# Can a qubit register hold any subset of values, or only specific subsets?

Say I have a quantum register containing 1 qubit. A qubit can hold either 0, 1 or both 0 and 1. In Dirac, one would write
0: $|0\rangle=\begin{bmatrix}1\\0\end{bmatrix}$

1: $|1\rangle=\begin{bmatrix}0\\1\end{bmatrix}$

0 and 1: $|+\rangle={1\over\sqrt 2}\begin{bmatrix}1\\1\end{bmatrix}$

If we were to extend our register to 2 qubits:
0: $|00\rangle = |0\rangle \otimes |0\rangle = \begin{bmatrix}1\\0\end{bmatrix} \otimes \begin{bmatrix}1\\0\end{bmatrix} = \begin{bmatrix}1\\0\\0\\0\end{bmatrix}$

1: $|01\rangle = |0\rangle \otimes |1\rangle = \begin{bmatrix}1\\0\end{bmatrix} \otimes \begin{bmatrix}0\\1\end{bmatrix} = \begin{bmatrix}0\\1\\0\\0\end{bmatrix}$

0 and 1: $|0+\rangle = |0\rangle \otimes |+\rangle = \begin{bmatrix}1\\0\end{bmatrix} \otimes {1\over\sqrt 2}\begin{bmatrix}1\\1\end{bmatrix} = {1\over\sqrt 2}\begin{bmatrix}1\\1\\0\\0\end{bmatrix}$

2: $|10\rangle = |1\rangle \otimes |0\rangle = \begin{bmatrix}0\\1\end{bmatrix} \otimes \begin{bmatrix}1\\0\end{bmatrix} = \begin{bmatrix}0\\0\\1\\0\end{bmatrix}$

3: $|11\rangle = |1\rangle \otimes |1\rangle = \begin{bmatrix}0\\1\end{bmatrix} \otimes \begin{bmatrix}0\\1\end{bmatrix} = \begin{bmatrix}0\\0\\0\\1\end{bmatrix}$

Now if I would want to load the decimal values 1 and 2 in the quantum register, I have two ways of reasoning, of which one appears to be flawed:
a)
This is in line with the reasoning above for 1 qubit, section "0 and 1":
I want to add values $|01\rangle$ and $|10\rangle$. My first qubit is a 0 for the decimal value 1, and my first qubit is a 1 for the decimal value 2. This means my first qubit is both 0 and 1, and therefor $|+\rangle$. My second qubit is a 1 for decimal value 1, and a zero for decimal value 2, which also implies $|+\rangle$. Hence, I need to load
$|++\rangle = |+\rangle \otimes |+\rangle = {1\over\sqrt 2}\begin{bmatrix}1\\1\end{bmatrix} \otimes {1\over\sqrt 2}\begin{bmatrix}1\\1\end{bmatrix} = {1\over\sqrt 4}\begin{bmatrix}1\\1\\1\\1\end{bmatrix}$
Which provides me with 1 and 2, but also 0 and 3.

b)
$c ( |01\rangle + |10\rangle) = c( \begin{bmatrix}0\\1\\0\\0\end{bmatrix} + \begin{bmatrix}0\\0\\1\\0\end{bmatrix}) = c \begin{bmatrix}0\\1\\1\\0\end{bmatrix}$, where $c$ is likely ${1\over\sqrt 2}$
Which provides me with 1 and 2 and only 1 and 2, and works out in mathematics, but fails in my current common sense.

My questions: Even though I see with math that b) holds and a) fails, can someone explain to me why this fails? Specifically, why is it that both individual qubits appear to be both 0 and 1, hence in superposition, but they are able to distinguish that as a total, only the decimal values 1 and 2 are present, and not 0,1,2 and 3 (which you would expect from 2 qubits in superposition).

• You've learned how tensor product works, that is great. Now you are making wild guesses how to add two numbers using qubits, but sorry, it simply does not make sense. – kludg Jul 23 '18 at 9:04

Mathematically, a pure single-qubit system can be described by any normalised vector in $\mathbb{C}^2$, i.e. $\left(\begin{array}{c} \alpha \\ \beta \end{array}\right)$ provided $|\alpha|^2+|\beta|^2=1$. Talking about "0 and 1" is already problematic. Clearly you're thinking about "an equal superposition of 0 and 1". Now, for two qubits, we can have any normalised state in $\mathbb{C}^4$, which clearly has many more possibilities than the tensor products of two single-qubit states (just count the number of parameters). So, sure, you can ask for "an equally weighted superposition of 01 and 10", and you'll get exactly what you had in (b): $(|01\rangle+|10\rangle)/\sqrt{2}$. If you try and create a normalised state with equal amplitudes for those two components using a state $|\psi\rangle\otimes|\phi\rangle$, you will necessarily have $|00\rangle$ and $|11\rangle$ components a well, so the state you've specified will be a superposition of all 4 possible basis states, not just the 2 that you want.
Let's also take it from the other perspective: let's say I've made $|+\rangle\otimes|+\rangle$. Now let's say that I measure in the computational basis and get the 0 answer. What's my state now? $|0\rangle\otimes|+\rangle$. If I measure the second qubit, I have a 50% change of getting the 0 answer. With a string 00 as my outcome, that certainly does not fulfil your requirement of being a bit 01 or 10.