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I do not understand how any state in Hilbert Space $\mathcal{H}=\mathcal{H}_A\otimes\mathcal{H}_B$ of dimension $\text{dim}(\mathcal{H}_A)\times\text{dim}(\mathcal{H}_B)$ can be decomposed in the Schmidt basis when the Schmidt basis has size $\text{min}\{\text{dim}(\mathcal{H}_A),\text{dim}(\mathcal{H}_A)$}.

Thank you in anticipation of your help.

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The Schimdt decomposition depends in the first place in the vector you want to "decompose", as far as I know. So you can't compare the raw dimensions of the spaces like that. – Learning is a mess Mar 29 '13 at 19:43
up vote 4 down vote accepted

This is because the Schmidt bases depend on the vector you want to decompose: given a vector $v\in\mathcal{H}$ you can find bases $\{u_i\}$ for $\mathcal{H}_A$ and $\{v_j\}$ for $\mathcal{H}_B$ such that $$v=\sum_k \lambda_k u_k\otimes v_k.\tag{1}$$ The bases $\{u_i\}$ and $\{v_j\}$ depend on $v$ in a nonlinear way (why? they're nonzero for $v=0$) and you can't expect linear algebraic arguments on dimension to hold quite as before. The way to phrase this is that the Schmidt bases of $v$ combine to form a basis $\{u_i\otimes v_j\}$ for $\mathcal{H}$ such that the off-diagonal coefficients of $v$ in the expansion $$v=\sum_{i,j} \lambda_{ij} \,u_i\otimes v_j\tag{2}$$ vanish. (You can still express all vectors in $\mathcal{H}$ in the form of eq. (2), of course, and the basis has size $\text{dim}(\mathcal{H}_A)\times\text{dim}(\mathcal{H}_B)$, as it should, but only a subspace of dimension $\text{min}\{\text{dim}(\mathcal{H}_A),\text{dim}(\mathcal{H}_B)\}$ can be written as in eq. (1). That's how the dimensions match up.)

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This follows from Singular Value Decomposition (go wikipedia it).

Given any state $\psi \in H = H_A \otimes H_B$, one can write it as \begin{align} \psi = \sum_{nm} W_{nm} \psi_n^A \psi_m^B, \end{align} where $\{ \psi_n^\xi \}$ form a basis for the subspace $H_\xi$. Let's assume that the dimensions of $H_A, H_B$ are finite (otherwise, your assertion is trivially true). Of course in general, $\dim(H_A) \neq \dim(H_B)$. Then the index $n$ in the sum runs from $1$ to $\dim(H_A)$ and $m$ from $1$ to $\dim(H_B)$.

Thus the coefficient matrix $W_{nm}$ is a $\dim(H_A) \times \dim(H_B)$ rectangular matrix in general. Singular value decomposition says it can be written as (in matrix form), \begin{align} W = U\Sigma V^\dagger, \end{align} where $U$ is a $\dim(H_A)$ unitary matrix and $V$ is a $\dim(H_B)$ unitary matrix, and $\Sigma$ is a rectangular, diagonal matrix of dimensions $\dim(H_A) \times \dim(H_B)$.

Since $\Sigma$ is diagonal, and is rectangular in general, there are only $\min(\dim(H_A), \dim(H_B))$ such singular values.

The $U$ matrix is a rotation of the $\psi_n^A$ basis, while $V^\dagger$ is a rotation of the $\psi_n^B$ basis. Thus the original state is written in the Schmidt form \begin{align} \psi = \sum_k^{\min(\dim(H_A),\dim(H_B))} \sigma_k \phi_k^A \phi_k^B, \end{align} where $\sigma_k$ are the $\min(\dim(H_A),\dim(H_B))$ singular values.

Thus the state only requires $\min(\dim(H_A),\dim(H_B))$ basis vectors (from either $A$ or $B$) to describe it.

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