What is the probability of measuring in the following circuit?

Question

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 $|10\rangle,|01\rangle$ is $0$ so I'm not sure what I did wrong.

Answer 1

Your computations are correct. If you look at the circuit you're creating, you can see the following:

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 $|1\rangle$.

Thus, in order to get the correct result, be sure not to put the measurement on the same wire when placing them. It should look like this:

The order of the measurements can be reversed, the only important thing is that they mustn't be stored on the same classical bit. With this circuit, I get the following probabilities:

Which is what we expect.