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The probability of measurement is the square of amplitude. After measurement, how to guess the original amplitude of state??

For example, in linear problem, we would like to know the exact solution, which has both of the positive and negative values.

In case of one qubit, the negative amplitude may be decided by estimating relative phase.

By measuring Z, X and Y basis, we can decide the $θ$ and $\phi$ in the state of Bloche sphere.
If $\phi >\pi$, we can guess that $|0\rangle $ and $|1\rangle $ have different sign of amplitude.
The method is available : How is it possible to guess what state the qubit was in by measuring it?

But how to estimate the negative amplitude of multiple qubits in one register?
What if the solution of value is all negative??

(Actually, I'd like to solve linear problem by quantum linear solver such as HHL algorithm )

Nyyni
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Honestly, I don't know how to find the signs of various amplitudes, but I wonder if it's really necessary. Let me explain better: the idea of solving linear equations and systems of linear equations using quantum computers comes from the fact that quantum computation inherently uses matrices, which also represent equations. The advantage lies in running your program once, having your result available more or less accurately in superposition. If you were to measure to obtain the amplitudes, you would likely lose this timing advantage. Additional operations to detect signs would further increase the execution time. For this reason, from what I know, in algorithms that provide a result in terms of superposition, no measurement should be performed, but rather, that result should be reused in a subsequent quantum algorithm.

Simona99
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As said in the other answer, HHL is best used as a part of more complex quantum algorithm, for example in machine learning. In other words, solution of a linear system is used as an input to subsequent calculation. Under such setting, you do not need to bother about signs or phases in case the results are complex numbers.

To get whole solution, i.e. including signs, real and imaginary parts, you need to perform a quantum state tomography. However, the tomography is exponentially complex in number of qubits. This effectively erases the speed up provided by HHL.

Martin Vesely
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