I have a quantum system that outputs a state similar to:

$|\psi\rangle = \frac{1}{2}(|00\rangle + |01\rangle + |10\rangle \pm |11\rangle)$.

So my question is: is there a way (by measurement or by adding gates) to figure out if the probability amplitude of $|11\rangle$ is positive or negative?

I need to know the sign because the sign will help me in the next steps in the system.

  • 1
    $\begingroup$ You are aware that these two states are fundamentally different? $\endgroup$ Apr 30, 2022 at 17:33
  • 1
    $\begingroup$ What do you mean by "the sign will help me in the next steps in the system"? $\endgroup$ Apr 30, 2022 at 21:09

2 Answers 2


There is a way that you can figure out the answer to your question using repeated measurements. Let's start with the first case where you have

$$ |\psi\rangle=|\psi_+\rangle=\frac{1}{2}\left(|00\rangle+|01\rangle+|10\rangle+|11\rangle\right) $$

This is actually a product state and can be written as

$$ |\psi_+\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle) \otimes \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle) $$

Applying an $H$ gate to the second qubit gives us

\begin{align*} (I \otimes H) |\psi_+\rangle&=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle) \otimes H\left(\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)\right) \\ &=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle) \otimes |0\rangle \\ &= \frac{1}{\sqrt{2}}(|00\rangle+|10\rangle) \end{align*}

So now if you measure your second qubit, you have 100% probability of obtaining $|0\rangle$. Now let's see what happens if we perform the same protocol for the other case. We start with

$$ |\psi\rangle=|\psi_-\rangle=\frac{1}{2}\left(|00\rangle+|01\rangle+|10|\rangle-|11\rangle\right) $$

This is actually an entangled state so it cannot be written as a product state. However we can still apply an $H$ gate to the second qubit:

\begin{align*} (I \otimes H) |\psi_-\rangle&=\frac{1}{2}\left(|0\rangle H|0\rangle+|0\rangle H|1\rangle+|1\rangle H|0\rangle-|1\rangle H|1\rangle\right) \\ &= \frac{1}{2}\left[|0\rangle \otimes \left(\frac{|0\rangle+|1\rangle}{\sqrt{2}}\right) + |0\rangle \otimes \left(\frac{|0\rangle-|1\rangle}{\sqrt{2}}\right) + |1\rangle \otimes \left(\frac{|0\rangle+|1\rangle}{\sqrt{2}}\right) - |1\rangle \otimes \left(\frac{|0\rangle-|1\rangle}{\sqrt{2}}\right)\right] \\ &= \frac{1}{2\sqrt{2}}\left(|00\rangle + |01\rangle + |00\rangle - |01\rangle +|10\rangle + |11\rangle - |10\rangle + |11\rangle \right) \\ &=\frac{1}{\sqrt{2}}\left(|00\rangle + |11\rangle\right) \end{align*}

So now if you measure your second qubit, you only have a 50% probability of obtaining $|0\rangle$. So in summary, you can repeatedly perform the experiment where you prepare $|\psi_\pm\rangle$, apply an $H$ gate and measure the second qubit and get a good approximation of the probability of measuring $|0\rangle$, which will tell you which of the two states ($|\psi_+\rangle$ or $|\psi_-\rangle$) you are producing.

This same method can be applied to general 2-qubits states i.e.

$$ |\psi_\pm\rangle=\alpha|00\rangle+ \beta|01\rangle+\gamma|10\rangle\pm\delta|11\rangle $$

But instead of 100% and 50%, you will get different probabilities.


I suggest you take a look at Algorithm 4.1 (vector state tomography) of Kerenidis and Prakash's A Quantum Interior Point Method for LPs and SDPs.

They provide a nice theorem to perform tomography on pure states. The second part of the algorithm estimates the sign of the amplitudes. I'd say you could adapt the second part of their routine to your specific case.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.