Von Neumann & Entanglement Entropy

1. 1
   Step 1:

   $$\rho=\sum_{i=1}^{d}\lambda_i\lvert e_i\rangle\langle e_i\rvert,\qquad \langle e_i\vert e_j\rangle=\delta_{ij}$$

   Spectral theorem for the Hermitian operator $\rho$; positivity gives $\lambda_i\ge0$ and $\mathrm{Tr}\rho=1$ gives $\sum_i\lambda_i=1$.
2. 2
   Step 2:

   $$\rho\log_2\rho=\eta(\rho)=\sum_{i=1}^{d}\eta(\lambda_i)\lvert e_i\rangle\langle e_i\rvert=\sum_{i=1}^{d}\lambda_i\log_2\lambda_i\,\lvert e_i\rangle\langle e_i\rvert$$

   Functional calculus with $\eta(x)=x\log_2x$. Applying $\eta$ in one go avoids forming $\log_2\rho$ on its own, which is undefined when $\rho$ has a zero eigenvalue.
3. 3
   Step 3:

   $$\mathrm{Tr}\big(\rho\log_2\rho\big)=\sum_{i=1}^{d}\lambda_i\log_2\lambda_i\,\mathrm{Tr}\lvert e_i\rangle\langle e_i\rvert=\sum_{i=1}^{d}\lambda_i\log_2\lambda_i\,\langle e_i\vert e_i\rangle$$

   Linearity of the trace and $\mathrm{Tr}\lvert a\rangle\langle b\rvert=\langle b\vert a\rangle$.
4. 4
   Step 4:

   $$S(\rho)=-\sum_{i=1}^{d}\lambda_i\log_2\lambda_i$$

   Each $\langle e_i\vert e_i\rangle=1$; multiply by $-1$.

1. 1
   Step 1:

   $$x\log_2x=\frac{x\ln x}{\ln2}$$

   Change of base, $\log_2x=\ln x/\ln2$; the constant factor does not affect the limit.
2. 2
   Step 2:

   $$\lim_{x\to0^+}x\ln x=\lim_{x\to0^+}\frac{\ln x}{1/x}$$

   Rewrite the indeterminate form $0\cdot(-\infty)$ as the quotient $\tfrac{-\infty}{+\infty}$, to which L'Hôpital's rule applies.
3. 3
   Step 3:

   $$=\lim_{x\to0^+}\frac{(\ln x)'}{(1/x)'}=\lim_{x\to0^+}\frac{1/x}{-1/x^2}$$

   L'Hôpital: numerator and denominator are differentiable on $(0,1)$, both diverge, and the derivative of the denominator, $-1/x^2$, never vanishes there.
4. 4
   Step 4:

   $$=\lim_{x\to0^+}(-x)=0$$

   Simplify $\frac{1/x}{-1/x^2}=-x$.
5. 5
   Step 5:

   $$\eta(x)=x\log_2x\ \text{on }(0,1],\quad \eta(0):=0\ \Rightarrow\ \eta\in C\big([0,1]\big)$$

   Defining $\eta(0)$ as the limit makes $\eta$ continuous on the closed interval, so $S$ is a continuous function of the eigenvalues and zero-probability eigenvectors contribute nothing.

1. 1
   Step 1:

   $$\rho_A=\sum_i\mu_i\lvert a_i\rangle\langle a_i\rvert,\quad \rho_B=\sum_j\nu_j\lvert b_j\rangle\langle b_j\rvert\ \Rightarrow\ \rho_A\otimes\rho_B=\sum_{i,j}\mu_i\nu_j\lvert a_ib_j\rangle\langle a_ib_j\rvert$$

   Kronecker product of the two spectral decompositions; $\{\lvert a_ib_j\rangle\}$ is an orthonormal basis of $\mathcal H_A\otimes\mathcal H_B$, so the eigenvalues are the products $\mu_i\nu_j$.
2. 2
   Step 2:

   $$S(\rho_A\otimes\rho_B)=-\sum_{i,j}\mu_i\nu_j\log_2(\mu_i\nu_j)=-\sum_{i,j}\mu_i\nu_j\big(\log_2\mu_i+\log_2\nu_j\big)$$

   Eigenvalue formula and $\log_2(ab)=\log_2a+\log_2b$ (terms with $\mu_i\nu_j=0$ vanish on both sides).
3. 3
   Step 3:

   $$=-\Big(\sum_j\nu_j\Big)\sum_i\mu_i\log_2\mu_i-\Big(\sum_i\mu_i\Big)\sum_j\nu_j\log_2\nu_j=S(\rho_A)+S(\rho_B)$$

   Factor the double sums and use $\sum_i\mu_i=\sum_j\nu_j=1$.

1. 1
   Step 1:

   $$R_Y(\theta)\lvert0\rangle=\cos\tfrac\theta2\lvert0\rangle+\sin\tfrac\theta2\lvert1\rangle$$

   First column of $R_Y(\theta)=\begin{pmatrix}\cos\frac\theta2&-\sin\frac\theta2\\\sin\frac\theta2&\cos\frac\theta2\end{pmatrix}$.
2. 2
   Step 2:

   $$\mathrm{CNOT}_{0\to1}\Big(\cos\tfrac\theta2\lvert00\rangle+\sin\tfrac\theta2\lvert10\rangle\Big)=\cos\tfrac\theta2\lvert00\rangle+\sin\tfrac\theta2\lvert11\rangle=:\lvert\psi_\theta\rangle$$

   Qubit 1 starts in $\lvert0\rangle$; CNOT flips it exactly when qubit 0 is $\lvert1\rangle$.
3. 3
   Step 3:

   $$\lvert\psi_\theta\rangle=\lambda_0\lvert0\rangle\lvert0\rangle+\lambda_1\lvert1\rangle\lvert1\rangle,\qquad \lambda_0=\cos\tfrac\theta2,\ \lambda_1=\sin\tfrac\theta2\quad(0<\theta<\pi)$$

   This is already a Schmidt decomposition: $\{\lvert0\rangle,\lvert1\rangle\}$ is orthonormal on both sides and the coefficients are positive for $0<\theta<\pi$ (otherwise absorb signs into the basis vectors).
4. 4
   Step 4:

   $$\rho_0=\mathrm{Tr}_1\lvert\psi_\theta\rangle\langle\psi_\theta\rvert=\begin{pmatrix}\cos^2\frac\theta2&0\\0&\sin^2\frac\theta2\end{pmatrix}$$

   By the theorem $\rho_A=\sum_k\lambda_k^2\lvert a_k\rangle\langle a_k\rvert$; the cross terms $\lvert0\rangle\langle1\rvert$ carry the factor $\langle1\vert0\rangle=0$ from qubit 1.
5. 5
   Step 5:

   $$E(\psi_\theta)=-\cos^2\tfrac\theta2\log_2\cos^2\tfrac\theta2-\sin^2\tfrac\theta2\log_2\sin^2\tfrac\theta2=h\big(\cos^2\tfrac\theta2\big)$$

   Eigenvalue formula with $\sin^2\tfrac\theta2=1-\cos^2\tfrac\theta2$, which is the definition of the binary entropy.
6. 6
   Step 6:

   $$\theta=\tfrac\pi3:\quad \cos^2\tfrac\pi6=\Big(\tfrac{\sqrt3}{2}\Big)^2=\tfrac34,\qquad \sin^2\tfrac\pi6=\tfrac14$$

   Values at $\theta/2=\pi/6$.
7. 7
   Step 7:

   $$h\big(\tfrac34\big)=-\tfrac34\log_2\tfrac34-\tfrac14\log_2\tfrac14=-\tfrac34\big(\log_23-2\big)-\tfrac14(-2)$$

   $\log_2\tfrac34=\log_23-\log_24=\log_23-2$ and $\log_2\tfrac14=-2$.
8. 8
   Step 8:

   $$=-\tfrac34\log_23+\tfrac32+\tfrac12=2-\tfrac34\log_23$$

   Collect the rational terms.
9. 9
   Step 9:

   $$=2-\tfrac34\times1.584962500721156=2-1.188721875540867=0.811278124459133$$

   $\log_23\approx1.584962500721156$.
10. 10
    Step 10:

    $$\theta=\tfrac\pi2:\ h\big(\tfrac12\big)=-2\cdot\tfrac12\log_2\tfrac12=1;\qquad \theta\in\{0,\pi\}:\ h(1)=h(0)=0$$

    $\theta=\pi/2$ gives the Bell state $\lvert\Phi^+\rangle$ with one full ebit; $\theta=0$ and $\theta=\pi$ give the product states $\lvert00\rangle$ and $\lvert11\rangle$.

Quick check

Quick check

1. 1
   Step 1:

   $$H_0:\ \lvert000\rangle\mapsto\tfrac1{\sqrt2}\big(\lvert000\rangle+\lvert100\rangle\big)\ \xrightarrow{\mathrm{CNOT}_{0\to1}}\ \tfrac1{\sqrt2}\big(\lvert000\rangle+\lvert110\rangle\big)\ \xrightarrow{\mathrm{CNOT}_{1\to2}}\ \tfrac1{\sqrt2}\big(\lvert000\rangle+\lvert111\rangle\big)$$

   Apply the three gates in order; each CNOT copies the computational value of its control into its target.
2. 2
   Step 2:

   $$\rho=\lvert\mathrm{GHZ}\rangle\langle\mathrm{GHZ}\rvert=\tfrac12\big(\lvert000\rangle\langle000\rvert+\lvert000\rangle\langle111\rvert+\lvert111\rangle\langle000\rvert+\lvert111\rangle\langle111\rvert\big)$$

   Expand the outer product; this is an $8\times8$ matrix with four nonzero entries equal to $\tfrac12$ at positions $(0,0),(0,7),(7,0),(7,7)$.
3. 3
   Step 3:

   $$\rho_{\{0\}}=\mathrm{Tr}_{12}\rho=\tfrac12\big(\lvert0\rangle\langle0\rvert\langle00\vert00\rangle+\lvert0\rangle\langle1\rvert\langle11\vert00\rangle+\lvert1\rangle\langle0\rvert\langle00\vert11\rangle+\lvert1\rangle\langle1\rvert\langle11\vert11\rangle\big)$$

   $\mathrm{Tr}_{12}(\lvert a\,bc\rangle\langle a'\,b'c'\rvert)=\lvert a\rangle\langle a'\rvert\,\langle b'c'\vert bc\rangle$, applied to each of the four terms.
4. 4
   Step 4:

   $$=\tfrac12\big(\lvert0\rangle\langle0\rvert+\lvert1\rangle\langle1\rvert\big)=\tfrac I2,\qquad S(\rho_{\{0\}})=h\big(\tfrac12\big)=1$$

   The coherences need $\langle11\vert00\rangle=0$ and vanish; the remaining state is maximally mixed.
5. 5
   Step 5:

   $$\rho_{\{0,1\}}=\mathrm{Tr}_{2}\rho=\tfrac12\big(\lvert00\rangle\langle00\rvert\langle0\vert0\rangle+\lvert00\rangle\langle11\rvert\langle1\vert0\rangle+\lvert11\rangle\langle00\rvert\langle0\vert1\rangle+\lvert11\rangle\langle11\rvert\langle1\vert1\rangle\big)$$

   Trace out only qubit 2: $\mathrm{Tr}_2(\lvert ab\,c\rangle\langle a'b'\,c'\rvert)=\lvert ab\rangle\langle a'b'\rvert\,\langle c'\vert c\rangle$.
6. 6
   Step 6:

   $$=\tfrac12\big(\lvert00\rangle\langle00\rvert+\lvert11\rangle\langle11\rvert\big)=\mathrm{diag}\big(\tfrac12,0,0,\tfrac12\big)$$

   Again the coherences vanish. In the ordered basis $\lvert00\rangle,\lvert01\rangle,\lvert10\rangle,\lvert11\rangle$ the matrix is diagonal.
7. 7
   Step 7:

   $$\mathrm{spec}\,\rho_{\{0,1\}}=\big\{\tfrac12,\tfrac12,0,0\big\}\ \Rightarrow\ S(\rho_{\{0,1\}})=-2\cdot\tfrac12\log_2\tfrac12-0-0=1$$

   A diagonal matrix displays its eigenvalues; zeros contribute nothing. Note $1<\log_24=2$: the pair is far from maximally mixed.
8. 8
   Step 8:

   $$\rho_{\{2\}}=\mathrm{Tr}_{01}\rho=\tfrac12\big(\lvert0\rangle\langle0\rvert+\lvert1\rangle\langle1\rvert\big)=\tfrac I2,\qquad S(\rho_{\{2\}})=1$$

   Same computation as for qubit 0, now with $\langle ab\vert a'b'\rangle$ supplied by qubits $0,1$.
9. 9
   Step 9:

   $$S(\rho_{\{0,1\}})=S(\rho_{\{2\}})=1$$

   Consistency check with the equal-marginal theorem: $\{0,1\}$ and $\{2\}$ are complementary parts of a pure state. The Schmidt form across this cut is $\tfrac1{\sqrt2}\lvert00\rangle\lvert0\rangle+\tfrac1{\sqrt2}\lvert11\rangle\lvert1\rangle$.

1. 1
   Step 1:

   $$\lvert\Phi^+\rangle\langle\Phi^+\rvert:\quad S(\rho_{AB})=0,\qquad \rho_A=\mathrm{Tr}_B\lvert\Phi^+\rangle\langle\Phi^+\rvert=\tfrac I2,\qquad S(\rho_A)=1$$

   The global state is pure; its marginal is maximally mixed (partial-trace lesson).
2. 2
   Step 2:

   $$\tfrac I4=\tfrac I2\otimes\tfrac I2:\quad S(\rho_{AB})=2,\qquad \rho_A=\mathrm{Tr}_B\tfrac I4=\tfrac I2\,\mathrm{Tr}\tfrac I2=\tfrac I2,\qquad S(\rho_A)=1$$

   Additivity gives $S(\tfrac I4)=1+1$; the partial trace of a product is the first factor times the trace of the second.
3. 3
   Step 3:

   $$S(\rho_{AB})-S(\rho_A)=\begin{cases}-1&\lvert\Phi^+\rangle\\+1&I/4\end{cases}$$

   For the Bell state the whole is less uncertain than the part, which is impossible for classical probability distributions (where $H(X,Y)\ge H(X)$). A negative $S(AB)-S(A)$ certifies entanglement; the maximally mixed state shows the classical ordering.