What is the Schmidt decomposition for the 00 + 01 state?

Question

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.

Adam Zalcman · Accepted Answer

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
$$
|psi rangle = frac{|0rangle|0rangle + |0rangle|1rangle}{sqrt{2}} = |0ranglefrac{|0rangle + |1rangle}{sqrt{2}} = |0rangle|+rangle,
$$
so
$$
lambda_0 = 1 quad lambda_1 = 0 
|phi_0^Arangle = |0rangle quad |phi_1^Arangle = |1rangle 
|phi_0^Brangle = |+rangle quad |phi_1^Brangle = |-rangle.
$$

What is the Schmidt decomposition for the 00 + 01 state?

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