Schmidt Decomposition

Every bipartite pure state, however tangled it looks, has a hidden diagonal form: a single sum that pairs one orthonormal vector on the left with one on the right. That canonical form is the Schmidt decomposition, and it falls straight out of the singular value decomposition of the coefficient matrix. Its weights — the Schmidt coefficients — are immune to anything either party can do locally, which is exactly why they, and the single number built from them (the entanglement entropy), are the measure of entanglement for pure states. This lesson turns the qualitative "entangled vs. not" of 1.4.1 into a quantitative dial.

Learning Objectives

After this lesson you will be able to:

1. State and prove the Schmidt decomposition theorem from the SVD of the coefficient matrix.
2. Read off the Schmidt coefficients, Schmidt rank, and Schmidt bases of a bipartite pure state.
3. Show the Schmidt spectrum is a local-unitary invariant, and use Schmidt rank as the separability test.
4. Define the entanglement entropy $E = -\sum_k\lambda_k\log_2\lambda_k$ and identify maximally entangled states by their flat spectrum.
5. Compute Schmidt decompositions numerically with numpy.linalg.svd and interpret the spectrum.

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Intuition

A bipartite state generally needs $d_A d_B$ amplitudes — a whole matrix $C$ of them. The Schmidt decomposition says you can always rotate the local bases (a new basis for $A$, a new basis for $B$) so that this matrix becomes diagonal: the state collapses to a single sum $\sum_k\sigma_k\lvert u_k\rangle\lvert v_k\rangle$ with at most $\min(d_A,d_B)$ terms. The number of terms (the Schmidt rank) and the diagonal weights (the Schmidt coefficients $\sigma_k$) are intrinsic — choosing different starting bases, or applying any gates locally, just relabels $\lvert u_k\rangle$ and $\lvert v_k\rangle$ but never touches the $\sigma_k$. A product state is the special case of one term ($\sigma=(1,0,\dots)$); a Bell state is the opposite extreme, a perfectly flat spectrum. So "how entangled is this pure state?" has a clean answer: look at how spread-out the Schmidt spectrum is, and the SVD hands it to you for free.

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Theory

Setup: the coefficient matrix

From 1.4.1, any pure state of a bipartite system $\mathcal{H}_A\otimes\mathcal{H}_B$ (dimensions $d_A,d_B$) expands in product bases $\{\lvert i\rangle_A\}$, $\{\lvert j\rangle_B\}$ as

$$\lvert\Psi\rangle = \sum_{i=1}^{d_A}\sum_{j=1}^{d_B} C_{ij}\,\lvert i\rangle_A\otimes\lvert j\rangle_B, \qquad C\in\mathbb{C}^{d_A\times d_B},$$

and normalization is $\langle\Psi\lvert\Psi\rangle = \sum_{ij}|C_{ij}|^2 = \lVert C\rVert_F^2 = 1$ (the Frobenius norm). The matrix $C$ holds everything about the state; the Schmidt decomposition is just its SVD, read as a statement about $\lvert\Psi\rangle$.

The theorem

Schmidt decomposition. For any pure state $\lvert\Psi\rangle\in\mathcal{H}_A\otimes\mathcal{H}_B$ there exist orthonormal sets $\{\lvert u_k\rangle\}\subset\mathcal{H}_A$ and $\{\lvert v_k\rangle\}\subset\mathcal{H}_B$ and unique reals $\sigma_k > 0$ with $\sum_k\sigma_k^2 = 1$ such that

$$\boxed{\ \lvert\Psi\rangle = \sum_{k=1}^{r}\sigma_k\,\lvert u_k\rangle\otimes\lvert v_k\rangle\ }$$

The $\sigma_k$ are the Schmidt coefficients, $r\le\min(d_A,d_B)$ is the Schmidt rank, and the $\{\lvert u_k\rangle\},\{\lvert v_k\rangle\}$ are the Schmidt bases.

Proof. Apply the SVD (0.1.6) to the coefficient matrix: $C = U\Sigma V^\dagger$, with $U$ ($d_A\times d_A$) and $V$ ($d_B\times d_B$) unitary and $\Sigma$ diagonal with entries $\sigma_k\ge0$ (the singular values). Componentwise $C_{ij} = \sum_k U_{ik}\,\sigma_k\,(V^\dagger)_{kj} = \sum_k\sigma_k\,U_{ik}\,\overline{V_{jk}}$. Substitute:

$$\lvert\Psi\rangle = \sum_{ij}\sum_k \sigma_k\,U_{ik}\,\overline{V_{jk}}\,\lvert i\rangle\lvert j\rangle = \sum_k \sigma_k\Big(\underbrace{\textstyle\sum_i U_{ik}\lvert i\rangle}_{\lvert u_k\rangle}\Big)\otimes \Big(\underbrace{\textstyle\sum_j \overline{V_{jk}}\lvert j\rangle}_{\lvert v_k\rangle}\Big).$$

Because $U$ is unitary its columns $\lvert u_k\rangle$ are orthonormal; likewise the conjugated columns of $V$ give orthonormal $\lvert v_k\rangle$. Keeping only the $r$ strictly positive $\sigma_k$ gives the boxed form. Normalization forces $\sum_k\sigma_k^2 = \lVert C\rVert_F^2 = 1$. The $\sigma_k$ are the singular values of $C$, which are unique; the Schmidt bases are unique up to phases (and up to rotation within any degenerate $\sigma_k$ subspace). $\blacksquare$

The decisive feature: it is a single sum over one index $k$, not a double sum over $i,j$. One left vector is matched to exactly one right vector.

Schmidt rank and the separability test

The Schmidt rank $r = \operatorname{rank}(C)$ (number of nonzero $\sigma_k$) is the sharp measure of whether a pure state is entangled — exactly the criterion promised in 1.4.1 and 1.1.4:

• $r = 1$: $\lvert\Psi\rangle = \sigma_1\lvert u_1\rangle\lvert v_1\rangle$ is a product state (separable). With $\sigma_1=1$ this is just $\lvert u_1\rangle\otimes\lvert v_1\rangle$.
• $r \ge 2$: entangled — no single product of local vectors can reproduce it.

The Schmidt spectrum is a local-unitary invariant

Suppose Alice applies $U_A$ and Bob applies $U_B$. The coefficient matrix transforms as $C \to U_A\,C\,U_B^{\top}$ (Exercise E3; the transpose comes from $U_B$ acting on the second factor). Left and right multiplication by unitaries does not change singular values (0.1.6). Hence:

$$\boxed{\ \text{Local unitaries }U_A\otimes U_B\text{ leave every Schmidt coefficient }\sigma_k\text{ unchanged.}\ }$$

They only rotate the Schmidt vectors $\lvert u_k\rangle\to U_A\lvert u_k\rangle$, $\lvert v_k\rangle\to U_B^*\lvert v_k\rangle$. This is why the spectrum measures entanglement: it is precisely the part of the state that survives quotienting out each party's choice of local basis. Anything that genuinely changes entanglement must be a joint (entangling) operation.

Reduced states and entanglement entropy

The Schmidt coefficients are the bridge to the density-matrix formalism of Course 1.5. Tracing out one side of $\lvert\Psi\rangle\langle\Psi\lvert$ (the partial trace, 1.5.2) gives diagonal reduced states in the Schmidt bases:

$$\rho_A = \sum_k \sigma_k^2\,\lvert u_k\rangle\langle u_k\lvert, \qquad \rho_B = \sum_k \sigma_k^2\,\lvert v_k\rangle\langle v_k\lvert .$$

So $\rho_A$ and $\rho_B$ share the same nonzero eigenvalues $\lambda_k := \sigma_k^2$ — a striking fact: two subsystems of very different sizes have identically-shaped spectra. Since $\{\lambda_k\}$ is a probability distribution ($\lambda_k\ge0$, $\sum_k\lambda_k=1$), we measure entanglement by its Shannon entropy (0.2.3):

$$\boxed{\ E(\Psi) = -\sum_k \lambda_k\log_2\lambda_k = -\sum_k \sigma_k^2\log_2\sigma_k^2\ }\quad\text{(the \textbf{entanglement entropy}, in ebits).}$$

It is symmetric ($E_A = E_B$, since both sides share $\{\lambda_k\}$), zero iff the state is a product ($r=1$), and — being a function of the LU-invariant spectrum — itself a local-unitary invariant.

Maximal entanglement = flat spectrum

$E$ is maximized when the distribution $\{\lambda_k\}$ is uniform. For Schmidt rank $r$, the maximum is $E = \log_2 r$, attained when all $\sigma_k = 1/\sqrt r$. A state of two $d$-dimensional systems is maximally entangled when $\sigma_k = 1/\sqrt d$ for all $k$, giving $E = \log_2 d$ and maximally mixed reduced states $\rho_A=\rho_B=I/d$. For two qubits ($d=2$) that is $E = 1$ ebit — the Bell states (1.4.2), whose flat spectrum $(\tfrac1{\sqrt2},\tfrac1{\sqrt2})$ is exactly what "maximally entangled" means. The entanglement entropy thus interpolates continuously from $0$ (product) to $\log_2 d$ (maximal): a genuine dial, not a yes/no flag.

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Worked Examples

Example 1 — The Bell state is already in Schmidt form

$\lvert\Phi^+\rangle = \tfrac1{\sqrt2}(\lvert00\rangle+\lvert11\rangle)$ has coefficient matrix C = \tfrac1{\sqrt2}\begin{psmallmatrix}1&0\\0&1\end{psmallmatrix} = \tfrac1{\sqrt2}I. Its singular values are both $\tfrac1{\sqrt2}$, so $\sigma = (\tfrac1{\sqrt2},\tfrac1{\sqrt2})$, Schmidt rank $r=2$, with Schmidt bases $\{\lvert0\rangle,\lvert1\rangle\}$ on each side. The spectrum is flat, so $\lambda = (\tfrac12,\tfrac12)$ and $E = -2\cdot\tfrac12\log_2\tfrac12 = 1$ ebit — maximally entangled, with $\rho_A = \rho_B = I/2$.

Example 2 — Entanglement as a continuous dial

$\lvert\Psi_\theta\rangle = \cos\theta\,\lvert00\rangle + \sin\theta\,\lvert11\rangle$ is already a Schmidt decomposition: $C = \operatorname{diag}(\cos\theta,\sin\theta)$, so $\sigma = (\cos\theta,\sin\theta)$ (for $\theta\in[0,\tfrac\pi4]$) and $\lambda = (\cos^2\theta,\sin^2\theta)$. The entanglement entropy is the binary entropy $E(\theta) = H_2(\cos^2\theta)$. At $\theta=0$ the state is $\lvert00\rangle$, a product ($r=1$, $E=0$); at $\theta=\tfrac\pi4$ it is $\lvert\Phi^+\rangle$, maximal ($E=1$); in between it is partially entangled. One parameter sweeps the whole range from separable to maximal.

Example 3 — One sign flip: product vs. maximal

Compare two states that look almost identical:

$$\lvert P\rangle = \tfrac12(\lvert00\rangle+\lvert01\rangle+\lvert10\rangle+\lvert11\rangle), \qquad \lvert M\rangle = \tfrac12(\lvert00\rangle+\lvert01\rangle+\lvert10\rangle-\lvert11\rangle).$$

Their coefficient matrices are C_P = \tfrac12\begin{psmallmatrix}1&1\\1&1\end{psmallmatrix} and C_M = \tfrac12\begin{psmallmatrix}1&1\\1&-1\end{psmallmatrix} = \tfrac1{\sqrt2}H. The rows of $C_P$ are identical, so $\operatorname{rank}(C_P)=1$: $\sigma=(1,0)$, and indeed $\lvert P\rangle = \lvert+\rangle\lvert+\rangle$ is a product state, $E=0$. But $H$ is unitary, so $C_M = \tfrac1{\sqrt2}H$ has both singular values equal to $\tfrac1{\sqrt2}$: $\sigma=(\tfrac1{\sqrt2},\tfrac1{\sqrt2})$, $r=2$, $E=1$ — $\lvert M\rangle$ is maximally entangled (it is $\lvert\Phi^+\rangle$ dressed by a local $H$). A single sign separates a product state from a Bell state, and only the Schmidt spectrum sees it at a glance.

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Hands-on (Python)

import numpy as np

def schmidt(state, dA, dB, tol=1e-12):
    """Schmidt decomposition of a bipartite pure state (big-endian, A⊗B).
    Returns (sigma, U_cols, Vh_rows) keeping only sigma_k > tol."""
    C = state.reshape(dA, dB)                 # coefficient matrix C_{ij}
    U, s, Vh = np.linalg.svd(C)               # C = U diag(s) Vh
    keep = s > tol
    return s[keep], U[:, keep], Vh[keep, :]   # |u_k> = U[:,k], |v_k> = Vh[k,:]

def entanglement_entropy(sigma):
    """E = -sum lambda log2 lambda,  lambda = sigma^2 (in ebits)."""
    lam = sigma**2
    lam = lam[lam > 0]
    return float(max(0.0, -np.sum(lam * np.log2(lam))))   # entropy is nonnegative

# Build a few 2-qubit states (big-endian: index 2i+j ↔ |i>|j>):
def two_qubit(c00, c01, c10, c11):
    v = np.array([c00, c01, c10, c11], dtype=complex)
    return v / np.linalg.norm(v)

bell    = two_qubit(1, 0, 0, 1)               # |Φ+>
prod    = two_qubit(1, 1, 1, 1)               # |+>|+>  (Example 3, |P>)
maximal = two_qubit(1, 1, 1, -1)              # one sign flip (Example 3, |M>)

for name, st in [("Bell", bell), ("|+>|+>", prod), ("sign-flip", maximal)]:
    s, _, _ = schmidt(st, 2, 2)
    print(f"{name:>10}: sigma={np.round(s,3)}  rank={len(s)}  E={entanglement_entropy(s):.3f} ebits")
#       Bell: sigma=[0.707 0.707]  rank=2  E=1.000   (flat -> maximal)
#     |+>|+>: sigma=[1.]           rank=1  E=0.000   (product; the second σ≈0 is dropped)
#  sign-flip: sigma=[0.707 0.707]  rank=2  E=1.000   (Bell up to a local H)

# The dial of Example 2: E(θ) = H2(cos^2 θ), peaking at θ = π/4.
for theta in (0, np.pi/8, np.pi/4):
    st = two_qubit(np.cos(theta), 0, 0, np.sin(theta))
    s, _, _ = schmidt(st, 2, 2)
    print(f"θ={theta:.3f}: sigma={np.round(s,3)}  E={entanglement_entropy(s):.4f}")
# θ=0.000: sigma=[1.]           E=0.0000   (product)
# θ=0.393: sigma=[0.924 0.383]  E=0.6009   (partial)
# θ=0.785: sigma=[0.707 0.707]  E=1.0000   (maximal)

# Reconstruction check: Σ_k σ_k |u_k>⊗|v_k> rebuilds the state exactly.
s, U, Vh = schmidt(maximal, 2, 2)
rebuilt = sum(sk * np.kron(U[:, k], Vh[k, :]) for k, sk in enumerate(s))
print("reconstruct OK:", np.allclose(rebuilt, maximal))      # True

# Local-unitary invariance: H ⊗ I leaves the Schmidt spectrum unchanged.
H = np.array([[1, 1], [1, -1]], dtype=complex) / np.sqrt(2)
I2 = np.eye(2)
st2 = np.kron(H, I2) @ bell
print("σ after H⊗I:", np.round(schmidt(st2, 2, 2)[0], 3))    # [0.707 0.707] — unchanged

Numerical Schmidt rank. The rank is the count of singular values above a tolerance, never an exact equality test — finite precision turns a "zero" coefficient into something like $10^{-16}$. This is the careful version of the schmidt_rank helper from 1.4.1: pick tol comfortably above machine noise (here $10^{-12}$). The same SVD also drives the matrix-product-state truncation used in tensor-network simulators (Term 5.3).

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Exercises

E1 (easy). Give the Schmidt coefficients, rank, and entanglement entropy of $\lvert\Psi_\theta\rangle = \cos\theta\lvert00\rangle+\sin\theta\lvert11\rangle$ for $\theta=0$, $\tfrac\pi6$, $\tfrac\pi4$.

Solution

It is already in Schmidt form: $\sigma=(\cos\theta,\sin\theta)$. At $\theta=0$: $\sigma=(1,0)$, rank $1$, $E=0$ (product). At $\theta=\tfrac\pi6$: $\sigma=(\tfrac{\sqrt3}2,\tfrac12)$, rank $2$, $E = H_2(\tfrac34) = -\tfrac34\log_2\tfrac34-\tfrac14\log_2\tfrac14\approx0.811$. At $\theta=\tfrac\pi4$: $\sigma=(\tfrac1{\sqrt2},\tfrac1{\sqrt2})$, rank $2$, $E=1$ (maximal). ∎

E2 (easy). Why must $\sum_k\sigma_k^2 = 1$ for any normalized state?

Solution

Normalization gives $1 = \langle\Psi\lvert\Psi\rangle = \sum_{ij}|C_{ij}|^2 = \lVert C\rVert_F^2$, and the squared Frobenius norm equals the sum of squared singular values, $\lVert C\rVert_F^2 = \sum_k\sigma_k^2$. So $\sum_k\sigma_k^2 = 1$, which is also why $\lambda_k=\sigma_k^2$ is a valid probability distribution. ∎

E3 (medium). Show that under local unitaries the coefficient matrix transforms as $C \to U_A\,C\,U_B^{\top}$, and conclude the Schmidt coefficients are unchanged.

Solution

$(U_A\otimes U_B)\lvert\Psi\rangle = \sum_{ij}C_{ij}(U_A\lvert i\rangle)\otimes(U_B\lvert j\rangle) = \sum_{ij}C_{ij}\sum_{i'}(U_A){i'i}\lvert i'\rangle\sum{j'}(U_B){j'j}\lvert j'\rangle$. The new coefficient of $\lvert i'\rangle\lvert j'\rangle$ is $\sum{ij}(U_A){i'i}C{ij}(U_B){j'j} = (U_A C,U_B^{\top}){i'j'}$. Singular values are invariant under left/right multiplication by unitaries ($U_B^{\top}$ is unitary since $U_B$ is), so every $\sigma_k$ — hence $E$ — is unchanged; only the Schmidt vectors rotate. ∎

E4 (medium). Without numerics, find the Schmidt rank and entanglement entropy of $\lvert M\rangle = \tfrac12(\lvert00\rangle+\lvert01\rangle+\lvert10\rangle-\lvert11\rangle)$.

Solution

C_M = \tfrac12\begin{psmallmatrix}1&1\\1&-1\end{psmallmatrix} = \tfrac1{\sqrt2}H with $H$ the (unitary) Hadamard. A unitary has all singular values $1$, so $C_M$ has both singular values $\tfrac1{\sqrt2}$: $\sigma=(\tfrac1{\sqrt2},\tfrac1{\sqrt2})$, rank $2$, flat spectrum, $E=1$ ebit. It is maximally entangled — a Bell state up to the local unitary $H$ on one qubit. ∎

E5 (medium). Prove that $\rho_A = \operatorname{Tr}_B\lvert\Psi\rangle\langle\Psi\lvert = \sum_k\sigma_k^2\lvert u_k\rangle\langle u_k\lvert$, and deduce that $\rho_A$ is pure iff $\lvert\Psi\rangle$ is a product state.

Solution

Write $\lvert\Psi\rangle=\sum_k\sigma_k\lvert u_k\rangle\lvert v_k\rangle$. Then $\lvert\Psi\rangle\langle\Psi\lvert =\sum_{k,l}\sigma_k\sigma_l\lvert u_k\rangle\langle u_l\lvert\otimes\lvert v_k\rangle\langle v_l\lvert$. The partial trace over $B$ replaces $\lvert v_k\rangle\langle v_l\lvert$ by $\langle v_l\lvert v_k\rangle=\delta_{kl}$ (orthonormality), leaving $\rho_A=\sum_k\sigma_k^2\lvert u_k\rangle\langle u_k\lvert$ — already diagonal, with eigenvalues $\lambda_k=\sigma_k^2$. $\rho_A$ is pure iff one eigenvalue is $1$ and the rest $0$, i.e. a single nonzero $\sigma_k$, i.e. Schmidt rank $1$, i.e. $\lvert\Psi\rangle$ is a product. ∎ (Partial trace: 1.5.2.)

E6 (hard). Show $r\le\min(d_A,d_B)$, and that for two $d$-level systems the maximum entanglement entropy is $\log_2 d$, attained exactly when $\sigma_k=1/\sqrt d$ for all $k$.

Solution

$r=\operatorname{rank}(C)$ and $C$ is $d_A\times d_B$, so $r\le\min(d_A,d_B)$; for two $d$-level systems $r\le d$. Entanglement entropy is the Shannon entropy $H(\{\lambda_k\})$ of a distribution on $r\le d$ outcomes, which is maximized by the uniform distribution and equals $\log_2 r\le\log_2 d$ (0.2.3). The bound $\log_2 d$ is reached only when $r=d$ and all $\lambda_k=1/d$, i.e. $\sigma_k=1/\sqrt d$ — the flat spectrum, giving $\rho_A=\rho_B=I/d$. ∎

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Checkpoint

1. State the Schmidt decomposition and the key structural fact that distinguishes it from a generic double sum.
2. How do you compute the Schmidt coefficients of a state in practice?
3. What does Schmidt rank tell you, and why is the Schmidt spectrum unchanged by local unitaries?
4. Define the entanglement entropy and give its value for a product state and for a two-qubit Bell state.
5. What is the Schmidt spectrum of a maximally entangled state of two $d$-level systems, and what are its reduced states?

Answers

1. $\lvert\Psi\rangle = \sum_k\sigma_k\lvert u_k\rangle\lvert v_k\rangle$ with orthonormal $\{\lvert u_k\rangle\}$, $\{\lvert v_k\rangle\}$ and $\sigma_k>0$, $\sum_k\sigma_k^2=1$. It is a single sum pairing one $A$-vector with one $B$-vector.
2. Form the coefficient matrix $C$ and take its singular values via the SVD; the $\sigma_k$ are those singular values.
3. Rank $1$ ⟺ product, rank $\ge2$ ⟺ entangled. Local unitaries send $C\to U_A C U_B^{\top}$, which preserves singular values.
4. $E = -\sum_k\sigma_k^2\log_2\sigma_k^2$; $E=0$ for a product state, $E=1$ ebit for a Bell state.
5. Flat: $\sigma_k=1/\sqrt d$ for all $k$, $E=\log_2 d$; reduced states are maximally mixed, $\rho_A=\rho_B=I/d$.

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Further Reading

• [NC] Nielsen & Chuang, §2.5 — the Schmidt decomposition and its consequences for entanglement.
• [Pre] Preskill, Ph219, Ch. 2 — Schmidt decomposition, purification, and the reduced density operator.
• [Wat] Watrous, The Theory of Quantum Information, Ch. 2 — operator-Schmidt and SVD viewpoint.
• [NC] §12.5.1 — entanglement entropy as the entanglement measure for pure bipartite states.

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