Formula Sheet 1 — Foundations: States, Gates, Measurement, Density Operators (L01–L06)
States & Bloch Sphere (L01–L02) ¶ Qubit: ∣ ψ ⟩ = α ∣ 0 ⟩ + β ∣ 1 ⟩ |\psi\rangle=\alpha|0\rangle+\beta|1\rangle ∣ ψ ⟩ = α ∣0 ⟩ + β ∣1 ⟩ ∣ α ∣ 2 + ∣ β ∣ 2 = 1 |\alpha|^2+|\beta|^2=1 ∣ α ∣ 2 + ∣ β ∣ 2 = 1 ∣ ψ ⟩ = cos θ 2 ∣ 0 ⟩ + e i ϕ sin θ 2 ∣ 1 ⟩ |\psi\rangle=\cos\frac\theta2|0\rangle+e^{i\phi}\sin\frac\theta2|1\rangle ∣ ψ ⟩ = cos 2 θ ∣0 ⟩ + e i ϕ sin 2 θ ∣1 ⟩ r ⃗ = ( sin θ cos ϕ , sin θ sin ϕ , cos θ ) \vec r=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta) r = ( sin θ cos ϕ , sin θ sin ϕ , cos θ ) ρ = 1 2 ( I + r ⃗ ⋅ σ ⃗ ) \rho=\tfrac12(I+\vec r\cdot\vec\sigma) ρ = 2 1 ( I + r ⋅ σ ) ⇔ ∣ r ⃗ ∣ = 1 \Leftrightarrow |\vec r|=1 ⇔ ∣ r ∣ = 1 n n n 2 n 2^n 2 n U ∣ ψ ⟩ ∣ 0 ⟩ = ∣ ψ ⟩ ∣ ψ ⟩ U|\psi\rangle|0\rangle=|\psi\rangle|\psi\rangle U ∣ ψ ⟩ ∣0 ⟩ = ∣ ψ ⟩ ∣ ψ ⟩ ψ \psi ψ
Common Gates (L04) ¶ X = ( 0 1 1 0 ) X=\begin{pmatrix}0&1\\1&0\end{pmatrix} X = ( 0 1 1 0 ) Y = ( 0 − i i 0 ) Y=\begin{pmatrix}0&-i\\i&0\end{pmatrix} Y = ( 0 i − i 0 ) Z = ( 1 0 0 − 1 ) Z=\begin{pmatrix}1&0\\0&-1\end{pmatrix} Z = ( 1 0 0 − 1 ) H = 1 2 ( 1 1 1 − 1 ) H=\tfrac1{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix} H = 2 1 ( 1 1 1 − 1 ) S = diag ( 1 , i ) S=\text{diag}(1,i) S = diag ( 1 , i ) T = diag ( 1 , e i π / 4 ) T=\text{diag}(1,e^{i\pi/4}) T = diag ( 1 , e iπ /4 ) R n ( θ ) = e − i θ n ^ ⋅ σ ⃗ / 2 = cos θ 2 I − i sin θ 2 n ^ ⋅ σ ⃗ R_n(\theta)=e^{-i\theta\, \hat n\cdot\vec\sigma/2}=\cos\frac\theta2 I - i\sin\frac\theta2\,\hat n\cdot\vec\sigma R n ( θ ) = e − i θ n ^ ⋅ σ /2 = cos 2 θ I − i sin 2 θ n ^ ⋅ σ ∣ a , b ⟩ = ∣ a , a ⊕ b ⟩ |a,b\rangle=|a, a\oplus b\rangle ∣ a , b ⟩ = ∣ a , a ⊕ b ⟩ = diag ( 1 , 1 , 1 , − 1 ) =\text{diag}(1,1,1,-1) = diag ( 1 , 1 , 1 , − 1 ) H X H = Z HXH=Z H X H = Z H Z H = X HZH=X H Z H = X X Z = − Z X XZ=-ZX XZ = − ZX H 2 = I H^2=I H 2 = I { H , T , CNOT } \{H,T,\text{CNOT}\} { H , T , CNOT } U = e i α R z ( β ) R y ( γ ) R z ( δ ) U=e^{i\alpha}R_z(\beta)R_y(\gamma)R_z(\delta) U = e i α R z ( β ) R y ( γ ) R z ( δ )
Measurement (L01, L03) ¶ Projective: p ( m ) = ⟨ ψ ∣ P m ∣ ψ ⟩ p(m)=\langle \psi| P_m|\psi\rangle p ( m ) = ⟨ ψ ∣ P m ∣ ψ ⟩ P m ∣ ψ ⟩ / p ( m ) P_m|\psi\rangle/\sqrt{p(m)} P m ∣ ψ ⟩ / p ( m ) E m ≥ 0 E_m\ge0 E m ≥ 0 ∑ m E m = I \sum_m E_m=I ∑ m E m = I p ( m ) = Tr ( E m ρ ) p(m)=\operatorname{Tr}(E_m\rho) p ( m ) = Tr ( E m ρ ) E m = K m † K m E_m=K_m^\dagger K_m E m = K m † K m P P P X : H X{:}\,H X : H Y : S † H Y{:}\,S^\dagger H Y : S † H Z Z Z ⟨ P ⟩ = p ( + 1 ) − p ( − 1 ) \langle P\rangle = p(+1)-p(-1) ⟨ P ⟩ = p ( + 1 ) − p ( − 1 )
Entanglement (L03) ¶ Bell states: ∣ Φ ± ⟩ = ∣ 00 ⟩ ± ∣ 11 ⟩ 2 |\Phi^\pm\rangle=\tfrac{|00\rangle\pm|11\rangle}{\sqrt2} ∣ Φ ± ⟩ = 2 ∣00 ⟩ ± ∣11 ⟩ ∣ Ψ ± ⟩ = ∣ 01 ⟩ ± ∣ 10 ⟩ 2 |\Psi^\pm\rangle=\tfrac{|01\rangle\pm|10\rangle}{\sqrt2} ∣ Ψ ± ⟩ = 2 ∣01 ⟩ ± ∣10 ⟩ ∣ Φ + ⟩ |\Phi^+\rangle ∣ Φ + ⟩ H H H → \to → ∣ ψ ⟩ A B = ∑ i λ i ∣ i A ⟩ ∣ i B ⟩ |\psi\rangle_{AB}=\sum_i\lambda_i|i_A\rangle|i_B\rangle ∣ ψ ⟩ A B = ∑ i λ i ∣ i A ⟩ ∣ i B ⟩ λ i ≥ 0 \lambda_i\ge0 λ i ≥ 0 ⇔ \Leftrightarrow ⇔ > 1 >1 > 1 ⇔ \Leftrightarrow ⇔ ρ A \rho_A ρ A ∣ S ∣ ≤ 2 |S|\le2 ∣ S ∣ ≤ 2 2 2 2\sqrt2 2 2 → \to → → \to →
Density Operators (L06) ¶ ρ = ∑ i p i ∣ ψ i ⟩ ⟨ ψ i ∣ \rho=\sum_i p_i|\psi_i\rangle\langle \psi_i| ρ = ∑ i p i ∣ ψ i ⟩ ⟨ ψ i ∣ ρ ≥ 0 \rho\ge0 ρ ≥ 0 Tr ρ = 1 \operatorname{Tr}\rho=1 Tr ρ = 1 Tr ρ 2 ≤ 1 \operatorname{Tr}\rho^2\le1 Tr ρ 2 ≤ 1 ρ A = Tr B ρ A B \rho_A=\operatorname{Tr}_{B}\rho_{AB} ρ A = Tr B ρ A B ρ ↦ U ρ U † \rho\mapsto U\rho U^\dagger ρ ↦ U ρ U † E ( ρ ) = ∑ i K i ρ K i † \mathcal E(\rho)=\sum_i K_i\rho K_i^\dagger E ( ρ ) = ∑ i K i ρ K i † ∑ i K i † K i = I \sum_iK_i^\dagger K_i=I ∑ i K i † K i = I F ( ρ , σ ) = ( Tr ρ σ ρ ) 2 F(\rho,\sigma)=\left(\operatorname{Tr}\sqrt{\sqrt\rho\,\sigma\sqrt\rho}\right)^2 F ( ρ , σ ) = ( Tr ρ σ ρ ) 2 ρ \rho ρ F = ⟨ ψ ∣ ρ ∣ ψ ⟩ F=\langle \psi|\rho|\psi\rangle F = ⟨ ψ ∣ ρ ∣ ψ ⟩
Standard Noise Channels (L06–L07) ¶ Bit flip: K 0 = 1 − p I K_0=\sqrt{1-p}\,I K 0 = 1 − p I K 1 = p X K_1=\sqrt p\,X K 1 = p X Z Z Z E ( ρ ) = ( 1 − p ) ρ + p 3 ( X ρ X + Y ρ Y + Z ρ Z ) \mathcal E(\rho)=(1-p)\rho+\tfrac p3(X\rho X+Y\rho Y+Z\rho Z) E ( ρ ) = ( 1 − p ) ρ + 3 p ( XρX + Y ρ Y + ZρZ ) K 0 = ( 1 0 0 1 − γ ) K_0=\begin{pmatrix}1&0\\0&\sqrt{1-\gamma}\end{pmatrix} K 0 = ( 1 0 0 1 − γ ) K 1 = ( 0 γ 0 0 ) K_1=\begin{pmatrix}0&\sqrt\gamma\\0&0\end{pmatrix} K 1 = ( 0 0 γ 0 ) γ = 1 − e − t / T 1 \gamma=1-e^{-t/T_1} γ = 1 − e − t / T 1 λ = 1 − e − t / T ϕ \lambda = 1-e^{-t/T_\phi} λ = 1 − e − t / T ϕ 1 T 2 = 1 2 T 1 + 1 T ϕ \frac{1}{T_2}=\frac{1}{2T_1}+\frac{1}{T_\phi} T 2 1 = 2 T 1 1 + T ϕ 1 T 2 ≤ 2 T 1 T_2\le 2T_1 T 2 ≤ 2 T 1
Handy Linear Algebra ¶ Spectral: Hermitian A = ∑ a a P a A=\sum_a a\,P_a A = ∑ a a P a f ( A ) = ∑ a f ( a ) P a f(A)=\sum_a f(a)P_a f ( A ) = ∑ a f ( a ) P a e i θ A e^{i\theta A} e i θ A A 2 = I A^2=I A 2 = I cos θ I + i sin θ A \cos\theta\, I + i\sin\theta\, A cos θ I + i sin θ A ( A ⊗ B ) ( C ⊗ D ) = A C ⊗ B D (A\otimes B)(C\otimes D)=AC\otimes BD ( A ⊗ B ) ( C ⊗ D ) = A C ⊗ B D Tr ( A ⊗ B ) = Tr A Tr B \operatorname{Tr}(A\otimes B)=\operatorname{Tr} A\,\operatorname{Tr} B Tr ( A ⊗ B ) = Tr A Tr B [ A , B ] = A B − B A [A,B]=AB-BA [ A , B ] = A B − B A [ σ a , σ b ] = 2 i ε a b c σ c [\sigma_a,\sigma_b]=2i\varepsilon_{abc}\sigma_c [ σ a , σ b ] = 2 i ε ab c σ c
Usage note. Everything on this sheet is fair game on quizzes without derivation; exams allow this sheet. Cross-references (L##) point to the lecture where each result is derived. Textbook: Nielsen & Chuang, Chs. 1–2, 8.