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I have been working on a question where I have to decompose this matrix in terms of Pauli Matrices: \begin{bmatrix}1&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&1\end{bmatrix}

I already have a solution but I don't understand the solution I've been given, this is the solution:

First there is a truth table:

+------+-----------+
|Input | Output    |
+------+-----------+
| |00> | |00>+|11> | = 1/2(|00>+|11>+|11>+|11>)
| |01> | 0         | = 1/2(|01>-|01>+|10>-|10>)
| |10> | 0         | = 1/2(|10>-|10>+|01>-|01>)
| |11> | |11>+|00> | = 1/2(|00>+|00>+|11>+|11>)
+------+-----------+

I understand the truth table, but I don't understand the things after the "=" and I also don't understand how the final answer is achieved. This is the final answer:

$$ \frac{1}2(I_1 \otimes I_2) + \frac{1}2(Z_1 \otimes Z_2) + \frac{1}2(X_1 \otimes X_2) - \frac{1}2(Y_1 \otimes Y_2) $$

Any help in understanding the solution would be really appreciated. Thank you!

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I'm really not sure what the truth table is trying to represent but let me present a solution that (at least I think) is fairly simple.

First we note that the Pauli matrices together with the identity form an orthogonal basis on the vector space of $2\times2$ matrices $M_2(\mathbb{C})$. Where orthogonal is taken with respect to the inner product, $$ \langle M, N\rangle = \mathrm{Tr}[M^* N]. $$ Moreover, taking tensor products of two elements of our basis we get a basis for $M_4(\mathbb{C})$. Note the fact that these operators form a basis is exactly why we can decompose a matrix into a sum of tensor products of the Pauli operators together with the identity. Now it is slightly easier if we normalize our basis elements. For example $\|X\otimes Y\| = |\langle X\otimes Y,X\otimes Y\rangle|^{1/2} = \mathrm{Tr}[X^2 \otimes Y^2]^{1/2} = \mathrm{Tr}[I \otimes I]^{1/2} = 2$. Note all of the elements in our basis have norm $2$ as they all square to the identity. Thus multiplying each operator by $\frac12$ we get an orthonormal basis.

Now we have an orthonormal basis, it is straightforward to find the coefficients in the expansion, like how we would do this in $\mathbb{C}^n$ we need only take the inner product of our matrix with each of the orthonormal basis elements. Formally, this is because we know $$ M = \frac{c_{00}}{2} I \otimes I + \frac{c_{01}}{2} I \otimes X + \frac{c_{02}}{2} I \otimes Y + \dots + \frac{c_{33}}{2} Z \otimes Z, $$ then for example $$ c_{13} = \mathrm{Tr}[\frac{1}{2}(X \otimes Z) M] = \langle \frac12 (X\otimes Z), M\rangle. $$

Performing these computations for the matrix you specify, you should get out the correct coefficients.

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  • $\begingroup$ Thank you. I was just wondering does the Tr represent trace? And also would I have to work out c_31 as well as c_13, and c_30 as well as c_03 (and so on..) because tensor product is not commutative? $\endgroup$ – Sire Feb 27 at 22:08
  • $\begingroup$ Also the truth table was supposed to be using Dirac notation $\endgroup$ – Sire Feb 27 at 22:10
  • $\begingroup$ @Sire Yes, $\mathrm{Tr}$ represents the trace and yes, in general you would need to work out each coefficient separately. $\endgroup$ – Rammus Feb 27 at 22:50
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I would like to add that this matrix is not unitary since an operator described by it retrurns same results for input combination $|00\rangle$ and $|11\rangle$, i.e. $\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$, and for inputs $|01\rangle$ and $|10\rangle$ state $|00\rangle$ is returned. Hence such operator cannot be implemented on a quantum computer.

However, it can be decomposed in Pauli terms as Pauli matrices and their tensor products are basis of a matrix space (this is provided in answer above).

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