The Qubit

1. 1
   Step 1:

   $$\lvert\psi\rangle\in\mathbb{C}^2,\qquad \langle\psi\vert\psi\rangle=1$$

   Postulate 1: the state of an isolated system is a unit vector of its Hilbert space, here $\mathbb{C}^2$.
2. 2
   Step 2:

   $$\lvert\psi\rangle=\alpha\lvert0\rangle+\beta\lvert1\rangle,\qquad \langle\psi\rvert=\alpha^*\langle0\rvert+\beta^*\langle1\rvert$$

   Expand in the computational basis; the bra is the conjugate transpose, so the coefficients are complex-conjugated.
3. 3
   Step 3:

   $$\langle\psi\vert\psi\rangle=\alpha^*\alpha\langle0\vert0\rangle+\alpha^*\beta\langle0\vert1\rangle+\beta^*\alpha\langle1\vert0\rangle+\beta^*\beta\langle1\vert1\rangle$$

   The inner product is linear in the ket and antilinear in the bra, giving four terms.
4. 4
   Step 4:

   $$=\lvert\alpha\rvert^2\cdot1+\alpha^*\beta\cdot0+\beta^*\alpha\cdot0+\lvert\beta\rvert^2\cdot1=\lvert\alpha\rvert^2+\lvert\beta\rvert^2$$

   Orthonormality $\langle i\vert j\rangle=\delta_{ij}$ kills the cross terms, and $z^*z=\lvert z\rvert^2$.
5. 5
   Step 5:

   $$P(0)=\langle\psi\vert P_0\vert\psi\rangle=\lvert\langle0\vert\psi\rangle\rvert^2=\lvert\alpha\rvert^2,\qquad P(1)=\lvert\langle1\vert\psi\rangle\rvert^2=\lvert\beta\rvert^2$$

   Postulate 3 (Born rule) for the $Z$ measurement with projectors $P_0=\lvert0\rangle\langle0\rvert$, $P_1=\lvert1\rangle\langle1\rvert$.
6. 6
   Step 6:

   $$P(0)+P(1)=\lvert\alpha\rvert^2+\lvert\beta\rvert^2=\langle\psi\vert\psi\rangle=1$$

   The two outcomes are exhaustive and exclusive, so their probabilities must sum to one; this is exactly the unit-norm condition of Postulate 1, so the two postulates are consistent.

1. 1
   Step 1:

   $$\lvert\tilde\psi\rangle=(1+i)\lvert0\rangle+2\lvert1\rangle$$

   An unnormalised vector: $\tilde\alpha=1+i$, $\tilde\beta=2$.
2. 2
   Step 2:

   $$\lvert\tilde\alpha\rvert^2=(1+i)^*(1+i)=(1-i)(1+i)=1-i^2=2$$

   Modulus squared of a complex number is the number times its conjugate; $i^2=-1$.
3. 3
   Step 3:

   $$\lvert\tilde\beta\rvert^2=2^2=4,\qquad \lVert\tilde\psi\rVert^2=2+4=6$$

   Add the squared moduli (the derivation above with general coefficients).
4. 4
   Step 4:

   $$\lvert\psi\rangle=\frac{\lvert\tilde\psi\rangle}{\sqrt6}=\frac{1+i}{\sqrt6}\lvert0\rangle+\frac{2}{\sqrt6}\lvert1\rangle$$

   Divide by the norm $\sqrt6$; the result has unit norm because $\lVert c\,v\rVert=\lvert c\rvert\,\lVert v\rVert$.
5. 5
   Step 5:

   $$\alpha=\frac{1}{\sqrt6}+\frac{i}{\sqrt6}\approx0.4082+0.4082\,i,\qquad \beta=\frac{2}{\sqrt6}\approx0.8165$$

   Real and imaginary parts, to four decimals.
6. 6
   Step 6:

   $$P(0)=\frac{2}{6}=\frac13,\qquad P(1)=\frac{4}{6}=\frac23,\qquad P(0)+P(1)=1$$

   Born rule on the normalised amplitudes; the check $P(0)+P(1)=1$ confirms the normalisation.

Quick check

Quick check

1. 1
   Step 1:

   $$N=\lvert0.6\rvert^2+\lvert0.8i\rvert^2=0.36+(-0.8i)(0.8i)=0.36+0.64=1$$

   The modulus squared of $0.8i$ is the number times its conjugate; $-i\cdot i=1$. The vector is already a unit vector.
2. 2
   Step 2:

   $$P(0)=\frac{\lvert\alpha\rvert^2}{N}=0.36,\qquad P(1)=\frac{\lvert\beta\rvert^2}{N}=0.64$$

   Born rule; dividing by $N$ is harmless here because $N=1$.
3. 3
   Step 3:

   $$\alpha=0.6\,e^{i\cdot0},\qquad\beta=0.8\,e^{i\pi/2}$$

   Polar forms: $0.6$ lies on the positive real axis and $0.8i$ on the positive imaginary axis.
4. 4
   Step 4:

   $$\cos\tfrac\theta2=\frac{\lvert\alpha\rvert}{\sqrt N}=0.6\ \Rightarrow\ \theta=2\arccos0.6\approx1.8546\ \ (\approx106.26^\circ)$$

   The theorem's $r_0=\cos\tfrac\theta2$ with $\theta\in[0,\pi]$; consistently $\sin\tfrac\theta2=\sqrt{1-0.36}=0.8=\lvert\beta\rvert$.
5. 5
   Step 5:

   $$\phi=\arg\beta-\arg\alpha=\tfrac\pi2-0=\tfrac\pi2$$

   The relative phase; the global phase is $\gamma=\arg\alpha=0$.
6. 6
   Step 6:

   $$\cos\tfrac\theta2\lvert0\rangle+e^{i\pi/2}\sin\tfrac\theta2\lvert1\rangle=0.6\lvert0\rangle+0.8i\lvert1\rangle$$

   Reassemble the canonical form: $e^{i\pi/2}=i$.

1. 1
   Step 1:

   $$N=\lvert2\rvert^2+\lvert1+2i\rvert^2=4+(1-2i)(1+2i)=4+1-4i^2=9$$

   Squared moduli as in the worked example of the first section; $-4i^2=4$.
2. 2
   Step 2:

   $$\lvert\psi\rangle=\frac{1}{\sqrt9}\big(2\lvert0\rangle+(1+2i)\lvert1\rangle\big)=\tfrac23\lvert0\rangle+\tfrac{1+2i}{3}\lvert1\rangle$$

   Divide by the norm $\sqrt N=3$: this is what the Normalise button does.
3. 3
   Step 3:

   $$P(0)=\tfrac49\approx0.4444,\qquad P(1)=\tfrac59\approx0.5556$$

   $\lvert\tfrac23\rvert^2=\tfrac49$ and $\lvert1+2i\rvert^2/9=\tfrac59$; the editor shows the same bars before and after normalising, because it always divides by $N$.
4. 4
   Step 4:

   $$\theta=2\arccos\tfrac23\approx1.6821,\qquad\phi=\arg(1+2i)-\arg2=\arctan2\approx1.1071$$

   $1+2i$ lies in the first quadrant, so its argument is $\arctan\tfrac21$; $\arg2=0$.

1. 1
   Step 1:

   $$K\sim\mathrm{Bin}(N,p),\qquad p=0.64,\quad N=1000$$

   $K$ counts the shots with outcome $1$: a sum of $N$ independent Bernoulli variables, each $1$ with probability $p$.
2. 2
   Step 2:

   $$\mathbb E[K]=Np=640,\qquad\mathrm{Var}(K)=Np(1-p)=1000\cdot0.64\cdot0.36=230.4$$

   A Bernoulli variable has mean $p$ and variance $p-p^2=p(1-p)$; means and (for independent terms) variances add.
3. 3
   Step 3:

   $$\hat f=\frac KN,\qquad\mathbb E[\hat f]=p,\qquad\sigma_{\hat f}=\sqrt{\frac{p(1-p)}{N}}=\sqrt{\frac{0.64\cdot0.36}{1000}}=\frac{0.48}{\sqrt{1000}}\approx0.0152$$

   Dividing a random variable by $N$ divides its standard deviation by $N$; $\sqrt{0.64\cdot0.36}=0.8\cdot0.6=0.48$.
4. 4
   Step 4:

   $$\hat f\in\big[p-1.96\,\sigma_{\hat f},\ p+1.96\,\sigma_{\hat f}\big]\approx[0.610,\ 0.670]\quad\text{in about }95\%\text{ of runs}$$

   Central limit theorem: with $Np(1-p)=230.4\gg1$ the binomial is close to a normal distribution, and $\pm1.96$ standard deviations hold $95\%$ of a normal law.
5. 5
   Step 5:

   $$\chi^2=\frac{(K_0-E_0)^2}{E_0}+\frac{(K-E_1)^2}{E_1}=\frac{(K-Np)^2}{Np(1-p)}=z^2,\qquad z=\frac{\hat f-p}{\sigma_{\hat f}}$$

   With $K_0=N-K$ and $E_0=N(1-p)$ the two deviations are equal up to sign, and $\tfrac1{Np}+\tfrac1{N(1-p)}=\tfrac{1}{Np(1-p)}$.
6. 6
   Step 6:

   $$\mathbb E[\chi^2]=1,\qquad P(\chi^2>3.84)\approx0.05$$

   Two bins whose counts must add up to $N$ leave $2-1=1$ degree of freedom; $z$ is approximately standard normal, so $\chi^2=z^2$ has mean $1$ and exceeds $1.96^2\approx3.84$ in about $5\%$ of runs.

1. 1
   Step 1:

   $$\langle+\rvert=\big(\lvert+\rangle\big)^\dagger=\tfrac{1}{\sqrt2}\begin{pmatrix}1&1\end{pmatrix}$$

   The bra is the conjugate transpose; the entries are real, so conjugation does nothing.
2. 2
   Step 2:

   $$\langle+\vert+i\rangle=\tfrac{1}{\sqrt2}\begin{pmatrix}1&1\end{pmatrix}\tfrac{1}{\sqrt2}\begin{pmatrix}1\\i\end{pmatrix}=\tfrac12(1\cdot1+1\cdot i)=\tfrac{1+i}{2}$$

   Row times column.
3. 3
   Step 3:

   $$\lvert\langle+\vert+i\rangle\rvert^2=\frac{(1-i)(1+i)}{4}=\frac{1-i^2}{4}=\frac{2}{4}=\frac12$$

   Modulus squared via the complex conjugate.
4. 4
   Step 4:

   $$\langle+i\rvert=\tfrac{1}{\sqrt2}\begin{pmatrix}1&-i\end{pmatrix},\qquad \langle+i\vert-i\rangle=\tfrac12\big(1\cdot1+(-i)(-i)\big)=\tfrac12(1+i^2)=0$$

   For comparison, within the $Y$ basis: conjugating $i\mapsto -i$ in the bra is essential; forgetting it would give $\tfrac12(1-i^2)=1$, the wrong answer.
5. 5
   Step 5:

   $$\langle0\vert\pm\rangle=\tfrac{1}{\sqrt2},\quad \langle0\vert\pm i\rangle=\tfrac{1}{\sqrt2},\quad \langle1\vert\pm\rangle=\pm\tfrac{1}{\sqrt2},\quad \langle1\vert\pm i\rangle=\pm\tfrac{i}{\sqrt2}$$

   Overlaps with the $Z$ basis read off the column vectors directly; every modulus squared is $\tfrac12$.
6. 6
   Step 6:

   $$\langle\pm\vert\pm' i\rangle=\tfrac12\big(1\pm(\pm' i)\big)\ \Rightarrow\ \lvert\langle\pm\vert\pm' i\rangle\rvert^2=\tfrac14(1+1)=\tfrac12$$

   All four $X$–$Y$ overlaps have the form $\tfrac12(1+\epsilon i)$ with $\epsilon=\pm1$, whose modulus squared is $\tfrac14\cdot2$.

Quick check