If I try to write the two-qubit state $$ |\psi \rangle = \frac{|0 \rangle |0 \rangle + |0 \rangle |1 \rangle}{\sqrt{2}}$$ as $$ |\psi \rangle = \lambda_0 |\phi_0 \rangle |\phi_0 \rangle + \lambda_1 |\phi_1 \rangle |\phi_1 \rangle $$ for states $|\phi_0 \rangle = a |0 \rangle + b |1 \rangle$ and $|\phi_1 \rangle = c |0 \rangle + d |1 \rangle$, and real constants $\lambda_0$ and $\lambda_1$, I arrive at some contradiction that the quantity $\lambda_0 a b + \lambda_1 c d$ must equal both $1/\sqrt{2}$ and $0$, which makes me think a Schmidt decomposition for this state doesn't exist. However, the Schmidt decomposition theorem states that it can be done for any pure state $|\psi \rangle$ of a composite system.

I must have misunderstood what the Schmidt decomposition actually is or I am doing things the wrong way, I would appreciate it if someone could enlighten me on this.


1 Answer 1


Note that the two bases in the Schmidt decomposition do not necessarily coincide (for reasons related to the fact that the two unitary matrices in the singular value decomposition do not necessarily coincide). It is therefore clearer to write

$$ |\psi \rangle = \lambda_0 |\phi_0^A \rangle |\phi_0^B \rangle + \lambda_1 |\phi_1^A \rangle |\phi_1^B \rangle $$

where superscript $A$ indicates the first qubit and superscript $B$ indicates the second qubit. Next, observe that

$$ |\psi \rangle = \frac{|0\rangle|0\rangle + |0\rangle|1\rangle}{\sqrt{2}} = |0\rangle\frac{|0\rangle + |1\rangle}{\sqrt{2}} = |0\rangle|+\rangle, $$


$$ \lambda_0 = 1 \quad \lambda_1 = 0 \\ |\phi_0^A\rangle = |0\rangle \quad |\phi_1^A\rangle = |1\rangle \\ |\phi_0^B\rangle = |+\rangle \quad |\phi_1^B\rangle = |-\rangle. $$

  • $\begingroup$ Oh, I was thinking that because the two Hilbert spaces were identical the two bases should coincide, I think it is due to the notation used in the material I was using to study. Now it makes sense, thank you! $\endgroup$
    – AdonaiH
    Jun 2, 2021 at 1:22

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