q4-harder

Question 4 (Harder): Quantum Operators (Translating a State Mapping to a Matrix)

Source: Chapter 02, slides 3-4, 7-9, 19, 21, 27, 32-33; Chapter 03, slides 11, 38-39; Chapter 04, slides 11-16

These problems go beyond the base Question 4 set: mappings where information is deliberately incomplete and must be recovered from orthonormality, mappings on 3 qubits with genuinely entangling truth tables, mappings given between two non-computational bases at once, and a mapping given as a symbolic formula rather than concrete numbers.

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Problem 1: recovering a missing output from orthonormality alone

Given: on a single qubit, only one output is specified,

$$|0⟩\to \frac{1}{\sqrt{3}}|0⟩+\sqrt{\frac{2}{3}}|1⟩,$$

and you are told the unknown operator $U$ is orthonormal (a valid gate). Find $U|1⟩$, then assemble the full matrix $U$.

Step 1, what orthonormality forces. $U^{†}U=I$ means the columns of $U$ must form an orthonormal set. Column 0 is already fixed as $\left(\frac{1}{\sqrt{3}},\sqrt{\frac{2}{3}}\right)$; this has norm

$${\left(\frac{1}{\sqrt{3}}\right)}^2+{\left(\sqrt{\frac{2}{3}}\right)}^2=\frac{1}{3}+\frac{2}{3}=1,$$

so it is already a valid unit vector. Column 1 must be a unit vector orthogonal to column 0, since distinct columns of an orthonormal matrix must have zero overlap.

Step 2, find a vector orthogonal to column 0. In ${\mathbb{C}}^2$, a real vector orthogonal to $\left(\frac{1}{\sqrt{3}},\sqrt{\frac{2}{3}}\right)$ is obtained by swapping the two entries and flipping one sign:

$$\left(-\sqrt{\frac{2}{3}},\frac{1}{\sqrt{3}}\right).$$

Check orthogonality directly:

$$\frac{1}{\sqrt{3}}\left(-\sqrt{\frac{2}{3}}\right)+\sqrt{\frac{2}{3}}\cdot \frac{1}{\sqrt{3}}=-\frac{\sqrt{2}}{3}+\frac{\sqrt{2}}{3}=0.✓$$

Its norm is $\frac{2}{3}+\frac{1}{3}=1$, so it is already unit length.

Step 3, the phase freedom. This orthogonal unit vector is only determined up to an overall phase $e^{i\gamma }$ — multiplying it by any $e^{i\gamma }$ keeps it unit length and orthogonal to column 0. Nothing in the problem statement fixes $\gamma$, so we take the simplest choice $\gamma =0$ (a real solution). A different piece of information (e.g. a stated Hermiticity requirement, or a second constraint on $U|1⟩$) would be needed to pin down $\gamma$ uniquely.

Step 4, assemble.

$$U|1⟩=-\sqrt{\frac{2}{3}}|0⟩+\frac{1}{\sqrt{3}}|1⟩,U=\left(\begin{array}{cc}\frac{1}{\sqrt{3}}& -\sqrt{\frac{2}{3}}\\ \sqrt{\frac{2}{3}}& \frac{1}{\sqrt{3}}\end{array}\right).$$

Step 5, verify. Both columns are unit length and mutually orthogonal (Steps 1-2), so $U^{†}U=I$ by construction. This $U$ has the form $\left(\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right)$ with $\cos \alpha =\frac{1}{\sqrt{3}}$, i.e. it is a real rotation matrix, matching the $R_Y(\theta )$ family from chapter 04 slide 14 with $\theta =2\alpha$.

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Problem 2: the Toffoli gate’s full $8\times 8$ matrix from its truth table

Given, on 3 qubits: $|x,y,z⟩\to |x,y,z\oplus (x\wedge y)⟩$ (the target bit flips only when both control bits $x,y$ are $1$).

Step 1, write out all 8 images. Using the ordering $|x,y,z⟩$ with $z$ the target:

$$|000⟩\to |000⟩,|001⟩\to |001⟩,|010⟩\to |010⟩,|011⟩\to |011⟩,$$ $$|100⟩\to |100⟩,|101⟩\to |101⟩,|110⟩\to |111⟩,|111⟩\to |110⟩.$$

Only the last two rows change: $x=y=1$ toggles $z$.

Step 2, write the outer-product sum. Six terms are identity-like ($|v⟩⟨v|$ for the six untouched basis states) plus two swapped terms:

$$U=\sum _{v\ne 110,111}|v⟩⟨v|+|111⟩⟨110|+|110⟩⟨111|.$$

Step 3, place entries. Order the basis by index $4x+2y+z$, positions $0$–$7$. Every position keeps a $1$ on the diagonal except positions $6$ ($110$) and $7$ ($111$), which swap:

$$U=\left(\begin{array}{cccccccc}1& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 0& 0& 1& 0\end{array}\right).$$

Step 4, verify. This is a permutation matrix (exactly one $1$ per row and column), so $U^{†}U=I$ automatically. This is the Toffoli gate, and it is non-Clifford (Question 7, Math Q6) precisely because it cannot be decomposed as a product of one- and two-qubit Clifford gates, even though the matrix itself is a simple $0/1$ permutation.

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Problem 3: iSWAP from its mapping (entangling, complex-valued, not a plain permutation)

Given: $|00⟩\to |00⟩$, $|01⟩\to i|10⟩$, $|10⟩\to i|01⟩$, $|11⟩\to |11⟩$.

Step 1, write the outer-product sum.

$$U=|00⟩⟨00|+i|10⟩⟨01|+i|01⟩⟨10|+|11⟩⟨11|.$$

Step 2, place columns. Column 0 ($⟨00|$): $(1,0,0,0)$. Column 1 ($⟨01|$): image is $i|10⟩$, so $(0,0,i,0)$. Column 2 ($⟨10|$): image is $i|01⟩$, so $(0,i,0,0)$. Column 3 ($⟨11|$): $(0,0,0,1)$.

Step 3, assemble.

$$U=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 0& i& 0\\ 0& i& 0& 0\\ 0& 0& 0& 1\end{array}\right).$$

Step 4, verify. Each column has exactly one nonzero entry, of modulus $|i|=1$ or $|1|=1$, and the nonzero entries of distinct columns sit in distinct rows (row $0$, row $2$, row $1$, row $3$ respectively), so all columns are mutually orthogonal and unit length: $U^{†}U=I$. Valid. Unlike Problem 2, this mapping genuinely entangles the computational basis states $|01⟩,|10⟩$ into each other with an extra factor of $i$, rather than merely permuting bit strings unchanged in value.

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Problem 4: change of basis between the Bell basis and the computational basis

Given, following the standard Bell-measurement convention: $|{\Phi }^{+}⟩\to |00⟩$, $|{\Phi }^{-}⟩\to |01⟩$, $|{\Psi }^{+}⟩\to |10⟩$, $|{\Psi }^{-}⟩\to |11⟩$, where

$$|{\Phi }^{\pm }⟩=\frac{|00⟩\pm |11⟩}{\sqrt{2}},|{\Psi }^{\pm }⟩=\frac{|01⟩\pm |10⟩}{\sqrt{2}}.$$

Unlike every problem in the base Question 4 file, each input ket here is itself a two-term superposition, so every bra in the outer-product sum expands into two pieces before columns can be read off.

Step 1, expand every bra.

$$⟨{\Phi }^{+}|=\frac{1}{\sqrt{2}}(⟨00|+⟨11|),⟨{\Phi }^{-}|=\frac{1}{\sqrt{2}}(⟨00|-⟨11|),$$ $$⟨{\Psi }^{+}|=\frac{1}{\sqrt{2}}(⟨01|+⟨10|),⟨{\Psi }^{-}|=\frac{1}{\sqrt{2}}(⟨01|-⟨10|).$$

Step 2, write the outer-product sum and distribute.

$$U=|00⟩⟨{\Phi }^{+}|+|01⟩⟨{\Phi }^{-}|+|10⟩⟨{\Psi }^{+}|+|11⟩⟨{\Psi }^{-}|$$ $$=\frac{1}{\sqrt{2}}|00⟩(⟨00|+⟨11|)+\frac{1}{\sqrt{2}}|01⟩(⟨00|-⟨11|)+\frac{1}{\sqrt{2}}|10⟩(⟨01|+⟨10|)+\frac{1}{\sqrt{2}}|11⟩(⟨01|-⟨10|).$$

Step 3, collect by input basis ket (column) rather than by term. Column $⟨00|$ receives a contribution from the first two terms: $\frac{1}{\sqrt{2}}|00⟩$ and $\frac{1}{\sqrt{2}}|01⟩$, so

$${\text{col}}_{00}=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0,0\right).$$

Column $⟨11|$ receives $\frac{1}{\sqrt{2}}|00⟩$ and $-\frac{1}{\sqrt{2}}|01⟩$:

$${\text{col}}_{11}=\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0,0\right).$$

Column $⟨01|$ receives $\frac{1}{\sqrt{2}}|10⟩$ and $\frac{1}{\sqrt{2}}|11⟩$:

$${\text{col}}_{01}=\left(0,0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right).$$

Column $⟨10|$ receives $\frac{1}{\sqrt{2}}|10⟩$ and $-\frac{1}{\sqrt{2}}|11⟩$:

$${\text{col}}_{10}=\left(0,0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right).$$

Step 4, assemble in the standard column order $\{00,01,10,11\}$.

$$U=\frac{1}{\sqrt{2}}\left(\begin{array}{cccc}1& 0& 0& 1\\ 1& 0& 0& -1\\ 0& 1& 1& 0\\ 0& 1& -1& 0\end{array}\right).$$

Step 5, verify orthonormality directly. ${\text{col}}_{00}\cdot {\text{col}}_{00}=\frac{1}{2}+\frac{1}{2}=1$; ${\text{col}}_{00}\cdot {\text{col}}_{11}=\frac{1}{2}-\frac{1}{2}=0$; ${\text{col}}_{01}\cdot {\text{col}}_{10}=\frac{1}{2}-\frac{1}{2}=0$; all cross terms between $\{00,11\}$-columns and $\{01,10\}$-columns vanish automatically since they have disjoint nonzero rows. So $U^{†}U=I$. Valid. This $U$ is exactly the Bell-basis-to-computational-basis change-of-basis matrix.

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Problem 5: a symbolic mapping, deriving $R_Z(\theta )$ as a formula and connecting it to $P(\lambda )$

Given, as a formula in the real parameter $\theta$ rather than concrete numbers:

$$|0⟩\to e^{-i\theta /2}|0⟩,|1⟩\to e^{i\theta /2}|1⟩.$$

Step 1, write the outer-product sum.

$$U(\theta )=e^{-i\theta /2}|0⟩⟨0|+e^{i\theta /2}|1⟩⟨1|.$$

Step 2, assemble.

$$U(\theta )=\left(\begin{array}{cc}{e}^{-i\theta /2}& 0\\ 0& e^{i\theta /2}\end{array}\right)=R_Z(\theta ).$$

Step 3, verify orthonormality for every $\theta$. $U(\theta {)}^{†}U(\theta )$ has diagonal entries $e^{i\theta /2}{e}^{-i\theta /2}=1$ and $e^{-i\theta /2}{e}^{i\theta /2}=1$, with off-diagonal entries $0$, so $U(\theta {)}^{†}U(\theta )=I$ for every real $\theta$: a whole one-parameter family of valid gates, not a single matrix.

Step 4, connect to $P(\lambda )$ by factoring out a global phase. Factor $e^{-i\theta /2}$ out of both diagonal entries:

$$R_Z(\theta )=e^{-i\theta /2}\left(\begin{array}{cc}1& 0\\ 0& e^{i\theta }\end{array}\right)=e^{-i\theta /2}P(\theta ).$$

Step 5, check the factoring is correct entry by entry. $(1,1)$ entry: $e^{-i\theta /2}\cdot 1=e^{-i\theta /2}$, matching $U(\theta )$‘s $(1,1)$ entry directly. $(2,2)$ entry: $e^{-i\theta /2}\cdot e^{i\theta }=e^{i\theta /2}$, matching $U(\theta )$‘s $(2,2)$ entry. Both match exactly.

Conclusion. $R_Z(\theta )$ and $P(\theta )$ act identically on any state up to the global phase factor $e^{-i\theta /2}$, so they are physically indistinguishable in isolation (Question 2, Why Q4) even though their matrices differ — but that global-phase factor stops being ignorable the moment $R_Z$ or $P$ is used as the controlled version of itself inside a larger circuit, since a controlled gate cannot discard a phase that only appears on one branch of the control.