from itertools import product import sympy as sp from sympy import Matrix, PermutationMatrix, sqrt from sympy.combinatorics import Permutation
For convenience we define two functions which map bit-strings to the corresponding bra and ket vectors.
def ket(bits: str) -> Matrix: """If bits is a string of '0' and '1' this returns |bits>.""" vec = [0] * 2**len(bits) vec[int(bits, 2)] = 1 return Matrix(vec) def bra(bits: str) -> Matrix: """If bits is a string of '0' and '1' this returns <bits|.""" return ket(bits).H
In the following we define the trace for arbitrary square matrices and the partial trace for square matrices whose dimension is a power of two. Note that the concept of the partial trace only makes sense if one can consider the matrix as element of some tensor product (i.e. the dimension is a product \(m_1\cdot m_2\cdot\ldots\)). We further restrict to \(m_i=2\) to keep the implementation and the interface simple and since probably nothing else is needed in the context of the book.
def trace(M: Matrix): # TODO: what is the correct return type? """Returns the sum of the diagonal elements. Requires a square matrix.""" dim = M.shape[0] assert dim == M.shape[1], "M must be square" result = 0 for i in range(dim): result += M[i, i] return result def ptrace(M: Matrix, bit_positions: list[int] = None) -> Matrix: """Partial Trace. The Matrix must have dimension 2^n. The bit_positions describe which bits (from the right) get traced out.""" dim = M.shape[0] assert dim == M.shape[1], "M must be square" assert dim.bit_count() == 1, "Dim of M must be 2^n for some n >= 0" bit_size = dim.bit_length() - 1 if bit_positions is not None: assert len(bit_positions) > 0, "bit_positions must not be empty" assert all(0 <= pos <= bit_size for pos in bit_positions), "Invalid bit_positions" if bit_positions is None: # shortcut return Matrix([[trace(M)]]) if len(bit_positions) == 1 and dim == 2: # corner case return Matrix([[trace(M)]]) elif len(bit_positions) > 1: ps = bit_positions.copy() ps.sort() # IMPORTANT! RM = M while len(ps) > 0: pos = ps.pop() RM = ptrace(RM, [pos]) return RM rpos = bit_size - 1 - bit_positions[0] mat = [[0]*(dim//2) for _ in range(dim//2)] for i, j in product(range(dim//2), range(dim//2)): ibits = f"{i:b}".zfill(bit_size - 1) jbits = f"{j:b}".zfill(bit_size - 1) i0 = int(ibits[:rpos] + "0" + ibits[rpos:], 2) i1 = int(ibits[:rpos] + "1" + ibits[rpos:], 2) j0 = int(jbits[:rpos] + "0" + jbits[rpos:], 2) j1 = int(jbits[:rpos] + "1" + jbits[rpos:], 2) mat[i][j] = M[i0, j0] + M[i1, j1] return Matrix(mat)
As a quick demo let us compute some partial traces of the following matrices:
A = Matrix([[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14, 15]]) B = Matrix([[8*i + j for j in range(8)] for i in range(8)])
Note that the matrix B has the same structure as as A, just bigger (it counts from 0
to 63).
A0 = Matrix([[5, 9], [21, 25]]) B0 = Matrix([[9, 13, 17, 21], [41, 45, 49, 53], [73, 77, 81, 85], [105, 109, 113, 117]]) A1 = Matrix([[10, 12], [18, 20]]) B1 = Matrix([[18, 20, 26, 28], [34, 36, 42, 44], [82, 84, 90, 92], [98, 100, 106, 108]]) B2 = Matrix([[36, 38, 40, 42], [52, 54, 56, 58], [68, 70, 72, 74], [84, 86, 88, 90]]) assert ptrace(A, [0]) == A0 assert ptrace(A, [1]) == A1 assert ptrace(B, [0]) == B0 assert ptrace(B, [1]) == B1 assert ptrace(B, [2]) == B2 "PASSED"
PASSED
Let us briefly outline a method of computing the partial trace with the help of standard
sympy functionality. I used this method to quickly calculate the asserted results in the
listing above - independently of ptrace.
For bit_positions == [0], it is easy to read out the results from the input
matrix. Essentially you only have to to view the matrix as a block-matrix of 2-by-2
matrices which you replace by their traces.
To reduce the case of e.g. bit_positions == [2] to bit_positions == [0] use the fact
that permuting the bit pattern like so 'xyz' -> 'zyx' induces a Permutation on the basis
vectors which corresponds to a permutation matrix \(M_1\). After doing the partial trace we
are left with 'zy' and have to permute this back to 'yz' (Matrix \(M_2\)). In other words we
have the identity
\[ \ptrace{i}{B} = M_2 \, \ptrace{0}{M_1 B M_1} \, M_2 . \]
Note that it is always guaranteed that \(M_1\) and \(M_2\) are their own inverses. Hence no adjoint operation is needed. The following code shows how simple this is:
m1 = Permutation(7)(1,4)(3,6) m2 = Permutation(3)(1,2) M1 = PermutationMatrix(m1).as_explicit() M2 = PermutationMatrix(m2).as_explicit() assert ptrace(B, [2]) == M2*ptrace(M1*B*M1, [0])*M2 "PASSED"
PASSED
Finally, ptrace(B, [0,1]) is the same as ptrace(ptrace(B, [1]), [0]).
assert ptrace(B, [0, 2]) == ptrace(ptrace(B, [2]), [0]) # easier to do 2 first, then 0 because ... assert ptrace(B, [0, 2]) == ptrace(ptrace(B, [0]), [1]) # ... otherwise needed to adjust indices assert ptrace(B, [2, 0]) == ptrace(ptrace(B, [2]), [0]) assert ptrace(B, [0, 1, 2]) == Matrix([[trace(B)]]) assert ptrace(B, [2, 1, 0]) == Matrix([[trace(B)]]) "PASSED"
PASSED
Let \(\rho\) be a density operator. Show that \(\trace{\rho^2}\leq1\), with equality if and only if \(\rho\) is a pure state.
By the characterization of density operators (Theorem 2.5) there are \(0 < p_i\leq 1\) and orthonormal \(\{|\psi_i\rangle\}\) such that:
\[ \rho = \sum_i p_i |\psi_i\rangle\langle\psi_i| . \]
Hence:
\[ \trace{\rho^2} = \trace{ \sum_i p_i^2 |\psi_i\rangle\langle\psi_i| } = \sum_i p_i^2 \leq \sum_i p_i = 1 , \]
which proves the first claim. Equality can clearly only hold if for all \(i\): \(p_i=1\). By the trace condition this is only possible if there is only one summand in the sum making up \(\rho\). This means that \(\rho\) is pure in that case. QED.
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.
Show that an arbitrary density matrix for a mixed state qubit may be written as
\[ \rho = \frac{I + \overrightarrow{r} \cdot \overrightarrow{\sigma} }{2} \]
where \(\overrightarrow{r}\) is a real three-dimensional vector such that \(\norm{\overrightarrow{r}}\leq1\). This vector is known as the Bloch vector for the state \(\rho\). Conversely any such \(\rho\) is also a density operator.
The matrix representation of \(\rho\) looks as follows:
\[ \rho = \frac{1}{2} \begin{bmatrix} 1 + z & x - iy \\ x + iy & 1 - z \end{bmatrix} = \begin{bmatrix} p & \frac{1}{2}\zeta^* \\ \frac{1}{2}\zeta & q \end{bmatrix} . \]
Let \(r=\norm{\overrightarrow{r}}\). We see that the matrices \(D\) which can be represented by \(\rho\) are characterized by the above matrix together with the constraints:
\[ p, q \geq 0,\; \zeta\in\CC \text{ with } p + q = 1 \text{ and } r = (p - q)^2 + |\zeta|^2 \leq 1 . \]
Now let \(|\psi\rangle=e^{i\gamma}(c|0\rangle+e^{i\varphi}s|1\rangle)\) be a pure state (\(c=\cos(\theta/2)\) and \(s=\sin(\theta/2)\)). The corresponding density matrix is:
\[ \,|\psi\rangle\langle\psi| = \begin{bmatrix} c^2 & e^{-i\varphi}cs \\ e^{i\varphi}cs & s^2 \end{bmatrix} . \]
Recall the following trigonometric identities: \(\cos\theta=c^2-s^2\) and \(\sin\theta=2cs\). This implies that the above constraint is satisfied. Hence pure states are in \(D\). In particular we have \(r=1\) for pure states.
On the other hand the same trigonometric identities also show that the constraint strengthened by \(r=1\) characterizes the pure states. Indeed, the correspondence between the two representations is given by \(x+iy=\zeta=e^{i\varphi}\sin\theta\) and \(z=p-q=\cos\theta\). This shows claims 3 and 4 (for the latter recall that the Bloch-Sphere is parameterized by \((x,y,z)=(\sin\theta\cos\varphi,\;\sin\theta\sin\varphi,\;\cos\theta)\)).
Clearly \(D\) is a convex set, hence all density matrices are in \(D\). This follows from the fact that density operators are the (closed) convex hull of the pure states. This shows the original version of claim 1 (we extended it a bit).
To show the rest of claim 1 it remains to prove that \(D\) contains only density operators.
That \(D\) contains only hermitian operators with trace=1 is clear. By exercise 4.5 we have \((\overrightarrow{r} \cdot \overrightarrow{\sigma})^2=\norm{\overrightarrow{r}}^2I\). Since \(\overrightarrow{r} \cdot \overrightarrow{\sigma}\) is hermitian and not a multiple of \(I\) we deduce that it has the two eigenvalues \(\pm\norm{\overrightarrow{r}}\). Hence (see section on notation for the meaning):
\[ 0 \leq \frac{1 - \norm{\overrightarrow{r}}}{2} \leq \rho \leq \frac{1 + \norm{\overrightarrow{r}}}{2} \leq 1 . \]
In particular, \(\rho\) is indeed a positive operator. Hence, \(\rho\) is a density matrix. This shows the rest of claim 1.
\(I/2\) is represented by \(\overrightarrow{r}=(0,0,0)\). It corresponds to the ensemble \(\{(1/2,|0\rangle),(1/2,|1\rangle)\}\).
Let \(\rho\) be a density operator. A minimal ensemble for \(\rho\) is an ensemble \(\{(p_i, |\psi_i\rangle)\}\) containing a number of elements equal to the rank of \(\rho\). Let \(|\psi\rangle\) be any state in the support of \(\rho\) . (The support of a Hermitian operator \(A\) is the vector space spanned by the eigenvectors of \(A\) with non-zero eigenvalues.) Show that there is a minimal ensemble for \(\rho\) that contains \(|\psi\rangle\), and moreover that in any such ensemble \(|\psi\rangle\) must appear with probability
\[ p = \frac{1}{\langle\psi|\rho^{-1}|\psi\rangle} , \]
where \(\rho^{-1}\) is defined to be the inverse of \(\rho\), when \(\rho\) is considered as an operator acting only on the support of \(\rho\). (This definition removes the problem that \(\rho\) may not have an inverse.)
Consider the representation of \(\rho\) by its eigenvectors (with non-zero eigenvalues):
\[ \rho = \sum_{i=1}^n \lambda_i |i\rangle\langle i| . \]
Since \(|\psi\rangle\) is in the support of \(\rho\) there are unique complex numbers \(v_1,\ldots,v_n\) with:
\[ \sum_{i=1}^n v_i \sqrt{\lambda_i} |i\rangle \in \RR |\psi\rangle \text{ and } \sum_{i=1}^n |v_i|^2 = 1 . \]
Now extend these numbers to a unitary matrix \(u\in\CC^{n\times n}\) such that \(u_{ij}=v_j\) (formally the existence follows from the Gram-Schmidt procedure). Let
\[ \sqrt{p_i} |\psi_i\rangle := \sum_{i=1}^n u_{ij} \sqrt{\lambda_j} |j\rangle . \]
By construction \(|\psi\rangle=|\psi_1\rangle\). By Theorem 2.6 (Unitary Freedom in the ensemble for density matrices) we found an ensemble containing \(|\psi\rangle\) so the existence is established. Conversely, the same Theorem states that any ensemble representing \(\rho\) and satisfying \(|\psi_1\rangle=|\psi_1\rangle\) must satisfy the above constraint for a matrix \(u\) whose first row is \(v\).
By the above contraint and the spectral theorem we have:
\[ \sqrt{p_i p_j} \langle\psi_i|\rho^{-1}|\psi_j\rangle = \sum_{kl} u_{ik}^\dagger \sqrt{\lambda_k} \langle k| \rho^{-1} u_{jl} \sqrt{\lambda_l} |l\rangle = \sum_{k} u_{jk} u_{ki}^* \lambda_k \langle k|\rho^{-1}|k\rangle = \delta_{ij} . \]
Setting \(i=j=1\) proves the claim. QED.
Remark: The proof shows that each minimal ensemble \((\sqrt{p_i}|\psi_i\rangle)\) is an orthonormal basis on the support of \(\rho\) with respect to the inner product given by \(\langle\varphi|\rho^{-1}|\psi\rangle\).
Suppose a composite of systems A and B is in the state \(|a\rangle|b\rangle\), where \(|a\rangle\) is a pure state of system A, and \(|b\rangle\) is a pure state of system B. Show that the reduced density operator of system A alone is a pure state.
The joint state is indeed a simple tensor product (i.e. a product state):
\[ \,|a\rangle |b\rangle = |a\rangle \otimes |b\rangle . \]
The corresponding density matrix is:
\[ \rho = |a\rangle\langle a| \otimes |b\rangle\langle b| . \]
The reduced density operator for A is obtained by taking the partial trace with respect to B:
\[ \rho_A = \ptrace{B}{\rho} = |a\rangle\langle a|. \]
This is indeed the pure state \(|a\rangle\). QED.
For each of the four Bell states, find the reduced density operator for each qubit.
The four Bell States are given by:
\[ \,|\beta_{xy}\rangle = \frac{1}{\sqrt{2}} \left( |0y\rangle + (-1)^{x} |1\overline{y}\rangle \right) , \]
where \(\overline{y}\) denotes the negation of \(y\). Using \(\trace{|i\rangle\langle j|}=\delta_{ij}\) we deduce:
\[ \ptrace{2}{|\beta_{xy}\rangle} = \frac{1}{2} (|0\rangle\langle0| + |1\rangle\langle1|) = \frac{1}{2} I . \]
and in the same way:
\[ \ptrace{1}{|\beta_{xy}\rangle} = \frac{1}{2} (|y\rangle\langle y| \,+\, |\overline{y}\rangle\langle\overline{y}|) = \frac{1}{2} I . \]
So in all 8 cases we get the same result: \(I/2\).
Remark: We see that knowledge of all partial traces is not sufficient to reconstruct the original state.
Extend the proof of the Schmidt decomposition to the case where A and B may have state spaces of different dimensionality.
Of course one may reuse the proof from the book by utilizing the general form of the singular value decompostion for non-square matrices.
Alternatively: Assume wlog \(\dim(H_A)\leq\dim(H_B)\). We already have a proof for the case of equal dimensions hence assume \(\dim(H_A)<\dim(H_B)\). Now we extend A by a direct sum with a Hilbert Space \(R\) such that:
\[ \dim(H_A \oplus R) = \dim(H_B) . \]
By the already proved special case of the theorem we get a Schmidt-Decomposition on \((H_A\oplus R)\otimes H_B\):
\[ \,|\psi\rangle = \sum_i \lambda_i \, |i_A\rangle |i_B\rangle . \]
We have to prove that \(|i_A\rangle\in H_A\) for all \(i\). Indeed, let \(|r\rangle\in R\) and consider:
\[ \langle i_A|r\rangle = \langle i_A, i_B | r, i_B \rangle = \sum_j \langle j_A, j_B | r, i_B \rangle = \langle \psi | r, i_B \rangle = 0 . \] This shows the claim. QED.
Suppose ABC is a three component quantum system. Show by example that there are quantum states of such systems which can not be written in the form
\[ \,|\psi\rangle = \sum_i \lambda_i |i_A, i_B, i_C\rangle \]
where \(\lambda_i\) are real numbers, and \(|i_A\rangle\), \(|i_B\rangle\), \(|i_C\rangle\) are orthonormal vectors of the respective systems.
Suppose a Schmidt-like decomposition is possible for a state with respect to ABC. Then the reduced density operators in A, B and C must have the same set of eigenvalues. Hence, to show that such a decomposition is not possible for a given state it suffices to show that the sets of eigenvalue are not all equal.
To spare us boring calculations lets introduce a nice function which takes a state vector of a three-qubit system ABC as input (this is already the simplest case where we can find counter examples). As output it reports the eigenvalues of the density matrix of each subsystem.
def get_eigenvals_of_ABC(state_vector: Matrix): """Calculate eigenvals of subsystems of three-qubit system (for Exercise 2.77).""" density = state_vector * state_vector.H DA = ptrace(density, [1, 2]) DB = ptrace(density, [0, 2]) DC = ptrace(density, [0, 1]) result = "" for S, D in [("A", DA), ("B", DB), ("C", DC)]: result += f"Eigenvalues in {S}: {D.eigenvals()}.\n" return result
The following code-snippit shows that
\[ \frac{1}{\sqrt{3}} (|000\rangle + |010\rangle + |011\rangle) \]
cannot be decomposed as required:
get_eigenvals_of_ABC((ket('000') + ket('010') + ket('011')) / sqrt(3))
Eigenvalues in A: {1/2 - sqrt(5)/6: 1, sqrt(5)/6 + 1/2: 1}.
Eigenvalues in B: {1/2 - sqrt(5)/6: 1, sqrt(5)/6 + 1/2: 1}.
Eigenvalues in C: {1: 1, 0: 1}.
QED.
Prove that a state \(|\psi\rangle\) of a composite system AB is a product state if and only if it has Schmidt number 1. Prove that \(|\psi\rangle\) is a product state if and only if \(\rho^A\) (and thus \(\rho^B\)) are pure states.
A product state is a state which can be represented by a simple tensor product \(|\alpha\rangle\otimes|\beta\rangle\). But this is by definition the same as having Schmidt-Rank 1.
For the second part consider the Schmidt-Decomposition:
\[ \, |\psi\rangle = \sum_{i=1}^r \lambda_i |i_A\rangle \otimes |i_B\rangle . \]
Hence the reduced state on system A is (take the partial trace on B):
\[ \rho^A = \sum_{i=1}^r \lambda_i^2 |i_A\rangle\langle i_A| . \]
Since \(\sum_i\lambda_i^2=1\) and \(\lambda_i>0\) this can only be a pure state (a projection operator, i.e. \(\rho^2=\rho\)) if \(r=1\). This is the same as \(|\psi\rangle\) being a product state. QED.
Consider a composite system consisting of two qubits. Find the Schmidt decompositions of the states
\[ \frac{|00\rangle + |11\rangle}{\sqrt{2}}; \quad \frac{|00\rangle + |01\rangle + |10\rangle + |11\rangle}{2}; \quad \text{and} \quad \frac{|00\rangle + |01\rangle + |10\rangle}{\sqrt{3}} . \]
The first state clearly is already represented as Schmidt-Decomposition.
The second term has Schmidt-Decomposition \(|+\rangle\otimes|+\rangle\), where
\(|+\rangle=(|0\rangle+|1\rangle)/\sqrt{2}\) is "the" positive eigenvector of the Pauli Operator X.
For the third one we have to calculate something. The proof of the existence of the Schmidt-Decomposition gives a procedure to actually calculate it. First let us find the Matrix \(A\in\CC^{2\times2}\) such that:
\[ \frac{|00\rangle + |01\rangle + |10\rangle}{\sqrt{3}} = \sum_{ij} A_{ij} |i\rangle \otimes |j\rangle . \]
Clearly the following is the only matrix satisfying the above equation:
\[ A = \frac{1}{\sqrt{3}} \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} . \]
We need to find the singular value decomposition of \(A\). Since \(A\) is hermitian this simplifies to find the spectral decomposition \(A=UDU^\dagger\).
One can use sympy to do this. The eigenvalues are:
\[ \lambda_{\pm} = \frac{\sqrt{3}}{6} \left( 1 \pm \sqrt{5} \right) . \]
"The" eigenvectors are:
\[ \,|\pm_A\rangle = \frac{1}{\sqrt{10 \pm 2\sqrt{5}}} \left((1 \pm \sqrt{5})|0\rangle + 2|1\rangle \right) . \]
With this we can write down the Schmidt-Decomposition:
\[ \frac{|00\rangle + |01\rangle + |10\rangle}{\sqrt{3}} = \lambda_+ |+_A\rangle|+_A\rangle + \lambda_- |-_A\rangle|-_A\rangle . \]
Note that the fact that the eigenvectors can be chosen with real entries simplifies things.
Suppose \(|\psi\rangle\) and \(|\varphi\rangle\) are two pure states of a composite quantum system with components A and B, with identical Schmidt coefficients. Show that there are unitary transformations \(U\) on system A and \(V\) on system B such that \(|\psi\rangle = (U \otimes V )|\varphi\rangle\).
That \(|\psi\rangle\) and \(|\varphi\rangle\) have the same Schmidt coefficients means that there are positive numbers \(\lambda_i\) such that
\begin{align*} \, |\psi\rangle &= \sum_{i=1}^r \lambda_i |i_{A,\psi}\rangle |i_{B,\psi}\rangle , \\ \, |\varphi\rangle &= \sum_{i=1}^r \lambda_i |i_{A,\varphi}\rangle |i_{B,\varphi}\rangle , \end{align*}for some orthonormal sets \((|i_{A,\psi}\rangle)\), \((|i_{A,\varphi}\rangle)\), \((|i_{B,\psi}\rangle)\), \((|i_{B,\varphi}\rangle)\). By the latter fact it is easy to find a unitary \(U\) which maps \(|i_{A,\varphi}\rangle\) to \(|i_{A,\psi}\rangle\) and similarly a \(V\) for the B-system. These two unitary operators satisfy the claim.
Let \(|AR_1\rangle\) and \(|AR_2\rangle\) be two purifications of a state \(\rho^A\) to a composite system AR. Prove that there exists a unitary transformation \(U_R\) acting on system R such that
\[ \,|AR_1\rangle = (I_A \otimes U_R ) |AR_2\rangle . \]
Let
\[ \rho^A = \sum_{i=1}^r \lambda_i^2 |i_A\rangle\langle i_A| \]
be a Spectral-Decomposition of \(\rho^A\) (in particular all the vectors involved form an orthonormal set). Without loss of generality we may assume that \(|AR_2\rangle\) adheres to the generic construction from the book, that is
\[ \, |AR_2\rangle = \sum_{i=1}^r |i_A\rangle \otimes |i_R\rangle \]
with an orthonormal set \((|i_R\rangle)\). A priori the Hilbert Space of \(R\) could have dimension smaller then \(r\). But in that case we could simply augment \(R\) by some additional dimensions.
Since \(|AR_1\rangle\) and \(|AR_2\rangle\) are purifications of the same state they have the same Schmidt Numbers (the \(\lambda_i\) from above). From Exercise 2.80 we deduce that there exist unitary matrices \(V_A\) and \(V_R\) such that
\[ \,|AR_1\rangle = (V_A \otimes V_R ) |AR_2\rangle . \]
To simplify notation let us write \(V=V_A\). Slightly abusing notation, let us write \(\ptrace{R}{|\psi\rangle}\) when we mean \(\ptrace{R}{|\psi\rangle\langle\psi|}\). We have
\[ \rho^A = \ptrace{R}{|AR_1\rangle} = \ptrace{R}{V\otimes V_R \, |AR_2\rangle} = V^\dagger \rho^A V . \]
This implies that \(V\) leaves the eigenspaces of \(\rho^A\) invariant (that is, an eigenvector of some eigenvalue is mapped to a possibly different eigenvector of the same eigenvalue).
For the moment let us assume that all \(\lambda_i\) are equal, that is \(\lambda_i=1/r\). Let \((v_{ij})\in\CC^{r\times r}\) be the matrix representation of \(V\) on the eigenspace of \(1/r\). We have:
\[ \, |AR_1\rangle = V \otimes V_R \, |AR_2\rangle = \frac{1}{r} \sum_{ij} v_{ji} |j_A\rangle \otimes V_R|i_R\rangle = \frac{1}{r} \sum_{j} |j_A\rangle \otimes \left( \sum_i v_{ji} V_R|i_{R}\rangle \right) . \]
Let \(|j_{R'}\rangle=\sum_{i}v_{ji}V_R|i_{R}\rangle\). These vectors form an orthonormal set since \((v_{ij})\) is a unitary matrix (and transposed unitary matrices are unitary too) and the \((|i_R\rangle)\) are orthonormal. Hence there is a unitary matrix \(U_R\) such that \(|j_{R'}\rangle=U|j_R\rangle\). This together with the previous equation implies the desired result:
\[ \, |AR_1\rangle = (I_A \otimes U_R) |AR_2\rangle . \]
The general case follows by decomposing \(|AR_1\rangle\) and \(|AR_2\rangle\) into direct sums over the eigenspaces of \(\rho^A\) and repeating the above reasoning for each one. Note that in the above reasoning for the special case the particular value of the trace was unimportant! QED.
Suppose \(\{p_i , |\psi_i\rangle\}\) is an ensemble of states generating a density matrix \(\rho\) for a quantum system A. Introduce a system R with orthonormal basis \(|i\rangle\).
Remark (which is missing in the book): In sub-exercise 3 one may assume that the space \(R\) is large enough to hold all the \(|i\rangle\) even if the purification does not need that many dimensions (two ensembles always have at least \(r\) (rank) members but might have more, see also Theorem 2.6).
This was essentially shown in the book right before Exercise 2.79! In the book it was implicitly assumed that the \(|\psi\rangle\) are orthonormal (by using the Schmidt-Decomposition of \(\rho\)). But the proof did not use this fact. QED.
Recall that:
\[ \,|AR\rangle = \sum_i \sqrt{p_i} |\psi_i\rangle |i\rangle . \]
The measurement of \(i\) as viewed from the whole system AR is represented by \(I\otimes|i\rangle\langle i|\) (see Box 2.6). Hence the probability to measure \(i\) is:
\[ p(i) = \langle AR|I\otimes|i\rangle\langle i| AR\rangle = p_i . \]
The post-measurement state is:
\[ \frac{1}{p(i)} I\otimes|i\rangle\langle i| AR\rangle = |\psi\rangle|i\rangle . \]
Restricted to A this is corresponds to \(|\psi\rangle\) (note that this does not directly follow from Postulate 4 but it is easy to show that product state behave nice under decomposition into the corresponding subsystems). QED.
Although it is not asked for, here is the state of R expressed as a density matrix (before measurement):
\[ \rho^R = \ptrace{A}{|AR\rangle\langle AR|} = \sum_{ij} \sqrt{p_i p_j} \langle\psi_i|\psi_j\rangle |i\rangle\langle i| . \]
It is nice to verify that the after measurement results restricted to R are consistent with the above:
\[ p(i) = p_i = \trace{|i\rangle\langle i| \rho^R} , \]
and
\[ \, |i\rangle = \frac{1}{p_i} |i\rangle\langle i| \rho^R |i\rangle\langle i| . \]
Unfortunately there is a notation clash here. Let us rename the orthonormal set used in part 1 to \((|i_0\rangle)\) and the corresponding purified vector to \(|AR_0\rangle\). By exercise 2.81 (freedom of purifications) there is a unitary operator \(U_R\) such that:
\[ |AR\rangle = I\otimes U_R \; |AR_0\rangle \]
Hence we may use the basis \((|i\rangle=U_R|i_0\rangle)\) to achieve the desired goal.
But note a subtle issue here: A priori \(U_R\) is not necessarily defined on each individual \(|i_0\rangle\). The reason is, that each density operator has ensembles with arbitrary many states (of course they are all linearly dependent to some \(r\) dimensional set due to Theorem 2.6). This implies that R0 can be arbitrary large if one takes the construction in sub-exercise 1. On the other hand one can just "add some dimensions" to R if necessary and extend \(U_R\) unitarily. QED.