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

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?

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.

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 $$d=\sin\theta$$.
• Just to be clear, are the exponential terms added to give an additional degree of freedom along another axis? Is the exponential a bit like its own rotation matrix embedded in those terms (in that it's secretly $sin$ and $cos$ terms summed)? Finally, what is the physical interpretation of this? When a polarisation detector rotates its polariser, how is this interpreted mathematically? Apr 21, 2020 at 17:21
• the phase is to take into account all possible unitary operations that can be performed on a qubit. You need it because for example $|0\rangle+|1\rangle$ is different from $|0\rangle+i |1\rangle$ (in terms of the polarisation, think diagonal vs circular polarisation). If you represent the qubit in the bloch sphere, than yes the unitaries are one-to-one with 3D rotations, but note that this nice representation only holds for one qubit (in higher dimensions it gets more complicated). Different orientations of a polariser will correspond to different measurement bases (here $|u_i\rangle$).