# Help finding mistake when modifying $T$ injection protocols

I am a little confused about where I am going wrong when computing the action of the following circuit:

My understanding is that the CNOT gate acts on the second qubit as a control and the first qubit as a target. However, we can flip the CNOT so that it is acting as a control on the first qubit and target on the second qubit by introducing $$H \otimes H$$ before and after the CNOT gate.

Now the circuit looks like:

I work through the circuit as follows:

Original states:

$$T|+\rangle = \frac{|0\rangle + w |1\rangle}{\sqrt{2}}$$ $$|\psi\rangle = \alpha |0\rangle + \beta |1\rangle$$

Action of $$H \otimes H$$ on the states:

$$H T|+\rangle = \frac{(1 + w)|0\rangle + (1-w)|1\rangle}{2}$$

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

Action of CNOT on the resulting states:

$$CNOT(1 \rightarrow 2): \frac{1}{2 \sqrt(2)}( (1+w)(\alpha + \beta)|00 \rangle + (1+w)(\alpha - \beta)|01 \rangle + (1-w)(\alpha + \beta)|10 \rangle + (1-w)(\alpha - \beta)|11 \rangle )$$

Action of $$H \otimes H$$ on the resulting state:

$$(H \otimes H)(T|+\rangle \otimes |\psi \rangle) = \frac{1}{\sqrt{2}} \alpha |00\rangle + \beta |01\rangle + w\alpha |10\rangle + w\beta |11\rangle$$

But then, if I measure this state, I do not get $$T|\psi\rangle$$ (when I measure the first qubit) or $$TS|\psi\rangle$$ (when I measure the second qubit) as the circuit suggests?

Instead I get $$\frac{1}{\sqrt{2}}(\alpha |0\rangle + \beta |1\rangle)$$ when I measure the first qubit and $$\frac{1}{\sqrt{2}}(w\alpha|0\rangle + w\beta|1\rangle)$$ when I measure in the second qubit.

Can anyone identify what I am doing wrong here?

• As the answer says (although expressed slightly differently) I think you've just expanded the two-qubit state before applying the controlled-not without ever applying the controlled-not, which will change $10\leftrightarrow 11$. Mar 25 at 13:23

When applying the CNOT(1 $$\rightarrow$$ 2) you got confused in the last row. The $$(1-\omega)|1\rangle$$ flips the second qubit so $$(\alpha+\beta)|0\rangle$$ turns into $$(\alpha+\beta)|1\rangle$$ (and the same for the other term). So the resulting state is $$(1-\omega)(\alpha - \beta)|10\rangle + (1-\omega)(\alpha+\beta)|11\rangle$$.

• Thank you, I understand my mistake. Although, where did the factor of $\frac{1}{\sqrt{2}}$ go? Apr 16 at 9:39

After the second $$H\otimes H$$ operator, once the mistake with the CNOT is corrected, the qubits should be in this state:

$$\frac{1}{\sqrt{2}}\left(\alpha\left|00\right\rangle + \beta\omega\left|01\right\rangle + \alpha\omega\left|10\right\rangle + \beta\left|11\right\rangle\right)$$

Once the first (leftmost) qubit is measured, the second qubit is either directly in state $$\alpha\left|0\right\rangle + \beta\omega\left|1\right\rangle$$ if $$|0\rangle$$ came out (in which case you do not need to apply the classically-controlled $$S$$ correction to the second qubit) or it is in state $$\alpha\omega\left|0\right\rangle + \beta\left|1\right\rangle=\alpha\left|0\right\rangle + \beta\omega^{-1}\left|1\right\rangle$$ and in that case you need to apply $$S$$ to retrieve the correct state.

• why isn't it in the state $\frac{1}{\sqrt{2}} (\alpha |00\rangle + \beta w |01 \rangle + \alpha w |10 \rangle + \beta |11 \rangle)$? Apr 16 at 9:38
• I omitted this constant out of habits, but you are certainly correct.
– AG47
Apr 16 at 9:42