Understanding bases in quantum mechanics

Question

For $l = 1$ the angular momentum operator $L_z$ has the eigenvalues $\hbar,0,-\hbar$ and the eigenstates are then $|1,1\rangle, |1,0\rangle, |1,-1\rangle$.

Now, we can calculate the matrix elements of the $L_x$ and $L_y$ operators in the basis of $L_z$ eigenstates which is given by: $$\begin{pmatrix}|1,1\rangle\\|1,0\rangle\\|1,-1\rangle\end{pmatrix}$$

Then by solving the eigenvalue equation $$L_x\mathbf{v} = \hbar m_x\mathbf{v},$$ for the eigenvalue $m_x = 1$, where $L_x$ is now a matrix, we can determine the eigenvector $\mathbf{v}$ which is the state in which the wave function collapsed and the value for $m_x=1$ was measured. As far as I understand, this is the $L_x$ eigenstate in the $L_z$ eigenstate basis.

Now, I have to calculate the probability of the eigenvalue $\hbar$ for $m_x=1$ being measured in the state $|1,1\rangle$. For the probability $P$ we can write: $$P(|1,1\rangle \rightarrow |\mathbf{v}\rangle) = |\langle \mathbf{v}|1,1\rangle|^2.$$

As far as I understand, in order to be able to calculate the inner product of these two states, I have to convert one to the basis of the other. So if $|1,1\rangle$ is given in an abstract basis and $\mathbf{v}$ is given in the basis of $L_z$ eigenstates, then I have to convert $|1,1\rangle$ to the $L_z$ eigenstate basis or vice versa.

The official solution does the following: $$\langle\mathbf{v}| = v_1\langle 1,1| + v_2\langle 1,0| + v_3\langle 1,-1|,$$ where $v_1, v_2, v_3$ are the components of $\mathbf{v}$.

I do not understand this step. What is being converted to what basis and how exactly? Is my understanding that in a scalar product two eigenstates need to be converted to the same basis correct? This would mean that $\mathbf{v}$ and $|\mathbf{v}\rangle$ are basically the same thing just in two different bases, the former in the basis of $L_z$ eigenstates and the latter in an abstract basis.

Answer 1

Your eigenstate $\vert 1,1\rangle_x$ of $\hat L_x$ with eigenvalue $+1$ will be a linear combination of the eigenstates $\vert 1,m\rangle_z$ of $\hat L_z$. In other words: \begin{align} \vert 1,1\rangle_x&=v_1\vert 1,1\rangle_z+ v_0\vert 1,0\rangle_z +v_{-1}\vert 1,-1\rangle_z\, , \\ _x\langle 1,1\vert&=v^*_1\,{_z\langle} 1,1\vert+ v^-_0\,{_z\langle} 1,0\vert +v^*_{-1}\,{_z\langle} 1,-1\vert\, , \end{align} and thus \begin{align} \vert _x\langle 1,1\vert 1,1\rangle_z= \vert v_1\vert^2 \end{align} and so forth. You can also express $\vert 1,1\rangle_x$ directly as \begin{align} (v_1,v_0,v_{-1})^\top \end{align} so that your inner product is then \begin{align} (v_1,v_0,v_{-1})^*\cdot\left(\begin{array}{c}1\\ 0 \\ 0\end{array}\right) =v_1^* \end{align} and thus the magnitude is $\vert v_1\vert^2$ as before.