7

If I have the $X$ gate acting on a qubit and the $\lambda_6$ gate acting on a qutrit, where $\lambda_6$ is a Gell-Mann matrix, the system is subjected to the Hamiltonian:

$\lambda_6X= \begin{pmatrix}0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 \end{pmatrix} $

In case anyone doubts this matrix, it can be generated with the following script (MATLAB/octave):

```
lambda6=[0 0 0; 0 0 1; 0 1 0];
X= [0 1; 1 0 ];
kron(lambda6,X)
```

However consider the alternative Hamiltonian:

$-\frac{1}{2}Z\lambda_1 + \frac{1}{2}\lambda_1 - \frac{1}{\sqrt{3}}X\lambda_8+\frac{1}{3}X$.

*This is the exact same Hamiltonian!*

The following script proves it:

```
lambda1=[0 1 0;1 0 0;0 0 0];
lambda8=[1 0 0;0 1 0;0 0 -2]/sqrt(3);
Z= [1 0; 0 -1 ];
round(-0.5*kron(Z,lambda1)+0.5*kron(eye(2),lambda1)-(1/sqrt(3))*kron(X,lambda8)+(1/3)*kron(X,eye(3)))
```

The "round" in the last line of code can be removed, but the format will be uglier because some of the 0's end up being around $10^{-16}$.

1) I thought the Pauli decomposition for two qubits is unique, why would the Pauli-GellMann decomposition of a qubit-qutrit be non-unique?

2) How would I obtain the decomposition $\lambda_6X$ from the above 6x6 matrix?

I think this is the correct answer: simply that the two decompositions are in different bases, which is what I alluded to in my comment to the other answer: in one case it acts on the qubit first then the qutrit, and in the other case it's the other way around (different bases). I might have become confused because until recently I was almost exclusively working with Hamiltonians that contained Z matrices (Ising models), and everything commutes there so this issue never came up. – user1271772 – 2018-05-29T19:54:43.573