Let ${\displaystyle H_{1}}$  and ${\displaystyle H_{2}}$  be Hilbert spaces of dimensions n and m respectively. Assume ${\displaystyle n\geq m}$ . For any vector ${\displaystyle w}$  in the tensor product ${\displaystyle H_{1}\otimes H_{2}}$ , there exist orthonormal sets ${\displaystyle \{u_{1},\ldots ,u_{m}\}\subset H_{1}}$  and ${\displaystyle \{v_{1},\ldots ,v_{m}\}\subset H_{2}}$  such that ${\textstyle w=\sum _{i=1}^{m}\alpha _{i}u_{i}\otimes v_{i}}$ , where the scalars ${\displaystyle \alpha _{i}}$  are real, non-negative, and unique up to re-ordering.

Proof edit

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.

Some properties of the Schmidt decomposition are of physical interest.

Spectrum of reduced states edit

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 edit

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 edit

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).

The Schmidt rank is defined for bipartite systems, namely quantum states

${\displaystyle |\psi \rangle \in H_{A}\otimes H_{B}}$ 

The concept of Schmidt rank can be extended to quantum systems made up of more than two subsystems.^[1]

Consider the tripartite quantum system:

${\displaystyle |\psi \rangle \in H_{A}\otimes H_{B}\otimes H_{C}}$ 

There are three ways to reduce this to a bipartite system by performing the partial trace with respect to ${\displaystyle H_{A},H_{B}}$  or ${\displaystyle H_{C}}$ 

${\displaystyle {\begin{cases}{\hat {\rho }}_{A}=Tr_{A}(|\psi \rangle \langle \psi |)\\{\hat {\rho }}_{B}=Tr_{B}(|\psi \rangle \langle \psi |)\\{\hat {\rho }}_{C}=Tr_{C}(|\psi \rangle \langle \psi |)\end{cases}}}$ 

Each of the systems obtained is a bipartite system and therefore can be characterized by one number (its Schmidt rank), respectively ${\displaystyle r_{A},r_{B}}$  and ${\displaystyle r_{C}}$ . These numbers capture the "amount of entanglement" in the bipartite system when respectively A, B or C are discarded. For these reasons the tripartite system can be described by a vector, namely the Schmidt-rank vector

${\displaystyle {\vec {r}}=(r_{A},r_{B},r_{C})}$ 

Multipartite systems edit

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] edit

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 ${\displaystyle (4,2,2)}$ .

• Singular value decomposition
• Purification of quantum state

• Pathak, Anirban (2013). Elements of Quantum Computation and Quantum Communication. London: Taylor & Francis. pp. 92–98. ISBN 978-1-4665-1791-2.