For example, one have measured some states like $|0\rangle$ in the computational basis for many times and got the approximate probability of getting 0 and 1 ($P(0)$ and $P(1)$). Then how does he calculate the off-diagonal elements of the density of the initial quantum state? The system is open and with noise.

  • $\begingroup$ you might also want to have a look at how quantum state tomography works $\endgroup$
    – glS
    Commented Aug 14, 2018 at 16:24

1 Answer 1


With the given measurements, you cannot: there is no observable difference between many different states such as $|\pm\rangle=(|0\rangle\pm|1\rangle)/\sqrt{2}$.

In order to determine what the state is completely, you need more measurements. If you're using projective measurements, you need two more. These would typically be projections onto the bases $$ |\pm\rangle \qquad\text{and}\qquad (|0\rangle\pm i|1\rangle)/\sqrt{2} $$ However, if you are willing to use POVMs, you can define a single measurement, comprising four measurement operators that does the job. One way to visualise these is in the Bloch-sphere: you need a set of axes that provides a basis in the three dimensional space. Three vectors cannot do this because they satisfy a completeness relation which constrains them to be in a plane. The most effective way to do this is to inscribe a regular tetrahedron inside the Bloch sphere and take the 4 corners as measurement bases.


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