# Evaluate the given quantum circuit The input state on the qubit is $$|1\rangle\otimes(2|0\rangle+|1\rangle)$$. I am assuming this means that the first qubit has the state $$|1\rangle$$, and the second qubit has the state $$2|0\rangle+|1\rangle$$. I could be wrong though. My question is, given the quantum circuit is shown below, what is the probability of the measurement outcomes, as well as the outcome measurement, and finally the new state after measurement.

If there is anything that needs to be further specified please let me know. Thank you.

The starting state is

$$2 | 1 0 \rangle + | 1 1 \rangle$$

After the $$H \otimes 1$$

$$\sqrt{2} | 0 0 \rangle - \sqrt{2} | 1 0 \rangle + 1/\sqrt{2} | 0 1 \rangle - 1/\sqrt{2} | 1 1 \rangle$$

where the first 2 terms come from the first term above and the last two from the last term.

Then the CNOT

$$\sqrt{2} | 0 0 \rangle - \sqrt{2} | 1 1 \rangle + 1/\sqrt{2} | 0 1 \rangle - 1/\sqrt{2} | 1 0 \rangle$$

You didn't normalize initially so it doesn't make sense to talk about probabilities right now. You would have to normalize by a factor of $$\sqrt{5}$$

Edit:

$$H | 1 \rangle = 1/\sqrt{2} | 0 \rangle - 1/\sqrt{2} | 1 \rangle$$

so

$$\begin{eqnarray*} H \otimes 1 | 10 \rangle &=& 1/\sqrt{2} | 00 \rangle - 1/\sqrt{2} | 10 \rangle\\ (H \otimes 1) 2 | 10 \rangle &=& 2/\sqrt{2} | 00 \rangle - 2/\sqrt{2} | 10 \rangle\\ &=& \sqrt{2} | 00 \rangle - \sqrt{2} | 10 \rangle\\ H \otimes 1 | 11 \rangle &=& 1/\sqrt{2} | 01 \rangle - 1/\sqrt{2} | 11 \rangle\\ \end{eqnarray*}$$

The normalization factor doesn't change with applying $$H \otimes 1$$ or $$CNOT$$. It is still $$\sqrt{5}$$. The normalized state is

$$\frac{\sqrt{2}}{\sqrt{5}} | 0 0 \rangle - \frac{\sqrt{2}}{\sqrt{5}} | 1 1 \rangle + 1/\sqrt{2*5} | 0 1 \rangle - 1/\sqrt{2*5} | 1 0 \rangle$$

as can be checked by:

$$\begin{eqnarray*} (\frac{\sqrt{2}}{\sqrt{5}})^2+(-\frac{\sqrt{2}}{\sqrt{5}})^2+(-1/\sqrt{2*5})^2+(1/\sqrt{2*5})^2 &=& \frac{2}{5}+\frac{2}{5}+\frac{1}{10}+\frac{1}{10}\\ &=& 1 \end{eqnarray*}$$

• Thank you for the response. after the Hadamard gate, how do you end up with the state given. More specifically, how do you get $\sqrt 2$ in the numerator? – Frank Schroer IV Jan 6 '19 at 2:47
• Also, do you have to normalize after applying the Hadamard gate, before applying the CNOT gate? – Frank Schroer IV Jan 6 '19 at 3:01