7
$\begingroup$

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 $10^{-16}$.

I thought the Pauli decomposition for two qubits is unique, why would the Pauli-GellMann decomposition of a qubit-qutrit be non-unique, and how would the decomposition $\lambda_6X$ from the above 6x6 matrix be obtained?

$\endgroup$
0

2 Answers 2

5
$\begingroup$

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 $6$.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ May 29, 2018 at 19:54
4
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ 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 :'( $\endgroup$ May 27, 2018 at 9:27
  • $\begingroup$ @user1271772 I'm not sure I understand: does this answer your question, after the typo was fixed? $\endgroup$
    – glS
    May 29, 2018 at 19:03
  • 1
    $\begingroup$ $\mathbb{C}^2 \otimes \mathbb{C}^3 \simeq \mathbb{C}^6 \simeq \mathbb{C}^3 \otimes \mathbb{C}^2$ but there is data in this isomorphism. Not canonical. Think with categories. $\endgroup$
    – AHusain
    May 29, 2018 at 20:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.