What is $\gamma$ in the qutrit?

Question

We know that the qubit is defined as follows $$\lvert\psi\rangle = \alpha\lvert 0\rangle + \beta\lvert 1\rangle$$ where $\alpha, \beta \in \mathbb{C}$. We can also rewrite the state of the qubit using trigonometry, $$\lvert\psi\rangle = \cos\biggl(\frac{\theta}{2}\biggr)\lvert 0\rangle + e^{i\phi}\sin\biggl(\frac{\theta}{2}\biggr)\lvert 1\rangle$$ So, upon further research we discover the qutrit, which is quite similar: $$\lvert\psi\rangle = \alpha\lvert 0\rangle + \beta\lvert 1\rangle + \gamma\lvert 2\rangle$$ where $\alpha, \beta, \gamma \in \mathbb{C}$.

Provided $\alpha$ and $\beta$ are still equal to their respective trig functions, what is $\gamma$ equal to?

In other words, how do we rewrite the state of a qutrit?

Answer 1

Actually, you've got to back up a bit and understand why the qubit state can be written in that form. The coefficients $\alpha$ and $\beta$ need to satisfy some constraints:

1. First, $\lvert\alpha\rvert^2 + \lvert\beta\rvert^2 = 1$ (for pure states), which immediately suggests using cosine and sine to represent the magnitudes. With a phase factor, any $\alpha$ and $\beta$ which satisfy this constraint can be written as $$\begin{align} \alpha &= e^{i\phi_\alpha}\cos\frac{\theta}{2} & \beta &= e^{i\phi_\beta}\sin\frac{\theta}{2} \end{align}$$
2. Then the overall phase factor is physically irrelevant, so you might as well represent all states that differ by an overall phase by one representative of that set. There are various conventions you can use to pick which state to use to represent the set, but the typical choice is to choose the one in which $\alpha$ is real. So you extract and throw away an overall phase of $e^{i\phi_{\alpha}}$, leaving $$\begin{align} \alpha &= \cos\frac{\theta}{2} & \beta &= e^{i(\phi_\beta - \phi_\alpha)}\sin\frac{\theta}{2} \end{align}$$ Then you can define $\phi$ as $\phi_\beta - \phi_\alpha$.

With a qutrit, the corresponding first constraint is $\lvert\alpha\rvert^2 + \lvert\beta\rvert^2 + \lvert\gamma\rvert^2 = 1$. Obviously, if you use the same trig expressions as before, that won't be true (unless $\gamma = 0$). So the expressions will necessarily be different. One straightforward way to parametrize the space is $$\begin{align} \alpha &= e^{i\phi_\alpha}\cos\frac{\theta}{2} & \beta &= e^{i\phi_\beta}\sin\frac{\theta}{2}\cos\frac{\varphi}{2} & \gamma &= e^{i\phi_\gamma}\sin\frac{\theta}{2}\sin\frac{\varphi}{2} \end{align}$$ and then again, you can extract an overall phase and come up with $$\begin{align} \alpha &= \cos\frac{\theta}{2} & \beta &= e^{i\phi_1}\sin\frac{\theta}{2}\cos\frac{\varphi}{2} & \gamma &= e^{i\phi_2}\sin\frac{\theta}{2}\sin\frac{\varphi}{2} \end{align}$$ Any (pure) qutrit state will be representable (or physically equivalent to one that is representable) in this form. However, I don't know if this is how it's actually done in practice.

Answer 2

A main reason to parametrize qubits as $$\lvert\psi\rangle = \cos\biggl(\frac{\theta}{2}\biggr)\lvert 0\rangle + e^{i\phi}\sin\biggl(\frac{\theta}{2}\biggr)\lvert 1\rangle$$ is that $\phi$ and $\theta$ correspond to the angles on the Bloch sphere.

For qutrits, no Bloch sphere representation exists; thus, while it of course still possible to find similar parametrizations for qutrit states (such as the one given in the answer of David Z), they are far less meaningful.