is as follows:

\begin{align*} E_0 &= \bra{0_E} U \ket{0_E} = X \sprod{0_E}{0_E} + Y \bra{0_E}X\ket{0_E} = X , \\ E_1 &= \bra{1_E} U \ket{0_E} = X \sprod{1_E}{0_E} + Y \bra{1_E}X\ket{0_E} = Y . \end{align*}

Hence

\[ \calE(\rho) = \frac{1}{2} \left(X \rho X + Y \rho Y \right) = \begin{bmatrix} \rho_{11} & 0 \\ 0 & \rho_{00} \end{bmatrix} . \]

Interpreting this in the same way as in exercise 8.4 we deduce that \(\calE\) corresponds to a spin-flip followed by a measurement (without obtaining the measurement outcome).

Exercise 8.6 (Composition of quantum operations)

Suppose \(\calE\) and \(\calF\) are quantum operations on the same quantum system. Show that the composition \(\calF\circ\calE\) is a quantum operation, in the sense that it has an operator-sum representation. State and prove an extension of this result to the case where \(\calE\) and \(\calF\) do not necessarily have the same input and output spaces.

Proof

Suppose

\begin{align*} \calE(\rho) &= \sum_i E_i \rho E_i^\dagger , \\ \calF(\rho) &= \sum_j F_j \rho F_j^\dagger . \end{align*}

Then

\begin{align*} \calF \circ \calE(\rho) = \sum_{ij} F_j E_i \rho E_i^\dagger F_j^\dagger . \end{align*}

This suggests to use all the \(F_jE_i\) as operation elements of the composition. That \(\calF\circ\calE\) is indeed a quantum operation now follows from

\[ \sum_{ij} E_i^\dagger F_j^\dagger F_j E_i = \sum_i E_i^\dagger E_i = I . \]

The proof directly generalizes (without any change) to the case where \(\calE:A\to\?B\) and \(\calF:B\to\?C\). QED.

Exercise 8.7

Suppose that instead of doing a projective measurement on the combined principal system and environment we had performed a general measurement described by measurement operators \(\{M_m\}\). Find operator-sum representations for the corresponding quantum operations \(\{\calE_m\}\) on the principal system, and show that the respective measurement probabilities are \(\trace{\calE_m(\rho)}\).

Solution

Let \(\sigma=\sum_jq_j\proj{j}\) be the state of the environment E. The generalized operations look as follows

\[ \calE_m(\rho) = \ptrace{E}{M_m U \rho\otimes\sigma U^\dagger M_m^\dagger} . \]

According to the posutulates the probability to measure \(m\), after an evolution with \(U\), is

\[ p(m) = \trace{M_m U \rho\otimes\sigma U^\dagger M_m^\dagger} = \trace{\calE_m(\rho)} . \]

Here we use the property \(\trace{\ptrace{E}{A}}=\trace{A}\) of the partial trace. The post-measurement state for the whole system is is

\[ \frac{M_m U \rho\otimes\sigma U^\dagger M_m^\dagger}{p(m)} , \]

and for the principal system it is

\[ \ptrace{E}{\frac{M_m U \rho\otimes\sigma U^\dagger M_m^\dagger}{p(m)}} = \frac{\calE_m(\rho)}{\trace{\calE_m(\rho)}} . \]

It is not hard to see that for any choice of an ONB \(\{e_k\}\) on E the operators

\[ E_{jk} = \sqrt{q_j} \bra{e_k} M_m U \ket{j} \]

are elements for the operation \(\calE_m\), which also satisfy \(\sum\?E_{jk}^\dagger\?E_{jk}\leq\?I\):

\[ \sum_{jk} E_{jk}^\dagger E_{jk} = \sum_{jk} q_j \bra{j}U^\dagger M_m^\dagger \ket{e_k} \bra{e_k} M_m U \ket{j} = \sum_j q_j \bra{j}U^\dagger M_m^\dagger M_m U \ket{j} \leq \sum_j q_j \bra{j}U^\dagger U \ket{j} = \sum_j q_j = I . \]

Exercise 8.8 (Non-trace-preserving quantum operations)

Explain how to construct a unitary operator for a system–environment model of a non-trace-preserving quantum operation, by introducing an extra operator, \(E_\infty\), into the set of operation elements \(E_k\), chosen so that when summing over the complete set of \(k\), including \(k=\infty\), one obtains \(\sum_kE_k^\dagger\?E_k=I\).

Solution

Let us define

\[ E_\infty = \sqrt{I - \sum_k E_k^\dagger E_k} . \]

Clearly \(E_\infty^\dagger=E_\infty\) and

\[ \sum_k' E_k^\dagger E_k = I . \]

Here and in the following we let \(\sum_k'\) denote the sum over all \(k\) including \(k=\infty\). On the other hand \(\sum_k\) excludes \(k=\infty\).

As in the book consider an environment described by a Hilbert space which has a basis \(\{\ket{e_k}\}\) (including \(k=\infty\)). Define \(U\) by

\[ U \ket{\psi} \ket{e_0} \equiv \sum_k' E_k \ket{\psi} \ket{e_k} \]

and unitary extension to the full system - exactly as in the book. Consider the following projection defined on E:

\[ P = I_E - \proj{e_\infty} . \]

We shall prove that

\[ \calE(\rho) = \ptrace{E}{PU \rho\otimes \proj{e_0} U^\dagger P} . \]

In fact, the RHS equals

\[ \ptrace{E}{\sum_{kl}' E_k \rho E_l^\dagger \otimes P \ket{e_k}\bra{e_l} P} = \ptrace{E}{\sum_{kl} E_k \rho E_l^\dagger \otimes \ket{e_k}\bra{e_l} } = \sum_{k} E_k \rho E_k^\dagger = \calE(\rho) \]

as desired.

Exercise 8.9 (Measurement model)

If we are given a set of quantum operations \(\{\calE_m\}\) such that \(\sum_m\calE_m\) is trace-preserving, then it is possible to construct a measurement model giving rise to this set of quantum operations. For each \(m\), let \(E_{mk}\) be a set of operation elements for \(\calE_m\). Introduce an environmental system, E, with an orthonormal basis \(\ket{m,k}\) in one-to-one correspondence with the set of indices for the operation elements. Analogously to the earlier construction, define an operator \(U\) such that

\[ U \ket{\psi} \ket{e_0} = \sum_{mk} E_{mk} \ket{\psi} \ket{m,k} . \]

Next, deﬁne projectors \(P_m=\sum_k\proj{m,k}\) on the environmental system, E. Show that performing \(U\) on \(\rho\otimes\proj{e_0}\), then measuring \(P_m\) gives \(m\) with probability \(\trace{\calE_m(\rho)}\), and the corresponding post-measurement state of the principal system is \(\calE_m(\rho)/\trace{\calE_m(\rho)}\).

Solution

That \(U\) can be extended to a unitary operator can be seen as in the other cases of constructing an environment model. We have

\begin{align*} \ptrace{E}{P_m U \rho \otimes \proj{e_0} U^\dagger P_m} &= \ptrace{E}{\sum_{m,k,l} E_{mk}\rho E_{ml}^\dagger \otimes \ket{m,k}\bra{m,l}} \\ &= \sum_{m,k} E_{mk}\rho E_{mk}^\dagger \\ &= \calE_m(\rho) . \end{align*}

Hence, using \(\trace{\ptrace{E}{\ldots}}=\trace{\ldots}\), the probability to measure \(m\) after applying \(U\) to \(\rho\otimes\proj{e_0}\) is

\[ p(m) = \trace{P_m U \rho \otimes \proj{e_0} U^\dagger P_m} = \trace{\calE_m(\rho)} . \]

The post-measurement state of the whole system is

\[ \frac{P_m U \rho \otimes \proj{e_0} U^\dagger P_m}{p(m)} . \]

Taking the partial trace of the environment yields the post-measurement state of the principal system

\[ \frac{\ptrace{E}{P_m U \rho \otimes \proj{e_0} U^\dagger P_m}}{p(m)} = \frac{\calE_m(\rho)}{\trace{\calE_m(\rho)}} . \]

Exercise 8.10

Give a proof of Theorem 8.3 based on the freedom in the operator-sum representation, as follows. Let \(\{E_j\}\) be a set of operation elements for \(\calE\). Define a matrix \(W_{jk}=\trace{E_j^\dagger\?E_k}\). Show that the matrix \(W\) is Hermitian and of rank at most \(d^2\), and thus there is unitary matrix \(u\) such that \(uWu^\dagger\) is diagonal with at most \(d^2\) non-zero entries. Use \(u\) to deﬁne a new set of at most \(d^2\) non-zero operation elements \(\{F_j\}\) for \(\calE\).

Proof 1

The statement is a direct consequence of the proof of Kraus's Theorem as given in the book (Theorem 8.1). Recall that the diagonalization of \(\sigma\)

\[ \sigma = \sum_{k=1}^M \proj{s_k} \]

is used to define the operation elements by

\[ E_k \ket{\psi} = \braket{\tilde{\psi}}{s_k} \]

That is, there are precisely \(M\) elements to be defined. How large can \(M\) be? Recall that \(\sigma\) is a density matrix on the Hilbert space \(\calH_R\otimes\calH_Q\) of the joint system \(RQ\). By assumption \(d=\dim(\calH_Q)\) and by construction \(\dim(\calH_R)=\dim(\calH_Q)=d\). Hence

\[ \dim(\calH_R\otimes\calH_Q) = d^2 . \]

Hence \(\sigma\in\CC^{d^2\times\?d^2}\) and therefore \(M\leq\?d^2\). QED.

Proof 2

The matrix \(W\) is hermitian because

\[ W_{kj}^* = \trace{E_k^\dagger E_j}^* = \trace{(E_k^\dagger E_j)^\dagger} = \trace{E_j^\dagger E_k} = W_{jk} . \]

Let \(\calH\) be the \(d\)-dimensional Hilbert space on which the \(E_k\) operate. Recall that

\[ A, B \mapsto \trace{A^\dagger B} \]

is a scalar product on the \(d^2\)-dimensional space of linear operators on \(\calH\) (see Hilbert-Schmidt space). It turns it into a Hilbert space. The claim that \(\rank{W}\leq\?d^2\) thus follows from the following lemma (with \(N=d^2\)).

Lemma

Let \(\{v_j\}_{j=1..M}\) be a bunch of vectors of an \(N\)-dimensional complex Hilbert space \(V\) and define

\[ W_{jk} = \braket{v_j}{v_k} . \]

Then \(\rank{W}\leq\?N\).

Proof

Let \(\{b_n\}_{n=1..N}\) be an ONB of \(V\). Hence there are unique numbers \(A_{jn}\in\CC\) such that

\[ v_j = \sum_{n=1}^N A_{jn} b_n . \]

Interpreted as a matrix we have \(A\in\CC^{M\times\?N}\), hence (the rank is always at most the minimum of the two dimensions)

\[ \rank{A} \leq \min(M, N) \leq N . \]

Consider

\[ W_{jk} = \braket{v_j}{v_k} = \sum_{nm} A_{jn}^* A_{km} \braket{b_n}{b_m} = \sum_n A_{jn}^* A_{kn} = (AA^\dagger)_{kj} \]

Hence

\[ W = \left(AA^\dagger \right)^\top \]

Recall that transposition and conjugation does not change the rank of a matrix. Moreover \(\rank{AB}\leq\min(\rank{A},\rank{B})\) for any two matrices \(A,B\). The last thing follows easily from the characterization of the rank as the dimension of the image of a linear mapping. Hence

\[ \rank{W} \leq \rank{A} \leq N . \]

QED.

Let \(M\) be the number of elements. Hence \(W\in\CC^{M\times\?M}\). Since \(W\) is hermitian there is a uniatry matrix \(u\in\CC^{M\times\?M}\) such that

\[ D = u W u^\dagger \]

is diagonal. Observe that

\[ D_{jk} = \sum_{mn} u_{jm} W_{mn} u_{kn}^* = \trace{\left(\sum_m u_{jm}^* E_m \right)^\dagger \left( \sum_n u_{kn}^* E_n \right)} . \]

Hence defining \(F_k=\sum_nu_{kn}^*E_n\) we have

\[ D_{jk} = \trace{F_j^\dagger F_k} . \]

We note two things:

Since \(D\) is diagonal the \(F_j\) are actual orthogonal with respect to the Hilbert-Schmidt scalar product.
Since \(\rank{D}=\rank{W}\leq\?d^2\) only at most \(d^2\) of the \(F_j\) are non-zero (\(\trace{F^\dagger\?F}=0\) implies \(F=0\)).

This concludes the proof. QED.

Exercise 8.11

Suppose \(\calE\) is a quantum operation mapping a \(d\)-dimensional input space to a \(d'\)-dimensional output space. Show that \(\calE\) can be described using a set of at most \(dd'\) operation elements \(\{E_k\}\).

Proof

With a minor adaption we can reduce this to one of the proofs of exercise 8.10. Let

\[ d_1 = \max(d, d') . \]

Then \(E_k\in\CC^{d\times\?d'}\). By padding with zeros we can interpret the \(E_k\) as elements of a \(dd'\)-dimensional subspace \(V\) of \(\CC^{d_1\times\?d_1}\). Note that this does not affect the definition of \(W\).

Now the proof of the special case \(d=d'\) goes through as is. The crucial thing to note is that the lemma has to be applied with \(N=dd'\) and the just defined Hilbert space \(V\). QED.

WIP Exercise 8.12

Why can we assume that \(O\) has determinant \(1\) in the decomposition (8.93)?

Remark: In a sense my solution below solves the exercise. But I have the vague feeling that it is not meant like that. Playing around with examples I have the impression that \(\det(M)\geq0\) just holds always. At the moment I can only prove this for a special case (see exercise 8.16). I do not even know if it is hard to prove. I have the feeling that it should not be so hard, but I still do not see it.

Maybe not a solution

As a real orthogonal matrix \(O\) has \(\det(O)=\pm1\). To see this recall that the determinant is the product of all eigenvalues which are complex number of modulus 1 for an orthogonal matrix (which is unitary if interpreted as a complex matrix). But the determinant of a real matrix is real, hence only \(\pm1\) is possible for their product.

If the determinant is already \(1\) we are done. Otherwise just replace \(O\) by \(-O\) and \(S\) by \(-S\). Of course this we destroy the property that \(S\) is positive.

Exercise 8.13

Show that unitary transformations correspond to rotations of the Bloch sphere.

Solution

Probably the idea is to prove it via the findings of the section (probably using the formulas for \(M_{jk}\) and \(c_k\)). Why else would this problem be in this chapter? But I have not succeeded in doing so.

On the other hand we already know that claim and used it in earlier chapters. Therefore I only sketch the proof. Essentially the problem is equivalent to exercise 4.6. To see this recall that every unitary operator is equal to some rotation \(R_{\hat{n}}(\theta)\) around some axis and some angle - up to a global phase. I have also made a small discussion for the (easy) step to generalize the statement of the exercise to density matrix in chapter 4.

Exercise 8.14

Show that \(\det(S)\) (From the polar decomposition of \(M=OS\)) need not be positive.

Remark

I assume that positive means strictly greater than zero here. Sometimes zero is included for positivity so it is always a bit unclear what is meant. There are two reasons to see it this way:

The polar decomposition explicitly states that there is an orthogonal matrix \(O\) and a positive matrix \(S\) (Moreover \(S\) is unique and \(O\) is unique on the support of \(S\)). Coincidentally here we already have a case were positive actually means that \(0\) is included, because it is defined as

\[ \forall\ket{\psi}\neq0: \; \bra{\psi} S \ket{\psi} \geq 0 . \]

In particular \(S=0\), or more generally that the kernel of \(S\) is non-trivial, is a possibility.

But positivity implies \(\det(S)\geq0\). Hence \(0\) can be the only violation of being "positive". Hence the strict interpretation of "positive" in this exercise. A loop hole would be that the exercise is somehow meant in some other way going away from the polar decomposition.
The second reason is that I believe that \(\det(M)\geq0\) always holds. See also my remark in exercise 8.12. If this is true I do not see any way to interpret the exercise in a way such that not positive means strictly negative.

Solution

For the bit-flip (\(E_0=\sqrt{p}I\), \(E_1=\sqrt{1-p}X\)) we have

\[ M = \diag(1, 2p-1, 2p-1) \]

(and \(c=0\)). See chapter 8.3.3 or look into my sage code. Clearly for \(p=1/2\) we have \(\det(M)=0\). In that case \(S=\sqrt{M^\dagger\?M}=M\) (and you could choose \(O=I\)). Hence also \(\det(S)=0\).

Exercise 8.15

Suppose a projective measurement is performed on a single qubit in the basis \(\ket{+}\), \(\ket{-}\), where \(\ket{\pm}=(\ket{0}\pm\ket{1})/\sqrt{2}\). In the event that we are ignorant of the result of the measurement, the density matrix evolves according to the equation

\[ \rho \mapsto \calE(\rho) = \proj{+}\rho\proj{+} + \proj{-}\rho\proj{-} . \]

Illustrate this transformation on the Bloch sphere.

Solution

We reduce this to a known problem by transforming the operation into a phase-flip. This has the advantage that we already know what it does in the Bloch-sphere, and even if not, it is an easier problem since we can work in the standard basis (and know for example how the matrices of the Pauli matrices look like in that basis).

Let \(H=(X+Z)/\sqrt{2}\) be the Hadamard transform (note that \(H^\dagger=H\) and \(H^2=1\)). Recall that it swaps \(X\) and \(Z\) as well their eigenspace projectors (\(\ket{\pm}\) are the eigenvectors of \(X\)). Hence

\[ \calE'(\rho) = H \calE(H\rho H) H = \proj{0}\rho\proj{0} + \proj{1}\rho\proj{1} . \]

Note that \(\calE'\) is just the phase flip and we already know what it is. In the Bloch sphere it corresponds to a projection onto the z-axis (see chapter 8.3.3):

\[ M' = \diag(0, 0, 1), \; c' = 0 . \]

In the Bloch-sphere interpretation of unitary operator \(H\) acts as a rotation around \((\hat{x}+\hat{z})/\sqrt{2}\) by 180°. It swaps \(\hat{x}\) and \(\hat{z}\) and negates \(\hat{y}\), that is, as a rotation in the Bloch sphere \(\rho\mapsto\?H\rho\?H\) is given by this matrix:

\[ T = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix} . \]

Hence the following describes \(\calE\):

\[ M = TM'T = \diag(1, 0, 0), \; c = Tc' = 0 . \]

Exercise 8.16

The graphical method for understanding single qubit quantum operations was derived for trace-preserving quantum operations. Find an explicit example of a non-trace-preserving quantum operation which cannot be described as a deformation of the Bloch sphere, followed by a rotation and a displacement.

Solution

Before delivering an example I find it instructive to discover what violations are even possible from dropping the trace preservation property. Hence the solution is divided into two parts.

What does still hold?

Let \(E\) be an arbitrary linear operator on \(\CC^2\) and let us consider the following quantum operation

\[ \calE(\rho) = E \rho E^\dagger . \]

Note that physicality (\(\trace{\calE(\rho)}\leq1\)) might be violated for some \(\rho\) but for this exercise physicality is not really relevant. It could always be restored by multiplication with a small \(\varepsilon\?>\?0\).

Note that from Kraus' theorem we know that an arbitrary single-qubit operation needs up to four operation elements (c.f. exercise 8.10). We try to generalize some of our findings at the end. But for now let us only consider the one-element case.

Normally \(\calE\) only acts on density operators, that is, operators of the form

\[ \rho = 1 + x X + y Y + z Z , \]

where \(x,y,z\in\RR\). For notational convenience we left out the factor \(1/2\). It does not really matter for our purposes. Note that the space of such matrices is a three-dimensional real-affine subspace of \(\CC^{2\times2}\equiv\CC^4\). We already know from the book (see also my calculations above) that trace-preserving operations map this space into an real-affine subspace of \(\CC^{2\times2}\). Fixing \(I\) as the origin of these spaces the resulting mapping is real-affine-linear (not sure how to say it correctly).

It simplifies matters to consider the larger (four-dimensional) real linear (without the affine) subspace \(S\) consisting of the following matrices:

\[ \rho = e + x X + y Y + z Z , \]

where \(e,x,y,z\in\RR\). Note that any matrix in \(\CC^{2\times2}\) can be written like that but with complex coefficients (indeed: \(I,X,Y,Z\) is an ONB with respect to the scalar product \(A,B\mapsto\trace{A^\dagger\?B}\)). In particular \(E\) can be written as

\[ E = sI + uX + vY + wZ \]

for \(s,u,v,w\in\CC\). Let us prove the following Lemma.

Lemma

Any quantum operation (trace-preserving or not) acts as a real-linear map from \(S\) to itself.

Proof

Let us conveniently write \(\rho=(e,x,y,z)\) in the basis \((I,X,Y,Z)\). We have

\[ X\rho Y = (-\ii z, y, x, \ii e) \text{ and } Y\rho X = (\ii z, y, x, -\ii e) . \]

Certainly you can see the pattern here and generalize it to the other two Pauli-operator pairs. Moreover

\[ X\rho X = (e, x, -y, -z) , \]

and analogously for the other two Pauli-operators. Hence

\[ \rho\mapsto\?uX\rho\?u^*X = (\chi e, \chi x, -\chi y, -\chi z) \]

where \(\chi=\abs{u}^2\). Hence this type of transformation is real-linear. On the other hand it is easy to see that

\[ uX \rho v^*Y + vY \rho u^* X = (-\beta z, \alpha y, \alpha x, \beta e) \]

where \(\alpha=2\Re(u^*v)\) and \(\beta=\ii\?uv^*-\ii\?u^*v=2\Im(u^*v)\). Hence this transformation is also real-linear. Observe that

\[ E \rho E^\dagger \]

can be decomposed into a sum of transformations similar to the proceeding ones. Hence it is real linear too. Since a quantum operation is a sum of such one-element operations it follows that any quantum operation is real linear. QED.

Clearly a map is trace-preserving if it maps the affine subspace described by \(e=1\) into itself (works for any affine subspace defined by setting \(e\) to a non-zero constant). Hence, we derived in a slightly different way the conclusion that trace-preserving quantum operations correspond to mappings within the Bloch-space.

It would be interesting to characterize what maps are possible. I believe they are characterized by \(\det(M)\geq0\) (where \(M:\RR^4\to\RR^4\) for the non-trace-preserving maps). At least for single-element operations we can use sage to verify that the determinant is always non-negative:

E = s*Id + u*X + v*Y + w*Z
M = make_qop1d_matrix_4d([E])
factor(det(M))

(s^2 - u^2 - v^2 - w^2)^2*(conjugate(s)^2 - conjugate(u)^2 - conjugate(v)^2 - conjugate(w)^2)^2

So for \(\calE(\rho)=E\rho\?E^\dagger\) with an arbitrary \(E=s+uX+vY+wZ\) we have

\[ \det(M) = \abs{s^2-u^2-v^2-w^2}^4 . \]

In principle the most general operation can be constructed from four single operation elements (see exercise 8.10). However the symbolic determinant as computed by sage is too complicated to see a pattern (or to use factor).

Example

We have seen in the first part of the solution that non-trace-preserving operation can at least be interpreted in \(\RR^4\). I also conjectured that this representation has non-negative determinant. For trace preserving operations this would imply that the determinant with respect to \(\RR^3\) is also positive.

The following is an example for a non-trace-preserving operation with negative determinant with respect to \(\RR^3\):

\[ E_0 = I, \quad E_1 = X + \ii Y . \]

Indeed, for \(\rho=(e,x,y,z)\) with respect to the Pauli basis \((I,X,Y,Z)\) (recall \(e=1\) for density matrices) we have

\[ \calE(\rho) = (3e-2z,x,y,2e-z) . \]

The following sage code "proves" it:

E0 = Id
E1 = X + i*Y
rho = e*Id + x*X + y*Y + z*Z

qop1d([E0, E1], rho)

(3*e - 2*z, x, y, 2*e - z)

If we project this into the three dimensional Bloch-space the corresponding matrix is \(\diag(1,1,-1)\), which has a negative determinant. Hence it cannot be represented as a deformation of the Bloch sphere, followed by a rotation and a displacement. Otherwise the determinant would be non-negative (as is the case in \(\RR^4\)).

Exercise 8.17

Verify (8.101) as follows. Define

\[ \calE(A) = \frac{A + XAX + YAY + ZAZ}{4} , \]

and show that

\[ \calE(I) = I; \quad \calE(X) = \calE(Y) = \calE(Z) = 0 . \]

Now use the Bloch sphere representation for single qubit density matrices to verify (8.101).

Solution

This is almost to simple to justify but let me try my best. The claim \(\calE(I)=I\) follows from \(X^2=Y^2=Z^2=I\). That \(\calE(N)=0\) for any Pauli matrix \(N\) follows from the fact that \(NMN=-M\) if \(M\) and \(N\) are different Pauli matrices and \(NMN=N\) otherwise.

Finally, that \(\calE(\rho)=I/2\) for any density matrix follows from linearity of \(\calE\) and the fact that any density matrix can be written as

\[ \rho = \frac{1}{2} (I + x X + y Y + z Z) \]

for \(x,y,z\in\RR\) (see exercise 2.72).

Exercise 8.18

For \(k\geq1\) show that \(\trace{\rho^k}\) is never increased by the action of the depolarizing channel.

Proof

Recall that in a suitable orthonormal basis every density matrix is represented by a diagonal matrix

\[ \diag(q_1,\ldots,q_d) \]

with \(\sum_jq_j=1\) and \(q_j\geq0\). Hence

\[ \trace{\rho^k} = f_k(q) \]

where

\[ f_k(q) = \sum_{j=1}^d q_j^k . \]

Observe that \(f_k\) is a strictly convex function, that is

\[ f_k(\lambda q + (1-\lambda) q') < \lambda f_k(q) + (1-\lambda) f_k(q') \]

for all \(\lambda\in(0,1)\) and \(q\neq\?q'\) (and \(\sum_jq_j=1\) and \(q_j\geq0\), and similarly for \(q'\)). This follows from the convexity of the scalar valued function \(x\mapsto\?x^k\), which in turn follows from the positivity of the second derivatives (on the domain of definition). By the method of Lagrange multipliers we easily see that

\[ q^\min = (1/d,\ldots,1/d) \]

is the unique minimum of \(f_k\). Now we are prepared to prove the claim. In fact:

\[ \trace{\calE(\rho)} = f_k(pq^\min + (1-p)q) \leq p f_k(q^\min) + (1-p) f_k(q) \leq f_k(q) = \trace{\rho} . \]

QED.

Exercise 8.19

Find an operator-sum representation for a generalized depolarizing channel acting on a \(d\)-dimensional Hilbert space.

Solution

We will define \(d^2\) operation elements

\[ E_{mn} = \frac{1}{d} P_m Q_n \text{ for } m,n \in \{0,1,\ldots,d-1\} \]

which implement the operation \(\calE(\rho)=I/d\). Hence by scaling with \(\sqrt{p}\) and adding an additional element \(\sqrt{1-p}I\) we can implement the depolarizing channel. In the following let \(\rho=\sum_{ij}\rho_{ij}\ket{i}\bra{j}\). For \(n\in\{0,\ldots,d-1\}\) define

\[ Q_n \ket{j} = e^{2\pi\ii nj/d} \ket{j} . \]

Observe that

\[ \sum_n Q_n \ket{j}\bra{k} Q_n^\dagger = d \delta_{jk} . \]

Hence

\[ \rho' = \frac{1}{d} \, \sum_n Q_n \rho Q_n^\dagger = \sum_j \rho_{jj} \proj{j} . \]

Next, for \(n\in\{0,\ldots,d-1\}\) define

\[ P_n\ket{j} = \sum_j \ket{n+j} , \]

where the sum \(n+j\) is interpreted modulo \(d\). Hence the \(P_n\) are cyclic permutations.

\[ \frac{1}{d} \sum_n P_n \rho' P_n^\dagger = \frac{1}{d} \sum_{jn} \rho_{jj} \ket{n+j}\braket{j}{j}\bra{n+j} = \frac{1}{d} I . \]

QED.

Exercise 8.20 (Circuit model for amplitude damping)

Show that the circuit in Figure 8.13 models the amplitude damping quantum operation, with \(\sin(\theta/2)^2=\gamma\).

                 ┌───┐
rho_in: ────■────┤ X ├
        ┌───┴───┐└─┬─┘
   |0>: ┤ Ry(θ) ├──■──
        └───────┘

Solution

Let \(B'\) be the action of the circuit:

\begin{align*} B' &= (I \otimes P_0 + X \otimes P_1) \cdot (P_0 \otimes I + P_1 \otimes R_y(\theta)) \\ &= P_0 \otimes P_0 + P_1 \otimes P_0 R_y(\theta) + XP_0 \otimes P_1 + XP_1 \otimes P_1 R_y(\theta) . \end{align*}

Let \(\ket{\psi}=a\ket{0}+b\ket{1}\). Then

\[ B' \ket{\psi}\ket{0} = a \ket{00} + bc\ket{10} + 0 + bs\ket{01} = a\ket{00} + b(c\ket{10} + s\ket{01}) , \]

where \(c=\cos(\theta/2)\) and \(s=\sin(\theta/2)\). Thus

\begin{align*} E_0 \ket{\psi} = \bra{0_E} B' \ket{\psi}\ket{0_E} &= a\ket{0} + bc\ket{1} \\ E_1 \ket{\psi} = \bra{1_E} B' \ket{\psi}\ket{0_E} &= bs\ket{0} \\ \end{align*}

In other words, using \(s=\sqrt{\gamma}\) and \(c=\sqrt{1-\gamma}\), we have

\begin{align*} E_0 &= \begin{bmatrix} 1 & 0 \\ 0 & \sqrt{1-\gamma} \end{bmatrix} , \\ E_1 &= \begin{bmatrix} 0 & \sqrt{\gamma} \\ 0 & 0 \end{bmatrix} . \end{align*}

This is as desired.

Exercise 8.21 (Amplitude damping of a harmonic oscillator)

Suppose that our principal system, a harmonic oscillator, interacts with an environment, modeled as another harmonic oscillator, through the Hamiltonian

\[ H = \chi (a^\dagger b + a b^\dagger) \]

where \(a\) and \(b\) are the annihilation operators for the respective harmonic oscillators, as deﬁned in Section 7.3.

Using \(U=\exp(-\ii\?H\tau)\), denoting the eigenstates of \(b^\dagger\?b\) as \(\ket{k_b}\), and selecting the vacuum state \(\ket{0_b}\) as the initial state of the environment, show that the operation elements \(E_k=\bra{k_b}U\ket{0_b}\) are found to be

\[ E_k = \sum_{n=k}^\infty \sqrt{\binom{n}{k}} \sqrt{(1-\gamma)^{n-k}\gamma^k} \ket{n-k}\ket{n} , \]

where \(\gamma=1-\cos(\chi\tau)^2\) is the probability of loosing a single quantum of energy, and states such as \(\ket{n}\) are eigenstates of \(a^\dagger\?a\).
Show that the operation elements \(E_k\) define a trace-preserving quantum operation.
I am not really sure why it is this formula for \(H\). From the text I would have expected \(H=\chi(a^\dagger\?b-ab^\dagger)\). But it doesn't matter for this exercise, both work.

Proof of part 1

Recall a lemma related to the Baker-Campbell-Hausdorff formula and Lie-Algebras:

\[ e^{\lambda G} A e^{-\lambda G} = \sum_n \frac{\lambda^n}{n!} [(G)^n, A] , \]

where \([(G)^n,A]\) is the \(n\)-fold commutator. We will apply it to \(G=a^\dagger\?b+ab^\dagger\) and \(A=a\). We have

\[ [G, a] = -b \text{ and } [G, b] = -a . \]

In particular \([(G)^2,A]=A\) for \(A\in\{a,b\}\). Hence (with \(\lambda=-\ii\chi\tau\), and abbreviating \(\theta=\chi\tau\))

\[ U a U^\dagger = \sum_n \frac{\theta^n}{n!} c_n \]

where

\[ c_n = \begin{cases} i^n a & \text{if } n \text{ is even.} \\ i^n b & \text{else.} \end{cases} . \]

Thus, using the Taylor expansion of \(\sin\) and \(\cos\), we get

\[ U a U^\dagger = \cos(\theta) a + \sin(\theta) b . \]

In the following let us abbreviate \(c=\cos(\theta)\) and \(s=\sin(\theta)\). Observe \(s^2=\gamma\) and \(c^2=1-\gamma\). Using the formula for \(UaU^\dagger\) we obtain

\[ U \ket{n_a, 0_b} = \frac{1}{\sqrt{n!}} \, U (a^\dagger)^n \ket{0_a, 0_b} = \frac{1}{\sqrt{n!}} \, (ca^\dagger+sb^\dagger)^n U \ket{0_a, 0_b} = \frac{1}{\sqrt{n!}} \, (ca^\dagger+sb^\dagger)^n \ket{0_a, 0_b} . \]

Since \(a\) and \(b\) commute we can use the binomial formula to obtain

\[ U \ket{n_a, 0_b} = \frac{1}{\sqrt{n!}} \sum_n \binom{n}{k} (ca^\dagger)^{n-k} (sb^\dagger)^k \ket{0_a,0_b} = \sum_{n\geq k} \sqrt{\binom{n}{k}} c^{n-k} s^k \ket{(n-k)_a,k_b} , \]

which is the claim of part 1. QED.

Proof of part 2

This can be proved in very abstract terms. We just have to use that \(E_k=\bra{k_b}U\ket{0_b}\) where \(U\) is unitary (on the combined system) and \(\{\ket{k_b}\}\) is an ONB on the environment. In fact (denoting by \(Q\) the principal system)

\[ \sum_k E_k^\dagger E_k = \sum_k \bra{0_b} U^\dagger \proj{k_b} U \ket{0_b} = \bra{0_b} U^\dagger U \ket{0_b} = I_Q . \]

This identity characterizes trace-preserving operations. QED.

Remark

It wouldn't be hard to directly verify the claim using the closed formula for \(E_k\) from this concrete example. Writing \(a_{nk}=\sqrt{\binom{n}{k}(1-\gamma)^{n-k}\gamma^k}\) the claim essentially follows from the binomial formula

\[ \sum_{k=0}^n a_{nk}^2 = ((1-\gamma) + \gamma)^n = 1 . \]

Exercise 8.22 (Amplitude damping of single qubit density matrix)

For the general single qubit state

\[ \rho = \begin{bmatrix} a & b \\ b^* & c \end{bmatrix} . \]

show that amplitude damping leads to

\[ \calE_{AD}(\rho) = \begin{bmatrix} 1 - (1-\gamma)(1-a) & b\sqrt{1-\gamma} \\ b^*\sqrt{1-\gamma} & c(1-\gamma) \end{bmatrix} . \]

Solution

The claim essentially follows from this small sage script

E0 = matrix.diagonal([1, sqrt(1-g)])
E1 = matrix([[0, sqrt(g)], [0, 0]])

a, c = SR.var('a c', domain='real')
b = SR.var('b', domain='complex')

rho = matrix([
    [a, b],
    [b.conjugate(), c],
])

A = simplify(E0*rho*E0.H + E1*rho*E1.H)

assert A == matrix([
    [               c*g + a, b*sqrt(1-g)],
    [sqrt(1-g)*conjugate(b), c*(1-g)]
])
"PASSED"

'PASSED'

The upper left corner looks different to what was stated in the exercise. But using \(a+c=1\) equality is easily verified.

Exercise 8.23 (Amplitude damping of dual-rail qubits)

Suppose that a single qubit state is represented by using two qubits, as

\[ \ket{\psi} = a \ket{01} + b \ket{10} . \]

Show that \(\calE_{AD}\otimes\calE_{AD}\) applied to this state gives a process which can be described by the operation elements

\begin{align*} E_0^{\mathrm{dr}} &= \sqrt{1-\gamma} I, \\ E_1^{\mathrm{dr}} &= \sqrt{\gamma} \ket{00}\bra{01}, \\ E_2^{\mathrm{dr}} &= \sqrt{\gamma} \ket{00}\bra{10}, \end{align*}

that is, either nothing (\(E_0^{\mathrm{dr}}\)) happens to the qubit, or the qubit is transformed (\(E_1^{\mathrm{dr}}\), \(E_2^{\mathrm{dr}}\)) into the state \(\ket{00}\), which is orthogonal to \(\ket{\psi}\). This is a simple error-detection code, and is also the basis for the robustness of the ‘dual-rail’ qubit discussed in Section 7.4.

Remark

The original exercise statement contained a small error. It claimed that only two operation elements are needed: \(E_0^{\mathrm{dr}}\) and \(E_1^{\mathrm{dr}}+E_2^{\mathrm{dr}}\). This is incorrect.

Moreover the formula for \(E_0^{\mathrm{dr}}\) needs a proper interpretation. But we will make it precise in the solution below.

Proof

Let us abbreviate \(c=\sqrt{1-\gamma}\) and \(s=\sqrt{\gamma}\). Recall the operation elements of \(\calE_{AD}\):

\[ E_0 = \begin{bmatrix} 1 & 0 \\ 0 & c \end{bmatrix}, \quad E_1 = \begin{bmatrix} 0 & s \\ 0 & 0 \end{bmatrix}. \]

A set of operation elements for \(\calE_{AD}\otimes\calE_{AD}\) is given by the four elements \(\{E_{ij}:=E_i\otimes\?E_j\}\) which we will write down in the following:

\[ E_{00} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & c & 0 & 0 \\ 0 & 0 & c & 0 \\ 0 & 0 & 0 & c^2 \end{bmatrix} \] and

\[ E_{01} = \begin{bmatrix} 0 & s & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & sc \\ 0 & 0 & 0 & 0 \end{bmatrix} \]

and

\[ E_{10} = \begin{bmatrix} 0 & 0 & s & 0 \\ 0 & 0 & 0 & sc \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]

and

\[ E_{11} = \begin{bmatrix} 0 & 0 & 0 & s^2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} . \]

Here the rows and columns correspond to the basis

\[ (\ket{00},\ket{01},\ket{10},\ket{11}) \]

in that order. Observe that all four operation elements leave the subspace spanned by the first three basis elements invariant. This is important since \(\ket{\psi}\) lives in that subspace. With respect to this subspace we have

\begin{align*} E_{00} &= \proj{00} + c(\proj{01} + \proj{10}) , \\ E_{01} &= s \ket{00}\bra{01} , \\ E_{10} &= s \ket{00}\bra{10} , \\ E_{11} &= 0 . \end{align*}

In particular we only need three elements and we have \(E_1^{\mathrm{dr}}=E_{01}\) and \(E_2^{\mathrm{dr}}=E_{10}\). Moreover \(E_{00}\) corresponds to \(cI\) if we understand \(I\) to be the identity on the subspace spanned by \(\ket{01}\) and \(\ket{10}\), which makes sense since \(\ket{\psi}\) even lives in that space. QED.

Exercise 8.24 (Spontaneous emission is amplitude damping)

A single atom coupled to a single mode of electromagnetic radiation undergoes spontaneous emission, as was described in Section 7.6.1. To see that this process is just amplitude damping, take the unitary operation resulting from the Jaynes–Cummings interaction, Equation (7.77), with detuning \(\delta=0\), and give the quantum operation resulting from tracing over the field.

Solution

The Jaynes-Cummings interaction is essentially given by

\[ H = \delta Z + g(a^\dagger \sigma_- + a \sigma_+) , \]

where \(Z\), \(\sigma_\pm=X\mp\ii\?Y\) act on the two-level atom and \(a\), \(a^\dagger\) on the field (see e.g. equation (7.71)). In the basis \((\ket{00},\ket{01},\ket{10})\) (the right qubit is the atom) the Hamiltonian looks as follows:

\[ H = \begin{bmatrix} \delta & 0 & 0 \\ 0 & -\delta & g \\ 0 & g & \delta \end{bmatrix} . \]

Let \(U=\exp(-\ii\?H\tau/2)\) and define \(c=\cos(g\tau)\), \(s=\sin(g\tau)\). For \(\delta=0\) we have

\[ U = \begin{bmatrix} 1 & 0 & 0 \\ 0 & c & -\ii s \\ 0 & -\ii s & c \end{bmatrix} . \]

Let \(\rho_E=\proj{0}\) be the state of the field - the environment - with zero photons. Then the following clearly describes spontaneous emission:

\[ \calE(\rho) = \ptrace{E}{U \rho_E\otimes\rho U^\dagger} . \]

We have to show that this is actually amplitude damping.

theta = SR.var('theta', domain='real')
b, b1 = SR.var('b b1', domain='complex')

c = cos(theta)
s = sin(theta)

# NOTE: we add a trivial action on |11>. The point of this is solely to be able to
# calculate the partial trace below. The function we use only accepts matrices with
# dimensions being a power of 2 (more precisely: Hilbert spaces described by qubits). If a
# state would actually be equal to |11> the Jaynes-Cummings interaction would not leave it
# invariant. But we can use this as a mathematical trick since the three-dimensional
# subspace we consider contains everything we need and is left invariant by the
# interaction.
U = matrix([
    [1,    0,    0, 0],
    [0,    c, -i*s, 0],
    [0, -i*s,    c, 0],
    [0,    0,    0, 1],
])

# An arbitrary mixed state (as input for the operation).
# b1 is actually conjugate(b) but this looks ugly and we do not need it.
rho = matrix([
    [p, b],
    [b1, 1-p],
])

M = U * kron(P0, rho) * U.H

# TODO: reimplement for sage
from chapter_2 import ptrace
from sympy import Matrix

M = Matrix(M)
T = ptrace(M, [1]) # bits are counted from the right.
T

Matrix([
[p - (p - 1)*sin(theta)**2,           b*cos(theta)],
[            b1*cos(theta), -(p - 1)*cos(theta)**2]])

Comparing this with the statement of exercise 8.22 we see that this is indeed amplitude damping with \(\gamma=s^2=\sin(g\tau)^2\).

Exercise 8.25

If we deﬁne the temperature \(T\) of a qubit by assuming that in equilibrium the probabilities of being in the \(\ket{0}\) and \(\ket{1}\) states satisfy a Boltzmann distribution, that is \(p_0=e^{-E_0/k_BT}/\mathcal{Z}\) and \(p_1=e^{-E_1/k_BT}/\mathcal{Z}\), where \(E_0\) is the energy of the state \(\ket{0}\), \(E_1\) the energy of the state \(\ket{1}\), and \(\mathcal{Z}=e^{-E_0/k_BT}+e^{-E_1/k_BT}\), what temperature describes the state \(\rho_\infty\)?

Solution

By convention we typically assume \(E_0\?<\?E_1\) (and we assume those values to be given constants). Note that this is equivalent to \(p_0\?>\?1/2\). The limit case \(E_0=E_1\) corresponds to \(p_0=1/2\). It is easy to see that

\[ p_0 = \frac{1}{1 + \exp(-(E_1-E_0)/k_BT)} \]

This can be used to get a formula for \(T\) in terms of \(p=p_0\):

\[ T = \frac{E_1 - E_0}{k_B \log(p/(1-p))} . \]

This is the temperature of a state \(\rho=p\proj{0}+(1-p)\proj{1}\). We have the following edge cases:

\[ T(p=1) = \lim_{p\to1} T(p) = 0, \quad T(p=1/2) = \lim_{p\to1/2} T(p) = \infty . \]

Exercise 8.26 (Circuit model for phase damping)

Show that the circuit in Figure 8.15 can be used to model the phase damping quantum operation, provided \(\theta\) is chosen appropriately.


rho_in: ────■────
        ┌───┴───┐
   |0>: ┤ Ry(θ) ├
        └───────┘

Proof

This is similar to exercise 8.20 - just simpler. Let \(B\) be the operation of the circuit:

\[ B = P_0 \otimes I + P_1 \otimes R_y(\theta) . \]

Let \(\ket{\psi}=a\ket{0}+b\ket{1}\). Then

\[ B\ket{\psi}\ket{0_E} = a \ket{00} b\ket{1} R_y(\theta)\ket{0} = a\ket{00} + b(c\ket{10} + s\ket{11}) , \]

where \(c=\cos(\theta)\) and \(s=\sin(\theta)\). From here we get

\begin{align*} E_0\ket{\psi} &= \bra{0_E} B \ket{\psi}\ket{0_E} = a \ket{0} + bc\ket{1} , \\ E_1\ket{\psi} &= \bra{1_E} B \ket{\psi}\ket{0_E} = bs \ket{1} . \end{align*}

In other words:

\[ E_0 = \begin{bmatrix} 1 & 0 \\ 0 & c \end{bmatrix}, \quad E_1 = \begin{bmatrix} 0 & 0 \\ 0 & s \end{bmatrix}, \]

which is phase damping (at least for e.g. \(\theta\in[0,\pi/2]\)) - as desired. QED.

Exercise 8.27 (Phase damping = phase flip channel)

Give the unitary transformation which relates the operation elements of (8.127)–(8.128) to those of (8.129)–(8.130); that is, find \(u\) such that \(\tilde{E}_k=\sum_ju_{kj}E_j\).

For reference:

\[ E_0 = \begin{bmatrix} 1 & 0 \\ 0 & \sqrt{1-\lambda} \end{bmatrix} , \; E_1 = \begin{bmatrix} 0 & 0 \\ 0 & \sqrt{\lambda} \end{bmatrix} . \]

and \(\tilde{E}_0=\sqrt{\alpha}I\), \(\tilde{E}_1=\sqrt{1-\alpha}Z\).

Solution 1

In the following I prove

\begin{align*} \tilde{E}_0 &= \sqrt{\alpha} E_0 + \sqrt{1-\alpha} E_1 , \\ \tilde{E}_1 &= \sqrt{1-\alpha} E_0 - \sqrt{\alpha} E_1 , \end{align*}

and briefly show how to come up with precisely this ansatz. Recall that the action of phase damping is given by

\[ \?\begin{bmatrix} p & b \\ b^* & q \end{bmatrix} \mapsto \begin{bmatrix} p & \sqrt{1-\lambda}\,b \\ \sqrt{1-\lambda}\,b^* & q \end{bmatrix} . \]

Phase flip is given by

\[ \?\begin{bmatrix} p & b \\ b^* & q \end{bmatrix} \mapsto \begin{bmatrix} p & (2\alpha-1)b \\ (2\alpha-1)b^* & q \end{bmatrix} . \]

Hence there is a bijection \([0,1]\to[1/2,1]\) given by

\[ \alpha=(1+\sqrt{1-\lambda})/2 \]

which maps between phase damping and "half of" all phase flips. This already shows that phase damping is just phase flip and we already have a formula for \(\alpha\). Hence there must be a unitary matrix \(u\) such that

\begin{align*} \tilde{E}_0 &= u_{00} E_0 + u_{01} E_1 , \\ \tilde{E}_1 &= u_{10} E_0 + u_{11} E_1 . \end{align*}

By inspecting the matrices it is clear that we must have \(u_{00}=\sqrt{\alpha}\) and \(u_{10}=\sqrt{1-\alpha}\). Since the matrix \(u\) is unitary we also must have \(\abs{u_{01}}=\sqrt{1-\alpha}\) and \(\abs{u_{11}}=\sqrt{\alpha}\). Since everything else is real we directly conclude that \(u_{10}\) and \(u_{11}\) have to be real too. By looking at the \(11\) entries of the operation elements we directly see that \(u_{01}\) must be positive. Hence by (e.g.) unitarity of \(u\) we deduce that \(u_{11}\) must be negative. Hence the only possible \(u\) is the one implicitly given in the above equation.

Since we already know that \(u\) must exist and we also showed that it is unique the mentioned equation follows.

Solution 2

This is not really a second solution but more like a consistency check for the first solution (we all make mistakes so it is best practice to double check things from time to time).

The claim we have to prove is equivalent to:

\begin{align*} \sqrt{\alpha} &= \sqrt{\alpha} \sqrt{1-\lambda} + \sqrt{1-\alpha} \sqrt{\lambda} , \\ -\sqrt{1-\alpha} &= \sqrt{1-\alpha} \sqrt{1-\lambda} - \sqrt{\alpha} \sqrt{\lambda} . \end{align*}

In principle there would be four equations (one for each diagonal entry and each phase-flip element) but the other two are just trivial. Let us abbreviate \(c=\sqrt{1-\lambda}\) and \(s=\sqrt{\lambda}\). Note that \(c,s\in[0,1]\) and \(c^2+s^2=1\). The first equation is equivalent to

\[ \sqrt{\alpha} = \sqrt{\alpha} c + \sqrt{1-\alpha} s . \]

Before we go on, recall this formula \(2\alpha=1+c\). It suffices to show that twice the square of the RHS is \(2\alpha=1+c\) (squaring only hides the sign but we already see that both sides are positive). Let us check this:

\begin{align*} 2(\sqrt{\alpha} \, c + \sqrt{1-\alpha} \, s)^2 &= (\sqrt{1+c} \, c + \sqrt{1-c} \, s)^2 \\ &= (1+c)c^2 + (1-c)s^2 + 2\sqrt{1-c^2} \, sc \\ &= 1 + c(c^2+s^2) \\ &= 1 + c . \end{align*}

This is as desired. The second equation

\[ -\sqrt{1-\alpha} = \sqrt{1-\alpha} c - \sqrt{\alpha} s . \]

can also be shown by showing that twice the square of the RHS is equal to \(2(1-\alpha)=1-c\) (the sign makes no problem since the RHS is at most \(\sqrt{1-\alpha}\) and this only if \(c=1\), \(s=0\) which implies \(\alpha=1\), a special case easily treated separately). Let us also check this:

\begin{align*} 2(\sqrt{1-\alpha} \, c - \sqrt{\alpha} \, s)^2 &= (\sqrt{1-c} \, c - \sqrt{1+c} \, s)^2 \\ &= (1-c)c^2 + (1+c)s^2 - 2\sqrt{1-c^2} \, sc \\ &= 1 - c(c^2+s^2) \\ &= 1 - c . \end{align*}

Exercise 8.28 (One `CNOT` phase damping model circuit)

Show that a single CNOT gate can be used as a model for phase damping, if we let the initial state of the environment be a mixed state, where the amount of damping is determined by the probability of the states in the mixture.

Solution

As an intermediate step let us find a solution involving a CZ gate instead of a CNOT gate. It is relatively intuitive how to do this if we recall that phase damping is just phase flip (exercise 8.27)


               rho_in: ─■─
                        │
p|0><0| + (1-p)|1><1|: ─■─

Here CZ gate is to be interpreted as having its control at the environment (it is symmetric of course but it helps to imagine it that way). The environment is in the mixed state

\[ \rho_E = p\proj{0} + (1-p)\proj{1} \]

which activates the CZ with probability \(1-p\) and does nothing otherwise. A similar construction was done for the deploarizing channel in figure 8.12. To rigorously prove that this actually implements the phase flip let \(U\) be the CZ gate and consider

\begin{align*} \ptrace{E}{U \rho\otimes\rho_E U^\dagger} &= p \, \ptrace{E}{U \rho\otimes P_0 U^\dagger} + (1-p) \, \ptrace{E}{U \rho\otimes P_1 U^\dagger} \\ &= p \, \ptrace{E}{\rho\otimes P_0} + (1-p) \, \ptrace{E}{(Z\rho Z)\otimes P_1} \\ &= p \, \rho + (1 - p) \, Z\rho Z . \end{align*}

Hence the above circuit implements phase flip. By exercise 8.27 we know that setting \(p=(1+\sqrt{1-\lambda})/2\) (c.f. solution of that exercise) is phase damping with parameter \(\lambda\).

Using the Hadamard gate and \(HXH=Z\) the circuit can be converted to a cicuit using a CNOT:


               rho_in: ───────■───────
                       ┌───┐┌─┴─┐┌───┐
p|0><0| + (1-p)|1><1|: ┤ H ├┤ X ├┤ H ├
                       └───┘└───┘└───┘

Note that we actually do not need the second Hadamard gate since it only acts on the environment after the interaction took place. That is, the following circuit implements the same operation:


               rho_in: ───────■──
                       ┌───┐┌─┴─┐
p|0><0| + (1-p)|1><1|: ┤ H ├┤ X ├
                       └───┘└───┘

The Hadamard gate maps between the eigenbasis \((\ket{0},\ket{1})\) of \(Z\) and the eigenbasis \((\ket{0},\ket{1})\) of \(X\). Hence the following circuit still implements phase damping:


               rho_in: ──■──
                       ┌─┴─┐
p|+><+| + (1-p)|-><-|: ┤ X ├
                       └───┘

That is, the environment is in the state \(p\proj{+}+(1-p)\proj{-}\) and the CNOT gate has its target there.

Exercise 8.29 (Unitality)

A quantum process \(\calE\) is unital if \(\calE(I)=I\). Show that the depolarizing and phase damping channels are unital, while amplitude damping is not.

Solution

This is a straightforward exercise. So let us mainly mention one thing. Some formulas like this one for the depolarizing channel

\[ \calE_{DC}(\rho) = p d\inv I + (1-p)\rho \]

only holds for density matrices. In that case we have to plug in \(d\inv\?I\) and not \(I\):

\[ \calE_{DC}(d\inv I) = (p + (1-p)) d\inv I = d\inv I . \]

For the phase damping let us show the claim on the (slightly) more general phase flip operation (exercise 8.27). We use the a formula based on operation elements where the above remark does not apply:

\[ \calE_{PF}(I) = p I^3 + (1-p) ZIZ = p I + (1-p) I = I . \]

Finally consider amplitude damping with \(E_0=\diag(1,c)\) and \(E_1=s\ket{0}\bra{1}\) where \(c,s\geq0\) with \(c^2+s^2=1\) (moreover we assume that \(c\neq1\) to avoid the identity operation which is indeed unital):

\[ \calE_{AD}(I) = E_0E_0^\dagger + E_1E_1^\dagger = \diag(1,c^2) + \diag(s^2, 0) = \diag(1+s^2,c^2) \neq I \]

Exercise 8.30 (\(T_2\leq2T_1\))

The \(T_2\) phase coherence relaxation rate is just the exponential decay rate of the off-diagonal elements in the qubit density matrix, while \(T_1\) is the decay rate of the diagonal elements (see Equation (7.144)). Amplitude damping has both nonzero \(T_1\) and \(T_2\) rates; show that for amplitude damping \(T_2=2T_1\). Also show that if amplitude and phase damping are both applied then \(T_2\leq2T_1\).

Proof

Recall the general equation for the relaxation from (7.144):

\[ \?\begin{bmatrix} (p-p_0) e^{-t/T_1} + p_0 & b e^{-t/T_2} \\ b^* e^{-t/T_2} & (p_0-p) e^{-t/T_1} + (1-p_0) \?\end{bmatrix} . \]

From exercise 8.22 we know

\[ \calE_{AD}(\rho) = \?\begin{bmatrix} 1 + (1-\gamma)(p-1) & b \sqrt{1-\gamma} \\ b^* \sqrt{1-\gamma} & (1-p) (1-\gamma) \?\end{bmatrix} . \]

This clearly corresponds to \(p_0=1\), \(T_2=2T_1\) and an unspecified \(t>0\). If we apply phase damping in addition we get

\[ \calE_{PD}\circ\calE_{AD}(\rho) = \?\begin{bmatrix} 1 + (1-\gamma)(p-1) & b \sqrt{(1-\lambda)(1-\gamma)} \\ b^* \sqrt{(1-\lambda)(1-\gamma)} & (1-p) (1-\gamma) \?\end{bmatrix} . \]

This clearly only makes the \(T_2\)-relaxation faster. In other words \(T_2\) gets smaller. QED.

TODO Exercise 8.31 (Exponential sensitivity to phase damping)

Using (8.126), show that the element \(\rho_{nm}=\bra{n}\rho\ket{m}\) in the density matrix of a harmonic oscillator decays exponentially as \(e^{-\lambda(n-m)^2}\) under the effect of phase damping, for some constant \(\lambda\).

Exercise 8.32

Explain how to extend quantum process tomography to the case of non-trace-preserving quantum operations, such as arise in the study of measurement.

Solution

The procedure explained in the book does not rely on the fact that the operation is trace-preserving. The only occasion it appeared was when it was mentioned that \(\chi\) has \(d^4-d^2\) free parameters. In the non-trace-preserving case the \(-d^2\) goes away.

Exercise 8.33 (Specifying a quantum process)

Suppose that one wished to completely specify an arbitrary single qubit operation \(\calE\) by describing how a set of points on the Bloch sphere \(\{\vec{r}_k\}\) transform under \(\calE\). Prove that the set must contain at least four points.

Proof

First of all recall that the bloch sphere description only applies to trace-preserving operations (exercise 8.16). Therefore let us add the additional resctriction that the operation has to be trace preserving.

The claim almost follows from the fact that trace-preserving operations act like affine maps in the (three-dimensional) Bloch-space. But it is well known that an affine map on a \(d\)-dimensional space needs exactly \(d+1\) affine-linearly independent points to be specified. I say "almost" because not every affine map corresponds to a quantum operation (at least I think so). But it still serves as good justification because we have already seen that quantum operations give rise to several non-trivial affine maps.

We show that every map (on the Bloch-space) of the form

\[ \vec{x} \mapsto RD \vec{x} + \vec{c} , \]

for any rotation matrix \(R\) (orthogonal and determinant plus one), diagonal matrix \(D\) with diagonal entries in \(0\?<\?d_i\?<\?1-\gamma\), and any vector \(\vec{c}\) with \(\norm{\vec{c}}\leq\gamma\) corresponds to a quantum operation. Here \(0\leq\gamma\?<\?1\) is an arbitrary parameter. There are more affine maps possible but this is sufficient for our purpose.

Note that e.g. mirroring at the \(z=0\) plane (a negative determinant map) and the mentioned affine maps together generate the group of all invertible affine maps (this is related to the polar decomposition of matrices). Hence it is not hard to see that even if we restrict the space of possible maps to maps of the above form we need at least four points.

Now let us justify the above claim. Any rotation \(\vec{x}\mapsto\?R\vec{x}\) is possible by unitary evolution (keyword: Pauli rotations). The depolarizing channel implements \(\vec{x}\mapsto\alpha\vec{x}\) for any \(\alpha\in(0,1)\). The phase flip can implement an operation which shrinks the x- and y-dimension by the same factor. In the same way a similar shrinkage of any pair of axes is possible. Hence a mapping \(\vec{x}\mapsto\?D\vec{x}\) is possible where the diagonal entries just have to satisfy \(0\?<\?d_i\?<\?1\).

Finally amplitude damping together with a rotation can be used to realize a non-trivial \(\vec{c}\). Note that amplitude damping also shrinks everything, but not more than by a factor \(1-\gamma\). This is the reason why we impose the condition \(0\?<\?d_i\?<\?1-\gamma\) on the realizability of the diagonal entries. QED.

Exercise 8.34 (Process tomography for two qubits)

Show that the \(\chi_2\) describing the black box operations on two qubits can be expressed as

\[ \chi_2 = \Lambda_2 \sigma \Lambda_2 , \]

where \(\Lambda_2=\Lambda\otimes\Lambda\), \(\Lambda\) is as deﬁned in Box 8.5, and \(\sigma\) is a block matrix of 16 measured density matrices,

\[ \sigma = P^\top \left( \sum_{mn} \rho_{mn}\otimes\rho'_{mn} \right) P \]

where \(\rho'_{nm}=\calE(\rho_{nm})\), \(\rho_{nm}=T_n\proj{00}T_m\), \(T_1=I\otimes\?I\), \(T_2=I\otimes\?X\), \(T_3=X\otimes\?I\), \(T_4=X\otimes\?X\), and \(P=I\otimes[(\rho_{00}+\rho_{12}+\rho_{21}+\rho_{33})\otimes\?I]\) is a permutation matrix.

Remarks

The formula for sigma is equivalent to the one in the book. I find it to be more concise. It follows from the fact that the \(\rho_{mn}\) are matrices with zeros every except at the position \((m,n)\).
I think that there is a tiny error in this exercise. I think one should use \(P\) to conjugate \(\Lambda_2\) so that the formula is as follows:

\[ \chi_2 = P \Lambda_2 P R' P \Lambda_2 P = P \Lambda_2 \sigma \Lambda_2 P , \]

where \(R'\) is the same as \(\sigma\) but without conjugating with \(P\).

Proof

I will slightly deviate from the exercise (but still solve it) to get a better insight into what is going on. I will change the exercise in the following way:

I consider \(d\geq1\) qubits - generalizing the statement (and actually making it easier to grasp).
I define a slightly different \(\Lambda_d\) avoiding the \(P\).

The task can be reduced to the case \(d=1\) which I already proved in the introduction to this chapter. I reuse the notation from there so please have a look. Let

\[ \calE(\rho) = \sum_{mn} \chi_{mn} K_m \rho K_n , \]

where \(K_m=\bigotimes_{x=0}^{d-1}N_{m_x}\). Our task is to find \(\chi\). Let

\[ \tilde{\Lambda}_d = \frac{1}{2^d} \sum_{a=0}^{2^d-1} \bigotimes_{x=0}^{d-1} L_{a_x} \otimes \bigotimes_{x=0}^{d-1} R_{a_x} . \]

Then

\[ \tilde{\Lambda}_d R' \tilde{\Lambda}_d = \frac{1}{4^d} \sum_{mnab} \chi_{mn} \left[\bigotimes_{x=0}^{d-1} L_{a_x} P_{i_xj_x} L_{b_x} \right] \otimes \left[\bigotimes_{x=0}^{d-1} R_{a_x} N_{m_x} P_{i_xj_x} N_{n_x} R_{b_x} \right] \]

But the RHS is just the tensorized version of a formula for the case \(d=1\). Hence the LHS must equal \(\chi\) (note that the tensor-product must be in the right order to not get a permuted version of \(\chi\)).

To see why the permuation matrix \(P\) is necessary in the context of the exercise let us again consider the case \(d=2\). Here we have

\begin{align*} \Lambda_2 &= \sum_a L_{a_1} \otimes R_{a_1} \otimes L_{a_0} \otimes R_{a_0} , \\ \tilde{\Lambda}_2 &= \sum_a L_{a_1} \otimes L_{a_0} \otimes R_{a_1} \otimes R_{a_0} . \end{align*}

Hence we see that only the two middle operators are swapped. This is the purpose of the permuation. We could also leave out the permuation and just replace \(\tilde{\Lambda}_2\) by \(\Lambda_2\), but in that case we would get a permuted version of \(\chi\). QED.

Exercise 8.35 (Process tomography example)

Consider a one qubit black box of unknown dynamics \(\calE_1\). Suppose that the following four density matrices are obtained from experimental measurements, performed according to Equations (8.173)–(8.176):

\begin{align*} \rho_1' &= \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} , \\ \rho_2' &= \begin{bmatrix} 0 & \sqrt{1-\gamma} \\ 0 & 0 \end{bmatrix} , \\ \rho_3' &= \begin{bmatrix} 0 & 0 \\ \sqrt{1-\gamma} & 0 \end{bmatrix} , \\ \rho_4' &= \begin{bmatrix} \gamma & 0 \\ 0 & 1-\gamma \end{bmatrix} . \end{align*}

where \(\gamma\) is a numerical parameter. From an independent study of each of these input–output relations, one could make several important observations: the ground state \(\ket{0}\) is left invariant by \(\calE_1\), the excited state \(\ket{1}\) partially decays to the ground state, and superposition states are damped. Determine the \(\chi\) matrix for this process.

Remark: I wonder why we should stop at computing the chi matrix. Anyway, let us extend the exercise by the requirement to also compute operation elements for this operation.

Solution 1

Let us first set up two routines to compute \(\beta\) and \(\lambda\). We specialize those to the single qubit case. Moreover we use the basis \((\rho_j)=(P_0, P_0X, XP_0,XP_0X)\) (where \(P_0=\proj{0}\)) for the space where the density matrices live.

def compute_beta_835(Es: list[matrix]):
    """Compute beta for single qubit and specific choice of basis of input matrix space."""
    rho = [P0, P0*X, X*P0, X*P0*X] # basis

    # As suggested in the text we flatten the space of 4x4 matrices to a 16-dimensional
    # vector. This turns beta into a 16x16 matrix.
    beta = matrix.zero(16)

    for m, n in product(range(4), range(4)):
        for j, k in product(range(4), range(4)):
            rho_prime = Es[m] * rho[j] * Es[n].H
            beta[j+4*k, m+4*n] = trace(rho[k].H * rho_prime)

    return beta


def compute_lambda_835(rho_prime):
    """Compute lambda for single qubit and specific choice of basis of input matrix
    space."""
    rho = [P0, P0*X, X*P0, X*P0*X] # basis

    lam = [None for _ in range(16)]

    for j in range(4):
        for k in range(4):
            lam[j+4*k] = trace(rho[k].H * rho_prime[j])

    return vector(lam)


def toMatrix_835(vec):
    """Convert a flattened 4x4 matrix back to matrix representation."""
    return matrix(SR, 4, 4, lambda m,n: vec[m+4*n])

The following code computes \(\chi\). As a basis for the operation elements we use the \(\tilde{E}_j\) as in box 8.5: \((I,X,-\ii\?Y,Z)\).

rho_prime_835 = [
    matrix([
        [1, 0],
        [0, 0],
    ]),
    matrix([
        [0, sqrt(1-g)],
        [0, 0],
    ]),
    matrix([
        [0, 0],
        [sqrt(1-g), 0],
    ]),
    matrix([
        [g, 0],
        [0, 1-g],
    ]),
]

# E-tilde as basis for beta as in box 8.5:
beta_835 = compute_beta_835([Id, X, -i*Y, Z])
kappa_835 = beta_835.pseudoinverse()
assert beta_835 == beta_835 * kappa_835 * beta_835, "kappa is pseudoinverse of beta"

lam_835 = compute_lambda_835(rho_prime_835)
chi_835 = toMatrix_835(kappa_835 * lam_835)

Let us print \(\chi\):

chi_835

[-1/4*g + 1/2*sqrt(-g + 1) + 1/2                               0                               0                           1/4*g]
[                              0                           1/4*g                          -1/4*g                               0]
[                              0                          -1/4*g                           1/4*g                               0]
[                          1/4*g                               0                               0 -1/4*g - 1/2*sqrt(-g + 1) + 1/2]

To obtain the operation elements we have to diagonalize \(\chi\).

D, U = chi_835.eigenmatrix_right()
print(f"D (diagonalized chi matrix) = \n{D}")
print(f"\nU (columns contain eigenvectors) = \n{U}")

D (diagonalized chi matrix) =
[-1/2*g + 1          0          0          0]
[         0          0          0          0]
[         0          0          0          0]
[         0          0          0      1/2*g]

U (columns contain eigenvectors) =
[                          1                           1                           0                           0]
[                          0                           0                           1                           1]
[                          0                           0                           1                          -1]
[-(g + 2*sqrt(-g + 1) - 2)/g  (g - 2*sqrt(-g + 1) - 2)/g                           0                           0]

There are only two non-zero eigenvalues of the chi-matrix:

\[ d_0 = 1 - \frac{g}{2}, \quad d_3 = \frac{g}{4} . \]

The corresponding normalized eigenvectors are

\[ u_0 = (1, 0, 0, r)/\sqrt{1+r^2}, \quad u_3 = (0, 1, -1, 0)/\sqrt{2} , \]

where \(r=g\inv(1-\sqrt{1-g})^2\). This implies the following operation elements:

\begin{align*} E_0 &= \sqrt{\frac{1-\frac{g}{2}}{1+r^2}} \, (I + rZ) , \\ E_3 &= \frac{\sqrt{g}}{2} \, (X + \ii Y) . \end{align*}

r = (1/g)*(1-sqrt(1-g))^2
op835 = make_operation([sqrt(1-g/2)*(Id+r*Z)/sqrt(1+r^2), sqrt(g)*(X+i*Y)/2])

Let us briefly verify that this operation indeed implements the desired operation:

rho = [P0, P0*X, X*P0, X*P0*X]
assert op835(rho[0]) == rho_prime_835[0]
assert op835(rho[1]) == rho_prime_835[1]
assert op835(rho[2]) == rho_prime_835[2]
assert op835(rho[3]) == rho_prime_835[3]
"PASSED"

'PASSED'

Solution 2

In box 8.5 another way to compute the chi-matrix was suggested. It is based on a matrix \(\Lambda\):

Lambda1 = (kron(Z, Id) + kron(X, X)) / 2

In addition we need the following utility functions. The first one computes the matrix

\[ R' := \begin{bmatrix} \rho_1' & \rho_2' \\ \rho_3' & \rho_4' \end{bmatrix} . \]

It utilizes the fact that this matrix is just \(\sum_j\rho_j\otimes\rho_j'\) for our specific choice of \((\rho_j)\). The second one just implements the formula (see (8.179))

\[ \chi = \Lambda R' \Lambda . \]

def compute_rho_prime_matrix_835(rho_primes):
    rhos = [P0, P0*X, X*P0, X*P0*X]
    lam = matrix.zero(4)

    for rho, rho_prime in zip(rhos, rho_primes):
        lam += kron(rho, rho_prime)

    return lam


def compute_chi_via_Lambda_835(rho_primes):
    return Lambda1 * compute_rho_prime_matrix_835(rho_primes) * Lambda1

So let us compute \(\chi\):

# Take rho_prime_835 from solution 1:
chi_835_2 = compute_chi_via_Lambda_835(rho_prime_835)

Let us do a quick check that we got the same thing as in the first solution. Note that this is only possible because we used the same bases for the operation elements (\((I,X,-\ii\?Y,Z)\)) and the matrix space where the density matrices live (\((P_0,P_0X,XP_0,XP_0X)\)). Otherwise the number might be completely different of course.

assert chi_835 == chi_835_2
"PASSED"

'PASSED'

The rest is the same as in the first solution.

Chapter 8

Table of Contents

Setup

Imports

Utility code

Bloch sphere representation

Derivation of the formula for \(M_{jk}\) and \(c_k\)

Sage code

Process tomography for single qubit

Exercises

Exercise 8.1 (Unitary evolution as a quantum operation)

Proof

Preparation

The actual proof

Exercise 8.2 (Measurement as a quantum operation)

Solution

Exercise 8.3

Proof

Exercise 8.4 (Measurement)

Solution

Exercise 8.5 (Spin flips)

Solution

Exercise 8.6 (Composition of quantum operations)

Proof

Exercise 8.7

Solution

Exercise 8.8 (Non-trace-preserving quantum operations)

Solution

Exercise 8.9 (Measurement model)

Solution

Exercise 8.10

Proof 1

Proof 2

Exercise 8.11

Proof

WIP Exercise 8.12

Maybe not a solution

Exercise 8.13

Solution

Exercise 8.14

Solution

Exercise 8.15

Solution

Exercise 8.16

Solution

What does still hold?

Example

Exercise 8.17

Solution

Exercise 8.18

Proof

Exercise 8.19

Solution

Exercise 8.20 (Circuit model for amplitude damping)

Solution

Exercise 8.21 (Amplitude damping of a harmonic oscillator)

Proof of part 1

Proof of part 2

Exercise 8.22 (Amplitude damping of single qubit density matrix)

Solution

Exercise 8.23 (Amplitude damping of dual-rail qubits)

Proof

Exercise 8.24 (Spontaneous emission is amplitude damping)

Solution

Exercise 8.25

Solution

Exercise 8.26 (Circuit model for phase damping)

Proof

Exercise 8.27 (Phase damping = phase flip channel)

Solution 1

Solution 2

Exercise 8.28 (One CNOT phase damping model circuit)

Solution

Exercise 8.29 (Unitality)

Solution

Exercise 8.30 (\(T_2\leq2T_1\))

Proof

TODO Exercise 8.31 (Exponential sensitivity to phase damping)

Exercise 8.32

Solution

Exercise 8.28 (One `CNOT` phase damping model circuit)