q6-clifford-gates-tracing

Question 6: Clifford Gates (Step-by-Step Tracing of Quantum States)

Source: Chapter 03, slides 3, 7-12, 28, 30-34, 38-39, 43-44

Task type for every problem below: you are given a circuit diagram, and you must find the quantum state after every single gate, one step at a time, ending at the final state.

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Problem 1

Circuit:

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

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). The $|10⟩$ term has control $1$, so its target flips $0\to 1$, giving $|11⟩$; the $|00⟩$ term has control $0$ and is untouched.

$$\frac{1}{\sqrt{2}}\left(|00⟩+|11⟩\right).$$

Final state.

$$\frac{1}{\sqrt{2}}\left(|00⟩+|11⟩\right).$$

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Problem 2

Circuit:

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

Step 1 (start). $|00⟩$.

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

Step 3 (after $Z$ on q0). $Z$ multiplies the amplitude by $-1$ whenever q0 is $|1⟩$; the $|10⟩$ term picks up $-1$, the $|00⟩$ term is untouched.

$$\frac{1}{\sqrt{2}}\left(|00⟩-|10⟩\right).$$

Step 4 (after controlled-NOT, control q0, target q1). The $|10⟩$ term (carrying coefficient $-1/\sqrt{2}$) has control $1$, so its target flips: $-|11⟩$; the $|00⟩$ term is untouched.

$$\frac{1}{\sqrt{2}}\left(|00⟩-|11⟩\right).$$

Final state.

$$\frac{1}{\sqrt{2}}\left(|00⟩-|11⟩\right).$$

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Problem 3

Circuit:

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

Step 1 (start). $|00⟩$.

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

Step 3 (after $S$ on q0). $S=\mathrm{diag}(1,i)$ multiplies the amplitude by $i$ whenever q0 is $|1⟩$; the $|10⟩$ term picks up $i$, the $|00⟩$ term is untouched.

$$\frac{1}{\sqrt{2}}\left(|00⟩+i|10⟩\right).$$

Step 4 (after controlled-NOT, control q0, target q1). The $|10⟩$ term (coefficient $i/\sqrt{2}$) has control $1$, so its target flips: $i|11⟩$; the $|00⟩$ term is untouched.

$$\frac{1}{\sqrt{2}}\left(|00⟩+i|11⟩\right).$$

Final state.

$$\frac{1}{\sqrt{2}}\left(|00⟩+i|11⟩\right).$$

This shows a relative phase picked up before entangling survives inside the entangled state.

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Problem 4

Circuit:

q0: |0> -----H--*--x--
                 |  |
q1: |0> --X------Z--x--

(here the x--x column denotes swap between q0 and q1, and the *--Z column is controlled-Z between q0 and q1)

Step 1 (start). $|00⟩$.

Step 2 (after $X$ on q1). $|01⟩$.

Step 3 (after $H$ on q0). q0 goes from $|0⟩$ to $\frac{|0⟩+|1⟩}{\sqrt{2}}$, q1 stays $|1⟩$.

$$\frac{1}{\sqrt{2}}\left(|01⟩+|11⟩\right).$$

Step 4 (after $CZ$, between q0 and q1). $CZ$ multiplies the amplitude by $-1$ only when both qubits are $|1⟩$; the $|11⟩$ term qualifies and picks up $-1$, the $|01⟩$ term (q0 $=0$) is untouched.

$$\frac{1}{\sqrt{2}}\left(|01⟩-|11⟩\right).$$

Step 5 (after swap on q0,q1). Swap exchanges the two digits in each term: $|01⟩\to |10⟩$, $|11⟩\to |11⟩$ (symmetric, unchanged).

$$\frac{1}{\sqrt{2}}\left(|10⟩-|11⟩\right)=\frac{1}{\sqrt{2}}|1⟩\otimes \left(|0⟩-|1⟩\right),$$

an unentangled product state.

Final state.

$$\frac{1}{\sqrt{2}}\left(|10⟩-|11⟩\right)=|1⟩\otimes |-⟩.$$

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Problem 5

Circuit:

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

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). The $|100⟩$ term has control (q0) equal to $1$, so q1 flips: $\frac{1}{\sqrt{2}}\left(|000⟩+|110⟩\right)$.

Step 4 (after controlled-NOT, control q1, target q2). The $|110⟩$ term now has control (q1) equal to $1$, so q2 flips: $\frac{1}{\sqrt{2}}\left(|000⟩+|111⟩\right)$. The $|000⟩$ term has $q1=0$, untouched.

Final state.

$$\frac{1}{\sqrt{2}}\left(|000⟩+|111⟩\right),$$

the GHZ state, built here by a chain of controlled-NOT gates (q0 controls q1, q1 controls q2) rather than fanning out directly from q0.

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Problem 6

Circuit:

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

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). $\frac{1}{\sqrt{2}}\left(|00⟩+|11⟩\right)$.

Step 4 (after $Z$ on q1). $Z$ multiplies the amplitude by $-1$ whenever q1 is $|1⟩$; the $|11⟩$ term picks up $-1$, $|00⟩$ is untouched.

$$\frac{1}{\sqrt{2}}\left(|00⟩-|11⟩\right).$$

Final state.

$$\frac{1}{\sqrt{2}}\left(|00⟩-|11⟩\right).$$

Comparison to Problem 2. Both circuits ($Z$ on q0 before entangling, versus $Z$ on q1 after entangling) reach the exact same final state, because after the controlled-NOT the only term with a $|1⟩$ on either wire is $|11⟩$, so a $Z$ on either wire lands on the same term.

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Problem 7

Circuit:

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

Step 1 (start). $|0⟩$.

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

Step 3 (after first $S$). $S$ multiplies the $|1⟩$ amplitude by $i$.

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

Step 4 (after second $S$). $S$ multiplies the $|1⟩$ amplitude by $i$ again: $i\cdot i=-1$.

$$\frac{1}{\sqrt{2}}\left(|0⟩-|1⟩\right).$$

Final state.

$$\frac{1}{\sqrt{2}}\left(|0⟩-|1⟩\right)=|-⟩.$$

This confirms $S$ applied twice equals $Z$ (i.e. $SS=Z$, since $Z|+⟩=|-⟩$ directly), but only becomes visible once the qubit already has a nonzero $|1⟩$ amplitude to act on. Applying $SS$ to a bare $|0⟩$ would leave it at $|0⟩$ with no visible effect at all, since $S$ never touches the $|0⟩$ amplitude.

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Problem 8

Circuit:

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

Step 1 (start). $|00⟩$.

Step 2 (after first controlled-NOT, control q0, target q1). Control q0 is $|0⟩$, so nothing changes.

$$|00⟩.$$

Step 3 (after $H$ on q0). $\frac{1}{\sqrt{2}}\left(|00⟩+|10⟩\right)$.

Step 4 (after second controlled-NOT, control q0, target q1). The $|10⟩$ term has control $1$, so q1 flips:

$$\frac{1}{\sqrt{2}}\left(|00⟩+|11⟩\right).$$

Final state.

$$\frac{1}{\sqrt{2}}\left(|00⟩+|11⟩\right).$$

Lesson. The first controlled-NOT in this circuit was a no-op (its control qubit was still $|0⟩$ at that point in the trace), so only the last two gates actually determined the outcome. Always check the control’s actual value at each step rather than assuming a gate has an effect just because it is drawn on the circuit.