The Schmidt decomposition

Intuition

Every pure state of two systems can be written as a single sum $\sum_{i}\sqrt{\lambda_{i}}\ket{a_{i}}\ket{b_{i}}$ , with orthonormal states on each side and positive weights adding to one — its Schmidt decomposition. It shows at a glance how entangled the state is: one term means a product, more terms mean entanglement. The two reduced states have the same nonzero eigenvalues, the weights $\lambda_{i}$ , so the two parts are exactly as mixed as each other, however different their dimensions. The number of terms, the Schmidt rank, is at most the smaller of the two dimensions.

Any table of numbers can be turned, by choosing new axes on each side, into a diagonal list of stretch factors. The Schmidt decomposition does that for the table of amplitudes of a two-part state: the stretch factors are how entangled it is.

The two Schmidt weights of $\cos\theta\ket{00}+\sin\theta\ket{11}$ against $\theta$ : $\cos^{2}\theta$ and . At either end one weight is 1 and the state is a product; where they cross, at , the weights are equal and the entanglement is the most two qubits can hold.

Every pure state of two parts can be written

\ket{\psi}=\sum_{i=1}^{r}\sqrt{\lambda_{i}}\,\ket{a_{i}}\ket{b_{i}},\qquad \lambda_{i}>0,\ \ \sum_{i}\lambda_{i}=1

Properties

The $\ket{a_{i}}$ are orthonormal in $\mathcal{H}_{A}$ and the $∣ b_{i}$ in ; the Schmidt rank is at most .

The Schmidt decomposition

\ket{\psi}\in\mathcal{H}_{A}\otimes\mathcal{H}_{B}\ \Longrightarrow\ \ket{\psi}=\sum_{i}\sqrt{\lambda_{i}}\,\ket{a_{i}}\ket{b_{i}}

Diagonalise the reduced state of $A$ and expand the whole state along its eigenvectors. The states of $B$ that go with them are forced to be orthogonal, because their overlaps are the elements of $\hat{\rho}_{A}$ in its own eigenbasis, which vanish off the diagonal. Their lengths are the square roots of the eigenvalues; normalising them gives the decomposition.

Proof steps

\hat{\rho}_{A}=\sum_{i}\lambda_{i}\ket{a_{i}}\bra{a_{i}}

$\hat{\rho}_{A}$ is Hermitian: diagonalise it by the spectral theorem.

\ket{\psi}=\sum_{i}\ket{a_{i}}\ket{\tilde b_{i}},\qquad \ket{\tilde b_{i}}=\sum_{b}\braket{a_{i}\,b|\psi}\ket{b}

Expand the state along the eigenbasis of $\hat{\rho}_{A}$ ; each $\ket{\tilde b_{i}}$ is a state of , not yet normalised.

\hat{\rho}_{A}=\operatorname{tr}_{B}\ket{\psi}\bra{\psi}=\sum_{i,j}\braket{\tilde b_{j}|\tilde b_{i}}\,\ket{a_{i}}\bra{a_{j}}

Trace out $B$ .

\braket{\tilde b_{j}|\tilde b_{i}}=\lambda_{i}\,\delta_{ij}

Compare with the diagonal form: the $\ket{\tilde b_{i}}$ are orthogonal, with squared lengths $\lambda_{i}$ .

\ket{\psi}=\sum_{\lambda_{i}>0}\sqrt{\lambda_{i}}\,\ket{a_{i}}\ket{b_{i}},\qquad \ket{b_{i}}=\frac{\ket{\tilde b_{i}}}{\sqrt{\lambda_{i}}}

Normalise the nonzero ones; the rest vanish.

Applications

Practice

One Term for a Product

A product state needs a single term in its Schmidt decomposition, with weight one. Any entangled state needs more.

Try it

What is the Schmidt rank of a product state?