q7-non-clifford-init

Question 7: Non-Clifford Gates (State Initialization with Clifford and Non-Clifford Gates)

Source: Chapter 04, slides 6-8, 11-16

Task type for every problem below: you are given a target quantum state that a Clifford-only circuit cannot reach, and you must draw a circuit using both Clifford gates ($H$, $S$, Pauli $X/Y/Z$, controlled-NOT) and at least one non-Clifford gate ($P(\lambda )$, $T$, or a continuous-angle rotation $R_Y(\theta )$) that prepares it from the all-zero basis state, then verify by tracing forward.

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Problem 1: $\frac{|0⟩+e^{i\pi /4}|1⟩}{\sqrt{2}}$, single qubit

Circuit:

q0: |0> --H--T--

Step 1 (start). $|0⟩$.

Step 2 (after $H$). $\frac{1}{\sqrt{2}}\left(|0⟩+|1⟩\right)$.

Step 3 (after $T$). $T=\mathrm{diag}\left(1,e^{i\pi /4}\right)$ multiplies the $|1⟩$ amplitude by $e^{i\pi /4}$, leaving the $|0⟩$ amplitude untouched.

$$\frac{1}{\sqrt{2}}\left(|0⟩+e^{i\pi /4}|1⟩\right),$$

the target state.

Why Clifford alone cannot do this. $H$ alone only ever produces real, equal-weight superpositions with phase $0$ or $\pi$ between the two terms; $T$ is what supplies the $\pi /4$ relative phase that neither $H$ nor $S$ (which only offers multiples of $\pi /2$) can produce.

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Problem 2: $\frac{|00⟩+e^{i\pi /4}|11⟩}{\sqrt{2}}$, two qubits

Circuit:

q0: |0> --H--*-----
             |
q1: |0> -----X--T--

Step 1 (start). $|00⟩$.

Step 2 (after $H$ on q0). $\frac{1}{\sqrt{2}}\left(|00⟩+|10⟩\right)$.

Step 3 (after controlled-NOT, control q0, target q1). $\frac{1}{\sqrt{2}}\left(|00⟩+|11⟩\right)$, the Bell state.

Step 4 (after $T$ on q1). The $|00⟩$ term has $q1=|0⟩$ and is untouched; the $|11⟩$ term has $q1=|1⟩$ and picks up $e^{i\pi /4}$.

Final state.

$$\frac{1}{\sqrt{2}}\left(|00⟩+e^{i\pi /4}|11⟩\right),$$

the target state.

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Problem 3: $\cos (\pi /8)|0⟩+\sin (\pi /8)|1⟩$, single qubit, arbitrary rotation angle

Circuit:

q0: |0> --RY(pi/4)--

Step 1 (start). $|0⟩$.

Step 2 (after $R_Y(\pi /4)$).

$$R_Y(\theta )|0⟩=\cos \left(\frac{\theta }{2}\right)|0⟩+\sin \left(\frac{\theta }{2}\right)|1⟩$$

for any real $\theta$; substituting $\theta =\pi /4$:

$$\cos \left(\frac{\pi }{8}\right)|0⟩+\sin \left(\frac{\pi }{8}\right)|1⟩,$$

the target state, directly in one gate.

Why this needs a non-Clifford gate. Clifford single-qubit gates only ever reach the six stabilizer states ($|0⟩$, $|1⟩$, $|+⟩$, $|-⟩$, $|+i⟩$, $|-i⟩$); $\pi /8$ is not one of the fixed Clifford angles (multiples of $\pi /2$), so only a continuously parameterized rotation gate like $R_Y$ can land exactly here.

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Problem 4: $\frac{|000⟩+e^{i\pi /4}|111⟩}{\sqrt{2}}$, GHZ state with a non-Clifford phase

Circuit:

q0: |0> --H--*-----*-----
             |     |
q1: |0> -----X-----|-----
                    |
q2: |0> ------------X--T--

Step 1 (start). $|000⟩$.

Step 2 (after $H$ on q0). $\frac{1}{\sqrt{2}}\left(|000⟩+|100⟩\right)$.

Step 3 (after controlled-NOT, control q0, target q1). $\frac{1}{\sqrt{2}}\left(|000⟩+|110⟩\right)$.

Step 4 (after controlled-NOT, control q0, target q2). $\frac{1}{\sqrt{2}}\left(|000⟩+|111⟩\right)$, the ordinary GHZ state.

Step 5 (after $T$ on q2). The $|000⟩$ term has $q2=|0⟩$ and is untouched; the $|111⟩$ term has $q2=|1⟩$ and picks up $e^{i\pi /4}$.

Final state.

$$\frac{1}{\sqrt{2}}\left(|000⟩+e^{i\pi /4}|111⟩\right),$$

the target state.

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Problem 5: $\frac{|0⟩+e^{i\pi /8}|1⟩}{\sqrt{2}}$, single qubit, general phase gate $P(\lambda )$

Circuit:

q0: |0> --H--P(pi/8)--

Step 1 (start). $|0⟩$.

Step 2 (after $H$). $\frac{1}{\sqrt{2}}\left(|0⟩+|1⟩\right)$.

Step 3 (after $P(\pi /8)$). $P(\lambda )=\mathrm{diag}\left(1,e^{i\lambda }\right)$ multiplies the $|1⟩$ amplitude by $e^{i\pi /8}$, leaving $|0⟩$ untouched.

Final state.

$$\frac{1}{\sqrt{2}}\left(|0⟩+e^{i\pi /8}|1⟩\right),$$

the target state. This is the same construction as Problem 1 but with an arbitrary angle $\lambda =\pi /8$ instead of the fixed $T=P(\pi /4)$, showing $T$ is just one fixed instance of the more general $P(\lambda )$ family.

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Problem 6: $\cos (\pi /8)|00⟩+\sin (\pi /8)|11⟩$, a non-maximally-entangled two-qubit state

Circuit:

q0: |0> --RY(pi/4)--*--
                     |
q1: |0> -------------X--

Step 1 (start). $|00⟩$.

Step 2 (after $R_Y(\pi /4)$ on q0). $\cos (\pi /8)|0⟩+\sin (\pi /8)|1⟩$ on q0, tensor $|0⟩$ on q1.

$$\cos (\pi /8)|00⟩+\sin (\pi /8)|10⟩.$$

Step 3 (after controlled-NOT, control q0, target q1). The $|10⟩$ term has control $1$, so q1 flips: $\sin (\pi /8)|11⟩$; the $|00⟩$ term has control $0$ and is untouched: $\cos (\pi /8)|00⟩$.

Final state.

$$\cos (\pi /8)|00⟩+\sin (\pi /8)|11⟩,$$

the target state.

Why this needs a non-Clifford gate. Clifford gates ($H$, controlled-NOT) alone only ever reach the maximally-entangled Bell states, with equal weight $1/\sqrt{2}$ on both terms; reaching an unequal weighting like $\cos (\pi /8)$ versus $\sin (\pi /8)$ requires a continuously variable rotation angle before the entangling controlled-NOT.

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Problem 7: Bell pair on q0,q1 tensor a non-Clifford single-qubit state on q2

Target: $\frac{|00⟩+|11⟩}{\sqrt{2}}$ (qubits 0,1) $\otimes$ $\left(\cos (\pi /8)|0⟩+\sin (\pi /8)|1⟩\right)$ (qubit 2).

Circuit:

q0: |0> --H--*----------------
             |
q1: |0> -----X----------------
q2: |0> --RY(pi/4)-------------

Step 1, structural observation. q0,q1 and q2 share no gate, so they can be traced independently and combined at the end by a tensor product, the same reasoning used for parallel sub-circuits in Question 5.

Step 2, q0,q1 branch. After $H$ then controlled-NOT: $\frac{1}{\sqrt{2}}\left(|00⟩+|11⟩\right)$, the ordinary Bell state.

Step 3, q2 branch. After $R_Y(\pi /4)$: $\cos (\pi /8)|0⟩+\sin (\pi /8)|1⟩$.

Final state.

$$\frac{1}{\sqrt{2}}{\left(|00⟩+|11⟩\right)}_{01}\otimes {\left(\cos (\pi /8)|0⟩+\sin (\pi /8)|1⟩\right)}_2,$$

the target state. The Bell pair needed only Clifford gates; only the q2 register needed the non-Clifford rotation.

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Problem 8: $\frac{|0⟩+e^{i3\pi /4}|1⟩}{\sqrt{2}}$, composing a Clifford phase gate with a non-Clifford phase gate

Circuit:

q0: |0> --H--S--T--

Step 1 (start). $|0⟩$.

Step 2 (after $H$). $\frac{1}{\sqrt{2}}\left(|0⟩+|1⟩\right)$.

Step 3 (after $S$). $S=\mathrm{diag}(1,i)$ multiplies the $|1⟩$ amplitude by $i=e^{i\pi /2}$, leaving $|0⟩$ untouched.

$$\frac{1}{\sqrt{2}}\left(|0⟩+e^{i\pi /2}|1⟩\right).$$

Step 4 (after $T$). $T$ multiplies the $|1⟩$ amplitude by $e^{i\pi /4}$ again, and phases on the same amplitude add:

$$e^{i\pi /2}\cdot e^{i\pi /4}=e^{i(\pi /2+\pi /4)}=e^{i3\pi /4}.$$

Final state.

$$\frac{1}{\sqrt{2}}\left(|0⟩+e^{i3\pi /4}|1⟩\right),$$

the target state.

Why this matters. This shows a Clifford phase gate ($S$, angle $\pi /2$) and a non-Clifford phase gate ($T$, angle $\pi /4$) applied to the same qubit combine by adding their angles, since both are diagonal in the same basis and commute with each other. The same $3\pi /4$ target could equally well have been reached by a single $P(3\pi /4)$ in place of $S$ and $T$ together.