Chapter 9

We have

\[ D((1,0), (1/2,1/2)) = \frac{1}{2} ( 1/2 + 1/2 ) = \frac{1}{2} , \]

and

\[ D((1/2,1/3,1/6), (3/4,1/8,1/8)) = \frac{1}{2} ( 1/4 + 5/24 + 1/24 ) = \frac{1}{4} . \]

\[ D((p,1-p), (q, 1-q)) = \frac{1}{2} (\abs{p-q} + \abs{1-p-1+q}) = \abs{p-q} . \]

QED.

We have

\[ F((1,0), (1/2,1/2)) = \sqrt{1\cdot1/2} + 0 = \sqrt{2\inv} \]

and

\[ F((1/2,1/3,1/6), (3/4,1/8,1/8)) = \sqrt{3/8} + \sqrt{1/24} + \sqrt{1/48} = \frac{1 + 4\sqrt{2}}{4\sqrt{3}} \approx 0.961 . \]

Let \(J\) be the index set of the probability distribution (I mean: when we write \(p_x\) the \(x\) is from \(J\)). For a set \(S\subset\?J\) let \(s\) be its 0-1-encoding, that is, \(s_x=1\) iff \(x\in\?S\). Clearly \(p(S)=p\cdot s\) if we write \(p\cdot\?s=\sum_xp_xs_x\). Let us calculate:

Here the \(\max_t\) runs over all \(t\) with \(t_x=\pm2\inv\) and \(\norm{p-q}_1=D(p,q)\) is the L1-norm. For the second equality we used that \(\sum_xp_x=\sum_xq_x=1\). Note that the inequality is just a basic version of Hölder's inequality (which can be seen directly of course). Equality clearly holds for

\[ t_x = \begin{cases} +2\inv & \text{for } p_x \geq q_x \\ -2\inv & \text{else.} \end{cases} . \]

Equivalently (using the bijection \(t=s-2\inv\)):

\[ s_x = \begin{cases} 1 & \text{for } p_x \geq q_x \\ 0 & \text{else.} \end{cases} . \]

Note that one could also invert the cases (and treat \(p_x=q_x\) arbitrarily) but this choice also shows that one can leave out the absolute values from the procedure. QED.

This statement was already proved in the solution of exercise 9.4.

For the commuting pair of density operators we have

\[ D_1 = \frac{1}{2} \left(\abs{\frac{3}{4}-\frac{2}{3}} + \abs{\frac{1}{4}-\frac{1}{3}}\right) = \frac{1}{12} . \]

Let us consider the second, more interesting example. To compute \(D_2=D(\rho,\sigma)\) we could compute the eigenvalues \(\lambda_i\) of \(\rho-\sigma\). We then have \(D_2=(\abs{\lambda_1}+\abs{\lambda_2})/2\).

plus = matrix((ket('0') + ket('1'))/sqrt(2))
minus = matrix((ket('0') - ket('1'))/sqrt(2))

rho = matrix([[3/4, 0], [0, 1/4]])
sigma = (2/3)*plus.H*plus + (1/3)*minus.H*minus

# To compute the distance we essentially have to compute the eigenvalues of rho - sigma
# (and add their absolute values).
A = rho - sigma
assert A.eigenvalues() == [-1/12*sqrt(13), 1/12*sqrt(13)]

# Just for fun: In this special case you can extract the relevant information from the
# A.H*A since this one is already diagonal.
A2 = A.H * A
assert A2 == (13/144) * matrix.diagonal([1, 1])
"PASSED"

'PASSED'

Hence

\[ D_2 = \frac{\sqrt{13}}{12} . \]

The exercise already outlines what to do. I just make it a little bit more precise. Let \(D=\diag(\ldots,d_i,\ldots)\) be a diagonalization of \(\rho-\sigma\):

\[ \rho - \sigma = U D U^\dagger \]

Now rewrite \(D\) in terms of two other diagonal matrices

\[ D = D_Q - D_S \]

where \((D_Q)_{ii}=d_i\) if \(d_i\?>0\) and \((D_S)_{ii}=-d_i\) if \(d_i\?<0\) (and zeros everywhere else). Now define

\[ Q = U D_Q U^\dagger, \; S = U D_S U^\dagger . \]

Clearly \(Q\) and \(S\) are two positive operators with \(\rho-\sigma=Q-S\) with orthogonal support. QED.

This directly follows from strong convexity with \(q_i=p_i\), using \(D(p,p)=0\). QED.

But in my opinion this result is actually misleading. The book emphasizes the fact that \(D\) is a metric, but actually it corresponds to a norm on the matrix space:

\[ \norm{\rho}_{S_1} = \trace{\abs{\rho}} . \]

We have \(D(\rho,\sigma)=2\inv\norm{\rho-\sigma}_{S_1}\). This is a so called Schatten norm. The Schatten classes are the natural analog of the \(L^p\)​-spaces for operators on Hilbert spaces. The above norm corresponds to \(p=1\). The Hilbert-Schmidt inner product corresponds to a Schatten norm too, here \(p=2\).

As for any norm we have the following more general result (of course one has to prove that the Schatten norm is actually a norm, but this is trivial based on what we already know about \(D\))

\[ D\left(\sum_i a_i \rho_i, \sum_i a_i \sigma_i \right) = 2\inv\norm{\sum_i a_i(\rho_i-\sigma_i)}_{S_1} \leq 2\inv\sum_i \abs{a_i} \norm{\rho_i-\sigma_i}_{S_1} = \sum_i \abs{a_i} D(\rho_i,\sigma_i) \]

Here the \(a_i\) could be any complex numbers. It doesn't even matter that the operators are density operators - it could be any matrices. And as said, this is just an abstract property of norms in general. The strong convexity result is different in the way that it really relies on special properties of this particular norm and the concrete arguments it is applied to (density operators, numbers which are probabilities).

Just for definiteness let me recall my favorite version of Schauder's fixed point theorem. There are various specializations and generalizations. In my oppinion the following version hits the sweet spot in terms of generality and simplicity. The main difference to the version mentioned in the exercise is that it gets rid of the totally irrelevant property that the sourrounding space is a Hilbert space. There are generalizations to more abstract settings but this might distract too much.

Theorem (Schauder's fixed point theorem)

Let \(V\) be a Banach space and \(K\subset\?V\) a compact and convex set. Let \(f:K\to\?K\) be continuous. Then \(f\) has a fixed point \(x=f(x)\).

In our setting we have \(f=\calE\), \(V=\CC^{d\times d}\) and \(K\) is the set of density matrices (c.f. Theorem 2.5):

\[ K = \{ \rho \in \CC^{d\times d} \; | \; \rho\geq0 , \trace{\rho} = 1 \} . \]

\(K\) is clearly a closed set (if \((\rho_n)\) are density matrices with a limit \(\rho_n\to\rho\) then the limit is also a density matrix). To show that it is compact it suffices to show that \(K\) is bounded (in finite dimensional vector spaces compact is the same as bounded and closed - Heine-Borel). Recall that the operator norm of a hermitian matrix is equal to the spectral radius - the maximum of the absolute values of its eigenvalues (search for the relation of the spectral radius to the operator norm on any decent resource on any of those topics, try e.g. wikipedia). Since the spectral radius is at most \(1\) the claim follows. Hence \(K\) is indeed compact.

Convexity is also a well-known property of \(K\). In fact, the density matrices are just the convex closure of the projection operators (this is just a mathematician's reformulation of the definition of density matrices as ensembles of pure states).

Since \(\calE\) is trace-preserving we have \(\calE(K)\subseteq\?K\).

By the representation by Kraus matrices a quantum operation extends to a linear operator on \(\CC^{d\times\?d}\). In finite dimensions linearity implies continuity (I didn't find a dedicated resource on that, just have a look into bounded operators). In particular \(\calE:K\to\?K\) is continuous.

The preconditions of Schauder's theorem are satisfied. Hence \(\calE\) has a fixed point. QED.

We already know from exercise 9.9 that there must be one fixed point \(\rho_0\). Suppose there was a second one \(\rho_1\neq\rho_0\). Then

\[ D(\rho_0,\rho_1) = D(\calE(\rho_0),\calE(\rho_1)) < D(\rho_0,\rho_1) . \]

The equality from the fixed point property. The inequality is the strict contractivity. But this inequality is a contradiction to the fact that \(d\?<\?d\) cannot hold for any real number \(d\)! Hence the claim that there was a second fixed point is wrong. QED.

Remark

The result is related to Banach's fixed point theorem. At least in finite dimensions a strictly contractive map is already uniformly contractive. That is, there exists a \(\lambda\?<1\) such that

\[ D(\calE(\rho),\calE(\sigma)) < \lambda D(\rho,\sigma) \]

for all \(\rho\), \(\sigma\). This can be easily seen by assuming the contrary and considering a sequence \((\rho_n,\sigma_n)\) which violates the inequality for \(\lambda=1-n\inv\) (also use the compactness of \(K\) to single out a converging sub-sequence). Hence Banach's fixed point theorem applies and even yields a procedure which converges exponentially fast to the unique fixed point.

By strong contractivity (or just basic properties of any norm - see my extensive commentary to exercise 9.8) we have

\[ D(\calE(\rho),\calE(\sigma)) \leq p D(\rho_0,\rho_0) + (1-p) D(\calE'(\rho),\calE'(\sigma)) \leq 0 + (1-p) D(\rho,\sigma) . \]

Hence the operation is strictly contractive. By exercise 9.10 there exists a unique fixed point. QED.

Of course one could use the fact that the depolarizing channel acts like

\[ (x,y,z) \mapsto (1-p)(x,y,z) \]

on the Bloch space together with the fact that this identification is an isometry - up to a factor of \(2\) (see equation (9.20)). But it also easily follows from just plugging stuff into other stuff:

\[ D(\calE(\rho),\calE(\sigma)) = 2\inv \norm{pI/2 + (1-p)\rho - pI/2 - (1-p)\sigma}_{S_1} = 2\inv (1-p) \norm{\rho-\sigma}_{S_1} = (1-p) D(\rho,\sigma) . \]

Of course it is also not hard to see that \(I/2\) is the fixed point and that the depolarizing channel contracts every other state towards it exponentially fast.

The bit flip channel is given by the following Kraus operators

\[ E_0 = \sqrt{p} \, I , \; E_1 = \sqrt{1-p} \, X . \]

The action on the Bloch sphere is easily seen to be (use \(XNX=-N\) for \(N\) being \(Y\) or \(Z\), and \(XXX=X\)):

\[ (x,y,z) \mapsto (x, (2p-1)y, (2p-1)z) . \]

Hence, if we are in the non-trivial case, \(p\neq1\), the set of fixed points is given by \(\{2\inv(1+xX)\,|\,x\in[-1,1]\}\) (we only consider density matrices). In the trivial case the bit flip is just the identity operation.

Recall that the spectral theorem implies \(f(UAU^\dagger)=Uf(A)U^\dagger\) for any continuous function \(f\) defined on the spectrum of \(A\) and any unitary \(U\). Actually this functional calculus is typically defined by a formula like

\[ f(A) = V f(D) V^\dagger \]

where \(D=V^\dagger\?AV\) is the diagonalization of \(A\) and \(f(D)\) is just applied entry-wise. So it is no surprise that the claimed formula \(\sqrt{UAU^\dagger}=U\sqrt{A}U^\dagger\) holds. But let us come back to the exercise:

For the first equality we applied the hint to \(\sqrt{\rho}\). For the third equality we applied it to \(\sqrt{\rho^{1/2}\sigma\rho^{1/2}}\). Finally we applied \(\trace{UAU^\dagger}=\trace{AU^\dagger\?U}=\trace{A}\), using the cyclicity property of the trace. QED.

To prove this we just need a minor modification of the proof of Theorem 9.4. So recall the proof given in the book. To give a bit of relevant context recall that

\[ \ket{\psi} = (U_R \otimes \sqrt{\rho} U_Q) \ket{m} \]

is a purification of \(\rho\) (and we have a similar formula for \(\ket{\varphi}\)). It is not hard to check that any unitary \(U_R\) and \(U_Q\) give rise to a purification. This is just a direct computation (hint: use \(\ptrace{R}{A}=\sum_i\bra{i_R}U_R^\dagger\?AU_R\ket{i_R}\) to compute the partial trace). That any purification on RQ has this form follows from exercise 2.81, but this is not really important here.

Fixing \(\ket{\psi}\) more or less means that we also fix \(U_R\), \(U_Q\). Actually, they are not really unique but we cannot easily tell which combinations are allowed so let us just assume they are fixed. On the other hand we are free to choose \(V_R\) and \(V_Q\) in any way because \(\ket{\varphi}\) can be chosen freely. Let

\[ U = V_Q V_R^\dagger U_R U_Q^\dagger , \]

as in the book. Note one subtle thing here: the operators must be treated as matrices here since they originally operate on different spaces (but this is explained in the book of course).

According to the book to obtain equality in \(F(\rho,\sigma)=\abs{\braket{\psi}{\varphi}}\) we have to accomplish that \(U=V^\dagger\) where \(\abs{\sqrt{\rho}\sqrt{\sigma}}V\) is the polar decomposition of \(\sqrt{\rho}\sqrt{\sigma}\). But this can easily be done by choosing e.g. \(V_R=I\) and

\[ V_Q = V^\dagger U_Q U_R^\dagger, \]

where \(V\) is expressed as a matrix with respect to the basis \((\ket{i_Q})_i\). QED.

This is just a short calculation

QED.

The first claim follows from the fact that \(\arccos:[0,1]\to[0,\pi/2]\) is a strictly decreasing bijection and that \(F\) has values in \([0,1]\). The second assertion follows from the fact that \(F(\rho,\sigma)=1\) iff \(\rho=\sigma\) by Uhlmann's theorem. QED.

This is a direct corollary of the monotonicity of the fidelity (Theorem 9.6) and the fact that \(\arccos:[0,1]\to[0,\pi/2]\) is decreasing (and \(F(\rho,\sigma)\in[0,1]\)). QED.

Follows from strict concavity with \(q_i=p_i\). QED.

Follows from exercise 9.19 with \(\sigma_i=\sigma\). QED.

The key to the proof is (9.60):

\[ F(\ket{\psi}, \rho)^2 = \bra{\psi} \rho \ket{\psi} . \]

Let us calculate:

In the first line we maximize over all projections. In the second line we set \(P=\proj{\psi}\). QED.

We have

\[ E(VU,\calF\circ\calE) = \max_{\rho} d(VU \rho U^\dagger V^\dagger, \calF(\calE(\rho))) . \]

Observe that

For the two equalities we used the assumed isometry property of the matric \(d\). The first inequality is just the triangle inequality. From this the claim directly follows. QED.

Let us recall the formula for \(\bar{F}\):

\[ \bar{F} = \sum_j p_j F(\rho_j, \calE(\rho_j))^2 . \]

Since \(0\leq\?F(\rho_j,\calE(\rho_j))\leq1\), and \(0\leq\?p_j\leq1\), and \(\sum_jp_j=1\) we see that \(\bar{F}=1\) is equivalent to

\[ \forall j: p_j > 0 \Rightarrow F(\rho_j, \calE(\rho_j)) = 1 . \]

On the other hand we already know that \(F(\rho,\sigma)=1\) is equivalent to \(\rho=\sigma\) (follows e.g. from Uhlmann's theorem). Hence \(\bar{F}=1\) is equivalent to

\[ \forall j: p_j > 0 \Rightarrow \rho_j = \calE(\rho_j) . \]

QED.