What does rotation of a basis mean in the Bell State?

Question

Forgive me for a simple question but I have been reading up on Bell's Theorem and its formulation in measuring a fully entangled state $$\frac{1}{\sqrt{2}}|00〉 + \frac{1}{\sqrt{2}}|11〉$$

and I understand that the probabilities of measuring the outcome of a single qubit for two different states are the same no matter the basis. However, most texts I read then have a paragraph similar to this:

Moreover the state of the second qubit is exactly equal to the outcome of the measurement — $|0〉$ if the measurement outcome is $0$, say. But now if the second qubit is measured in a basis rotated by $\theta$, then the probability that the outcome is also $0$ is exactly $\cos^2(θ)$."

Why is this true? In particular, we must remember that the states being described here are just shorthands for vectors in $\mathbb{C}^4$ and I'm not even sure what "rotating" a basis by an angle means here... To be clear, $|00〉= [0,0,0,0]^{T}$ and $|11〉= [1,1,1,1]^{T}$ so how do you rotate a basis or set of vectors?

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EDIT: I just want to add that I understand the implied argument. If $|v〉,|v^{\perp}〉$ are two vectors that make up the measurement basis and say the outcome of measuring the first qubit was $|v〉$, then it is implied that the second measurement's basis consists of the vectors "$|v_\theta〉,|v^{\perp}_\theta〉$" so that $$|v〉= |v_\theta〉\cos(\theta)+|v^{\perp}_\theta〉\sin(\theta)$$ $$|v^{\perp}〉= -|v_\theta〉\sin(\theta)+|v^{\perp}_\theta〉\cos(\theta)$$ and then we proceed by considering the probabilities of measuring the state $|v_\theta〉$ etc.

But what really are these vectors $|v_\theta〉,|v^{\perp}_\theta〉$ when their coordinates are complex numbers in a 4D space?

An obvious definition for trying to generalise rotations to this situation would be to define them as the vectors which make those equations true but the problems I see with this are:

1. Mathematically, we don't know such vectors exist without some explicit construction (and there are supposedly problems with considering rotations around axes in 4D so this isn't as esoteric a complaint as it might sound).

2. Such a definition is completely devoid of physical insight. Why would this definition correspond with anything to do with actual rotations of a polarising filter in an actual experiment?

Again, I apologise if this shows a deep misunderstanding of the physics etc. and any major clearings-up would be greatly appreciated.

Answer 1

If you measured the first qubit and found, say, the state $|0\rangle$, then you know that the second qubit is also in the state $|0\rangle$. To measure in a rotated basis means to measure in a basis of the form $$ |u_1\rangle = \cos\theta |0\rangle + e^{i\varphi}\sin\theta|1\rangle,\qquad |u_2\rangle = -e^{-i\varphi}\sin\theta |0\rangle + \cos\theta|1\rangle. $$ Equivalently, it means to measure in the computational basis after applying a rotation operation, which you can write in the form $$U(\theta,\varphi)=\begin{pmatrix}\cos\theta & -e^{-i\varphi}\sin\theta \\ e^{i\varphi}\sin\theta & \cos\theta\end{pmatrix}.$$ Note that this unitary maps $|0\rangle\mapsto|u_1\rangle$ and $|1\rangle\mapsto|u_2\rangle$.

Measuring $|0\rangle$ in this new basis will give you the outcome $|u_1\rangle$ with probability $|\langle 0|u_1\rangle|^2=\cos^2\theta$, and the outcome $|u_2\rangle$ with probability $\sin^2\theta$.

Finally, let me point out that parametrising unitary operations with angles is not really necessary. In general, a two-dimensional unitary (technically, an $SU(2)$ matrix) has the form $$\begin{pmatrix}a & b \\ -b^* & a^*\end{pmatrix}$$ with $a,b\in\mathbb C$ such that $|a|^2+|b|^2=1$. It is then common to parametrise these parametrise these components using angles, remembering that if $c^2+d^2=1$ and $c,d\in\mathbb R$ then there is always some $\theta\in\mathbb R$ such that $c=\cos\theta$ and $d=\sin\theta$.