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This is an important result in Quantum Computing because it means that the Hamiltonian of a Quantum System can be encoded as a sequence of real numbers and their corresponding Pauli Operator.

How do we prove this result?

(Note I am assuming that the Hermitian is of size $2^n \times 2^n$, and so by Pauli Operator I mean n-fold tensor products of the 2x2 Pauli Matrices)

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Consider an arbitrary $2\times 2$ matrix: $$ M=\left(\begin{array}{cc} a & b \\ c & d \end{array}\right) $$ What does it mean to impose that $M$ is Hermitian, $M=M^\dagger$? $$ a=a^\star, b=c^\star,d=d^\star $$ By that token, $a$ and $d$ are real and, if we write $c=x+iy$ for real $x,y$, then $b=x+iy$. Thus, we can write the matrix as $$ M=\left(\begin{array}{cc} a & x-iy \\ x+iy & d \end{array}\right). $$ You can readily verify that this is the same as $$ \frac{a+d}{2}I+xX+yY+\frac{a-d}{2}Z. $$ All four coefficients are real.

To think about the general case of $n$ qubits, realise that we can use a basis of $x\in\{0,1\}^n$. There are diagonal terms like $$ a|x\rangle\langle x| $$ which you can show how to construct from Paulis: $$ \bigotimes_{i=1}^n\frac{I+(-1)^{x_i}Z}{2} $$ Then there are terms $(a+ib)|x\rangle\langle y|+(a-ib)|y\rangle\langle x|)$. For these, use the construction $$ |0\rangle\langle 1|=\frac12(X+iY) $$ for qubits $i$ where $x_i\neq y_i$, and the same construction as for the diagonal case whenever $x_i=y_i$. It should quickly be clear that we can use just tensor products of Paulis. It might take a little more care to be clear that the imaginary components all vanish.

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  • $\begingroup$ Ah yes, I did mean the more general case! Will edit the question to make that clear. Sorry about that and thanks for the proof for the 2x2 case! $\endgroup$ May 12 '21 at 13:41

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