Partial Trace & Reduced States

1. 1
   Step 1:

   $$\mathrm{Tr}\big((M\otimes I)\rho_{AB}\big)=\sum_{a,b}\langle a\,b\vert(M\otimes I)\rho_{AB}\vert a\,b\rangle$$

   Trace in the product basis $\{\lvert ab\rangle\}$ of the composite space.
2. 2
   Step 2:

   $$=\sum_{a,b}\sum_{a'',b''}\langle ab\vert M\otimes I\vert a''b''\rangle\,\langle a''b''\vert\rho_{AB}\vert ab\rangle=\sum_{a,b}\sum_{a''}M_{aa''}\,\rho_{(a'',b),(a,b)}$$

   Insert the resolution of the identity $\sum_{a'',b''}\lvert a''b''\rangle\langle a''b''\rvert$; the Kronecker index formula gives $\langle ab\vert M\otimes I\vert a''b''\rangle=M_{aa''}\delta_{bb''}$.
3. 3
   Step 3:

   $$=\sum_{a,a''}M_{aa''}\Big(\sum_b\rho_{(a'',b),(a,b)}\Big)=\sum_{a,a''}M_{aa''}(\rho_A)_{a''a}=\mathrm{Tr}(M\rho_A)$$

   Do the $b$ sum first and recognise the component form of $\rho_A$; the last expression is the trace of the matrix product $M\rho_A$. This proves existence.
4. 4
   Step 4:

   $$\mathrm{Tr}(M\sigma)=\mathrm{Tr}(M\rho_A)\ \ \forall M\ \Rightarrow\ \mathrm{Tr}\big(M(\sigma-\rho_A)\big)=0\ \ \forall M$$

   For uniqueness, subtract the two identities (linearity of the trace).
5. 5
   Step 5:

   $$M=\lvert a'\rangle\langle a\rvert:\quad\mathrm{Tr}\big(\lvert a'\rangle\langle a\rvert\,\Delta\big)=\langle a\vert\Delta\vert a'\rangle=\Delta_{aa'}=0$$

   Test against each element of the Hilbert–Schmidt basis $\{\lvert a'\rangle\langle a\rvert\}$, writing $\Delta=\sigma-\rho_A$ and using the trace identity $\mathrm{Tr}(\lvert x\rangle\langle y\rvert\Delta)=\langle y\vert\Delta\vert x\rangle$.
6. 6
   Step 6:

   $$\Delta_{aa'}=0\ \ \forall a,a'\ \Rightarrow\ \sigma=\rho_A$$

   All matrix entries of the difference vanish. Since $\{\lvert a\rangle\langle a'\rvert\}$ is an orthonormal basis for $\langle X,Y\rangle=\mathrm{Tr}(X^\dagger Y)$, testing on it is equivalent to testing on all $M$.

∎

1. 1
   Step 1:

   $$\overline{(\rho_A)_{a'a}}=\sum_b\overline{\rho_{(a',b),(a,b)}}=\sum_b\rho_{(a,b),(a',b)}=(\rho_A)_{aa'}$$

   Hermiticity: conjugate the component formula and use $\overline{\rho_{xy}}=\rho_{yx}$ for $\rho_{AB}$.
2. 2
   Step 2:

   $$\langle\varphi\vert\rho_A\vert\varphi\rangle=\mathrm{Tr}\big(\lvert\varphi\rangle\langle\varphi\rvert\rho_A\big)=\mathrm{Tr}\big((\lvert\varphi\rangle\langle\varphi\rvert\otimes I)\rho_{AB}\big)=\sum_b\langle\varphi\,b\vert\rho_{AB}\vert\varphi\,b\rangle\ge0$$

   Positivity: trace identity, the uniqueness theorem with $M=\lvert\varphi\rangle\langle\varphi\rvert$, and positivity of $\rho_{AB}$ on each vector $\lvert\varphi\rangle\lvert b\rangle$.
3. 3
   Step 3:

   $$\mathrm{Tr}\,\rho_A=\mathrm{Tr}(I_A\rho_A)=\mathrm{Tr}\big((I_A\otimes I_B)\rho_{AB}\big)=\mathrm{Tr}\,\rho_{AB}=1$$

   Unit trace: the uniqueness theorem with $M=I_A$.
4. 4
   Step 4:

   $$\mathrm{Tr}(M\sigma)=\mathrm{Tr}(H_1\sigma)+i\,\mathrm{Tr}(H_2\sigma)$$

   Linearity: agreement on all Hermitian $M$ implies agreement on all $M$, so local observable statistics already pin down $\rho_A$.

∎

1. 1
   Step 1:

   $$\lvert\psi\rangle=\sum_{a,b}C_{ab}\lvert a\rangle\otimes\lvert b\rangle,\qquad C\in\mathbb C^{d_A\times d_B},\ C_{ab}=\psi_{(a,b)}$$

   Expand in the product basis and arrange the amplitudes as a $d_A\times d_B$ matrix (for two qubits: reshape $(\psi_0,\psi_1,\psi_2,\psi_3)$ row-major into $2\times2$).
2. 2
   Step 2:

   $$C=U\Sigma V^\dagger,\qquad C_{ab}=\sum_{k=1}^rU_{ak}\,s_k\,\overline{V_{bk}},\qquad s_1\ge\dots\ge s_r>0$$

   Singular value decomposition: $U$ ($d_A\times d_A$) and $V$ ($d_B\times d_B$) unitary, $\Sigma$ diagonal with the non-negative singular values; only the $r=\mathrm{rank}\,C$ non-zero ones contribute.
3. 3
   Step 3:

   $$\lvert\psi\rangle=\sum_{k=1}^rs_k\Big(\sum_aU_{ak}\lvert a\rangle\Big)\otimes\Big(\sum_b\overline{V_{bk}}\lvert b\rangle\Big)=\sum_{k=1}^rs_k\lvert a_k\rangle\otimes\lvert b_k\rangle$$

   Substitute and regroup using bilinearity of $\otimes$; define $\lvert a_k\rangle=\sum_aU_{ak}\lvert a\rangle$ and $\lvert b_k\rangle=\sum_b\overline{V_{bk}}\lvert b\rangle$.
4. 4
   Step 4:

   $$\langle a_k\vert a_l\rangle=\sum_a\overline{U_{ak}}U_{al}=(U^\dagger U)_{kl}=\delta_{kl},\qquad\langle b_k\vert b_l\rangle=\sum_bV_{bk}\overline{V_{bl}}=\overline{(V^\dagger V)_{kl}}=\delta_{kl}$$

   Orthonormality follows from the unitarity of $U$ and $V$ (their columns are orthonormal).
5. 5
   Step 5:

   $$1=\langle\psi\vert\psi\rangle=\sum_{k,l}s_ks_l\langle a_k\vert a_l\rangle\langle b_k\vert b_l\rangle=\sum_ks_k^2$$

   Normalisation; the inner product of product vectors factorises. Set $\lambda_k=s_k$.
6. 6
   Step 6:

   $$(\rho_A)_{aa'}=\sum_b\psi_{(a,b)}\overline{\psi_{(a',b)}}=\sum_bC_{ab}\overline{C_{a'b}}=(CC^\dagger)_{aa'}$$

   Component formula of the partial trace for $\rho=\lvert\psi\rangle\langle\psi\rvert$, whose entries are $\psi_{(a,b)}\overline{\psi_{(a',b')}}$.
7. 7
   Step 7:

   $$\rho_A=CC^\dagger=U\Sigma V^\dagger V\Sigma U^\dagger=U\Sigma^2U^\dagger=\sum_k\lambda_k^2\lvert a_k\rangle\langle a_k\rvert$$

   Insert the SVD; $V^\dagger V=I$. The columns of $U$ are the eigenvectors $\lvert a_k\rangle$, with eigenvalues $\lambda_k^2$. The same computation with $(\rho_B)_{bb'}=\sum_aC_{ab}\overline{C_{ab'}}=(C^T\overline C)_{bb'}$ gives $\rho_B=\sum_k\lambda_k^2\lvert b_k\rangle\langle b_k\rvert$.

∎

1. 1
   Step 1:

   $$\text{(i)}\Rightarrow\text{(iii)}:\ \ \rho_A=\mathrm{Tr}_B\big(\lvert\alpha\rangle\langle\alpha\rvert\otimes\lvert\beta\rangle\langle\beta\rvert\big)=\lvert\alpha\rangle\langle\alpha\rvert\,\langle\beta\vert\beta\rangle=\lvert\alpha\rangle\langle\alpha\rvert$$

   Definition of $\mathrm{Tr}_B$ on a product operator, $\mathrm{Tr}_B(X\otimes Y)=X\,\mathrm{Tr}\,Y$, and normalisation of $\lvert\beta\rangle$. A rank-one projector is pure.
2. 2
   Step 2:

   $$\text{(iii)}\Rightarrow\text{(ii)}:\ \ 1=\mathrm{Tr}\,\rho_A^2=\sum_{k=1}^r\lambda_k^4\quad\text{with}\quad\sum_{k=1}^r\lambda_k^2=1$$

   The eigenvalues of $\rho_A$ are the $\lambda_k^2$ (Schmidt theorem), so its purity is $\sum_k(\lambda_k^2)^2$.
3. 3
   Step 3:

   $$\sum_k\lambda_k^4=1=\Big(\sum_k\lambda_k^2\Big)^2\ \Rightarrow\ \lambda_k^2\lambda_l^2=0\ \ (k\ne l)\ \Rightarrow\ r=1$$

   Purity theorem of the previous lesson: purity $1$ forces exactly one non-zero eigenvalue. Since every Schmidt coefficient is strictly positive, there is only one of them.
4. 4
   Step 4:

   $$\text{(ii)}\Rightarrow\text{(i)}:\ \ r=1,\ \lambda_1=1\ \Rightarrow\ \lvert\psi\rangle=\lvert a_1\rangle\otimes\lvert b_1\rangle$$

   The Schmidt sum has a single term and $\lambda_1^2=1$: a product of unit vectors.
5. 5
   Step 5:

   $$\mathrm{spec}_{\neq0}(\rho_A)=\{\lambda_k^2\}_{k=1}^r=\mathrm{spec}_{\neq0}(\rho_B)$$

   Both marginals were diagonalised with the same weights in the Schmidt theorem, which gives the final statement.

∎