Worked Problems in Physics

Nielsen and Chuang Exercise 2.76

Posted in Nielsen/Chuang by rpmuller on June 6, 2010

Exercise 2.76 Extend the proof of the Schmidt decomposition to the case where {A} and {B} may have spaces of different dimensions.

We start with the state

\displaystyle  \left|\psi\right>=\sum_{jk}a_{jk}\left|j\right>\left|k\right>.

Here the matrix of values {a_{jk}} can be non-square rectangular. We can again do a SVD, yielding

\displaystyle  \left|\psi\right>=\sum_{ijk}u_{ji}d_{ii}v_{ik}\left|j\right>\left|k\right>.

The dimensions of {j} and {k} allow that the products {\left|i_A\right>=u_{ji}\left|j\right>} and {\left|i_B\right>=v_{ik}\left|k\right>} still make sense, and the diagonal index {i=\min(j,k)}.

Nielsen and Chuang Exercise 2.75

Posted in Nielsen/Chuang by rpmuller on June 1, 2010

Exercise 2.75 For each of the four Bell states, find the reduced density operator for each qubit.

\displaystyle  \rho^{00}=\frac{1}{2}\left(\left|00\right>\left<00\right|+\left|00\right>\left<11\right|+\left|11\right>\left<00\right|+\left|11\right>\left<11\right|\right)

\displaystyle  \rho^{00}_1 = \rho^{00}_2=\frac{1}{2}\left( \left|0\right>\left<0\right| + \left|1\right>\left<1\right| \right) = I/2.

\displaystyle  \rho^{01} = \frac{1}{2}\left(\left|00\right>\left<00\right|-\left|00\right>\left<11\right|-\left|11\right>\left<00\right|+\left|11\right>\left<11\right|\right)

\displaystyle  \rho^{01}_1 = \rho^{01}_2 = I/2.

\displaystyle  \rho^{10} = \frac{1}{2}\left(\left|10\right>\left<10\right|+\left|10\right>\left<01\right|+\left|01\right>\left<10\right|+\left|01\right>\left<01\right|\right)

\displaystyle  \rho^{10}_1 = \rho^{10}_2 = I/2.

\displaystyle  \rho^{11} = \frac{1}{2}\left(\left|10\right>\left<10\right|-\left|10\right>\left<01\right|-\left|01\right>\left<10\right|+\left|01\right>\left<01\right|\right)

\displaystyle  \rho^{11}_1 = \rho^{11}_2 = I/2.

Nielsen and Chuang Exercise 2.74

Posted in Nielsen/Chuang by rpmuller on June 1, 2010

Exercise 2.74 Suppose a composite of systems {A} and {B} is in the state {\left|a\right>\left|b\right>}, where {\left|a\right>} is a pure state of system {A}, and {\left|b\right>} is a pure state of system {B}. Show that the reduced density operator of system {A} alone is a pure state.

\displaystyle \rho = \left|a\right>\left|b\right>\left<a\right|\left<b\right|,

\displaystyle \mathrm{tr}_B(\rho) = \left|a\right>\left<a\right|\left<b|b\right>= \left|a\right>\left<a\right|,

and is thus a pure state.

Help Needed with Nielsen and Chuang Exercise 2.73

Posted in Nielsen/Chuang by rpmuller on June 1, 2010

Exercise 2.73 Given density matrix {\rho}, the minimal ensemble {\{p_k,\left|k\right>\}} contains a number of elements equal to the rank of {\rho}. Let {\left|i\right>} be any state in the support of {\rho}, where the support is defined as being spanned by the eigenvectors of {\rho} with nonzero eigenvalues. Show there exists a minimal ensemble for {\rho} that contains {\left|i\right>}, and that {\left|i\right>} appears with probability

\displaystyle  p_i = \frac{1}{\left<i|\rho^{-1}|i\right>}

where {\rho^{-1}} is the inverse that acts only on the support of {\rho}.

The density matrix {\rho} has the eigendecomposition {\rho=p_k\left|k\right>\left<k\right|}. We are given state {\left|i\right>} in the support of {\rho}, meaning that it is spanned by the eigenvectors with nonzero eigenvalues. This means that we can expand

\displaystyle \left|i\right> = \sum_{k'}\left<k'|i\right>\left|k'\right>= \sum_{k'}C_{k'i}\left|k'\right>

where the primed summation is only over those vectors in the support of {\rho}. Among other things, this means that the rank of {\left|i\right>} is no greater than {\rho}. Since we can expand {\left|i\right>} in the eigenvectors of {\rho}, this also means that {\left|i\right>} we can create a minimal ensemble for {\rho} than contains {\left|i\right>}.

Given {\rho} with rank {m}, consider the {m\times m} matrix {\sum_{k'}p_k\left|k\right>\left<k\right|}. We can write the projection of {\rho} onto {\left|i\right>} via

\displaystyle  \rho\left|i\right>=\sum_{kl'} C_{li}p_k\left|k\right>\left<k|l\right>=\sum_{k'}C_{ki}p_k\left|k\right>.

I need help here, if anyone has any suggestions. Just can’t push the equations home.

Nielsen and Chuang Exercise 2.72

Posted in Nielsen/Chuang by rpmuller on May 15, 2010

Exercise 2.72 Bloch sphere for mixed states. The Bloch sphere picture for pure states of a single qubit was introduced in Section 1.2. This description has an important generalization to mixed states as follows.

  1. Show that an arbitrary density matrix for a mixed state qubit may be written as

    \displaystyle \rho = \frac{I+\vec{r}\cdot\vec{\sigma}}{2},

    where {\vec{r}} is a real three-dimensional vector such that {||\vec{r}||\leq 1}. This vector is known as the Bloch vector for the state {\rho}.

  2. What is the Bloch vctor representation for the state {\rho=I/2}?
  3. Show that a state {\rho} is pure iff {||\vec{r}||=1}.
  4. Show that for pure states the description of the Bloch vector we have given coincides with that of Section 1.2.

Given that

\displaystyle  \left|\psi\right>=\cos\frac\theta 2\left|0\right>+e^{i\phi}\sin\frac\theta 2\left|1\right>,

we can write, for a pure state,

\displaystyle  \rho=\left|\psi\right>\left<\psi\right| = \left[\begin{array}{cc} \cos^2(\theta/2) & e^{-i\phi}\cos(\theta/2)\sin(\theta/2) \\ e^{i\phi}\cos(\theta/2)\sin(\theta/2) & \sin^2(\theta/2) \end{array} \right].

Massaging this a little bit

\displaystyle  \rho= \left[\begin{array}{cc} \cos^2(\theta/2) & \cos(\phi)\cos(\theta/2)\sin(\theta/2) - i\sin(\phi)\cos(\theta/2)\sin(\theta/2) \\ \cos(\phi)\cos(\theta/2)\sin(\theta/2) + i\sin(\phi)\cos(\theta/2)\sin(\theta/2) & 1-\cos^2(\theta/2) \end{array} \right].

We want this to have the form

\displaystyle  \rho= \frac{1}{2}\left[\begin{array}{cc} 1+r_3 & r_1-ir_2 \\ r_1+ir_2 & 1-r_3 \end{array} \right].

For this to work,

\displaystyle  2\cos^2(\theta/2) = 1+r_3,

\displaystyle r_3 = 2\cos^2(\theta/2)-1,

\displaystyle r_1 = 2\cos(\phi)\cos(\theta/2)\sin(\theta/2),

\displaystyle r_2 = 2\sin(\phi)\cos(\theta/2)\sin(\theta/2).

Looking at the norm of {r},

\displaystyle |r|^2 = 4\cos^4(\theta/2)-4\cos^2(\theta/2)+1 + 4\cos^2(\phi)\cos^2(\theta/2)\sin^2(\theta/2) + 4\sin^2(\phi)\cos^2(\theta/2)\sin^2(\theta/2)

\displaystyle =4\cos^4(\theta/2)-4\cos^2(\theta/2)+1+4\cos^2(\theta/2)\sin^2(\theta/2)(\cos^2(\phi)+\sin^2(\phi))

\displaystyle =4\cos^4(\theta/2)-4\cos^2(\theta/2)+1+4\cos^2(\theta/2)\sin^2(\theta/2)

\displaystyle =4\cos^2(\theta/2)(\cos^2(\theta/2)-1+\sin^2(\theta/2))+1

\displaystyle =4\cos^2(\theta/2)(\cos^2(\theta/2)-\cos^2(\theta/2))+1=1.

Thus, we have shown that a pure state corresponds to the proper form, with {|r|=1}, which completes goal 3 and goal 4.

Suppose we have a mixed state density matrix {[\rho=\sum_k p_k\left|k\right>\left<k\right|,} with {\sum_kp_k=1}. Such a state corresponds to

\displaystyle \rho=\frac{I+\sum_kp_k\vec{r_k}\cdot\vec\sigma}{2}.

We now have a linear combination of unit vectors {\sum_kp_k\vec{r_k}}, again with {\sum_kp_k=1}. This sum will correspond to some other vector {r}, with {0 \leq |r| \leq 1}. This completes goal 1 of the exercise.

The Bloch vector for the state {\rho=I/2} (goal 2) corresponds to {r=0}.

Nielsen and Chuang Exercise 2.71

Posted in Nielsen/Chuang by rpmuller on April 30, 2010

Exercise 2.71 Let {\rho} be a density operator. Show that {\mathrm{tr}(\rho^2)\leq 1}, with equality iff {\rho} is a pure state.

A density matrix is defined by {\rho=\sum_jp_j\left|j\right>\left<j\right|}, where {p_j} are nonnegative and {\sum_jp_j=1}. Its square is given by {\rho^2=\sum_{jk}p_jp_k\left|j\right>\left<j|k\right>\left<k\right|=\sum_jp_j^2\left|j\right>\left<j\right|}. Given the above constraints on {p_j}, {\sum_jp_j^2\leq 1}. If {\rho} is a pure state, {\rho^2=\rho}, and thus {p_j^2=p_j} for all {j}, meaning that {\sum_jp_j^2=1}.

Nielsen and Chuang Exercise 2.70

Posted in Nielsen/Chuang by rpmuller on April 15, 2010

Exercise 2.70 Suppose {E} is any positive operator acting on Alice’s qubit. Show that {\left<\psi|E\otimes I|\psi\right>} takes the same value when {\left|\psi\right>} is any of the four Bell states. Suppose a malevolent third part (‘Eve’) intercepts Alice’s qubit on the way to Bob in the superdense coding protocol. Can Eve infer anything about which of the four possible bit strings 00, 01, 10, 11 Alice is trying to send? If so, how so, or if not, why not?

For the first part,

  1. {\left<00+11|E\otimes I|00+11\right> = \left<0|E|0\right>\left<1|1\right> 		+ \left<0|E|1\right>\left<1|0\right>+ \left<1|E|0\right>\left<0|1\right> 		+ \left<1|E|1\right>\left<0|0\right> = \left<0|E|0\right>+\left<1|E|1\right>}
  2. {\left<00-11|E\otimes I|00-11\right> = \left<0|E|0\right>+\left<1|E|1\right>}
  3. {\left<01+10|E\otimes I|01+10\right> = \left<0|E|0\right>+\left<1|E|1\right>}
  4. {\left<01-10|E\otimes I|01-10\right> = \left<0|E|0\right>+\left<1|E|1\right>}

If Eve eavesdrops, she can only measure one qubit, and, as we show above, this yields the same result for all four states. However, if Bob has both states, he can do a measurement {E_1\otimes E_2} on both qubits

\displaystyle \left<00+11|E_1\otimes E_2|00+11\right> = \left<0|E_1|0\right>\left<1|E_2|1\right> 	+ \left<0|E_1|1\right>\left<1|E_2|0\right>

\displaystyle + \left<1|E_1|0\right>\left<0|E_2|1\right> + \left<1|E_1|1\right>\left<0|E_2|0\right>

which will not be the same for all four qubits.

Nielsen and Chuang Exercise 2.69

Posted in Nielsen/Chuang by rpmuller on April 15, 2010

Exercise 2.69 Verify that the Bell basis forms an orthonormal basis for the two qubit state space.

The Bell basis is given by

  1. {\Psi_{00}= (\left|00\right>+\left|11\right>)/\sqrt{2}}
  2. {\Psi_{01}= (\left|00\right>-\left|11\right>)/\sqrt{2}}
  3. {\Psi_{10}= (\left|10\right>+\left|01\right>)/\sqrt{2}}
  4. {\Psi_{11}= (\left|01\right>-\left|10\right>)/\sqrt{2}}

We note that, for the two electron states

  1. {\left<00|00\right>=1}
  2. {\left<00|11\right>=0}
  3. {\left<00|10\right>=0}
  4. {\left<00|01\right>=0}
  5. {\left<11|11\right>=1}
  6. {\left<11|10\right>=0}
  7. {\left<11|01\right>=0}
  8. {\left<10|10\right>=1}
  9. {\left<10|01\right>=0}
  10. {\left<01|01\right>=1}

from which it is easy to show that {\left<i|j\right>=\delta_{ij}} for {i} and {j} among the Bell states, and thus that the states form an orthonormal basis.

Nielsen and Chuang Exercise 2.68

Posted in Nielsen/Chuang by rpmuller on April 13, 2010

Exercise 2.68 Prove that {(\left|00\right>+\left|11\right>)/\sqrt{2}\neq\left|a\right>\left|b\right>} for all single qubit states {\left|a\right>} and {\left|b\right>}.

Suppose that {\left|a\right>=c\left|0\right>+d\left|1\right>} and {\left|b\right>=e\left|0\right>+f\left|1\right>}. Then

\displaystyle \left|ab\right>=ce\left|00\right>+cf\left|01\right>+de\left|10\right>+df\left|11\right>.

For this to be equal to the desired state, {cf=0}, which means either {c=0} or {f=0}. However, if {c=0}, then {ce\neq 1/\sqrt{2}}, and if {f=0}, then {df\neq 1/\sqrt{2}}. Which makes it impossible for the product of two states to be equal to {(\left|00\right>+\left|11\right>)/\sqrt{2}}.

Nielsen and Chuang Exercise 2.67

Posted in Nielsen/Chuang by rpmuller on April 13, 2010

Exercise 2.67 Suppose {V} is a Hilbert space with a subspace {W}. Suppose {U: W\rightarrow V} is a linear operator which preserves inner products, that is, for any {\left|w_1\right>} and {\left|w_2\right>} in {W},

\displaystyle  \left<w_1|U^\dagger U|w_2\right> = \left<w_1|w_2\right>.

Prove that there exists a unitary operator {U':V\rightarrow V} which extends {U}. That is, {U'\left|w\right>=U\left|w\right>} for all {\left|w\right>} in {W}, but {U'} is defined on the entire space {V}. Usually we omit the prime symbol {'} and just write {U} to denote the extension.

This is yet another of these exercised whose point I suspect that I am missing. However, I note that the composite operator {U'=U\otimes I}, where the identity is defined only on {V \perp W}, preserves inner products and is unitary provided that {U} is unitary.