Proof

The Schmidt decomposition is essentially a restatement of the singular value decomposition in a different context. Fix orthonormal bases ${\displaystyle \{e_{1},\ldots ,e_{n}\}\subset H_{1}}$ and ${\displaystyle \{f_{1},\ldots ,f_{m}\}\subset H_{2}}$. We can identify an elementary tensor ${\displaystyle e_{i}\otimes f_{j}}$ with the matrix ${\displaystyle e_{i}f_{j}^{\mathsf {T}}}$, where ${\displaystyle f_{j}^{\mathsf {T}}}$ is the transpose of ${\displaystyle f_{j}}$. A general element of the tensor product

${\displaystyle w=\sum _{1\leq i\leq n,1\leq j\leq m}\beta _{ij}e_{i}\otimes f_{j}}$

can then be viewed as the n × m matrix

${\displaystyle \;M_{w}=(\beta _{ij}).}$

By the singular value decomposition, there exist an n × n unitary U, m × m unitary V, and a positive semidefinite diagonal m × m matrix Σ such that

${\displaystyle M_{w}=U{\begin{bmatrix}\Sigma \\0\end{bmatrix}}V^{*}.}$

Write ${\displaystyle U={\begin{bmatrix}U_{1}&U_{2}\end{bmatrix}}}$ where ${\displaystyle U_{1}}$ is n × m and we have

${\displaystyle \;M_{w}=U_{1}\Sigma V^{*}.}$

Let ${\displaystyle \{u_{1},\ldots ,u_{m}\}}$ be the m column vectors of ${\displaystyle U_{1}}$, ${\displaystyle \{v_{1},\ldots ,v_{m}\}}$ the column vectors of ${\displaystyle {\overline {V}}}$, and ${\displaystyle \alpha _{1},\ldots ,\alpha _{m}}$ the diagonal elements of Σ. The previous expression is then

${\displaystyle M_{w}=\sum _{k=1}^{m}\alpha _{k}u_{k}v_{k}^{\mathsf {T}},}$

Then

${\displaystyle w=\sum _{k=1}^{m}\alpha _{k}u_{k}\otimes v_{k},}$

which proves the claim.

Spectrum of reduced states

Consider a vector ${\displaystyle w}$ of the tensor product

${\displaystyle H_{1}\otimes H_{2}}$

in the form of Schmidt decomposition

${\displaystyle w=\sum _{i=1}^{m}\alpha _{i}u_{i}\otimes v_{i}.}$

Form the rank 1 matrix ${\displaystyle \rho =ww^{*}}$. Then the partial trace of ${\displaystyle \rho }$, with respect to either system A or B, is a diagonal matrix whose non-zero diagonal elements are ${\displaystyle |\alpha _{i}|^{2}}$. In other words, the Schmidt decomposition shows that the reduced states of ${\displaystyle \rho }$ on either subsystem have the same spectrum.

Schmidt rank and entanglement

The strictly positive values ${\displaystyle \alpha _{i}}$ in the Schmidt decomposition of ${\displaystyle w}$ are its Schmidt coefficients, or Schmidt numbers. The total number of Schmidt coefficients of ${\displaystyle w}$, counted with multiplicity, is called its Schmidt rank.

If ${\displaystyle w}$ can be expressed as a product

${\displaystyle u\otimes v}$

then ${\displaystyle w}$ is called a separable state. Otherwise, ${\displaystyle w}$ is said to be an entangled state. From the Schmidt decomposition, we can see that ${\displaystyle w}$ is entangled if and only if ${\displaystyle w}$ has Schmidt rank strictly greater than 1. Therefore, two subsystems that partition a pure state are entangled if and only if their reduced states are mixed states.

Von Neumann entropy

A consequence of the above comments is that, for pure states, the von Neumann entropy of the reduced states is a well-defined measure of entanglement. For the von Neumann entropy of both reduced states of ${\displaystyle \rho }$ is ${\textstyle -\sum _{i}|\alpha _{i}|^{2}\log \left(|\alpha _{i}|^{2}\right)}$, and this is zero if and only if ${\displaystyle \rho }$ is a product state (not entangled).

Multipartite systems

The concept of Schmidt-rank vector can be likewise extended to systems made up of more than three subsystems through the use of tensors.

Example ^[2]

Take the tripartite quantum state ${\displaystyle |\psi _{4,2,2}\rangle ={\frac {1}{2}}{\big (}|0,0,0\rangle +|1,0,1\rangle +|2,1,0\rangle +|3,1,1\rangle {\big )}}$

This kind of system is made possible by encoding the value of a qudit into the orbital angular momentum (OAM) of a photon rather than its spin, since the latter can only take two values.

The Schmidt-rank vector for this quantum state is ${\displaystyle (4,2,2)}$.