# What is the probability of measuring in the following circuit?

So given the circuit and $$|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$$ above I would like to see what the probabilities of each possible measurement are.

To do this I tried expanding the circuit as follows:

1. after first Hadammard and CNOT we get $$(\alpha|0\rangle+ \beta|1\rangle) \otimes \frac{1}{\sqrt{2}}(|00\rangle \otimes |11\rangle)$$

2. Then we apply the second CNOT to get: $$\frac{1}{\sqrt{2}}(CNOT \otimes I)(\alpha|000\rangle + \alpha|011\rangle + \beta|100\rangle + \beta|111\rangle) = \frac{1}{\sqrt{2}}(\alpha|000\rangle + \alpha|011\rangle + \beta|110\rangle + \beta|101\rangle)$$

3. Now apply the final Hadamard to get: $$(H\otimes H \otimes I)\frac{1}{\sqrt{2}}(\alpha|000\rangle + \alpha|011\rangle + \beta|110\rangle + \beta|101\rangle) = \frac{1}{\sqrt{2}}(\alpha|+00\rangle + \alpha|+11\rangle + \beta|-10\rangle + \beta|-01\rangle) = \frac{1}{2}(\alpha|000\rangle + \alpha|100\rangle + \alpha|011\rangle + \alpha |111\rangle + \beta|010\rangle - \beta|110\rangle + \beta|001\rangle - \beta|101\rangle)$$

So now we have $$Pr[000]=Pr[011]=Pr[111] = \frac{1}{4}\alpha^2, Pr[010]=Pr[001]=Pr[110]=Pr[101] = \frac{1}{4}\beta^2$$

So now if we measure $$00$$ on the first two qubits:

$$Pr[00] = Pr[000] + Pr[001] = \frac{1}{4}$$.

And by similar logic $$Pr[11]=PR[01]=Pr[10]=\frac{1}{4}$$

But then when I implement this circuit in IBM quantum I get that probability of $$|10\rangle,|11\rangle$$ is $$0$$ so I'm not sure what I did wrong.

In particular, note the 0 at the bottom of your measurements. This means that both measurements are stored within the first bit of the result. Thus, the last bit will always be 0, since it stored no measurement. This also explains why you're seeing 00 and 10 with equal probability: the first qubit indeed has $$50\%$$ chance of being in state $$|0\rangle$$ and $$50\%$$ chance of being in state $$|1\rangle$$.