Weak measurement quantum circuit

[whatisreality]

Homework Statement

The circuit in the attached photo has a one-qubit operation $U$ which is controlled by the first qubit. The box with the meter and arrow denotes a measurement. If $U=H$ what is the probability of finding 0 or 1 in the final measurement of the second qubit?

Homework Equations

The Attempt at a Solution

I can write the general state $|\phi\rangle = \alpha |0\rangle + \beta |1\rangle$ so then the input I think is:
$(\alpha |0\rangle + \beta |1\rangle)|0\rangle$
Applying the Hadamard gate I'm not sure about, only because of the control bit - usually though control means only act if the first qubit is 1. So after that gate the state would be
$\alpha |00\rangle + \frac{1}{\sqrt{2}} \beta |1\rangle (|0\rangle + |1\rangle)$
So would the probability of the second qubit being zero be $\alpha^2 + \frac{\beta^2}{2}$? I'm not sure that H is really applied only when the first qubit is 1, could it also be that it's applied when the first qubit is zero?

 

[mark726]

Hi there,

Thank you for your post. Your approach is correct – the state after the Hadamard gate is applied would be:

$$(\alpha |0\rangle + \beta |1\rangle)\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$$

Now, the control bit indicates that the Hadamard gate will only be applied if the first qubit is 1. In this case, the first qubit is |0⟩ so the Hadamard gate will not be applied. Therefore, the state after the Hadamard gate is still:

$$(\alpha |0\rangle + \beta |1\rangle)\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$$

The probability of the second qubit being |0⟩ after measurement would be:

$$|\alpha|^2\frac{1}{2} + |\beta|^2\frac{1}{2} = \frac{1}{2}(|\alpha|^2 + |\beta|^2)$$

This is because the probability of measuring the first qubit as |0⟩ is 1, so the state after measurement would be:

$$|0\rangle(\alpha |0\rangle + \beta |1\rangle)\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) = (\alpha |00\rangle + \beta |10\rangle)\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$$

The probability of measuring the second qubit as |0⟩ is then:

$$|\alpha|^2\frac{1}{2} + |\beta|^2\frac{1}{2} = \frac{1}{2}(|\alpha|^2 + |\beta|^2)$$

I hope this helps clarify your understanding. Let me know if you have any further questions.