Preliminaries: Let $M$ be any $4\times 4$ unitary matrix. Now, define the matrices $A$ and $B$ by
$$A=M^\dagger\left[ \begin{pmatrix}
\cos(\theta) & -e^{i\lambda}\sin(\theta)\\
e^{i\phi}\sin(\theta) & e^{i(\lambda+\phi)}\cos(\theta)
\end{pmatrix}\otimes \begin{pmatrix}
1 & 0\\
0 &1
\end{pmatrix}\right]M$$
$$B=M^\dagger\left[\begin{pmatrix}
1 & 0\\
0 &1
\end{pmatrix}\otimes \begin{pmatrix}
\cos(\theta) & -e^{i\lambda}\sin(\theta)\\
e^{i\phi}\sin(\theta) & e^{i(\lambda+\phi)}\cos(\theta)
\end{pmatrix}\right]M$$
We may then define $f:\mathbb{R}^6\to \mathbb{R}$ by
$$f(\theta,\lambda,\phi,\delta,\alpha,\beta)=\prod_{C\in\{A,B\}}\left(\left|\begin{pmatrix}
0\\
0\\
1\\
0
\end{pmatrix}^TC\begin{pmatrix}
\cos(\delta)e^{i\alpha}\\
\sin(\delta)e^{i\beta}\\\
0\\
0
\end{pmatrix}\right|+\left|\begin{pmatrix}
0\\
0\\
0\\
1
\end{pmatrix}^TC\begin{pmatrix}
\cos(\delta)e^{i\alpha}\\
\sin(\delta)e^{i\beta}\\\
0\\
0
\end{pmatrix}\right|\right)$$
Now, what are the zeros of this function? It is easy to show that any element of the set
$$S=\{(k_1\pi ,\lambda, 2\pi k_2-\lambda,\delta,\alpha,\beta):k_1,k_2\in\mathbb{Z}\text{ and }\lambda,\delta,\alpha,\beta\in\mathbb{R}\}$$
is a solution to $f(\theta,\lambda,\phi,\delta,\alpha,\beta)=0$. This is because
$$\begin{pmatrix}
\cos(\theta) & -e^{i\lambda}\sin(\theta)\\
e^{i\phi}\sin(\theta) & e^{i(\lambda+\phi)}\cos(\theta)
\end{pmatrix}\Bigg|_{\theta=k_1 \pi,\phi=2\pi k_2-\lambda}=(-1)^{k_1}\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}$$
and the matrix $A$ collapses down to
$$A=M^\dagger[(-1)^{k_1}I_4]M=(-1)^{k_1}M^\dagger M=(-1)^{k_1}I_4$$
(where $I_4$ is the $4\times 4$ identity matrix), which in turn means all the inner products are zero.
My question: Is there some $4\times 4$ unitary matrix $M$ such that $S$ is all the solutions to the equation $f(\theta,\lambda,\phi,\delta,\alpha,\beta)=0$?
Motivation: If I could prove this question in the negative (there are no such matrices $M$), then I could show that at least $3$ qubits are required for an error detection code I am developing for a quantum computer. I won't bore you with all the details (unless someone wants to know more), but my problem has simplified to the linear algebra question above.
Work: So far, I have tried a lot of different ways of proving the above without much success. For any set matrix $M$, I can always find solutions that are not in $S$. However, it does seem like almost all the degrees of freedom are required because there are some $M$ where the solution seems to lie on a line in $\mathbb{R}^6$ space (basically it depends on the difference between $\alpha$ and $\beta$). That is, it does not form a family of solutions on a plane/hyper-plane in the space. My current work is to try to use different generators of $U(4)$ and see if I can say something if I decompose $M$ into constituent parts (Here are two different papers I have been using for said generators). Unfortunately, this has not simplified my problem to the point where I can get a satisfactory solution.
Best Answer
Not a full solution, an idea how to simplify the equation.
Vectors $\lvert e_i\rangle=\{(1,0,0,0)^{T}, (0,1,0,0)^{T}, (0,0,1,0)^{T}, (0,0,0,1)^{T}\}$ form a basis. $f$ is zero if for some vector $\lvert a\rangle=e^{i\alpha}\cos{\delta}\lvert e_1\rangle+e^{i\beta}\sin{\delta}\lvert e_2\rangle$: $$\tag{1} \langle e_3\rvert A\lvert a\rangle=\langle e_4\rvert A\lvert a\rangle=0 \; or \; \langle e_3\rvert B\lvert a\rangle=\langle e_4\rvert B\lvert a\rangle=0 $$ If we introduce matrix components $A_{ik}=\langle e_i \rvert A\lvert e_k\rangle$, equation (1) can be rewritten as: $$ A_{31}e^{i\alpha}\cos{\delta}+A_{32}e^{i\beta}\sin{\delta}=A_{41}e^{i\alpha}\cos{\delta}+A_{42}e^{i\beta}\sin{\delta}=0 $$ (or similar equation for $B$, which is omitted); which is satisfied if: $$ F(\theta,\lambda,\phi)\equiv A_{42}A_{31}-A_{41}A_{32}=0 $$ This is a single complex equation for 3 real parameters: $\theta,\lambda,\phi$. This way we can analyze an equation with just 3 parameters, which should be easier.