Ex10 8

$\def\abs#1{|#1|}\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\def\tr{\mathord{\mbox{tr}}}\mathbf{Exercise\ 10.8}$

The Schmidt Decomposition.

Every $m\times n$ matrix $M$, with $m \leq n$, has a singular value decomposition $M = UDV$ where $D$ is an $m\times n$ diagonal matrix with non-negative real entries, and $U$ and $V$ are $m\times m$ and $n\times n$ unitary matrices.

Let $\ket\psi \in A\otimes B$, where $A$ has dimension $m$ and $B$ has dimension $n$, with $m \leq n$. Let $\{ \ket i \}$ be a basis for $A$ and $\{ \ket j \}$ be a basis for $B$, then for some choice of $m_{ij}\in\bf C$

(1)

\begin{align} \ket\psi = \sum_{i = 0}^{m - 1}\sum_{j = 0}^{n - 1} a_{ij}\ket i\ket j. \end{align}

Let $M$ be the $m\times n$ matrix with entries $a_{ij}$. Use the singular value decomposition (SVD) for $M$ to find sets of orthonormal unit vectors $\{ \ket{\alpha_i}\} \in A$ and $\{ \ket{\beta_j}\} \in B$ such that

(2)

\begin{align} \ket\psi = \sum_{i = 0}^{m - 1} \lambda_i\ket{\alpha_i}\ket{\beta_j} \end{align}

where $\lambda_i$ is non-negative. The $\lambda_i$ are called the {\it Schmidt coefficients}\index{Schmidt coefficients}, and $K$, the number of $\lambda_i$, is called the {\it Schmidt rank}\index{Schmidt rank} or {\it Schmidt number}\index{Schmidt number} of $\ket\psi$.

The Schmidt Decomposition

Anders B Madsen 01 Feb 2024 20:47

Let $a_{ij} \in {\Bbb C}$ be the coefficients such that

(1)

\begin{align} \ket{\psi} = \sum_{i=0}^{m-1} \sum_{j=0}^{n-1} a_{ij} \ket i \ket j\,. \end{align}

Let $M$ be the $m \times n$ matrix with entries $a_{ij}$. Then by the singular value decomposition, $M = UDV$,

(2)

\begin{align} a_{ij} = \sum_{k=0}^{m-1} \sum_{\ell=0}^{n-1} U_{ik}D_{k\ell}V_{\ell j} = \sum_{k=0}^{m-1} U_{ik}\lambda_kV_{kj} \,, \end{align}

where we are assuming $m \leq n$ and where $\lambda_k$ are the non-negative diagonal entries of $D$.

Now rewrite $\ket{\psi}$ as

(3)

\begin{align} \ket{\psi} &= \sum_{i=0}^{m-1} \sum_{j=0}^{n-1} \sum_{k=0}^{m-1} U_{ik}\lambda_kV_{kj} \ket i \ket j \\ &= \sum_{k=0}^{m-1} \lambda_k \ket{\alpha_k} \ket{\beta_k}\,, \end{align}

where $\ket{\alpha_k} = U\ket k$ and $\ket{\beta_k} = V^T\ket k$. $\{\ket{\alpha_k}\}$ and $\{\ket{\beta_k}\}$ are orthonormal bases for $A$ and $B$, respectively, since $U$ and $V$ are unitary.

Furthermore, $\sum \lambda_k^2 = 1$, since the sum is normalised.

Last edited on 04 Feb 2024 13:15 by Anders B Madsen Show more