# How do I decompose the given $4\times 4$ matrix in terms of Pauli matrices? [duplicate]

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!

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.

• 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? – Sire Feb 27 at 22:08
• Also the truth table was supposed to be using Dirac notation – Sire Feb 27 at 22:10
• @Sire Yes, $\mathrm{Tr}$ represents the trace and yes, in general you would need to work out each coefficient separately. – Rammus Feb 27 at 22:50

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).