I tried to measure quantum state with a quantum state tomography. However, I encountered a situation when two different quantum states had the same density matrix. In particular, these states were $\frac{1}{\sqrt{2}}|0\rangle + \frac{1+i}{2}|1\rangle$ and $\frac{1-i}{2}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle$.

My question is: Is there any possibility how to distinguish these two states (or any other two states with the same density matrix) with quantum state tomography?

  • 2
    $\begingroup$ Both of those states are same, they only differ by a global phase! $\endgroup$
    – Hemant
    Nov 10, 2019 at 23:15

2 Answers 2


There isn't. A density matrix encodes all the knowledge available about a state, therefore if two states are described by the same density matrix, they are indistinguishable.

Ket vectors differing by only a global phase have always the same density matrix, and represent the same physical state.

  • 1
    $\begingroup$ Right, that. I wrote a similar answer in parallel. :-) $\endgroup$ Nov 10, 2019 at 23:58
  • $\begingroup$ I see now, thanks. $\endgroup$ Nov 11, 2019 at 5:47

The question presupposes a misconception that the vector form of a state $|\psi\rangle$ exists independently of its density operator form $|\psi\rangle\langle\psi|$, which is often described as secondary. In reality, the density operator of a state is all that truly exists --- and even then, it only exists as statistical information. In fact, you can construct quantum mechanics using density operators $\rho$ as the primary model of states. The formalism for this is similar to the Heisenberg picture of quantum mechanics, in which operators evolve over time. In a Heisenberg-type picture, you don't strictly need Hilbert spaces at all at first, you can define everything with abstract "operators". You can introduce a Hilbert space $\mathcal{H}$ later as an algebra representation of all of the operators. You can also introduce vector states $|\psi\rangle$ later as a simplified formalism for operator states that happen to have the form $\rho = |\psi\rangle\langle\psi|$.


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