Exercise 2.64

I would provide two solutions to this problem which are equivalent with each other in some sense.

Solution One

Suppose we have \(m\) linearly independent vectors \(|\psi_1\rangle, |\psi_2\rangle, \dotsc, |\psi_m\rangle\), and they can be considered as a basis of vector space \(V\). Then, any subset of \(\{|\psi_j\rangle\}_{j=1}^{m}\) is also linearly independent and can be considered as a basis of a subspace. We could construct the POVM based on above property.

• For \(i = 1, \dotsc, m\), if define \(P_i\) as a projector to subspace \({\rm span}\{|\psi_j\rangle\}_{j=1, \dotsc, m\text{ and } j\neq i}\), then

  (1)\[ E_{i} = a_i(I - P_i), a_i > 0 \]

  Here \(I - P_i\) is an orthogonal complement of project \(P_i\) operator, and it is a projector to the subspace spanned by only \(|\psi_i\rangle\) according to the main text. Then the probability of getting \(E_i\) for state \(|\psi_i\rangle\), i.e., \(\langle \psi_i|E_i|\psi_i\rangle\) is non-zero, while the probability of getting outcome \(E_i\) if given state \(|\psi_j\rangle\) where \(i\neq j\) is zero since \(|\psi_j\rangle\) is not in the subspace spanned by only \(|\psi_i\rangle\). Thus, if outcome \(E_i\) occurs, Bob knows with certainty that he was given the state \(|\psi_i\rangle\).

  To construct projector \(P_i\), one can apply the Gram-Schmidt procedure to convert an ordinary basis \(\{|\psi_j\rangle\}_{j=1, \dotsc, m\text{ and } j\neq i}\) into an orthonormal basis \(\{|\phi_{ik}\rangle\}_{k=1, \dotsc, m-1}\). Then \(P_i = \sum_k |\phi_{ik}\rangle\langle \phi_{ik}|\).

• For \(i = m+1\), we define

  (2)\[ E_{m+1} = I - \sum_{i=1}^{m} E_i \]

  The last POVM operator is constructed to make sure the sum of all POVM operators is identity.

To illustrate that eq. (1) and eq. (2) give a valid POVM, we still need to verify that \(E_{i}\) is positive operator.

• For \(i = 1,\dotsc, m\), since projector \(P_i\) is Hermitian, then we can compute

  (3)\[ (I - P)^{\dagger}(I - P) = (I - P)(I - P) = I - P - P + P^2 = I - P \]

  Note that \((I - P)^{\dagger}(I - P)\) is positive due to Exer2.25, so \(I- P\) is also positive and \(E_i\) is positive for \(a_i> 0\).

• For \(i = m+1\), consider an arbitrary state \(|v\rangle\), then

  (4)\[ \langle v|E_{m+1}|v\rangle = \langle v|v\rangle - \sum_{i=1}^{m}\langle v|E_i|v\rangle \]

  and for positive \(E_i\) we have \(\langle v|E_i|v\rangle \geq 0\). By setting \(a_i\) as a small enough number, we can always have \(\langle v|E_{m+1}|v\rangle\geq 0\) and \(E_{m+1}\) can be positive.

Solution Two

We could also consider a POVM in the following form.

• For \(i = 1, \dotsc, m\), let

  (5)\[ E_i = a_i|v_i\rangle\langle v_i|, a_i > 0 \]

  where \(\langle v_i|\psi_j\rangle = 0\) for \(i\neq j\) and \(\langle v_i|\psi_i\rangle \neq 0\). From the definition, the probability of getting \(E_i\) for state \(|\psi_i\rangle\), i.e., \(\langle \psi_i|E_i|\psi_i\rangle = |\langle v_i|\psi_i\rangle|\) is non-zero, while the probability of getting outcome \(E_i\) if given state \(|\psi_j\rangle\) where \(i\neq j\) is zero. Thus, if outcome \(E_i\) occurs, Bob knows with certainty that he was given the state \(|\psi_i\rangle\).

  To construct \(|v_i\rangle\), one can first apply the Gram-Schmidt procedure to convert an ordinary basis \(\{|\psi_j\rangle\}_{j=1, \dotsc, m\text{ and } j\neq i}\) into an orthonormal basis \(\{|\phi_{ik}\rangle\}_{k=1, \dotsc, m-1}\), then perform the \(m-\)th step of Gram-Schmidt procedure to create \(|v_i\rangle\) from \(\{|\phi_{ik}\rangle\}_{k=1, \dotsc, m-1}\) and from \(|\psi_i\rangle\). Note that \(|v_i\rangle\) is orthonormal to \(\{|\phi_{ik}\rangle\}_{k=1, \dotsc, m-1}\) so \(I - P_i = |v_i\rangle\langle v_i|\) and the two solutions are equivalent with each other.

• For \(i = m+1\), we define

  (6)\[ E_{m+1} = I - \sum_{i=1}^{m} E_i\tag{4}\label{16} \]

  The last POVM operator is constructed to make sure the sum of all POVM operators is identity.

To prove that eq. (5) and eq. (6) gives a valid POVM measurement, we need to verify each \(E_i\) is positive operator. Consider an arbitrary state \(|\phi\rangle\), then

• For \(i = 1,\dotsc, m\),

  (7)\[ \langle \phi|E_i|\phi\rangle = a_i\langle \phi|v_i\rangle\langle v_i|\phi\rangle = a_i |\langle \phi|v_i\rangle|^2 \geq 0 \]

  which shows \(E_i\) is positive operator.

• For \(i = m+1\),

  (8)\[ \begin{align} \langle \phi|E_{m+1}|\phi\rangle &= \langle \phi|\phi\rangle - \sum_{i=1}^{m} a_i\langle \phi|v_i\rangle\langle v_i|\phi\rangle = \langle \phi|\phi\rangle -\sum_{i=1}^{m} a_i|\langle \phi|v_i\rangle|^2 \end{align} \]

  By setting \(a_i\) as a small enough number, we can always have \(\langle v|E_{m+1}|v\rangle\geq 0\) and \(E_{m+1}\) can be positive.