Can quantum tomography helps to reconstruct the state? How is this possible with arbitrary quantum state? For example if I have a $$|\psi\rangle= (0.24506+0.9633i)|0\rangle + (0.0046238+0.10943i)|1\rangle$$ if I measure it will give either 1 or 0, How can I reconstruct a state using tomography using Qiskit?


1 Answer 1


If you have just one copy of the (unknown) state, you cannot perform QST to reconstruct the state. For this, you will need many copies.

As a matter of fact, there is Holevo's theorem which, loosely speaking, states that you cannot extract more than one bit of information out of a single qubit. To describe the state of a qubit you need considerably more bits, so that's impossible (unfortunately - this is one of the most restrictive theorems in the whole of quantum computing).

If you are able to prepare many different copies of the state (not necessarily at the same time), then you can perform state tomography. To learn how to do this with qiskit, check for instance their excellent tutorial on QST.

  • $\begingroup$ Hello, Thankyou for responding. How can I make copies of quantum state are there any techniques to do this? $\endgroup$
    – John Jones
    Mar 31, 2021 at 18:02
  • $\begingroup$ Do you know what the state is? Then you can work out how to prepare it from, say, the $|0\rangle$ state, and redo this over and over. If you don't know what the state is, there is another important restrictive theorem; the no-cloning theorem, which states exactly that you cannot do this (for any arbitrary state). $\endgroup$
    – JSdJ
    Mar 31, 2021 at 18:07
  • $\begingroup$ I was seeign this link they are saying you can see state vector after quantum state is teleported using tomography. can we do this? $\endgroup$
    – John Jones
    Mar 31, 2021 at 18:17
  • $\begingroup$ That's only for a very restrictive use case - where you know that the state is in either state from a basis, but no other option. Then, you can perform a measurement in that basis to check which of the two basis states it is - a 'normal' measurement is just that for the $Z:= \{|0\rangle,|1\rangle\}$ basis. $\endgroup$
    – JSdJ
    Mar 31, 2021 at 18:36
  • $\begingroup$ That means if I take $$|\psi\rangle= (0.24506+0.9633i)|0\rangle + (0.0046238+0.10943i)|1\rangle$$ and teleport form Alice to Bob, I can reconstruct the state vector $|\psi\rangle$ on the bob side using tomography? I was totally confused. Is that what they were discussing link? $\endgroup$
    – John Jones
    Mar 31, 2021 at 18:53

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