Output of Bell State Measurement

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 $$|00rangle$$ and $$|01rangle$$ whereas in the Bloch sphere it showed $$|00rangle$$ and $$|11rangle$$ with an angle of 180 degrees. What is the reason behind this as the output states should be $$|00rangle$$ and $$|11rangle$$ but then it is showing $$|00rangle$$ and $$|01rangle$$ in the histogram and it showed me the same when I ran that circuit on the quantum computer that is $$|00rangle$$ and $$|01rangle$$.
This is the photo of the circuit, histogram, and Bloch Sphere.

Quantum Computing Asked by Ved Dharkar on December 28, 2020

You probably did something like this:

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

Answered by KAJ226 on December 28, 2020

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

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

After measurement on simulator I get these results

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];


Answered by Martin Vesely on December 28, 2020

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