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I was studying VQLS in https://qiskit.org/textbook/ch-paper-implementations/vqls.html and run into the following normalization during the cost calculation. It says if $ |\Phi\rangle$ has a small norm. but, how come? I thought all vectors are unit vectors. Could you explain why we need the normalization factor? I think I missed something. Thanks

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This is due to how the $\mathbf{A}$ matrix was defined; from that same tutorial page we have:

$$\tag{1} \mathbf{A} = \sum_{n} c_n A_n $$

where each $A_n$ is unitary and $c_n$ is complex (and in the original VQLS paper they further impose $\lVert {\mathbf{A}}\rVert<1$ and bounded condition number) but $\mathbf{A}$ is never required to be unitary. Therefore, L2 norm isn't necessarily preserved under the action of this matrix and so a state like $|\Phi\rangle =\mathbf{A}|\psi\rangle$ may not be normalized.

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  • $\begingroup$ Thanks. That was a missing piece. $\endgroup$ Aug 27 at 10:52
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In a similar way to how the global phase difference of a state makes no physical difference, neither does amplitude of a state. We normalise states to have unit magnitude for mathematical convenience in the same way we don't carry around an $e^{i\phi}$ factor for arbitrary $\phi$ with all our states. This is because having unit vectors means we don't need to include normalisation in the Born rule. Here adding the normalisation is a way of making the cost function independent of amplitude so that you can find the global minimum of the cost function with respect to $|\Phi\rangle$ as opposed to having to find the minimum under the constraint that $|\Phi\rangle$ has unit magnitude using Lagrange multipliers. Doing so makes the minimisation mildly easier because you can just differentiate the cost with respect to the coefficients of $|\Phi\rangle$ given some orthonormal basis. That said both will give the same answer providing you normalise at the end.

In fact, in physics, there are a few situations such as infinite plane waves where one does not normalise the magnitude of the state to one. That said I do not know of any such situations in quantum computing as far as my education in the subject goes.

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  • $\begingroup$ @ Chris Long , Thanks but I am confused that $\langle\Psi|\Psi\rangle$ is 1 by definition isn't it? Quantum state is always normalized 1 and unitary transformation doesn't change the norm. $\endgroup$ Aug 25 at 14:01
  • $\begingroup$ No we set $\langle\Psi|\Psi\rangle=1$ becaus eit makes the probability rules simpler, you can use $\langle\Psi|\Psi\rangle=a$ for any real positive non-zero $a$ and unitaries will still not change the norm and all of quantum mechanics still works just you need to divide all the probabilities you calculated by $a$ so it is just easier to set $a=1$ but in this case it is easier to let $a$ vary in the minimisation and then set $a=1$ after. $\endgroup$
    – Chris Long
    Aug 25 at 14:20

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