Why is the decomposition of a qubit-qutrit Hamiltonian in terms of Pauli and Gell-Mann matrices not unique?

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?

user1271772

Posted 2018-05-27T05:41:03.770

Reputation: 8 162

Answers

5

You get two decompositions for your matrix (let's call it $A$) because you are using two different operatorial bases.

In the first case you are considering the matrix as acting in a space of dimension $3\times 2$, that is, using the operatorial basis $\{\lambda_i\sigma_j\}_{ij}\equiv\{\lambda_i\otimes\sigma_j\}_{ij}$.

In other words, you are computing the coefficients $c_{ij}=\operatorname{tr}((\lambda_i\otimes \sigma_j) A)$, finding $c_{61}$ to be the only not vanishing term. This decomposition will be unique, because $\operatorname{tr}\big[ (\lambda_i\sigma_j)(\lambda_k\sigma_l) \big]=N_{ij} \delta_{ik}\delta_{jl}$.

On the other hand, the second decomposition is obtained thinking of $A$ as a matrix in a space of dimensions $2\times 3$, that is, by decomposing it using the operatorial basis $\{\sigma_i\lambda_j\}_{ij}\equiv\{\sigma_i\otimes\lambda_j\}_{ij}$. This gives you new coefficients $d_{ij}\equiv\operatorname{tr}((\sigma_i \otimes\lambda_j) A)$, which do not have to be (and indeed are not) the same as the $c_{ij}$.

There is no paradox because $\{\sigma_i\otimes\lambda_j\}_{ij}$ and $\{\lambda_i\otimes\sigma_j\}_{ij}$ are two entirely different operatorial bases for a space of dimension $6$.

glS

Posted 2018-05-27T05:41:03.770

Reputation: 12 247

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

4

This looks essentially similar to the property of non-commutativity of the Kronecker product: $X\otimes \lambda_6\neq \lambda_6\otimes X$:

$$X\otimes\lambda_6 = \begin{pmatrix}0&1 \\1&0\end{pmatrix}\otimes \begin{pmatrix}0&0&0 \\0&0&1 \\0&1&0\end{pmatrix} = \begin{pmatrix}0&0&0&0&0&0 \\ 0&0&0&0&0&1\\ 0&0&0&0&1&0\\ 0&0&0&0&0&0\\ 0&0&1&0&0&0\\ 0&1&0&0&0&0 \end{pmatrix}$$

Unsurprisingly, you can't decompose $-\frac{1}{2}Z\lambda_1 + \frac{1}{2}I_2\lambda_1 - \frac{1}{\sqrt{3}}X\lambda_8+\frac{1}{3}XI_3 = \lambda_6X$ into $X\lambda_6$.

However, as both matrices are square, they are 'permutation similar', so that $X\otimes \lambda_6=P^T\left(\lambda_6\otimes X\right)P$ for some permutation matrix $P$

In other words, to answer part 1, for a given permutation/ordering, the decomposition is unique, but when the ordering is changed, the matrix/Hamiltonian undergoes a rotation $\left(P^T = P^{-1}\right)$, which also changes the decomposition.

It becomes clear what can be used to decompose a matrix of this form by splitting it into sub-matrices: by writing $$X\lambda_6 = \begin{pmatrix}A&B\\C&D\end{pmatrix},$$ where each sub-matrix $A, B, C$ and $D$ is a $3\times 3$ matrix, it becomes clear that $A=D=0$ and $B=C=\lambda_6$, which verifies $$X\lambda_6 = \begin{pmatrix}0&\lambda_6\\\lambda_6&0\end{pmatrix} = X\otimes \lambda_6$$

Performing the rotation/permuting and applying the same idea gives $$M=\begin{pmatrix}0&0&0&0&0&0 \\ 0&0&0&0&0&0\\ 0&0&0&0&0&1\\ 0&0&0&0&1&0\\ 0&0&0&1&0&0\\ 0&0&1&0&0&0 \end{pmatrix} = \begin{pmatrix}A&B\\C&D\end{pmatrix},$$ which gives that $$A=0,\quad B=C=\begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix},\quad D=\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}=\lambda_1$$

It follows that $B=C=\frac{1}{3}I_3-\frac{1}{\sqrt{3}}\lambda_8$, giving $$M=\begin{pmatrix}0&\frac{1}{3}I_3-\frac{1}{\sqrt{3}}\lambda_8\\\frac{1}{3}I_3-\frac{1}{\sqrt{3}}\lambda_8&\lambda_1\end{pmatrix}=\frac{1}{2}\left(I-Z\right)\otimes\lambda_1 + X\otimes\left(\frac{1}{3}I_3-\frac{1}{\sqrt{3}}\lambda_8\right).$$

Changing the order of the decomposition: $$M=\begin{pmatrix}A&&B&&C\\D&&E&&F\\G&&H&&J\end{pmatrix},$$ which gives $A=B=C=D=E=G=J=0$ and $F=H=X$, in turn giving $$M=\begin{pmatrix}0&&0&&0\\0&&0&&X\\0&&X&&0\end{pmatrix}=\lambda_6\otimes X$$

Mithrandir24601

Posted 2018-05-27T05:41:03.770

Reputation: 3 226

I guess this answers the question: $\lambda_6 X$ acts on the qutrir first then the qubit, whereas the other expression acts on the qubit first then the qutrir, but I still don't get why there's two decompositions because working with only qubits I've never seen something like this. I hate to edit the question after you did all this work, but the way it's written (which I apologize you already spent time answering) is wrong, because as you said, $X\lambda_6$ is not the matrix I have there :'( – user1271772 – 2018-05-27T09:27:55.763

@user1271772 I'm not sure I understand: does this answer your question, after the typo was fixed? – glS – 2018-05-29T19:03:21.570

1$\mathbb{C}^2 \otimes \mathbb{C}^3 \simeq \mathbb{C}^6 \simeq \mathbb{C}^3 \otimes \mathbb{C}^2$ but there is data in this isomorphism. Not canonical. Think with categories. – AHusain – 2018-05-29T20:38:20.367