# Understanding bases in quantum mechanics

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.

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.