Unitarity of CKM matrix?

Question

The CKM matrix is known to be unitary.

Now, is it just found experimentally?

Is the following idea correct?

We know that the weak force eigenstate $ \left( \begin{array}{c} d' \\ s' \\ b' \end{array} \right) $ are related to the physical state $ \left( \begin{array}{c} d \\ s \\ b \end{array} \right) $ via the Cabibbo-Kobayashi-Maskawa (CKM) matrix : $$ \left( \begin{array}{c} d' \\ s' \\ b' \end{array} \right) = V_{CKM}\left( \begin{array}{c} d \\ s \\ b \end{array} \right)$$

Orthogonality of the weak force eigenstates AND of the physical states dictates that: $$ ( d' s' b' ) \left( \begin{array}{c} d' \\ s' \\ b' \end{array} \right) = ( d s b ) \left( \begin{array}{c} d \\ s \\ b \end{array} \right) = 1 $$

SO:

$$1 = ( d' s' b' ) \left( \begin{array}{c} d' \\ s' \\ b' \end{array} \right) = V_{CKM}^{\dagger} V_{CMK} ( d s b ) \left( \begin{array}{c} d \\ s \\ b \end{array} \right) = V_{CKM}^{\dagger} V_{CMK}$$

Therefore $V_{CKM}^{\dagger} V_{CMK} = 1$

Answer 1

Your last equation does not make sense. On the left had side you have a scalar 1, while on the right hand side there is supposed to be the unit matrix. However, it is $$ 1=\begin{pmatrix}d' & s' & b'\end{pmatrix}\begin{pmatrix}d' \\ s' \\ b'\end{pmatrix} = \begin{pmatrix}d & s & b\end{pmatrix}V_{CKM}^\dagger V_{CKM}\begin{pmatrix}d \\ s \\ b\end{pmatrix}. $$ Now, since the $\begin{pmatrix}d \\ s \\ b\end{pmatrix}$ form an orthonormal basis, $V_{CKM}^\dagger V_{CKM}$ has to be the unit matrix. In general, if for an orthonormal basis $\{v_i\}$ of a vector space $V$, there is a Matrix $M$ such that $$v_i^\dagger M v_i =1$$ for all $i$, then $M$ must be the unit matrix.

In general, a change of bases between two orthonormal bases is always unitary.