# Unitary transformations that make a 2-qubit system non-observable

Apology in advance if this question is not entirely sound, I am just beginning to grasp q-computation. My question is the following:

Consider a 2-qubit system. Suppose your initial state is a superposition state

$$s_i=a|01\rangle+b|10\rangle$$

Is there a 4 by 4 matrix that represents a quantum gate that yields the state

$$0|00\rangle+0|01\rangle+0|10\rangle+0|11\rangle$$?

If this is possible, can this be generalized to $$n$$ q-bits, and $$n$$ amplitudes to provide an initial state like $$s_i$$ above?

There are a few reasons there isn't a quantum gate that can map to an "all 0s".

Most simply, that output state has no probability of measuring any state, which doesn't make much sense because if we have a pair of qubits and shine a laser on them (or whatever the readout process is), we have to get some information about them.

On a higher level, quantum gates are described by unitary matrices, which importantly means they 1) are reversible and 2) preserve norms.

For 1), if there's some gate that maps the initial state to 0, then it couldn't be reversible.

For 2), when we talk about a state $$|s\rangle = a|01\rangle + b|10\rangle$$, we typically associate $$a,b$$ with complex probability amplitudes such that the probability of measuring $$|01\rangle$$ is $$|a|^2$$ and $$|10\rangle$$ is $$|b|^2$$. Since we have to get some kind of result when we make a measurements $$|a|^2 + |b|^2 = 1$$. If we have some quantum unitary gate $$U$$ acting on $$|s\rangle$$, we always have some $$U|s\rangle = c|00\rangle + d|01\rangle + e|10\rangle + f|11\rangle$$ such that $$|c|^2 + |d|^2 + |e|^2 +|f|^2 = 1$$.

• Right, thank you. This makes me think of another question, which I guess ought to go into a different post. Or maybe I ask here. Are there unitary transformations that map, say the state $s_i$ (assuming now it has $n$ qbits ) to just one of the basis vectors? So you start with $a|00>+b|01>+c|10>+d|11>$ and end up with say $e|00>$. By the way, how do you write the right bracket in TeX?
– EGME
Oct 29, 2022 at 15:36
• \rangle gives you pretty right brackets in TeX (and \langle is the same for left brackets). If you know your state $|s\rangle$ which you want to map to $|00\rangle$, you can do that by performing a change of basis from some basis you create starting with $|s\rangle$ (by Gram-Schmidt or some other procedure) to the standard computational basis. Thing to be wary of is experimentally, implementing such a unitary to do this for a large number qubits gets very resource intensive quickly Oct 29, 2022 at 19:32
• Thanks! One more question (but if you think I should post it, then please do say). Is it possible to design a quantum transformation to determine $N$-linear independence over the integers of some set of reals (namely, if you have a set $T$ of reals, you want to check if $\sum_T c_t t=0$ where $|c_t|\le N, c_t\in \mathbb{Z}$. I am studying Shor's paper right now, haven't quite yet gotten a full handle on it.
– EGME
Oct 30, 2022 at 6:47