1
$\begingroup$

What is the procedure to generate a random qutrit quantum state?

That is, I want to generate a $3 \times 1$ vector $u$ such that $u$ is generated by a Haar random $3 \times 3$ unitary $U$ multiplied with the $3 \times 1$ column vector $$ \begin{pmatrix} 1 \\ 0 \\ 0\\ \end{pmatrix}. $$

$\endgroup$
0

1 Answer 1

2
$\begingroup$

There are two ways I can see (so made this an answer!):

  1. Generate a random unitary matrix and take the first column (or row)

For this, you're looking for CircularUnitaryMatrixDistribution[3] for the Haar distribution on unitary matrices. To get an actual matrix from this distribution, use

RandomVariate[CircularUnitaryMatrixDistribution[3]]

Check out the Background & Context section of the documentation:

Probabilistically, the circular unitary matrix distribution represents a uniform distribution over the unitary square matrices, while mathematically it is a so-called Haar measure on the unitary group $U(n)$.

Note that for efficiency, you can simply take the First element instead of dotting with {1,0,0}.

  1. Generate a random point on the 6D sphere

The unit sphere in $\mathbb{C}^3$ under the standard (physical) inner product can be viewed as the unit sphere in $\mathbb{R}^6$.

Then we have

Complex @@@ Partition[RandomPoint[Sphere[6]], 2]

as a random (representative of a) qutrit directly.

(One potential issue is that quantum mechanical states are not actually in correspondence with points on the unit sphere in $\mathbb{C}^3$, but with orbits of such points under the action of multiplication by phase. However, since each orbit has the same induced measure, choosing a random point on the sphere is equivalent to choosing a random orbit.)

$\endgroup$

Your Answer

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

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