5

All tensor products of $n$ Pauli operators $\{I,X,Y,Z\}$ (that is $4^n$ combinations) form an orthogonal basis for the vector space of $2^n \times 2^n$ complex matrices. Hence, for every matrix there is a unique decomposition as a linear combination of tensor products of Pauli unitaries. Same is true if we fix some other unitary basis. If we not fix the ...


3

As you say, start by expanding $e^{-iS\Delta t}=\cos(\Delta t)I-i\sin(\Delta t)S$, so you'd be calculating $$ (\cos(\Delta t)I-i\sin(\Delta t)S)\rho\otimes\sigma(\cos(\Delta t)I+i\sin(\Delta t)S). $$ If you multiply out all the terms, then the $\cos^2(\Delta t)$ comes from the two $I$ terms, leaving you with $\rho\otimes\sigma$. If you trace out the first ...


2

Replace in the diagram in exercise 4.12 the triangles with the corresponding bending wires. It seems that the resulting diagram is like the second identity in 4.9, but with two added straight horizontal wires. (The diagram is rotated, but that doesn't matter.) Apply this identity. The resulting diagram is a cap with added straight wires. It is permitted to ...


2

In this paper, the authors used Knot theory to define what they call 'Path Model Representation'. In a later section they convert this representation to qubits by switching to binary.


1

Following your attempt we have $$ \begin{align*} | \phi \rangle &= \sum_i \phi_i | a_i^{\phi} \rangle | b_i^{\phi} \rangle\\ | \gamma \rangle &= \sum_i \gamma_i | a_i^{\gamma} \rangle | b_i^{\gamma} \rangle, \end{align*} $$ $$ \rho \equiv \operatorname{tr}_B(| \psi \rangle \langle \psi |) = |\alpha|^2 \rho_{\phi \phi} + |\beta|^2 \rho_{\gamma \...


1

Here is the solution. The trick is to use "the only connectivity matters" rule. The swap rule of 4.9 helps us reorder the inputs, which then makes it topologically equivalent to the next diagram (match the first and second wires of the states).


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