2
$\begingroup$

I was trying to see the output of Bell State Measurement in IBM Quantum Experience but in the simulation down the circuit in the histogram, it showed me that the possible qubit will be $|00\rangle$ and $|01\rangle$ whereas in the Bloch sphere it showed $|00\rangle$ and $|11\rangle$ with an angle of 180 degrees. What is the reason behind this as the output states should be $|00\rangle$ and $|11\rangle$ but then it is showing $|00\rangle$ and $|01\rangle$ in the histogram and it showed me the same when I ran that circuit on the quantum computer that is $|00\rangle$ and $|01\rangle$. This is the photo of the circuit, histogram, and Bloch Sphere.

$\endgroup$
1
  • 1
    $\begingroup$ Hi @VED DHARKAR. Did you probably forget to attach the photo? $\endgroup$ Dec 11 '20 at 8:15
2
$\begingroup$

I tried to prepare Bell state with circuit described by $\text{CNOT} (H \otimes I)$ and the result is state $\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$ as expected.

Here is my circuit in the composer (including the code):

enter image description here

After measurement on simulator I get these results

enter image description here

Please make sure that you measure correctly, i.e. each qubit to separate classical bit:

measure q[0] -> c[0];
measure q[1] -> c[1];
$\endgroup$
1
$\begingroup$

You probably did something like this:

enter image description here

Since you are only measuring the top qubit, and a Bell state $|\psi \rangle = \dfrac{|00\rangle + |11\rangle}{\sqrt{2}}$ has $1/2$ probability of the first qubit in the state $|0\rangle$ and $1/2$ probability of it being in the state $|1\rangle$. Hence, you got the above probabilities readout plot.

If you also put in the second measurement like @MartinVesely did, then you will get the right answer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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