# Changing basis

1. Apr 18, 2010

Hi,

I'm reading Shankar's Principles of QM and I find it not very clear on how exactly should I change basis of operator. I know how to change basis of a vector so can I treat the columns of operator matrix as vectors and change them? Or is it something else?

2. Apr 26, 2010

### jasonRF

It is something a little different. Let $$v_1$$ denote a vector represented in basis 1. Then to represent this same vector in terms of a different basis, basis 2, we need to find a matrix $$T_{1:2}$$ that maps any vector representation from basis 1 to basis 2. Thus if we let $$v_2$$ denote that vector represented in basis 2, then
$$v_2 = T_{1:2} \, v_1$$.
This means that
$$v_1 = T_{1:2}^{-1} \, v_2$$
so the matrix that maps a vector representation from basis 2 to basis one is
$$T_{2:1}=T_{1:2}^{-1}$$.

Now, if we have a matrix representation of an operator in basis 1, say $$A_1$$, then it takes a vector represented in basis 1 and maps it to a different vector represented in basis 1. For our example let
$$y_1 = A_1 v_1$$.
So if we want to represent y in basis 2 we have,
$$y_2 = T_{1:2} y_1 = T_{1:2} A_1 v_1 = T_{1:2} A_1 T_{2:1} v_2$$.
Hence, if we want to represent the operator in basis 2, the matrix representation must be
$$A_2 = T_{1:2} A_1 T_{2:1} = T_{1:2} A_1 T^{-1}_{1:2}$$,
and we have
$$y_2 = A_2 v_2$$
as required. If you think about what is happening, it should be easy to remember.

Note that most linear algebra books will cover this.

jason

Last edited: Apr 26, 2010
3. Apr 28, 2010