I am interested in constructing a general qutrit computational basis $\{|0\rangle,|1\rangle,|2\rangle\}_{(\theta,\varphi,\phi,\psi)}$, that is, writing the most general $U(3)$ transformation matrix (in terms of trigonometric functions). My aim is to start form a certain computational basis (say the $\lambda_3$ eigenvectors) and then construct a general one, where $|0(\theta,\varphi,\phi,\psi)\rangle$ is given by

$$|0(\theta,\varphi,\phi,\psi)\rangle = \cos\frac{\theta}{2}|0\rangle +e^{i\phi}\sin\frac{\theta}{2}\cos\frac{\varphi}{2}|1\rangle +e^{i\psi}\sin\frac{\theta}{2}\sin\frac{\varphi}{2}|2\rangle,$$ as stated for a general pure qutrit state in a previous question (What is $\gamma$ in the qutrit?). I also read arXiv:quant-ph/9910001, a reference on Qutrit Entanglement, which uses a similar expression for a general pure qutrit state.

I have already found $$|1(\theta,\varphi,\phi,\psi)\rangle = -\sin\frac{\theta}{2}|0\rangle +e^{i\phi}\cos\frac{\theta}{2}\cos\frac{\varphi}{2}|1\rangle +e^{i\psi}\cos\frac{\theta}{2}\sin\frac{\varphi}{2}|2\rangle,$$ which is orthogonal to $|0(\theta,\varphi,\phi,\psi)\rangle$. For $|2(\theta,\varphi,\phi,\psi)\rangle$, I would need to determine a ket that is both orthogonal to $|0(\theta,\varphi,\phi,\psi)\rangle$ and $|1(\theta,\varphi,\phi,\psi)\rangle$. Something like $|0(\theta,\varphi,\phi,\psi)\rangle\times|1(\theta,\varphi,\phi,\psi)\rangle$, if this was in $\mathbb{R}^3$ and not in a Hilbert space.

With this in mind, I computed the following $$|2(\theta,\varphi,\phi,\psi)\rangle = e^{i\phi}\sin\frac{\varphi}{2}|1\rangle +e^{i\psi}\cos\frac{\varphi}{2}|2\rangle.$$

  • $\begingroup$ as you observed yourself, the problem is essentially that of finding a parametrization for the full set of orthonormal bases in $\mathbb C^3$. The problem with your approach is fundamental: in general, there is no algebraic way to parametrize the hyperplane orthogonal to a given vector. It is only possible in some dimensions, and has to do with the parallelizability of $S^n$; see math.stackexchange.com/a/4216712/173147 for some related discussions. In your calculation, you made an assumption on the form of $|1\rangle$.. but there's no good way to not make such an assumption $\endgroup$
    – glS
    Commented Oct 25, 2021 at 7:11

1 Answer 1


You have serious problems: your qutrits aren't right.

A general U(3) transformation can be factored in a sequence of U(2) transformations so that $$ U=U(2)_{23}\times U(2)_{12}\times U(2)_{23}\tag{1} $$ where $U(2)_{ij}$ is a $U(2)$ transformation mixing basis vectors $i$ and $j$.

You can choose your first basis state to be $U_{23}$-invariant, i.e. \begin{align} \vert 0\rangle \mapsto \left(\begin{array}{c} 1 \\ 0 \\ 0\end{array}\right) \end{align} but not your second and third basis vectors. The action of a full $U(3)$ transformation on these will depend on $9$ parameters, so your second and third qutrit will depend on as many parameters as that.

Now there's an overall phase in U(3) so we might as well coset out this overall phase and work with $SU(3)$, which has similar factorization as Eq.(1). In this case the $U(2)_{ij}$ are in fact either full or specialized SU(2) transformation and details can be found in this paper. More details on the factorization go back to this arXiv paper by some of the same authors.

This is not the only factorization (although it's simple): see also

M. Reck, A. Zeilinger, H. J. Bernstein, and P. Bertani, Experimental Realization of any Discrete Unitary Operator, Phys. Rev. Lett. 73, 58 (1994)

W. R. Clements, P. C. Humphreys, B. J. Metcalf, W. S. Kolthammer, and I. A. Walmsley, Optimal design for universal multiport interferometers, Optica 3, 1460 (2016).

N. J. Russell, L. Chakhmakhchyan, J. L. O’Brien, and A. Laing, Direct dialling of Haar random unitary matrices, New J. Phys. 19, 033007 (2017).

but they will all lead to the same conclusion: your second and third basis states will in general need $8$ (in SU(3)) or $9$ (in U(3)) parameters.

Of course in the case of a qubit, the Euler factorization $R_zR_yR_z$ allows you go get rid of the rightmost $R_z$ since both $\vert 0\rangle$ and $\vert 1\rangle$ just pick up a phase under $R_z$, but this no longer holds for $U(3)$.

  • $\begingroup$ Thanks for your answer and for including several references. I already suspected that I was getting away with it too simply. $\endgroup$ Commented Oct 23, 2021 at 13:25

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