Update week5.do.txt

Author: mhjensen

doc/src/week5/week5.do.txt

@@ -334,123 +334,6 @@ S=-\mathrm{Tr}[\rho\log_2{\rho}].
 !et
 This is the so-called Von Neumann entropy. How did we arrive at this expression? 
 
-
-
-
-!split
-===== Schmidt decomposition =====
-If we cannot write the density matrix in this form, we say the system
-$AB$ is entangled. In order to see this, we can use the so-called
-Schmidt decomposition, which is essentially an application of the
-singular-value decomposition.
-
-
-!split
-===== Pure states and Schmidt decomposition =====
-
-The Schmidt decomposition allows us to define a pure state in a
-bipartite Hilbert space composed of systems $A$ and $B$ as
-
-!bt
-\[
-\vert\psi\rangle=\sum_{i=0}^{d-1}\sigma_i\vert i\rangle_A\vert i\rangle_B,
-\]
-!et
-where the amplitudes $\sigma_i$ are real and positive and their
-squared values sum up to one, $\sum_i\sigma_i^2=1$. The states $\vert
-i\rangle_A$ and $\vert i\rangle_B$ form orthornormal bases for systems
-$A$ and $B$ respectively, the amplitudes $\lambda_i$ are the so-called
-Schmidt coefficients and the Schmidt rank $d$ is equal to the number
-of Schmidt coefficients and is smaller or equal to the minimum
-dimensionality of system $A$ and system $B$, that is $d\leq
-\mathrm{min}(\mathrm{dim}(A), \mathrm{dim}(B))$.
-
-
-!split
-===== Proof of Schmidt decomposition =====
-
-The proof for the above decomposition is based on the singular-value
-decomposition. To see this, assume that we have two orthonormal bases
-sets for systems $A$ and $B$, respectively. That is we have two ONBs
-$\vert i\rangle_A$ and $\vert j\rangle_B$. We can always construct a
-product state (a pure state) as
-
-!bt
-\[ \vert\psi \rangle = \sum_{ij} c_{ij}\vert i\rangle_A\vert
-j\rangle_B,
-\]
-!et
-where the coefficients $c_{ij}$ are the overlap coefficients which
-belong to a matrix $\bm{C}$.
-
-
-
-!split
-===== Further parts of proof =====
-
-If we now assume that the
-dimensionalities of the two subsystems $A$ and $B$ are the same $d$,
-we can always rewrite the matrix $\bm{C}$ in terms of a singular-value
-decomposition with unitary/orthogonal matrices $\bm{U}$ and $\bm{V}$
-of dimension $d\times d$ and a matrix $\bm{\Sigma}$ which contains the
-(diagonal) singular values $\sigma_0\leq \sigma_1 \leq \dots 0$ as
-
-!bt
-\[
-\bm{C}=\bm{U}\bm{\Sigma}\bm{V}^{\dagger}.
-\]
-!et
-
-!split
-===== SVD parts in proof =====
-
-This means we can rewrite the coefficients $c_{ij}$ in terms of the singular-value decomposition
-!bt
-\[
-c_{ij}=\sum_k u_{ik}\sigma_kv_{kj},
-\]
-!et
-and inserting this in the definition of the pure state $\vert \psi\rangle$ we have
-
-!bt
-\[
-\vert\psi \rangle = \sum_{ij} \left(\sum_k u_{ik}\sigma_kv_{kj} \right)\vert i\rangle_A\vert j\rangle_B.
-\]
-!et
-
-!split
-===== Slight rewrite =====
-We rewrite the last equation as
-
-!bt
-\[
-\vert\psi \rangle = \sum_{k}\sigma_k \left(\sum_i u_{ik}\vert i\rangle_A\right)\otimes\left(\sum_jv_{kj}\vert j\rangle_B\right),
-\]
-!et
-which we identify simply as, since the matrices $\bm{U}$ and $\bm{V}$ represent unitary transformations,
-!bt
-\[
-\vert\psi \rangle = \sum_{k}\sigma_k \vert k\rangle_A\vert k\rangle_B.
-\]
-!et
-
-
-!split
-===== Different dimensionalities =====
-
-It is straight forward to prove this relation in case systems $A$ and
-$B$ have different dimensionalities.  Once we know the Schmidt
-decomposition of a state, we can immmediately say whether it is
-entangled or not. If a state $\psi$ has is entangled, then its Schmidt
-decomposition has more than one term. Stated differently, the state is
-entangled if the so-called Schmidt rank is is greater than one.  There
-is another important property of the Schmidt decomposition which is
-related to the properties of the density matrices and their trace
-operations and the entropies. In order to introduce these concepts let
-us look at the two-qubit Hamiltonian described here.
-
-
-
 !split
 ===== Two-qubit system and calculation of density matrices and exercise =====
 
@@ -656,6 +539,123 @@ plt.show
 
 
 
+
+!split
+===== Schmidt decomposition =====
+If we cannot write the density matrix in this form, we say the system
+$AB$ is entangled. In order to see this, we can use the so-called
+Schmidt decomposition, which is essentially an application of the
+singular-value decomposition.
+
+
+!split
+===== Pure states and Schmidt decomposition =====
+
+The Schmidt decomposition allows us to define a pure state in a
+bipartite Hilbert space composed of systems $A$ and $B$ as
+
+!bt
+\[
+\vert\psi\rangle=\sum_{i=0}^{d-1}\sigma_i\vert i\rangle_A\vert i\rangle_B,
+\]
+!et
+where the amplitudes $\sigma_i$ are real and positive and their
+squared values sum up to one, $\sum_i\sigma_i^2=1$. The states $\vert
+i\rangle_A$ and $\vert i\rangle_B$ form orthornormal bases for systems
+$A$ and $B$ respectively, the amplitudes $\lambda_i$ are the so-called
+Schmidt coefficients and the Schmidt rank $d$ is equal to the number
+of Schmidt coefficients and is smaller or equal to the minimum
+dimensionality of system $A$ and system $B$, that is $d\leq
+\mathrm{min}(\mathrm{dim}(A), \mathrm{dim}(B))$.
+
+
+!split
+===== Proof of Schmidt decomposition =====
+
+The proof for the above decomposition is based on the singular-value
+decomposition. To see this, assume that we have two orthonormal bases
+sets for systems $A$ and $B$, respectively. That is we have two ONBs
+$\vert i\rangle_A$ and $\vert j\rangle_B$. We can always construct a
+product state (a pure state) as
+
+!bt
+\[ \vert\psi \rangle = \sum_{ij} c_{ij}\vert i\rangle_A\vert
+j\rangle_B,
+\]
+!et
+where the coefficients $c_{ij}$ are the overlap coefficients which
+belong to a matrix $\bm{C}$.
+
+
+
+!split
+===== Further parts of proof =====
+
+If we now assume that the
+dimensionalities of the two subsystems $A$ and $B$ are the same $d$,
+we can always rewrite the matrix $\bm{C}$ in terms of a singular-value
+decomposition with unitary/orthogonal matrices $\bm{U}$ and $\bm{V}$
+of dimension $d\times d$ and a matrix $\bm{\Sigma}$ which contains the
+(diagonal) singular values $\sigma_0\leq \sigma_1 \leq \dots 0$ as
+
+!bt
+\[
+\bm{C}=\bm{U}\bm{\Sigma}\bm{V}^{\dagger}.
+\]
+!et
+
+!split
+===== SVD parts in proof =====
+
+This means we can rewrite the coefficients $c_{ij}$ in terms of the singular-value decomposition
+!bt
+\[
+c_{ij}=\sum_k u_{ik}\sigma_kv_{kj},
+\]
+!et
+and inserting this in the definition of the pure state $\vert \psi\rangle$ we have
+
+!bt
+\[
+\vert\psi \rangle = \sum_{ij} \left(\sum_k u_{ik}\sigma_kv_{kj} \right)\vert i\rangle_A\vert j\rangle_B.
+\]
+!et
+
+!split
+===== Slight rewrite =====
+We rewrite the last equation as
+
+!bt
+\[
+\vert\psi \rangle = \sum_{k}\sigma_k \left(\sum_i u_{ik}\vert i\rangle_A\right)\otimes\left(\sum_jv_{kj}\vert j\rangle_B\right),
+\]
+!et
+which we identify simply as, since the matrices $\bm{U}$ and $\bm{V}$ represent unitary transformations,
+!bt
+\[
+\vert\psi \rangle = \sum_{k}\sigma_k \vert k\rangle_A\vert k\rangle_B.
+\]
+!et
+
+
+!split
+===== Different dimensionalities =====
+
+It is straight forward to prove this relation in case systems $A$ and
+$B$ have different dimensionalities.  Once we know the Schmidt
+decomposition of a state, we can immmediately say whether it is
+entangled or not. If a state $\psi$ has is entangled, then its Schmidt
+decomposition has more than one term. Stated differently, the state is
+entangled if the so-called Schmidt rank is is greater than one.  There
+is another important property of the Schmidt decomposition which is
+related to the properties of the density matrices and their trace
+operations and the entropies. In order to introduce these concepts let
+us look at the two-qubit Hamiltonian described here.
+
+
+
+
+
 !split
 ===== Exercise: Two-qubit Hamiltonian =====