# Change of basis

1. Jan 26, 2015

### aaaa202

We are given the vectors la> = (1,0) and lb> = (0,1) and then a Hamiltonian H which is a 2x2 matrix with 2 on the diagonal entires and zero elsewhere. I am asked to now represent H in the basis of the vectors la'> = 1/sqrt(2)(1,1) and lb'> = 1/sqrt(2)(1,-1), which are also eigenvectors of H because of the degeneracy of the eigenvalues of H.
I always get confused with these basis-change problems. First of all: What does it even mean that a matrix is represented in a particular basis? Why is H necessarily represented in the basis (1,0) and (0,1)?
Secondly I know that the basis change equation is:
H' = SHS^-1 (1)
But my QM also tells me that the elements of H' can be found by taking the inner products <a'lHla'>, <a'lHlb'> etc. How is this equivalent to (1)?

2. Jan 27, 2015

### Fredrik

Staff Emeritus
See this FAQ post: https://www.physicsforums.com/threads/matrix-representations-of-linear-transformations.694922/ [Broken]

Last edited by a moderator: May 7, 2017
3. Jan 27, 2015

### aaaa202

Okay but Im still not sure why for an orthonormal basis the basis change equation:
B = SAS^-1
Reduces to just calculating B using the equations B_ij = <bil A lbj> , where b1,b2,b3... are the new basis vectors of the new basis expressed in the old basis.

4. Jan 27, 2015

### Fredrik

Staff Emeritus
I didn't have time to answer that before. Let U be the linear operator that takes the old basis vectors to the new, i.e. $e_i'=Ue_i$. I'll start by proving that this U must be unitary (assuming that both bases are orthonormal). For all i,j, we have
$$\langle e_i,e_j\rangle =\delta_{ij}=\langle e_i',e_j'\rangle =\langle Ue_i,Ue_j\rangle =\langle e_i,U^\dagger U e_j\rangle,$$ and therefore
$$\langle e_i,(I-U^\dagger U)e_j\rangle =0.$$ This implies that for all j and all vectors x, we have
$$\langle x,(I-U^\dagger U)e_j\rangle =\left\langle\sum_{i=1}^2 x_i e_i,(I-U^\dagger U)e_j\right\rangle =\sum_{i=1}^2 (x_i)^* \langle e_i,(I-U^\dagger U)e_j\rangle =0.$$ In particular, this holds when $x=(I-U^\dagger U)e_j$. That implies that for all j, we have $(I-U^\dagger U)e_j=0$. This implies that $U^\dagger U=I$.

Now if we define $S=U^{-1}$ and use that $U^\dagger=U^{-1}$, we have
\begin{align}
H'_{ij}=\langle e_i',He_j'\rangle = \langle Ue_i,HUe_j\rangle =\langle e_i,U^\dagger HUe_j\rangle =(U^\dagger HU)_{ij} =(SHS^{-1})_{ij}
\end{align} In other words, the components of $H$ in the primed basis are the same as the components of $SHS^{-1}$ in the unprimed basis.

Last edited: Jan 27, 2015